Split (2)

train · ~239k rows (showing the first 191k)

id stringlengths 1 6	url stringlengths 16 1.82k	content stringlengths 37 9.64M
188300	https://www.youtube.com/watch?v=fDMBKPmiCKg	Work-Energy with a spring example Rhett Allain 50300 subscribers 15 likes Description 5558 views Posted: 17 Mar 2011 Suppose a ball is dropped from some height onto a spring. How much will the spring be compressed? If you like this, check out my ebook "Just Enough Physics" available at Amazon. 2 comments Transcript: okay let me do a quick example using work energy and the energy stored in the spring so i'm just going to make up something so suppose i take a ball of mass m and i drop it over a spring and so the ball is going to fall and compress the spring and so this spring let's just say has a natural length an uncompressed length of l 0 and a spring constant of k and let's say this ball starts a distance h above the top of the spring so the question is how much compressed for the spring i didn't write it all the way out because i'm lazy okay so when we have a problem like this we need to think okay first am i using momentum principle or work energy and you should say say it say it work energy because we're dealing with distance and not time so if we deal with work energy principle the next thing we need to do is say what's my system i could do several things in this case i'm going to say the system is the ball plus the spring plus the earth and so if i include the if i include the spring in there let me pull okay sorry for the interruption um so that's my system now if i include the spring in the system then i can have spring potential energy if i include the ball and the earth in the system then i can have gravitational potential energy okay so that's just going to make it a little bit easier so that i don't have to worry about the work done by gravity i don't work worry about the work done by the spring okay so that's my system now the next thing i need to do is say i need to i need to pick two positions in space that i can use work energy principle for so i'll have it start up here and then i'll have an end down here with the compressed spring and the ball right there that's two okay we'll call that um i maybe maybe i picked these things poorly but uh so let me just go ahead and say uh why one is going to be the way i use my poorly worded system oh i'm going to do this i'm tricky i'll call this y equals 0. so y 1 is going to be h and then y 2 is going to be negative we'll call it s how much is compressed tricky you would you don't have to do that you can pick anywhere you want for the origin okay so now i can use my work energy principle and if that is my system there's no work done because gravity is part of the system the spring is part of the system there's nothing else so i have change in kinetic plus change in spring potential plus change in gravitational potential equals zero so let's just let's just do it 0 equals k2 minus k1 plus us 2 minus us1 plus ug2 minus ug1 okay so now i can start putting in some values here except that i know some things are zero because what's the kinetic energy when it starts if i release it from rest zero what's the kinetic energy at the lowest point when it hits when it when the spring stops it keyword stops zero um what is the spring potential energy at two that's something but at one at this position up here how much is the spring compressed or stretched it's not so i have that okay so now i can just put in my values i'll do it right here so i have 0 equals u s 2 is going to be one half k s squared u g 2 is going to be negative g2 right negative m s let me call it gs and then i have minus ug1 minus m g h now what i want to solve for i want to solve for s so in that case how would you solve this i didn't give you numbers it doesn't matter if i didn't give you any numbers then you would have to use a quadratic equation here so because i have zero equals a s squared plus b s oh sorry it's bs plus c so s equals uh negative negative b plus or minus the square root of b squared minus plus no b squared minus 4ac over 2a that's right right 2a b squared ah you know you forget these things sometimes when you get older yeah that's right negative b plus that's right okay so then you just b would be that value with a negative uh c would be that with a negative and that would be that you plug it in and that's how you do it
188301	https://math.stackexchange.com/questions/502496/about-the-ratio-of-the-areas-of-a-convex-pentagon-and-the-inner-pentagon-made-by	Skip to main content About the ratio of the areas of a convex pentagon and the inner pentagon made by the five diagonals Ask Question Asked Modified 6 years, 7 months ago Viewed 2k times This question shows research effort; it is useful and clear 24 Save this question. Show activity on this post. I've thought about the following question for a month, but I'm facing difficulty. Question : Letting S′ be the area of the inner pentagon made by the five diagonals of a convex pentagon whose area is S, then find the max of S′S. It seems that a regular pentagon and its affine images would give the max. However, I don't have any good idea without tedious calculations. Can anyone help? Update : I crossposted to MO. euclidean-geometry polygons Share CC BY-SA 3.0 Follow this question to receive notifications edited Apr 13, 2017 at 12:58 CommunityBot 1 asked Sep 23, 2013 at 14:48 mathlovemathlove 153k1010 gold badges125125 silver badges310310 bronze badges 6 Have you used similar triangles to compare the side-length ratios, and in particular, the perimeter ratios? If you can get a perimeter ratio, I believe a good start would be to consider that the area ratio would be some degree-2 form of the perimeter ratio. – abiessu Commented Sep 23, 2013 at 14:55 Except that in a non-regular pentagon S, S′ is often a different shape (i.e., non-similar to S). Hmmm... – abiessu Commented Sep 23, 2013 at 15:03 @abiessu: Well, I can't get what you mean, sorry. – mathlove Commented Sep 23, 2013 at 15:22 No worries, I drew my own non-regular pentagon and began to realize some of the difficulty you are having. If it is possible to know the perimeter p of S and the perimeter p′ of S′, then I would expect that S′S is similar to (or the same as) p′2p2. – abiessu Commented Sep 23, 2013 at 15:24 1 I'm reminded of "Feynman's Triangle" or the " 1/7 th triangle theorem. Construct Cevians from vertices of a triangle to points on the opposite sides which are in the ratio of one third the distance to a neighbor vertex. Take it in a clockwise or anticlockwise but consistent sense. The ratio of the area of the inner to outer triangle is 1/7,Feynman worked on the problem during a dinner conversation, I suppose he got the right answer. – Alan Commented Sep 30, 2013 at 19:32 \| Show 1 more comment 5 Answers 5 Reset to default This answer is useful 1 Save this answer. Show activity on this post. Denote by zi (0≤i≤4) the vertices of the large pentagon Z and by wi the vertices of the small pentagon W. WLOG we may assume z0=(0,0),z2=(1,0),z3=(0,1) and w3=(a,0),w4=(b,0),w1=(0,d),w2=(0,c) with 0<a<b<1,0<c<d<1,d<ca<1b . One then obtains z1=(ba(1−c)a−bc,c(a−b)a−bc),z4=(a(c−d)c−ad,dc(1−a)c−ad),w0=(b(1−d)1−bd,d(1−b)1−bd) . This means that the moduli space M of such configurations up to affinity has dimension 4. The values a=c=3−5√2≐0.381966 and b=d=5√−12≐0.618034 correspond to regular pentagons Z and W. Further computation then gives 2area(Z)=1+(b−a)ca−bc−a(d−c)ad−c,2area(W)=bd(2−b−d)−ac(1−bd)1−bd ,(1) and we are interested in the quantity t(a,b,c,d):=area(W)area(Z) . For regular pentagons Z and W one has t=7−35√2≐0.145898. The expressions (1) are complicated enough to forbid calculating partial derivatives of t. On the other hand it is easy to see that infMt(a,b,c,d)=0. As for the maximum of t I simulated 107 random such configurations with Mathematica and obtained the following optimal quintuple (a,b,c,d,t): (0.383542,0.619248,0.381882,0.618277,0.14589) . This supports the conjecture that t is maximal for regular pentagons. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jun 1, 2017 at 6:56 answered May 30, 2017 at 12:10 Christian BlatterChristian Blatter 232k1414 gold badges203203 silver badges489489 bronze badges 2 Very nice. But I think the sign of the last term of 2 area(Z) should a minus. – T L Davis Commented Jun 1, 2017 at 0:43 @TLDavis: Thank you for noting the slip. It came about when copying the Mathematica output. – Christian Blatter Commented Jun 1, 2017 at 6:57 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. A complete solution is available at On the polygon determined by the short diagonals of a convex polygon, Jacqueline Cho, Dan Ismailescu, Yiwon Kim, Andrew Woojong Lee Share CC BY-SA 4.0 Follow this answer to receive notifications answered Dec 30, 2018 at 16:42 user84909user84909 41222 silver badges88 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Here is a partial answer. Say point C of the large pentagon ABCDE is free to move, so long as the area does not change. We are curious what happens to the red area. So C can move parallel to BD. As it does so, the red area increases on one side and decreases on the other side, due to CA and CE moving. Consider CE, and the change in red area due only to CE as C moves by ϵ (parallel to BD). The point where CE crosses BD moves by kϵ, where k is the constant ratio m/n of the altitude m from E to BD to the altitude n from E to C's locus line. Indeed, the change in area is monotonic in the distance s from B′ (the intersection of AD and CE) to BD, which itself is monotonic in the position of C. The area of the red pentagon is affected in this way both by the change in area due to CE and by that due to CA, and therefore it is maximized when C is somewhere in the middle, and it is strictly concave in C's position. Due to the symmetry of the regular pentagon, we can see the red area is at a local maximum in this case. The question asks us to prove this is the global maximum, but I would guess an even stronger result, that this is the only local maximum, and even that the red area is strictly concave in the positions of the vertices (subject to constant area of the large pentagon). If we could show that single-vertex area-preserving moves can transform any pentagon into any "nearby" one, with no vertex ever traveling far from its original position, then that would probably finish the proof of total strict concavity. It seems to be true, requiring O(1/ϵ) moves. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 4, 2016 at 10:07 MattMatt 9,72355 gold badges3434 silver badges4848 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Not a complete answer, and surely not elegant, but here's an approach. The idea is to consider "polygonal conics" in the following sense. Take the cone K given by K:x2+y2=z2 and slice it with the plane z=1. We get a circle γ, and let's inscribe in this circle a regular pentagon, P (The idea actually works with a regular n-gon with n=2k+1... should it be called n-oddgon?). The diagonals of this pentagon define a smaller pentagon P′. Let also γ′ be the circle in which P′ is itself inscribed in. Let also C,C′,S,S′ be the areas of γ,γ′,P,P′ respectively. We have C′C=S′S since all these areas depend only on the radiuses of γ and γ′. Let's now slice the initial cone and the cone K′ generated by the circle γ′ with a general plane (x−x0)cosθ+(z−z0)sinθ=0 (Informally, it's the plane passing through (x0,0,z0) and its normal vector lies in the plane y=0 and makes an angle θ with the x-axis). We get two nested conics. Considering only the cases when these conics are ellipses, we can compute the ratio of their areas (area of the inner ellipse over area of the outer one) and one finds that the maximum of this value occurs when both are circles. When only the inner conic is an ellipse, the ratio is zero. Considering now the "pentagonal cones" obtained by extruding the original nested pentagons along the two cones K and K′ and reasoning in a similar way as before, we have that the ratio between the areas of the nested "pentagonal conics" has a maximum when the pentagons are regular. A problem with this approach is that I don't have a clue of what class of pentagons we are dealing with, as it may be too narrow to be of any real interest for the problem. UPDATE: in the case of a pentagon, since five points in general position determine a conic, I guess we have covered all the possible cases. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Apr 5, 2016 at 5:24 answered Apr 4, 2016 at 7:12 marco trevimarco trevi 3,35811 gold badge2020 silver badges4040 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. The area ratio S′/S is always k2 where k=1−sin(π/10)1+cos(π/5). k is the ratio of the perimeter of the inner pentagon to the outer pentagon. To see this, note that the inner pentagon appears to to be similar the outer pentagon but rotated by 180 degrees. Assume the outer pentagon is inscribed in a unit circle. Label the outer regular pentagon vertices as ABCDE with A at the top. Label the inner pentagon vertices as abcde but starting with a at the bottom. The triangles ACD and Acd are similar. It’s easily seen that the altitude of the Acd triangle is 1−sin(π/10) and that of the ACD triangle is 1+cos(π/5). Thus, the ratio of cd to CD is k. Since cd and CD are sides of regular pentagons, the perimeter ratio must be k and the area ratio must be k2. I leave the proof to others, but I assert that the following are invariant under any affine transformation: parallel lines remain parallel; the ratio of line sub-segments (P,X) and (X,Q) is constant if line segment (P,Q) is intersected at X by another line segment (p,q); similar polygons remain similar; and, as a result, the perimeter and area ratios of similar polygons are constant. The results have been checked with a MATLAB script that generates random affine transformations. Might add that these results should hold for other regular polygons except for the value of k. For example, k=1/3 for a hexagon. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 17, 2017 at 1:13 answered Feb 17, 2017 at 0:55 T L DavisT L Davis 46844 silver badges88 bronze badges 3 1 This is wrong. The ratio is irrational if the large pentagon is regular, and is rational if the vertices of the large pentagon have integer coordinates. – Christian Blatter Commented May 28, 2017 at 13:18 @ChristianBlatter: I guess I misread the question. My answer only applies to affine transformations of a regular pentagon. But, a quick check with MATLAB doesn't seem to bear out your claim about integer coordinates. – T L Davis Commented May 29, 2017 at 16:28 It is obvious that the "small pentagon" of a pentagon with integer coordinates has rational coordinates, and that a pentagon whose vertices have rational coordinates has rational area. – Christian Blatter Commented May 29, 2017 at 18:00 Add a comment \| You must log in to answer this question. 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188302	https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-spring-2015/mit6_042js15_session21.pdf	“mcs” — 2015/5/18 — 1:43 — page 419 — #427 11.9 Connectivity Deﬁnition 11.9.1. Two vertices are connected in a graph when there is a path that begins at one and ends at the other. By convention, every vertex is connected to itself by a path of length zero. A graph is connected when every pair of vertices are connected. 11.9.1 Connected Components Being connected is usually a good property for a graph to have. For example, it could mean that it is possible to get from any node to any other node, or that it is possible to communicate between any pair of nodes, depending on the application. “mcs” — 2015/5/18 — 1:43 — page 420 — #428 420 Chapter 11 Simple Graphs But not all graphs are connected. For example, the graph where nodes represent cities and edges represent highways might be connected for North American cities, but would surely not be connected if you also included cities in Australia. The same is true for communication networks like the internet—in order to be protected from viruses that spread on the internet, some government networks are completely isolated from the internet. Figure 11.16 One graph with 3 connected components. Another example is shown in Figure 11.16, which looks like a picture of three graphs, but is intended to be a picture of one graph. This graph consists of three pieces (subgraphs). Each piece by itself is connected, but there are no paths be-tween vertices in different pieces. These connected pieces of a graph are called its connected components. Deﬁnition 11.9.2. A connected component of a graph is a subgraph consisting of some vertex and every node and edge that is connected to that vertex. So, a graph is connected iff it has exactly one connected component. At the other extreme, the empty graph on n vertices has n connected components. 11.9.2 Odd Cycles and 2-Colorability We have already seen that determining the chromatic number of a graph is a chal-lenging problem. There is one special case where this problem is very easy, namely, when the graph is 2-colorable. Theorem 11.9.3. The following graph properties are equivalent: 1. The graph contains an odd length cycle. 2. The graph is not 2-colorable. 3. The graph contains an odd length closed walk. “mcs” — 2015/5/18 — 1:43 — page 421 — #429 11.9. Connectivity 421 In other words, if a graph has any one of the three properties above, then it has all of the properties. We will show the following implications among these properties: 1. IMPLIES 2. IMPLIES 3. IMPLIES 1: So each of these properties implies the other two, which means they all are equiva-lent. 1 IMPLIES 2 Proof. This follows from equation 11.3. ⌅ 2 IMPLIES 3 If we prove this implication for connected graphs, then it will hold for an arbitrary graph because it will hold for each connected component. So we can assume that G is connected. Proof. Pick an arbitrary vertex r of G. Since G is connected, for every node u 2 V.G/, there will be a walk wu starting at u and ending at r. Assign colors to vertices of G as follows: color.u/ D ( black; if jwuj is even; white; otherwise: Now since G is not colorable, this can’t be a valid coloring. So there must be an edge between two nodes u and v with the same color. But in that case wu reverse.wv/ hv—ui is a closed walk starting and b ending at u, and b its length is jwuj C jwvj C 1 which is odd. ⌅ 3 IMPLIES 1 Proof. Since there is an odd length closed walk, the WOP implies there is an odd length closed walk w of minimum length. We claim w must be a cycle. To show this, assume to the contrary that w is not a cycle, so there is a repeat vertex occurrence besides the start and end. There are then two cases to consider depending on whether the additional repeat is different from, or the same as, the start vertex. In the ﬁrst case, the start vertex has an extra occurrence. That is, w D f b x r “mcs” — 2015/5/18 — 1:43 — page 422 — #430 422 Chapter 11 Simple Graphs for some positive length walks f and r that begin and end at x. Since jwj D jfj C jrj is odd, exactly one of f and r must have odd length, and that one will be an odd length closed walk shorter than w, a contradiction. In the second case, w D f y g y r where f is a walk from x to y for some y ¤ x, and r is a walk from y to x, and jgj > 0. Now g cannot have odd b length b or it would be an odd-length closed walk shorter than w. So g has even length. That implies that fb y r must be an odd-length closed walk shorter than w, again a contradiction. This completes the proof of Theorem 11.9.3. ⌅ Theorem 11.9.3 turns out to be useful, since bipartite graphs come up fairly often in practice. We’ll see examples when we talk about planar graphs in Chapter 12. 11.9.3 k-connected Graphs If we think of a graph as modeling cables in a telephone network, or oil pipelines, or electrical power lines, then we not only want connectivity, but we want connec-tivity that survives component failure. So more generally, we want to deﬁne how strongly two vertices are connected. One measure of connection strength is how many links must fail before connectedness fails. In particular, two vertices are k-edge connected when it takes at least k “edge-failures” to disconnect them. More precisely: Deﬁnition 11.9.4. Two vertices in a graph are k-edge connected when they remain connected in every subgraph obtained by deleting up to k # 1 edges. A graph is k-edge connected when it has more than one vertex, and pair of distinct vertices in the graph are k- connected. Notice that according to Deﬁnition 11.9.4, if a graph is k-connected, it is also j -connected for j k. This convenient convention implies that two vertices are connected according to deﬁnition 11.9.1 iff they are 1-edge connected according to Deﬁnition 11.9.4. From now on we’ll drop the “edge” modiﬁer and just say “k-connected.”9 9There is a corresponding deﬁnition of k-vertex connectedness based on deleting vertices rather than edges. Graph theory texts usually use “k-connected” as shorthand for “k-vertex connected.” But edge-connectedness will be enough for us. “mcs” — 2015/5/18 — 1:43 — page 423 — #431 11.9. Connectivity 423 For example, in the graph in ﬁgure 11.15, vertices c and e are 3-connected, b and e are 2-connected, g and e are 1 connected, and no vertices are 4-connected. The graph as a whole is only 1-connected. A complete graph, kn, is .n # 1/-connected. Every cycle is 2-connected. The idea of a cut edge is a useful way to explain 2-connectivity. Deﬁnition 11.9.5. If two vertices are connected in a graph G, but not connected when an edge e is removed, then e is called a cut edge of G. So a graph with more than one vertex is 2-connected iff it is connected and has no cut edges. The following Lemma is another immediate consequence of the deﬁnition: Lemma 11.9.6. An edge is a cut edge iff it is not on a cycle. More generally, if two vertices are connected by k edge-disjoint paths—that is, no edge occurs in two paths—then they must be k-connected, since at least one edge will have to be removed from each of the paths before they could disconnect. A fundamental fact, whose ingenious proof we omit, is Menger’s theorem which conﬁrms that the converse is also true: if two vertices are k-connected, then there are k edge-disjoint paths connecting them. It takes some ingenuity to prove this just for the case k D 2. 11.9.4 The Minimum Number of Edges in a Connected Graph The following theorem says that a graph with few edges must have many connected components. Theorem 11.9.7. Every graph, G, has at least jV.G/j # jE.G/j connected com-ponents. Of course for Theorem 11.9.7 to be of any use, there must be fewer edges than vertices. Proof. We use induction on the number, k, of edges. Let P.k/ be the proposition that every graph, G, with k edges has at least jV.G/j # k connected com-ponents. Base case (k D 0): In a graph with 0 edges, each vertex is itself a connected component, and so there are exactly jV.G/j D jV.G/j # 0 connected components. So P.0/ holds. “mcs” — 2015/5/18 — 1:43 — page 424 — #432 424 Chapter 11 Simple Graphs Inductive step: Let Ge be the graph that results from removing an edge, e 2 E.G/. So Ge has k edges, and by the induction hypothesis P.k/, we may assume that Ge has at least jV.G/j # k-connected components. Now add back the edge e to obtain the original graph G. If the endpoints of e were in the same connected component of Ge, then G has the same sets of connected vertices as Ge, so G has at least jV.G/j # k > jV.G/j # .k C 1/ components. Alternatively, if the endpoints of e were in different connected components of Ge, then these two components are merged into one component in G, while all other components remain unchanged, so that G has one fewer connected component than Ge. That is, G has at least .jV.G/j # k/ # 1 D jV.G/j # .k C 1/ connected components. So in either case, G has at least jV.G/j # .k C 1/ components, as claimed. This completes the inductive step and hence the entire proof by induction. ⌅ Corollary 11.9.8. Every connected graph with n vertices has at least n # 1 edges. A couple of points about the proof of Theorem 11.9.7 are worth noticing. First, we used induction on the number of edges in the graph. This is very common in proofs involving graphs, as is induction on the number of vertices. When you’re presented with a graph problem, these two approaches should be among the ﬁrst you consider. The second point is more subtle. Notice that in the inductive step, we took an arbitrary .kC1/-edge graph, threw out an edge so that we could apply the induction assumption, and then put the edge back. You’ll see this shrink-down, grow-back process very often in the inductive steps of proofs related to graphs. This might seem like needless effort: why not start with an k-edge graph and add one more to get an .k C 1/-edge graph? That would work ﬁne in this case, but opens the door to a nasty logical error called buildup error, illustrated in Problem 11.48. 11.10 Forests & Trees We’ve already made good use of digraphs without cycles, but simple graphs without cycles are arguably the most important graphs in computer science. 11.10.1 Leaves, Parents & Children Deﬁnition 11.10.1. An acyclic graph is called a forest. A connected acyclic graph is called a tree. “mcs” — 2015/5/18 — 1:43 — page 425 — #433 11.10. Forests & Trees 425 Figure 11.17 A 6-node forest consisting of 2 component trees. b f i d j c e h g Figure 11.18 A 9-node tree with 5 leaves. The graph shown in Figure 11.17 is a forest. Each of its connected components is by deﬁnition a tree. One of the ﬁrst things you will notice about trees is that they tend to have a lot of nodes with degree one. Such nodes are called leaves. Deﬁnition 11.10.2. A degree 1 node in a forest is called a leaf. The forest in Figure 11.17 has 4 leaves. The tree in Figure 11.18 has 5 leaves. Trees are a fundamental data structure in computer science. For example, in-formation is often stored in tree-like data structures, and the execution of many recursive programs can be modeled as the traversal of a tree. In such cases, it is often useful to arrange the nodes in levels, where the node at the top level is iden-tiﬁed as the root and where every edge joins a parent to a child one level below. Figure 11.19 shows the tree of Figure 11.18 redrawn in this way. Node d is a child of node e and the parent of nodes b and c. 11.10.2 Properties Trees have many unique properties. We have listed some of them in the following theorem. Theorem 11.10.3. Every tree has the following properties: 1. Every connected subgraph is a tree. 2. There is a unique path between every pair of vertices. “mcs” — 2015/5/18 — 1:43 — page 426 — #434 426 Chapter 11 Simple Graphs f e h c d g i j b Figure 11.19 The tree from Figure 11.18 redrawn with node e as the root and the other nodes arranged in levels. 3. Adding an edge between nonadjacent nodes in a tree creates a graph with a cycle. 4. Removing any edge disconnects the graph. That is, every edge is a cut edge. 5. If the tree has at least two vertices, then it has at least two leaves. 6. The number of vertices in a tree is one larger than the number of edges. Proof. 1. A cycle in a subgraph is also a cycle in the whole graph, so any sub-graph of an acyclic graph must also be acyclic. If the subgraph is also con-nected, then by deﬁnition, it is a tree. 2. Since a tree is connected, there is at least one path between every pair of ver-tices. Suppose for the purposes of contradiction, that there are two different paths between some pair of vertices. Then there are two distinct paths p ¤ q between the same two vertices with minimum total length jpj C jqj. If these paths shared a vertex, w, other than at the start and end of the paths, then the parts of p and q from start to w, or the parts of p and q from w to the end, must be distinct paths between the same vertices with total length less than jpj C jqj, contradicting the minimality of this sum. Therefore, p and q have no vertices in common besides their endpoints, and so p reverse.q/ is a cycle. 3. An additional edge b hu—vi together with the unique path between u and v forms a cycle. “mcs” — 2015/5/18 — 1:43 — page 427 — #435 11.10. Forests & Trees 427 4. Suppose that we remove edge hu—vi. Since the tree contained a unique path between u and v, that path must have been hu—vi. Therefore, when that edge is removed, no path remains, and so the graph is not connected. 5. Since the tree has at least two vertices, the longest path in the tree will have different endpoints u and v. We claim u is a leaf. This follows because, since by deﬁnition of endpoint, u is incident to at most one edge on the path. Also, if u was incident to an edge not on the path, then the path could be lengthened by adding that edge, contradicting the fact that the path was as long as possible. It follows that u is incident only to a single edge, that is u is a leaf. The same hold for v. 6. We use induction on the proposition P.n/ WWD there are n # 1 edges in any n-vertex tree: Base case (n D 1): P.1/ is true since a tree with 1 node has 0 edges and 1 # 1 D 0. Inductive step: Now suppose that P.n/ is true and consider an .nC1/-vertex tree, T . Let v be a leaf of the tree. You can verify that deleting a vertex of degree 1 (and its incident edge) from any connected graph leaves a connected subgraph. So by Theorem 11.10.3.1, deleting v and its incident edge gives a smaller tree, and this smaller tree has n # 1 edges by induction. If we re-attach the vertex, v, and its incident edge, we ﬁnd that T has n D .nC1/#1 edges. Hence, P.n C 1/ is true, and the induction proof is complete. ⌅ Various subsets of properties in Theorem 11.10.3 provide alternative characteri-zations of trees. For example, Lemma 11.10.4. A graph G is a tree iff G is a forest and jV.G/j D jE.G/j C 1. The proof is an easy consequence of Theorem 11.9.7.6 (Problem 11.55). 11.10.3 Spanning Trees Trees are everywhere. In fact, every connected graph contains a subgraph that is a tree with the same vertices as the graph. This is called a spanning tree for the graph. For example, Figure 11.20 is a connected graph with a spanning tree highlighted. Deﬁnition 11.10.5. Deﬁne a spanning subgraph of a graph, G, to be a subgraph containing all the vertices of G. “mcs” — 2015/5/18 — 1:43 — page 428 — #436 428 Chapter 11 Simple Graphs Figure 11.20 A graph where the edges of a spanning tree have been thickened. Theorem 11.10.6. Every connected graph contains a spanning tree. Proof. Suppose G is a connected graph, so the graph G itself is a connected, span-ning subgraph. So by WOP, G must have a minimum-edge connected, spanning subgraph, T . We claim T is a spanning tree. Since T is a connected, spanning subgraph by deﬁnition, all we have to show is that T is acyclic. But suppose to the contrary that T contained a cycle C. By Lemma 11.9.6, an edge e of C will not be a cut edge, so removing it would leave a connected, spanning subgraph that was smaller than T , contradicting the minimality to T . ⌅ 11.10.4 Minimum Weight Spanning Trees Spanning trees are interesting because they connect all the nodes of a graph using the smallest possible number of edges. For example the spanning tree for the 6-node graph shown in Figure 11.20 has 5 edges. In many applications, there are numerical costs or weights associated with the edges of the graph. For example, suppose the nodes of a graph represent buildings and edges represent connections between them. The cost of a connection may vary a lot from one pair of buildings or towns to another. Another example is where the nodes represent cities and the weight of an edge is the distance between them: the weight of the Los Angeles/New York City edge is much higher than the weight of the NYC/Boston edge. The weight of a graph is simply deﬁned to be the sum of the weights of its edges. For example, the weight of the spanning tree shown in Figure 11.21 is 19. Deﬁnition 11.10.7. A minimum weight spanning tree (MST) of an edge-weighted graph G is a spanning tree of G with the smallest possible sum of edge weights. Is the spanning tree shown in Figure 11.21(a) an MST of the weighted graph shown in Figure 11.21(b)? It actually isn’t, since the tree shown in Figure 11.22 is also a spanning tree of the graph shown in Figure 11.21(b), and this spanning tree has weight 17. “mcs” — 2015/5/18 — 1:43 — page 429 — #437 11.10. Forests & Trees 429 2 4 3 3 4 4 3 3 4 4 2 2 5 2 2 2 8 8 (a) (b) Figure 11.21 A spanning tree (a) with weight 19 for a graph (b). 2 3 3 4 2 2 8 Figure 11.22 An MST with weight 17 for the graph in Figure 11.21(b). “mcs” — 2015/5/18 — 1:43 — page 430 — #438 430 Chapter 11 Simple Graphs What about the tree shown in Figure 11.22? It seems to be an MST, but how do we prove it? In general, how do we ﬁnd an MST for a connected graph G? We could try enumerating all subtrees of G, but that approach would be hopeless for large graphs. There actually are many good ways to ﬁnd MST’s based on a property of some subgraphs of G called pre-MST’s. Deﬁnition 11.10.8. A pre-MST for a graph G is a spanning subgraph of G that is also a subgraph of some MST of G. So a pre-MST will necessarily be a forest. For example, the empty graph with the same vertices as G is guaranteed to be a pre-MST of G, and so is any actual MST of G. If e is an edge of G and S is a spanning subgraph, we’ll write S C e for the spanning subgraph with edges E.S/ [ feg. Deﬁnition 11.10.9. If F is a pre-MST and e is a new edge, that is e 2 E.G/ # E.F /, then e extends F when F C e is also a pre-MST. So being a pre-MST is contrived to be an invariant under addition of extending edges, by the deﬁnition of extension. The standard methods for ﬁnding MST’s all start with the empty spanning forest and build up to an MST by adding one extending edge after another. Since the empty spanning forest is a pre-MST, and being a pre-MST is, by deﬁnition, in-variant under extensions, every forest built in this way will be a pre-MST. But no spanning tree can be a subgraph of a different spanning tree. So when the pre-MST ﬁnally grows enough to become a tree, it will be an MST. By Lemma 11.10.4, this happens after exactly jV.G/j # 1 edge extensions. So the problem of ﬁnding MST’s reduces to the question of how to tell if an edge is an extending edge. Here’s how: Deﬁnition 11.10.10. Let F be a pre-MST, and color the vertices in each connected component of F either all black or all white. At least one component of each color is required. Call this a solid coloring of F . A gray edge of a solid coloring is an edge of G with different colored endpoints. Any path in G from a white vertex to a black vertex obviously must include a gray edge, so for any solid coloring, there is guaranteed to be at least one gray edge. In fact, there will have to be at least as many gray edges as there are components with the same color. Here’s the punchline: Lemma 11.10.11. An edge extends a pre-MST F if it is a minimum weight gray edge in some solid coloring of F . “mcs” — 2015/5/18 — 1:43 — page 431 — #439 11.10. Forests & Trees 431 2 4 3 3 2 2 8 Figure 11.23 A spanning tree found by Algorithm 1. So to extend a pre-MST, choose any solid coloring, ﬁnd the gray edges, and among them choose one with minimum weight. Each of these steps is easy to do, so it is easy to keep extending and arrive at an MST. For example, here are three known algorithms that are explained by Lemma 11.10.11: Algorithm 1. [Prim] Grow a tree one edge at a time by adding a minimum weight edge among the edges that have exactly one endpoint in the tree. This is the algorithm that comes from coloring the growing tree white and all the vertices not in the tree black. Then the gray edges are the ones with exactly one endpoint in the tree. Algorithm 2. [Kruskal] Grow a forest one edge at a time by adding a minimum weight edge among the edges with endpoints in different connected components. An edge does not create a cycle iff it connects different components. The edge chosen by Kruskal’s algorithm will be the minimum weight gray edge when the components it connects are assigned different colors. For example, in the weighted graph we have been considering, we might run Algorithm 1 as follows. Start by choosing one of the weight 1 edges, since this is the smallest weight in the graph. Suppose we chose the weight 1 edge on the bottom of the triangle of weight 1 edges in our graph. This edge is incident to the same vertex as two weight 1 edges, a weight 4 edge, a weight 7 edge, and a weight 3 edge. We would then choose the incident edge of minimum weight. In this case, one of the two weight 1 edges. At this point, we cannot choose the third weight 1 edge: it won’t be gray because its endpoints are both in the tree, and so are both colored white. But we can continue by choosing a weight 2 edge. We might end up with the spanning tree shown in Figure 11.23, which has weight 17, the smallest we’ve seen so far. “mcs” — 2015/5/18 — 1:43 — page 432 — #440 432 Chapter 11 Simple Graphs Now suppose we instead ran Algorithm 2 on our graph. We might again choose the weight 1 edge on the bottom of the triangle of weight 1 edges in our graph. Now, instead of choosing one of the weight 1 edges it touches, we might choose the weight 1 edge on the top of the graph. This edge still has minimum weight, and will be gray if we simply color its endpoints differently, so Algorithm 2 can choose it. We would then choose one of the remaining weight 1 edges. Note that neither causes us to form a cycle. Continuing the algorithm, we could end up with the same spanning tree in Figure 11.23, though this will depend on the tie breaking rules used to choose among gray edges with the same minimum weight. For example, if the weight of every edge in G is one, then all spanning trees are MST’s with weight jV.G/j # 1, and both of these algorithms can arrive at each of these spanning trees by suitable tie-breaking. The coloring that explains Algorithm 1 also justiﬁes a more ﬂexible algorithm which has Algorithm 1 as a special case: Algorithm 3. Grow a forest one edge at a time by picking any component and adding a minimum weight edge among the edges leaving that component. This algorithm allows components that are not too close to grow in parallel and independently, which is great for “distributed” computation where separate proces-sors share the work with limited communication between processors. These are examples of greedy approaches to optimization. Sometimes greediness works and sometimes it doesn’t. The good news is that it does work to ﬁnd the MST. Therefore, we can be sure that the MST for our example graph has weight 17, since it was produced by Algorithm 2. Furthermore we have a fast algorithm for ﬁnding a minimum weight spanning tree for any graph. Ok, to wrap up this story, all that’s left is the proof that minimal gray edges are extending edges. This might sound like a chore, but it just uses the same reasoning we used to be sure there would be a gray edge when you need it. Proof. (of Lemma 11.10.11) Let F be a pre-MST that is a subgraph of some MST M of G, and suppose e is a minimum weight gray edge under some solid coloring of F . We want to show that F C e is also a pre-MST. If e happens to be an edge of M, then F C e remains a subgraph of M, and so is a pre-MST. The other case is when e is not an edge of M. In that case, M C e will be a connected, spanning subgraph. Also M has a path p between the different colored endpoints of e, so M C e has a cycle consisting of e together with p. Now p has both a black endpoint and a white one, so it must contain some gray edge g ¤ e. The trick is to remove g from M C e to obtain a subgraph M C e # g. Since gray “mcs” — 2015/5/18 — 1:43 — page 433 — #441 11.11. References 433 edges by deﬁnition are not edges of F , the graph M C e # g contains F C e. We claim that M C e # g is an MST, which proves the claim that e extends F . To prove this claim, note that M Ce is a connected, spanning subgraph, and g is on a cycle of M C e, so by Lemma 11.9.6, removing g won’t disconnect anything. Therefore, M Ce#g is still a connected, spanning subgraph. Moreover, M Ce#g has the same number of edges as M, so Lemma 11.10.4 implies that it must be a spanning tree. Finally, since e is minimum weight among gray edges, w.M C e # g/ D w.M/ C w.e/ # w.g/ w.M/: This means that M C e # g is a spanning tree whose weight is at most that of an MST, which implies that M C e # g is also an MST. ⌅ Another interesting fact falls out of the proof of Lemma 11.10.11: Corollary 11.10.12. If all edges in a weighted graph have distinct weights, then the graph has a unique MST. The proof of Corollary 11.10.12 is left to Problem 11.70. MIT OpenCourseWare 6.042J / 18.062J Mathematics for Computer Science Spring 2015 For information about citing these materials or our Terms of Use, visit:
188303	https://www.theproductionacademy.com/blog/ohms-law	Ohm’s Law + Wattage — The Production Academy Sign up Calculating power needs is critical in production. Written by Scott Adamson Last week, we covered three foundational topics of electricity — current, voltage, and resistance. Once we understand what these terms mean, it’s equally important to understand how they interact with each other. This relationship is called Ohm’s Law. It was defined in the early 19th-century by German physicist and mathematician Georg Ohm and demonstrates how changing one of these aspects within our electrical system will affect the others. Ohm’s Law is a formula used to calculate the relationship between voltage, which in electrical terms is represented by the letter E, referencing an old-school term electromotive force, current, which is represented by the letter I, representing intensity, and resistance, which is represented by the letter R or by the Greek letter Omega (Ω). For our purposes, we’ll use V for voltage, A for current (amperes), and Ω for resistance. Ohm's law states: voltage = current • resistance. This relationship can be represented as a triangle. In this triangle, voltage = current • resistance. We can also see that current = voltage ÷ resistance. And we can calculate resistance, which is voltage ÷ current. When we talk about more complicated electronic circuits, this equation is particularly important because we’ll need to calculate these values along any point in the circuit. But for our purposes as live sound engineers, we’re talking about power distribution, which means we’re looking at fixed voltages — either 120V in the United States or 240V in Europe. Plus, when we’re on stage, we’re not often calculating electrical resistance. So while this doesn’t really come into play in our day-to-day workflow, it’s still important to have a solid understanding of how these topics work together. Get real-world live sound mixing tips straight to your inbox. Another important topic in electricity is Wattage. This is how we measure the full electrical power harnessed by a circuit, measured in Watts. Wattage is found with a simple equation: Watts = voltage • current, or W = VA. When we’re talking about audio, most of the power we need will be for amplifiers. This is what takes an audio signal and boosts it so it has enough power to drive speakers. Every amplifier has a power rating, measured in Watts. This is the amount of power it sends to the speaker at full volume, but this doesn’t mean that the amplifier is going to need that much power from our electrical system. By knowing how much current each of our production devices draw, we can figure out what can fit on each circuit and how much power we need over to make the show happen. Learn how to use this equation to calculate power needs in our Show Power course. Joel BlairOhms, wattage, power, electricity
188304	https://alevels.ai/revise/computer-science-ocr/notes/boolean-algebra/karnaugh-maps	Sign In - A-levels AI Sign in to Alevels.ai Welcome back! Please sign in to continue AppleGoogle Don’t have an account?Sign up
188305	https://www.learnelectronicsindia.com/post/power-factor-and-its-importance?srsltid=AfmBOoo-5W-5OuA5GwOo2SWVV0ckvi_bm2axsiMdzmJ-Bsh1wM2vxuzU	Power factor and its importance top of page A to Z Platforms for Electronics Engineers Sign in Home Buy Electronics Components Reels Order Electronics Projects Services Order Customised Projects PCB Design Buy Project Codes Buy Circuit Diagram Courses Course KITs Training & Internships Career Guidance Find Jobs Interview Q&A Job Guidance Mock Interview Free Courses & Tutorials Blogs Order Electronic Projects Power factor and its importance Kiranmai Chigurupati Jun 28, 2021 5 min read Updated: Jan 3 Rated NaN out of 5 stars. What is power factor? Power factor (PF) is defined as the ratio of real power to apparent power. Real power is measured in kW and apparent power is measured in kVA. Apparent power is also known as Demand. The power factor is a measure of how efficiently you are using electricity. The beer analogy To understand the concept of the power factor, let's see the beer analogy. As you can see, in the beer mug above some of the bear is getting wasted due to foam spill. Liquid beer -This represents the active power or real power or working power which is measured in kW. Foam - This represents the reactive power that is measured in kVAR. Mug -Mug as the whole represents the apparent power that is measured in kVA. If a circuit is 100 % efficient, demand would be equal to the real power available. A strain will be placed on the utility system when demand is greater than the power available. Demand is calculated by taking the average load during a certain time interval. The lower the power factor, higher the current and hence the operating cost. Power factor calculation Power factor is the cosine of the angle between voltage and current. S = P + jQ ; where S - Demand / Apparent power P - Real power Q - Reactive power Power factor - (P/S) 100 In the case of DC circuits, both inductive and capacitive reactance is zero because of zero frequency. Thus the power factor of DC circuits is always unity. In the case of AC circuits, power factors exist since the frequency is not zero. P = VI ( In DC circuits) P = VI COSØ( In AC circuits) PF = COSØ Resistive load: For pure resistive circuits, Q = 0 S = P S/P = 1 Thus unity power factor. Phasor diagram of R - Load Power factor phasor diagram of R-load Ø = 0 ; PF = COSØ = 1 Inductive load: In the case of Inductive loads (motors, generators, transformers - any load that has windings has inductance), the inductance of the windings does not allow the sudden change in current, so the current will lag behind the applied voltage and the power factor in such circuits is lagging power factor. Phasor diagram of L- Load Power factor phasor diagram of L-load P = VI Cos (90) = 0 PF = Cos (90) = 0 Thus power factor of the pure inductive circuit is zero. Capacitive load: In the case of capacitive loads, the capacitance does not allow the sudden change in voltage. Thus, the current leads the capacitor voltage. The power factor in such circuits is the leading power factor. Phasor diagram of C - Load Power factor phasor diagram of C-load P = VI Cos(-90) = 0 PF = Cos(-90) = 0 Thus power factor of the pure capacitive circuit is zero. Similarly power factor can be calculated for different types of loads viz., RL, RC, LC, RLC. Importance of power factor: Power factor is very crucial for economic operation and quality transmission of the power system. The current required to deliver the same power increases with a decrease in power factor. For example, Pure resistive load (unity PF) of 50A becomes 100A for a power factor of 0.5 (for same load and voltage). Poor factor indicates the inefficient usage of available power Low power factor results in increased cross-section and thus the equipment size. Reduction in the amount of available useful power. Lower power factor causes heat damage to insulation and other equipment. Hence greater the power factor, the higher the efficiency, the lower the losses, and the lower the operating cost. Every utility aims for a higher power factor since higher efficiency indicates better utilization of generating and transmission systems. Greater power factor also indicates lower current -> lesser cross-section -> lesser resistance -> lesser I^2R losses and thus indicates long-lasting equipment with low operating cost and high efficiency. Power factor correction: To overcome all the above-mentioned issues, it is very important to correct the power factor (i.e., making sure that the power factor is closer to unity, usually greater than 0.95). Linear loads such as induction motors and transformers with low power factors can make use of passive components such as capacitors or inductors. Non-linear loads such as arc furnaces and rectifiers distort the current drawn from the system. In such cases, active or passive power factor correction can be used to counteract the distortion and raise the power factor. A higher power factor is often desirable since it improves the voltage regulation at the load. The apparent power can be reduced by installing reactive elements which supply or absorb reactive power near the load. For example, the inductive effect of motors can be diluted by installing capacitor banks and vice versa. Benefits of power factor correction: Voltage regulation is improved. I^2R losses are reduced. which will increase the life span of insulation and other components. Electricity charges are reduced. Rating of the cables is reduced. Voltage dip problems are minimized. Reactive power penalties paid by industries for operating with lower power factors will be reduced. Further, you can learn in detail about How to calculate power factor. See also: 5G Technology in India! How to interface a DC motor with Arduino UNO using TinkerCAD? Controlling stepper motor with Joystick Sensor and Arduino Content Written by- Name - Kiranmai Chigurupati Portfolio Link - click here Order Electronics Projects Want us to guide you through your project or make the project for you? 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188306	https://byuistats.github.io/BYUI_M221_Book/Lesson10.html	Math 221 Home Unit 1 Lesson 1: Course Intro & Probability Campus Student Version Online Student Version World Wide Block Student Version Lesson 2: The Statistical Process Lesson 3: Describing Quantitative Data (Shape & Center) Lesson 4: Describing Quantitative Data (Spread) Lesson 5: Normal Distributions Lesson 6: The Central Limit Theorem Lesson 7: Calculating Probabilities involving the Sample Mean Lesson 8: Unit 1 Review Unit 2 Lesson 9: Hypothesis Test for One Mean (Sigma Known) Lesson 10: Confidence Interval for One Mean (Sigma Known) Lesson 11: Inference for One Mean (Sigma Unknown) Lesson 12: Inference for the Mean of Differences (Two Dependent Samples) Lesson 13: Inference for the Difference of Means (Two Independent Samples) Lesson 14: Inference using ANOVA (Several Independent Samples) Lesson 15: Unit 2 Review Unit 3 Lesson 16: Describing Categorical Data Lesson 17: Inference for One Proportion Lesson 18: Inference for Two Proportions Lesson 19: Inference for Independence of Categorical Data Lesson 20: Unit 3 Review Unit 4 Lesson 21: Describing Bivariate Data Lesson 22: Simple Linear Regression Lesson 23: Inference for Bivariate Data Lesson 24: Unit 4 Review Toolbox Data Lesson 10: Inference for One Mean with Sigma Known (Confidence Interval) Lesson Outcomes Political Polls Background Point Estimators Interval Estimators The Margin of Error Putting It All Together: Confidence Intervals for (\mu) when (\sigma) is Known A Little More Precision Factors Affecting the Width of the Confidence Interval Interpretation of Confidence Intervals Requirements Example: Costs of CABG Surgery Sample Size Calculations Rounding Up Example: Grade Inflation Comments Summary Navigation Optional Videos for this Lesson Part 1 Part 2 Part 3 Part 4 Part 5 Part 6 Lesson Outcomes By the end of this lesson, you should be able to: Recognize when a one mean (sigma known) confidence interval is appropriate Calculate the sample size required to achieve a specified margin of error and level of confidence Explain the meaning of a level of confidence Create a confidence interval for a single mean with σ known using the following steps: Find the point estimate Calculate the margin of error for the given level of confidence Calculate a confidence interval from the point estimate and the margin of error Interpret the confidence interval Check the requirements for the confidence interval Explain how the margin of error is affected by the sample size and level of confidence Political Polls During an election in the United States, many polls are conducted to determine the attitudes of likely voters. Poll results are usually reported as percentages. For example, a poll might state that 49% favor the Republican candidate and 51% favor the Democratic candidate. Polls always include a margin of error. The margin of error gives an estimate of the variability in the responses. A common value for the margin of error in political polls is 3%. When we consider the margin of error, we estimate that the true proportion of people who favor the Republican candidate is 49% ± 3%, or in other words, between 46% and 52%. It is not obvious whether the Republican candidate is favored by more or less than 50% of the voters. In this case, the political race is too close to know who might win. In this reading, we will explore the margin of error and its role in estimating a parameter. Background Point Estimators We have learned about several statistics. Remember, a statistic is any number computed based on data. The sample statistics we have discussed are used to estimate population parameters. \| \| Sample Statistic \| Population Parameter \| --- \| \| (\bar x) \| (\mu) \| \| Standard Deviation \| (s) \| (\sigma) \| \| Variance \| (s^2) \| (\sigma^2) \| \| (\vdots) \| (\vdots) \| (\vdots) \| The statistics above are called point estimators because they are just one number (one point on a number line) that is used to estimate a parameter. Parameters are generally unknown values. Think about the mean. If (\mu) is unknown, how do we know if (\bar x) is close to it? The short answer is that we will never know for sure if (\bar x) is close to (\mu). This does not mean that we are helpless. The laws of probability and the normal distribution provide a way for us to create a range of plausible values for a parameter (e.g. (\mu)) based on a statistic (e.g. (\bar x)). Interval Estimators A point estimator gives one specific value as an estimate of a parameter. An interval estimator is a range of plausible values for a parameter. We can create an interval estimate by starting with a point estimate and adding or subtracting the margin of error. In the political poll mentioned above, the point estimate for the support of the Republican candidate is 49%. By adding and subtracting the margin of error, we get the interval estimate: 46% to 52%. A confidence interval is a commonly used interval estimator. In this reading, we will explore how to create a confidence interval for the mean when (\sigma) is known. The Margin of Error Properties of Bell-shaped Curves The following questions will help you review your understanding of the normal distribution. Answer the following questions: Fill in the blank in the following sentence.“The 68-95-99.7% rule only applies for distributions that are _________.” Show/Hide Solution) The 68-95-99.7% rule only applies for distributions that are bell-shaped. In the past, some students have answered that the data must be normally distributed. Actually, the 68-95-99.7% rule will work for any distribution that is mound-shaped and symmetrical. As long as the data are unimodal and not skewed left or skewed right, this rule works well. Approximately what percentage of data from a bell-shaped distribution will lie within two standard deviations of the mean? Show/Hide Solution) Using the 68-95-99.7% rule, about 95% of the observations will lie within two standard deviations of the mean. The Distribution of the Sample Mean We learned in the reading Distribution of Sample Means & The Central Limit Theorem about the characteristics of the sample mean, (\bar x). Specifically, if the population from which the data are drawn is (i) approximately normal or (ii) if the sample size is large, then (\bar x) will be approximately normally distributed. Furthermore, if the original population has mean (\mu) and standard deviation (\sigma), then the sampling distribution of (\bar x) will have mean (\mu) and standard deviation (\sigma/\sqrt{n}). So, if either condition (i) or (ii) is met, then we can consider the sample mean (\bar x) as a normal random variable with mean (\mu) and standard deviation (\sigma/\sqrt{n}). According to the 68-95-99.7% rule for symmetric bell-shaped distributions, about 95% of the time, the sample mean ((\bar x)) will lie within two standard deviations of the population mean ((\mu)). This is an important concept. Make sure that you understand the logic above before you continue reading. How Far is (\bar x) from (\mu), or in other words, How Far is (\mu) from (\bar x)? Assuming that (\bar x) is approximately normally distributed, about 95% of the time, it will be within two standard deviations of (\mu). Remember, the standard deviation of (\bar x) is (\frac{\sigma}{\sqrt{n}}). For the variable (\bar x), two standard deviations would be equal to (2 \frac{\sigma}{\sqrt{n}}). If we collect a random sample from a population and (\bar x) is normally distributed, then about 95% of the time the sample mean (\bar x) will be less than (2 \frac{\sigma}{\sqrt{n}}) units away from the population mean (\mu). Notice that this is true, whether or not we know (\mu). We can express this as a probability statement: [ P\left( \mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95 ] Here is the magic: If (\bar x) is within 2 standard deviations of (\mu), then (\mu) is within 2 standard deviations of (\bar x). It may seem silly to state it, but this is very important. Click here if you love math) Starting with [\mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}}] We subtract (\mu) from all parts of the inequality: [\mu - 2 \frac{\sigma}{\sqrt{n}} - \mu < \bar X - \mu < \mu + 2 \frac{\sigma}{\sqrt{n}} - \mu] Which reduces to [- 2 \frac{\sigma}{\sqrt{n}} < \bar X - \mu < 2 \frac{\sigma}{\sqrt{n}}] Now, subtract (\bar X) from all the terms: [- 2 \frac{\sigma}{\sqrt{n}} - \bar X < \bar X - \mu -\bar X < 2 \frac{\sigma}{\sqrt{n}} -\bar X] And this simplifies to [- 2 \frac{\sigma}{\sqrt{n}}-\bar X < -\mu < 2 \frac{\sigma}{\sqrt{n}}-\bar X] Multiplying all the terms by (-1), we get [2 \frac{\sigma}{\sqrt{n}}+\bar X > \mu > -2 \frac{\sigma}{\sqrt{n}}+\bar X] Rewriting this reversing the order of the terms, we get [\bar X - 2 \frac{\sigma}{\sqrt{n}} < \mu < \bar X + 2 \frac{\sigma}{\sqrt{n}}] So, the statement [ P\left( \mu - 2 \frac{\sigma}{\sqrt{n}} < \bar X < \mu + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95 ] is equivalent to the statement [ P\left( \bar X - 2 \frac{\sigma}{\sqrt{n}} < \mu < \bar X + 2 \frac{\sigma}{\sqrt{n}} \right) \approx 0.95 ] Putting It All Together: Confidence Intervals for (\mu) when (\sigma) is Known Notice that if we know (\sigma), then with approximately 95% confidence, (\mu) will be between the following two values: [\left( \bar X - 2 \frac{\sigma}{\sqrt{n}}, ~ \bar X + 2 \frac{\sigma}{\sqrt{n}} \right)] This equation gives a confidence interval for (\mu). A confidence interval is actually a point estimate (\left( \bar x \right)) plus or minus the margin of error (\left( 2 \frac{\sigma}{\sqrt{n}} \right)). We use the letter (m) to denote the margin of error: [m = 2 \frac{\sigma}{\sqrt{n}}] Using this definition for (m), our confidence interval can be written as [( \bar x - m, ~ \bar x + m )] or [\bar x \pm m] This pattern will be repeated throughout the course as we create confidence intervals of the form: [ \left( \begin{align} point~ ~ ~ & & margin~ & & point~ ~ ~ & & margin~ \ estimate & ~ ~- & of~error & ~ ~, & estimate & ~ ~+ & of~error \end{align} \right) ] or [(\text{point estimate}) \pm (\text{margin of error})] A Little More Precision What (z) Corresponds to 95% of the Area? The 68-95-99.7% Rule for bell-shaped distribution is just a quick approximation. This is useful for estimation, but more precision is usually required. Answer the following question: For a standard normal distribution, between what two (z)-scores will 95% of the data fall? In other words, find the values (-z) and (z) such that the area between them is equal to 0.95. Use the Normal Probability Applet. Click here for a hint) To find the value of (z^), given a specific confidence level, do the following: Open the applet at Normal Probability Applet Shade the area between the two red lines (the middle portion of the graph) only and enter the decimal value for the desired percentage (e.g. 0.95 for 95%) in the Area box. The value of (z^) will be displayed below. Two numbers will be given along the horizontal axis. The only difference in these numbers is their sign: one is positive and one is negative. The positive value is the desired (z^). Show/Hide Solution) 95% of the area under the standard normal curve is between (z=-1.96) and (z=1.96) For normally distributed data, 95% of the observations will fall within 1.96 standard deviations of the mean. The actual formula for a 95% confidence interval for (\mu) (when (\sigma) is known) is: [ \left( \bar x - 1.96 \frac{\sigma}{\sqrt{n}}, \bar x + 1.96 \frac{\sigma}{\sqrt{n}} \right) ] Please notice that the number 2 was used as an approximation for the actual value of 1.96. When computing a 95% confidence interval for a mean with (\sigma) known, please use 1.96 in the equation. Worked Example: Rolling a Die 25 Times We will compute a 95% confidence interval for the mean of (n=25) rolls of a fair die. A die was rolled 25 times. The values rolled were: [ 3 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 3 ~ ~ ~ 4 ~ ~ ~ 2 ~ ~ ~ 2 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 5 ] The mean of these values is (\bar x = 3.04). It is a fact that for the outcome of a six-sided die, (\sigma=\sqrt{\frac{35}{12}} \approx 1.7078). Applying the formula for a 95% confidence interval, [ \left( \bar x - z^ \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^ \frac{\sigma}{\sqrt{n}} \right) ] we get: [ \left( 3.04 - 1.96 \frac{1.7078}{\sqrt{25}}, ~ 3.04 + 1.96 \frac{1.7078}{\sqrt{25}} \right) ] or [ (2.37, ~ 3.71) ] What about Other Confidence Levels? Most of the time, researchers report 95% confidence intervals. The number 95% is called the confidence level. Sometimes it is desirable to use a level of confidence that is different than 95%. In that case, we need to use a number besides 1.96 in the calculations. Suppose we want to create a 90% confidence interval. What value would we put in the blanks in the confidence interval formula below? [ \left( \bar x - \underline{~ ~ ~ ? ~ ~ ~} \frac{\sigma}{\sqrt{n}}, ~ \bar x + \underline{~ ~ ~ ? ~ ~ ~} \frac{\sigma}{\sqrt{n}} \right) ] Another way to state this question is to ask, between what two (z)-scores will 90% of the data in a standard normal distribution fall? We need to find the values of (-z) and (z) such that the area between them is equal to 0.90. Again, we use the Normal Probability Applet. So, for a 90% confidence interval, the value of (z) would be 1.645. Note: We use an asterisk (()) to indicate that (z) was not computed from data, but was determined based on a chosen confidence level, 90%. The formula for a 90% confidence interval for a mean when (\sigma) is known is: [ \left( \bar x - 1.645 \frac{\sigma}{\sqrt{n}}, ~ \bar x + 1.645 \frac{\sigma}{\sqrt{n}} \right) ] When (\sigma) is known, we use (z^ = 1.96) to create 95% confidence intervals, and we use (z^ = 1.645) to create 90% confidence intervals. Formula for the Confidence Interval for (\mu) ((\sigma) Known) With this notation, the confidence interval formula generalizes to: [ \left( \bar x - z^ \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^ \frac{\sigma}{\sqrt{n}} \right) ] where (z^) is determined by the level of confidence. If you want a 90% confidence interval, then (z^) is the number of standard deviations on either side of the mean that you must go to capture 90% of the data. Answer the following questions: If (\bar x) follows a normal distribution and (\sigma) is known, what is the equation for a 99% confidence interval for the true population mean? Show/Hide Solution) For a 99% confidence level, (z^ = 2.576). The equation for the 99% confidence interval becomes: [ \left( \bar x - 2.576 \frac{\sigma}{\sqrt{n}}, ~ \bar x + 2.576 \frac{\sigma}{\sqrt{n}} \right) ] Apply the equation from the previous problem to find a 99% confidence interval for the mean value rolled on the die. Use the data given above. For convenience, they are reproduced here: [ 3 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 3 ~ ~ ~ 4 ~ ~ ~ 2 ~ ~ ~ 2 ~ ~ ~ 5 ~ ~ ~ 6 ~ ~ ~ 2 ~ ~ ~ 6 ~ ~ ~ 4 ~ ~ ~ 1 ~ ~ ~ 1 ~ ~ ~ 5 ] Show/Hide Solution) If we were making a 99% confidence interval, we follow the same procedure as for a 95% confidence interval, only we use (z^=2.576). The 99% confidence interval for the true mean is: [ \left( 3.04 - 2.576 \frac{1.7078}{\sqrt{25}}, ~ 3.04 + 2.576 \frac{1.7078}{\sqrt{25}} \right) ] which simplifies to: [ (2.16, ~ 3.92) ] The most commonly used values for (z^) are: \| ConfidenceLevel \| (z^) \| --- \| \| 90% \| 1.645 \| \| 95% \| 1.960 \| \| 99% \| 2.576 \| \| \| \| Any other values that you need can be determined using the Normal Probability Applet. Factors Affecting the Width of the Confidence Interval Recall that the formula to compute the confidence interval for a mean, where (\sigma) is known is: [ \left( \bar x - z^ \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^ \frac{\sigma}{\sqrt{n}} \right) ] Answer the following questions: What would happen to the confidence interval if the sample size (n) was increased, but the other values were still the same? Show/Hide Solution) Since the sample size (n) is in the denominator, if (n) is increased, (\sqrt{n}) would increase, and the fraction (\frac{\sigma}{\sqrt{n}}) would decrease, which would lead to a smaller margin of error. In other words, if the sample size is increased, the width of a confidence interval will decrease–it will become narrower. What would happen to the confidence interval if the confidence level were increased, say from 95% to 99%? Show/Hide Solution) Increasing the confidence level will lead to larger values of (z^). If (z^) is increased, then the margin of error (z^ \frac{\sigma}{\sqrt{n}}) would increase. This would lead to a wider confidence interval. Interpretation of Confidence Intervals How do we interpret confidence intervals? What do they really mean? (Image source: Flickr/ICMA Photos) Consider a coin with two sides: one called “heads” and the other called “tails”. Imagine that you flipped this coin, but you have not looked at it yet. What is the probability that the coin shows heads? Strangely enough, the answer is, it depends! If the head is facing up, then the probability that the coin shows heads is 1. If the head is facing down, then the probability that it shows heads is 0. The coin has been tossed. There is no randomness left in the process. So, either the head is facing up or it is facing down. So, the probability that the coin shows heads is either 1 or 0. (We just don’t know which.) The fact that we do not know the outcome does not change it or make it random. Before we toss the coin, the probability that the coin will show heads is (\frac{1}{2}=0.5). After we toss the coin, the probability that we get heads is 1 or 0. Transferring this reasoning to confidence intervals, we get a similar result. Once we have collected data on something, there is no randomness in the system. Any confidence interval that is created using that data will either contain the true parameter ((\mu)) or it will not. After collecting data, the probability that a specific confidence interval will contain (\mu) is either 1 or 0. The correct interpretation of a 95% confidence interval is to say, “We are 95% confident that the true mean lies within the lower and upper bounds of the confidence interval.” Consider the 95% confidence interval for the true mean of 25 rolls of a fair die. We found the 95% confidence interval to be: ((2.37, 3.71)). When we interpret this confidence interval, we say, “We are 95% confident that the true mean is between 2.37 and 3.71.” The word, “confident” implies that if we repeated this process many, many times, 95% of the confidence intervals we would get would contain the true mean (\mu). It does not imply anything about whether or not one specific confidence interval will contain the true mean. We do not say that “there is a 95% probability (or chance) that the true mean is between 2.37 and 3.71.” The probability that the true mean (\mu) is between 2.37 and 3.71 is either 1 or 0. Requirements There are three requirements that need to be checked when computing a confidence interval for a mean with (\sigma) known: A simple random sample was drawn from the population (\bar x) is normally distributed (\sigma) is assumed to be known The requirement of normality is satisfied if (a) the raw data are normally distributed or (b) the sample size is large. This procedure is robust to moderate departures from normality. Even if the requirement that (\bar x) is normally distributed is not satisfied perfectly, it is usually okay to conduct the test. In practice, we never really know (\sigma). This procedure is primarily used to help you understand the idea of confidence intervals. When (\sigma) is unknown, we use a slightly different computation. Example: Costs of CABG Surgery (Photo credit: Vanderbilt Photo/Neil Brake) This reading focuses on an important aspect of designing a study: determining the sample size. It is important for health care administrators to know the mean hospital costs for patients who have coronary artery bypass graft (CABG) surgery. A large hospital is planning a study to determine their mean costs for patients who have CABG surgery. A study will be conducted in which the charts of patients who had CABG surgery will be sampled, and their hospital costs will be recorded. For budgetary reasons, the hospital administrators do not want to collect a sample that is too large. However, if the sample size is not large enough, the confidence interval will be too wide to be useful as a planning tool. After a discussion among the senior administration, they have determined that they want to estimate the mean hospital costs of CABG surgery within $2000 (i.e., plus or minus $2000.) In other words, they want the confidence interval for the true mean to have a margin of error of $2000 dollars. Recall the equation for the confidence interval is: [ \left( \bar x - z^ \frac{\sigma}{\sqrt{n}}, \bar x + z^ \frac{\sigma}{\sqrt{n}} \right) ] The part of the equation that is added to and subtracted from (\bar x) is called the margin of error. We will denote the margin of error by the letter (m). [ m=z^ \frac{\sigma}{\sqrt{n}} ] To use this formula, the parameter (\sigma) must be given to us. It is the true standard deviation of the data you are observing. If you do not know (\sigma), you can estimate it using the standard deviation reported in a previous study or by conducting a pilot study. A study published by another hospital reported that the standard deviation of the costs for CABG surgery was $28,705. In the following questions, you will compute the margin of error, (m), for a future study of the hospital costs for CABG surgery. The hospital administrators want to use a 95% level of confidence. Assume the standard deviation can be estimated to be (\sigma = \$28,705). Answer the following questions: If the hospital administrators want to be 95% confident in the results, what should the value of (z^) be? Show/Hide Solution) [ \begin{align} z^=1.96 \end{align} ] If the hospital collects a sample of (n=100) patients’ costs, what would the margin of error be? Show/Hide Solution) [ \begin{array}{1cl} m &= z^ \frac{\sigma}{\sqrt{n}} \ &= 1.96 \cdot \frac{28705}{\sqrt{100}} \ &= \$5626.18 \end{array} ] If the hospital collects a sample of (n=1000) patients’ costs, what would the margin of error be? Show/Hide Solution) [ \begin{array}{1cl} m &= z^ \frac{\sigma}{\sqrt{n}} \ &= 1.96 \cdot \frac{28705}{\sqrt{1000}} \ &= \$1779.15 \end{array} ] Choose any other value for the sample size, (n). Find the margin of error for your sample size. Show/Hide Solution) Answers will vary. Repeat Question 11 until you find the sample size that will yield a margin of error as close to $2000 as possible, without going over. Show/Hide Solution) [ n = 792 ] Sample Size Calculations The process you followed in Questions 8 - 12 is effective, but tedious. There must be an easier way! The trick is to solve the margin of error equation (m = z^ \frac{\sigma}{\sqrt{n}}) for (n). Click here if you love math) [m = z^ \frac{\sigma}{\sqrt{n}}] Multiply both sides of the equation by (\sqrt{n}): [m \cdot \sqrt{n} = z^ \frac{\sigma}{\sqrt{n}} \cdot \sqrt{n}] Which reduces to: [m \cdot \sqrt{n} = z^ \cdot \sigma] Divide both sides of the equation by (m). Cancelling the (m)’s on the left hand side: [\sqrt{n} = \frac{z^ \cdot \sigma}{m}] Squaring both sides: [\left( \sqrt{n} \right)^2 = \left( \frac{z^ \cdot \sigma}{m} \right)^2] Which simplifies to the desired result: [n = \left( \frac{z^ \cdot \sigma}{m} \right)^2] This gives us the sample size formula, which tells the number of observations required in order to obtain a specified margin of error: [ n = \left( \frac{z^\sigma}{m} \right)^2 ] Once the level of confidence is selected, (z^) is automatically determined. In practice, the most common level of confidence is 95%, which means that (z^) would equal 1.96. If you need a reminder on finding the value of (z^), click here. Answer the following question: In Question 12, you used the process of guess-and-check to find the sample size. For this question, use the sample size formula to compute the sample size required to estimate the mean cost of CABG surgery, (\mu), within $2000 with 95% confidence. Recall that in a previous study, the standard deviation was found to be (\sigma = \$28,705). Show/Hide Solution) [ \begin{align} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{1.96 \cdot 28705}{2000} \right)^2 \ &= 791.3 \end{align} ] Rounding Up In Question 13, you computed the sample size required to get a margin of error of $2000. Notice that the result was 791.3, which is not a whole number. What does that mean? Does that suggest that you will survey only a fraction of the last patient’s costs? Of course not! When doing sample size calculations, if the answer is not a whole number, you always round up to the next highest whole number. This will allow you to get a sample size that is large enough to attain your desired margin of error. If you want to estimate the mean cost of CABG surgery within $2000 with 95% confidence, you will need to survey the files of 792 patients. When doing sample size calculations, you always round up. Answer the following questions: Answer this question without doing any computations. If the hospital administrators wanted to estimate the mean cost of CABG surgery with a margin of error of $1000 at the 95% confidence level, should the sample size be larger or smaller than (n=792)? Show/Hide Solution) In question 5, we determined that the sample size required to have a margin of error of $2000 is 792.The margin of error is in the denominator of the sample size formula. If we want to reduce the margin of error, we need to increase the sample size. The sample size should be larger than (n=792). Find the sample size required to estimate the mean cost of CABG surgery with a margin of error of $1000 and with a 95% confidence level. Show/Hide Solution) [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{1.96 \cdot 28705}{1000} \right)^2 \ &= 3165.4 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=3166). What happened to the sample size required when the margin of error is cut in half from $2000 to $1000? Show/Hide Solution) If we divide the two sample sizes, we get: [ \frac{3166}{792} \approx 4 ] In order to reduce the margin of error by half, we need to quadruple the sample size. Remember the sample size required to have a margin of error of $2000 at the 95% level of confidence is (n=792). If we wanted to estimate the mean cost with the same margin of error at the 99% level of confidence, would the sample size be larger or smaller? Show/Hide Solution) The value of (z^) would increase from 1.96 to 2.576, which would increase the required sample size. Find the sample size required to estimate the mean cost of CABG surgery with a margin of error of $2000 and at the 99% confidence level. Show/Hide Solution) [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{2.576 \cdot 28705}{2000} \right)^2 \ &= 1366.9 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=1367). If we increase the confidence level, what happens to the required sample size? Show/Hide Solution) If we increase the confidence level the required sample size also increases. Example: Grade Inflation The administration at BYU-Idaho is concerned about the possibility of grade inflation in their courses. Grades in a course are coded on a scale from 0 to 4, as given in the following table: \| A \| A- \| B+ \| B \| B- \| C+ \| C \| C- \| D+ \| D \| D- \| F \| \| 4 \| 3.7 \| 3.3 \| 3 \| 2.7 \| 2.3 \| 2 \| 1.7 \| 1.3 \| 1 \| 0.7 \| 0 \| Historical records indicate that the standard deviation of the numerical value for the grades students’ receive in a course is (\sigma = 0.68). Due to the confidential nature of grades, the administration will not release the current value of the mean grade earned on campus. An old reference indicates that the true mean value of the grades earned at BYU-Idaho students is 3.15. It would be interesting to estimate the current mean grade point average at BYU-Idaho. You have been asked to help study the issue of grade inflation at BYU-Idaho. A certain number of individual grades will need to be sampled to estimate the mean grade earned at BYU-Idaho. Answer the following questions: What sample size would be required to estimate the true mean grade earned at BYU-Idaho with 95% confidence and a margin of error of 0.2 grade points? Show/Hide Solution) [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{1.96 \cdot 0.68}{0.2} \right)^2 \ &= 44.4 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=45). What sample size would be required to estimate the true mean grade earned at BYU-Idaho with 98% confidence and a margin of error of 0.2 grade points? Show/Hide Solution) Using the Normal Probability Applet, we find that (z^ = 2.3263) for a 98% confidence level. [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{2.3263 \cdot 0.68}{0.2} \right)^2 \ &= 62.6 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=63). What sample size would be required to estimate the true mean grade earned at BYU-Idaho with 95% confidence and a margin of error of 0.1 grade points? Show/Hide Solution) [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{1.96 \cdot 0.68}{0.1} \right)^2 \ &= 177.6 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=178). What sample size would be required to estimate the true mean grade earned at BYU-Idaho with 98% confidence and a margin of error of 0.1 grade points? Show/Hide Solution) [ \begin{array}{1cl} n &= \left( \frac{z^\sigma}{m} \right)^2 \ &= \left( \frac{2.326 \cdot 0.68}{0.1} \right)^2 \ &= 250.2 \end{array} ] We always round up in sample size calculations, so the required sample size is (n=251). Comments What confidence level should be used? The most common choice for the confidence level is 95%. Other values are occasionally used, but this tends to meet the needs of most researchers. How do we find the value of (\sigma)? In a statistics class, this number will be given to you in the problem. In real life, you will need to estimate this value based on either ((i)) a previous study, ((ii)) the results of a pilot study, or ((iii)) your best professional judgement (an educated guess). Sometimes it is not possible to use data from a previous study or to easily conduct a pilot study. In these cases, the only thing a researcher can do is to use their understanding of the process to make a good guess of the value of (\sigma). Ultimately, sample size calculations are nothing more than a good estimate of the actual sample size that will be required. In many cases, budgets will limit the sample size that can be drawn. A researcher must assess whether a smaller-than-ideal sample will suffice for their needs. This is one of the difficult issues that arises in the actual practice of statistics. Summary The margin of error gives an estimate of the variability of responses. It is calculated as (\displaystyle{m=z^\frac{\sigma}{\sqrt{n}}}) where (z^) represents a calculated z-score corresponding to a particular confidence level. A confidence interval is an interval estimator used to give a range of plausible values for a parameter. The width of a confidence interval depends on the chosen confidence level (and its corresponding value of (z^)) as well as the sample size ((n)). This is the equation for calculating confidence intervals: [\displaystyle{\left(\bar x-z^\frac{\sigma}{\sqrt{n}},~\bar x+z^\frac{\sigma}{\sqrt{n}}\right)}] The sample size formula allows us to estimate the number of observations required to obtain a specific margin of error. (\displaystyle{n=\left(\frac{z^\sigma}{m}\right)^2}) Navigation \| Previous Reading \| This Reading \| Next Reading \| \| Lesson 9: Inference for One Mean: Sigma Known (Hypothesis Test) \| Lesson 10: Inference for One Mean: Sigma Known (Confidence Interval) \| Lesson 11: Inference for One Mean: Sigma Unknown \| Copyright © 2020 Brigham Young University-Idaho. All rights reserved.
188307	https://byjus.com/chemistry/nomenclature-saturated-hydrocarbons/	Nomenclature is a system of terms or rules that are used for forming these terms or names in a distinct field of science and arts. In simple terms it is an assignment of names to organic compounds. Saturated hydrocarbons are organic compounds that consist of carbon and hydrogen single bonds. In these compounds, there is the maximum number of hydrogen atom present for every carbon atoms. For example alkanes. The saturated hydrocarbons are named according to the following rules: Longest Chain Rule: The parent chain of the compound is considered the longest chain of carbon atoms. Lowest Set of Locants: The numbering of the carbon atoms starts from the end which gives the lowest number to the carbon atom carrying the substituent. Presence of Same Substituent More Than Once: Prefixes such as di, tri, etc are given to the substituents which are present twice, thrice respectively on the parent chain. Naming Different Substituents: If more than one substituent is present then the substituents are arranged in their alphabetical order. Naming Different Substituents At Equivalent Positions: If two different substituents are present in the same position from the two ends then the substituents are named so that the substituent which comes first in the alphabetical order gets the lowest number. Naming The Complex Substituents: Naming of the complex substituent is done when the substituent on the parent chain has a branched structure (i.e complex structure). These substituents are named as a substituted alkyl group and the carbon atom of this substituent attached to the parent chain is numbered 1. The name of this type of substituent is written in brackets. Let us understand it with the help of an example: In this case, we have 9 carbon atoms in the straight chain. 5th Carbon atom from both the ends of the straight-chain, consists of substituents having 3 carbon chains. On the first two carbon atoms of the substituent group, there is one additional carbon atom attached. Now if we consider this as a new parent chain, it has a substituent which has one additional carbon each. For naming them we will firstly number the parent chain. In this case, we have 9 carbon atoms in a straight chain which is also the parent chain. Then we find that the substituent is in the fifth position. Now taking the substituent we will observe that we have 3 substituent carbons and out of these three, two substituents have additional carbons attached. We find that the longest chain in this can be the first four carbon atom chains but this is wrong as the last carbon is not attached to the parent chain. So we will consider only three carbon atom chains as the main chain. Thus it can be named propane and in the first and second position, we have methyl group. We can write the name as 1-2 Dimethyl propane, but it will be written as 1-2 Dimethyl propyl as it is a substituent group. Now taking the substituent with the parent chain we will get 5-(1-2-Dimethyl Propyl) and as the parent chain has 9 carbon atoms so, it will be named nonane. Thus, the final name of the compound will be 5-(1-2-Dimethyl Propyl)nonane. Watch the video to understand the nomenclature of saturated hydrocarbons and get a deeper knowledge about the concept of nomenclature. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below Request OTP on Voice Call Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Request OTP on Voice Call Website Post My Comment So good for 🙏children’s and makes students mind so💪💪💪💪 Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
188308	https://algebrica.org/identities-using-reference-angles/	Updates The Loop Explore Glossary Life is an unknow About Us Home Search Results Search in glossary Identities Using Reference Angles Antonio Lupetti Updated: 09:02 PM • 15 May 2025 Verified Content • Reviewed for accuracy and reliability. 391 views Send Share this post with a friend and contribute to making knowledge accessible to everyone. WhatsApp Telegram Email The method of reference angles refers to a set of trigonometric identities that allow one to express trigonometric functions of non-acute angles, represented on the unit circle, in terms of the corresponding acute angle in the first quadrant of the Cartesian coordinate system. In other words, any trigonometric function sine, cosine, tangent, or cotangent, with an argument of the form: [ \pi \pm \alpha,\quad \frac{\pi}{2} \pm \alpha,\quad \pi \pm \alpha,\quad \frac{3\pi}{2} \pm \alpha,\quad 2\pi – \alpha, \quad \ldots ] can be rewritten as a function of (\alpha) by appropriately adjusting the sign. This technique simplifies the evaluation of trigonometric functions at arbitrary angles without relying on a calculator. it also strengthens your understanding of symmetry and periodicity within the unit circle. Consider an angle of the form: [ \frac{\pi}{2} + \alpha \tag{1} ] In Cartesian coordinates, this angle is located in the second quadrant. A geometric analysis shows that: [ \sin\left(\frac{\pi}{2} + \alpha\right) = \cos\alpha,\quad \cos\left(\frac{\pi}{2} + \alpha\right) = -\sin\alpha ] In other words, the sine of the angle (\alpha) (the vertical segment in the first quadrant) is equal in length to the cosine of the angle (\frac{\pi}{2} + \alpha) (the horizontal segment in the second quadrant), except that the latter has a negative sign. Since from the definition of tangent and cotangent we have: [ \tan\alpha = \frac{\sin\alpha}{\cos\alpha}\quad \text{and}\quad \cot\alpha = \frac{\cos\alpha}{\sin\alpha} ] Thus we obtain: [ \tan\left(\frac{\pi}{2} + \alpha\right) = \frac{\cos\alpha}{-\sin\alpha} = -\cot\alpha ] [ \cot\left(\frac{\pi}{2} + \alpha\right) = \frac{-\sin\alpha}{\cos\alpha} = -\tan\alpha ] For an angle of the form: [ \frac{\pi}{2} – \alpha \tag{2} ] In Cartesian coordinates, this angle is located in the first quadrant. We have: [ \sin\left(\frac{\pi}{2} – \alpha\right) = \cos\alpha,\quad \cos\left(\frac{\pi}{2} – \alpha\right) = \sin\alpha ] Thus, the tangent and cotangent functions are: [ \tan\left(\frac{\pi}{2} – \alpha\right) = \frac{\cos\alpha}{\sin\alpha} = \cot\alpha ] [ \cot\left(\frac{\pi}{2} – \alpha\right) = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha ] Consider the angle [ \pi + \alpha \tag{3} ] which is located in the third quadrant. The identities become: [ \sin(\pi + \alpha) = -\sin\alpha,\quad \cos(\pi + \alpha) = -\cos\alpha ] Therefore: [ \tan(\pi + \alpha) = \frac{-\sin\alpha}{- \cos\alpha} = \tan\alpha ] [\cot(\pi + \alpha) = \frac{-\cos\alpha}{-\sin\alpha} = \cot\alpha ] Consider the angle [ \pi – \alpha \tag{4} ] which is situated in the second quadrant. The identities are: [ \sin(\pi – \alpha) = \sin\alpha,\quad \cos(\pi – \alpha) = -\cos\alpha ] The tangent and cotangent functions become: [ \tan(\pi – \alpha) = \frac{\sin\alpha}{- \cos\alpha} = -\tan\alpha, ] [ \cot(\pi – \alpha) = \frac{-\cos\alpha}{\sin\alpha} = -\cot\alpha ] For an angle of the form [ \frac{3\pi}{2} + \alpha \tag{5} ] located in the fourth quadrant, we have: [ \sin\left(\frac{3\pi}{2} + \alpha\right) = -\cos\alpha, \quad \cos\left(\frac{3\pi}{2} + \alpha\right) = \sin\alpha ] The tangent and cotangent functions are: [ \tan\left(\frac{3\pi}{2} + \alpha\right) = \frac{-\cos\alpha}{\sin\alpha} = -\cot\alpha ] [ \cot\left(\frac{3\pi}{2} + \alpha\right) = \frac{\sin\alpha}{-\cos\alpha} = -\tan\alpha ] Consider the angle [ \frac{3\pi}{2} – \alpha \tag{6} ] which is in the fourth quadrant. We have: [ \sin\left(\frac{3\pi}{2} – \alpha\right) = -\cos\alpha, \quad \cos\left(\frac{3\pi}{2} – \alpha\right) = -\sin\alpha ] The tangent and cotangent become: [ \tan\left(\frac{3\pi}{2} – \alpha\right) = \frac{-\cos\alpha}{- \sin\alpha} = \cot\alpha ] [ \cot\left(\frac{3\pi}{2} – \alpha\right) = \frac{-\sin\alpha}{- \cos\alpha} = \tan\alpha ] Finally, for an angle of the form [ 2\pi – \alpha = -\alpha \tag{7} ] which represents the reduction of (-\alpha) using periodicity, we have: [ \sin(2\pi – \alpha) = -\sin\alpha,\quad \cos(2\pi – \alpha) = \cos\alpha ] The corresponding tangent and cotangent functions are: [ \tan(2\pi – \alpha) = \frac{-\sin\alpha}{\cos\alpha} = -\tan\alpha ] [ \cot(2\pi – \alpha) = \frac{\cos\alpha}{-\sin\alpha} = -\cot\alpha ] Continuing on this topic 672Unit Circle 2.8kSine and Cosine 837Tangent and Cotangent 478Secant and Cosecant 381Arcsine and Arccosine 374Arctangent and Arccotangent 393Trigonometric Identities 208Pythagorean Theorem 499Pythagorean Identity 452Right Triangle Trigonometry 481Law of Sines 312Law of Cosines Previous Next Insights, analysis, and handpicked content for STEM students and tech professionals. 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188309	https://www.nagwa.com/en/explainers/450158479087/	Lesson Explainer: Solving a System of Two Equations Using a Matrix Inverse \| Nagwa Lesson Explainer: Solving a System of Two Equations Using a Matrix Inverse \| Nagwa Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Classes My Messages My Reports My Wallet My Classes My Messages My Reports Lesson Explainer: Solving a System of Two Equations Using a Matrix Inverse Mathematics • First Year of Secondary School Join Nagwa Classes Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher! Check Available Classes Next Session: Saturday 20 September 2025 • 1:00pm Remaining Seats: 12 Try This In this explainer, we will learn how to solve a system of two linear equations using the inverse of the matrix of coefficients. We can solve a system of two linear equations, which are also called simultaneous equations, using the substitution or elimination methods, so it is fair to ask why we need to learn a different method to solve the same system. In fact, using a matrix inverse to solve a system of two linear equations is more involved than the previous two methods, which further justifies this question. We are studying this method as a model to understand the relationship between a system of linear equations and matrices. Since the system of two linear equations is the simplest model that relates a system of equations to matrices, it makes sense to start here. The method we will learn in this explainer can be used for a system containing a larger number of linear equations and unknown variables, although we will not discuss larger systems here. While it is not too difficult to solve a system of two linear equations without using matrices, it is more challenging to do this when we have three or more equations involved. Understanding the relationship between a system of linear equations and matrices lets us organize the given system of equations into a concise matrix equation, which can be solved using a method analogous to what we will discuss here. Before we discuss how to solve simultaneous equations using matrices, we need to understand how to solve a matrix equation. Recall the inverse matrix. Definition: Inverse Matrices Given a square matrix , the inverse matrix is a square matrix of the same order satisfying where is the identity matrix of the same order. If such a matrix exists, we say that matrix is invertible. Consider a matrix equation , where and are known and matrices, respectively, and is an unknown matrix. We further assume that is an invertible matrix. We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication is well defined. Since matrix is invertible, there exists an inverse matrix . Multiplying from the left by on both sides of the equation , we obtain On the left-hand side of the equation, we know that , where is the multiplicative identity. Hence, Substituting this expression to the left-hand side of equation (1), we can write Both and are known matrices; hence, this gives the solution to the matrix equation . How To: Solving Matrix Equations Let be an invertible matrix and be a matrix such that the multiplication is defined. Matrix satisfying the equation is given by This method gives us a way to solve any matrix equation of the form if matrix is invertible. However, this method cannot be used when is not invertible. This could happen if is not a square matrix or if is square and . In such cases, the matrix equation has either an infinite number of solutions or no solution. For a simple example, we can think of the case where , where is a zero matrix. We know that is not invertible since . The matrix equation has no solution if is a nonzero matrix, since the multiplication of a zero matrix by any matrix results in a zero matrix. On the other hand, if is a zero matrix of the natural order resulting from this matrix multiplication, any matrix satisfies the equation . This means that this matrix equation has an infinite number of solutions. In our first example, we will solve a matrix equation using the inverse matrix. Example 1: Solving Equations of Matrices Using Their Inverses Given that what is the value of ? Answer In this example, we are given a matrix equation. Matrix is an unknown matrix. If we find this matrix, we can find the value of . The example does not give us what matrix is, but it gives us the inverse of this matrix . Recall that the inverse of a square matrix , if it exists, is a matrix satisfying where is the identity matrix. We can multiply from the left by on both sides of the given equation to obtain We know that , which is a multiplicative identity, so we can neglect the factor and fill in the provided expression for on the right-hand side to write Computing this matrix multiplication, we obtain This leads to the unknown matrix. We know that a pair of matrices are equal if each pair of corresponding entries in the matrices are equal. Hence, this leads to In particular, the example asks for the value of , which is In the previous example, we solved a matrix equation when we were given the inverse matrix . If we are not provided the expression for , we can find the inverse matrix by using the following formula, as long as . Formula: Inverse of a 2 × 2 Matrix Let such that . Then, where . If , matrix is not invertible. Let us consider an example where we solve a matrix equation by first finding the inverse of a matrix. Example 2: Solving Equations of Matrices Using Their Inverses Given that determine the values of and . Answer In this example, we are given a matrix equation. Matrix is an unknown matrix. If we find this matrix, we can find the values of and . We recall that, given matrices and , an unknown matrix satisfying the equation is given by if the inverse matrix exists and the matrix multiplication can be defined. In our given example, this matrix corresponds to the matrix . Hence, this can be written as if the inverse matrix exists and the matrix multiplication is well defined. Hence, we need to begin by finding the inverse of this matrix, if it exists. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that Applying this formula to the given matrix, Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a matrix: Hence, using the determinant of the matrix we found earlier, Substituting this expression into equation (2), We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. Computing this matrix multiplication, we obtain Finally, computing the scalar multiplication, This leads to the unknown matrix. We know that a pair of matrices are equal if each pair of corresponding entries in the matrices are equal. Hence, this leads to So far, we have considered a few examples where we solved matrix equations using the inverse matrix. Let us turn our attention to the system of two linear equations. How To: Representing Systems of Two Equations as Matrix Equations Consider a system of equations given by for some known constants , , , , , . We can write this system of two equations as one matrix equation If we carry out the matrix multiplication on the left-hand side of the matrix equation, this is the same as Equating the corresponding entries of the matrices on both sides of this equation leads back to the system of two linear equations. Hence, this matrix equation is equivalent to the system of two linear equations. Since we can write the system of equations as a matrix equation, we can solve this system using the matrix inverse. We can see that the coefficients of and in the system of equations became the matrix in the matrix equation. This is called the coefficient matrix, because its entries come from the coefficients of the simultaneous equations. When writing down the coefficient matrix, we need to be careful with the order of the entries. Since the variable matrix has as its first entry, the coefficients of go in the first column. Hence, the same coefficient matrix is used even if the first equation in the system is written as . Rather than following the order of the coefficients written in the given equation, we need to consider which variable it is the coefficient of. We also note that the column matrix on the right-hand side of the matrix equation contains the constant terms from the right-hand sides of the simultaneous equations. The order of these constants must be consistent with the coefficient matrix. Since the coefficients of the first equation, , are written in the first row of the coefficient matrix, the constant from this equation must also appear in the first row of this matrix equation. Just like we discussed when solving matrix equations, this also means that the system of equations has either no solution or an infinite number of solutions when the coefficient matrix is not invertible. In the next example, we will write a pair of simultaneous equations into a matrix equation and then solve the matrix equation using the matrix inverse. Example 3: Solving a Pair of Simultaneous Equations Using Matrices Consider the simultaneous equations Express the given simultaneous equations as a matrix equation. Write down the inverse of the coefficient matrix. Multiply through by the inverse, on the left-hand side, to solve the matrix equation. Answer Part 1 Recall that a pair of simultaneous equations given by can be written as the matrix equation Here, it is important to note that the coefficient matrix is the coefficient of the simultaneous equations in the order given in the variable matrix . This means that the first column of the coefficient matrix contains the coefficients of variable , while the second column contains the coefficients of variable . In particular, we should first notice that and are written in the opposite order. We can rearrange this pair of simultaneous equations to say Then, we can write Part 2 In this part, we need to find the inverse of the coefficient matrix. In the previous part, we found the coefficient matrix . We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that Applying this formula to the coefficient matrix, Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a matrix: Hence, using the determinant of the matrix we found earlier, Part 3 In this part, we need to multiply through by the inverse on the left-hand side and solve the matrix equation. We begin with the matrix equation Multiplying both sides of the equation by the inverse of the coefficient matrix, we have On the left-hand side of this equation, the inverse of the coefficient matrix is multiplied by the coefficient matrix. Recall that, for any invertible matrix , we have where is the identity matrix. This means Since is the identity matrix, which is multiplied by the variable matrix, we can neglect this term. This leads to We can now substitute the inverse matrix from the previous part: Computing this matrix multiplication, we obtain Finally, computing the scalar multiplication, Hence, the solution to the matrix equation is In the previous example, we solved the matrix equation corresponding to a given pair of simultaneous equations. While we did not explicitly verify this, it can be shown that the values we found for and satisfy the given simultaneous equations. In the next problem, we will solve a pair of simultaneous equations by using matrices. Example 4: Solving a System of Two Equations Using Matrices Use matrices to solve the system Answer In this example, we need to solve the system of two linear equations by using matrices. We know that we can write a system of two linear equations into an equivalent matrix equation. Let us recall this process. Given the system of equations we can write an equivalent matrix equation Here, it is important to note that the coefficient matrix is the coefficient of the simultaneous equations in the order given in the variable matrix . This means that the first column of the coefficient matrix contains the coefficients of variable , while the second column contains the coefficients of variable . We note that the variable in the first equation is only accompanied by a negative sign. This indicates that the coefficient of in this equation is . Also, the variable in the second equation does not display any coefficients meaning that its coefficient is equal to 1. We can rewrite this pair of simultaneous equations with this information: Then, we can write the matrix equation We can solve this matrix equation by multiplying from the left the inverse of the coefficient matrix if it exists. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that Applying this formula to the coefficient matrix, Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a matrix: Hence, using the determinant of the matrix we found earlier, Recall that, for any invertible matrix , we have where is the identity matrix. This means that we will be able to remove the coefficient matrix from the left-hand side of equation (3) by multiplying from the left the inverse of the coefficient matrix. Multiplying both sides of the equation by the inverse of the coefficient matrix, we have We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication on the right-hand side of the equation above is well defined. Computing this matrix multiplication, we obtain Finally, computing the scalar multiplication, This gives us the solution of the matrix equation We know that a pair of matrices are equal if each pair of corresponding entries are equal. Hence, we obtain the solution to the given system of equations In our final example, we will solve a real-world problem involving a system of two equations using matrices. Example 5: Solving Two Equations with Two Unknowns Using Matrices The length of a rectangle is 6 cm more than twice its width, and twice its length is 39 cm more than its width. Given this, use matrices to determine the perimeter of the rectangle. Answer In this example, we need to find the perimeter of the rectangle whose length and width are related according to the given description. We know that the perimeter of a rectangle is given by twice the sum of its length and width. Let us begin by denoting the length and width of the rectangle by unknown constants and respectively. Then, Let us begin by writing down the relationship between and in the form of equations. First, we are given that the length of a rectangle is 6 cm more than twice its width. This can be written as Second, we are given that twice the length is 39 cm more than its width. This can be written as Let us rearrange these equations so that the left-hand sides of the equations contain the variables and the right-hand sides contain the constants: Now, we will solve this system of equations by using matrices. Recall that the system of equations can be written as the matrix equation Hence, we can write our system of equations as We can solve this matrix equation by multiplying from the left the inverse of the coefficient matrix if it exists. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that Applying this formula to the coefficient matrix, Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a matrix: Hence, using the determinant of the matrix we found earlier, Recall that, for any invertible matrix , we have where is the identity matrix. This means that we will be able to remove the coefficient matrix from the left-hand side of equation (4) by multiplying from the left the inverse of the coefficient matrix. Multiplying both sides of the equation by the inverse of the coefficient matrix, we have We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication on the right-hand side of the equation above is well defined. Computing this matrix multiplication, we obtain Finally, computing the scalar multiplication, We know that a pair of matrices are equal if each pair of corresponding entries are equal. Hence, we obtain the solution to the given system of equations This leads to the perimeter of the rectangle: Lastly, we note that matrices and their multiplicative inverses can be applied in the field of cryptography. In this context, the aim is to encode a message before it is transmitted, to prevent it from being understood if it is intercepted during transmission. The message can then be decoded once it reaches its desired destination. As an illustration, we might want to transmit the message “help”. We start by representing each letter as a number, using the following simple alphabet table. a: 1 b: 2 c: 3 d: 4 e: 5 f: 6 g: 7 h: 8 i: 9 j: 10 k: 11 l: 12 m: 13 n: 14 o: 15 p: 16 q: 17 r: 18 s: 19 t: 20 u: 21 v: 22 w: 23 x: 24 y: 25 z: 26 Then, taking our message two letters at a time, we write each two-letter section as a matrix: Next, to encode our message, we choose an invertible matrix as our coding matrix and multiply each of the above matrices on the left by . For instance, choosing , our message encodes as Therefore, we transmit the message . Now, recall that for a matrix with , its multiplicative inverse is given by where . In this case, we have , so Once the message has been received, to decode it we must multiply each of these matrices on the left by . This will “undo” the effect of and give us back the original pair of matrices . We can then read off the message “help” from these four numbers by looking up the corresponding letters in the alphabet table. Let us finish by recapping a few important concepts from the explainer. Key Points Let be an invertible matrix and be a matrix such that the multiplication is defined. Matrix satisfying the equation is given by If matrix is not invertible, the matrix equation has either an infinite number of solutions or no solution. Consider a system of equations given by for some known constants , , , , , . We can write this system of two equations as one matrix equation This equation can then be solved using the matrix inverse: Lesson Menu Lesson Lesson Plan Lesson Presentation Lesson Video Lesson Explainer Lesson Playlist Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! 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188310	https://www.reddit.com/r/learnmath/comments/2rs5o4/why_is_unit_vector_is_represented_as_cosx_sinx/	Why is unit vector is represented as (cosx, sinx)? : r/learnmath Skip to main contentWhy is unit vector is represented as (cosx, sinx)? : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •11 yr. ago WannaFuckFredDurst Why is unit vector is represented as (cosx, sinx)? RESOLVED Share Related Answers Section Related Answers Effective strategies for mastering algebra Tips for improving mental math skills Exploring real-world uses of number theory Comparing different methods of integration Best practices for preparing for math exams New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of January 8, 2015 Reddit reReddit: Top posts of January 2015 Reddit reReddit: Top posts of 2015 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
188311	https://nptel.ac.in/courses/106104028	Initiatives Learn about NPTELLocal ChaptersIndustry AssociatesTranslationGATE PreparationCSRInternshipPreDoc FellowshipFaculty InitiativesWorkshops Courses NPTEL CoursesMedical CoursesSWAYAM & NPTELDomainsNPTEL+ NPTEL STARS ----------- Coordinators ------------ NPTEL Testimonials ------------------ Help Videos ----------- More NPTEL BlogCourses on YTAbout usNOC Semester InformationCertification courses offered by IndustryCareersMerchandiseSpecial Lecture SeriesInternational NPTEL LearnersFAQDocumentsBooksLink to old site Log in Course Details Theory of Computation Lecture-01 What is theory of computation? Set membership problem, basic notions like alphabet, strings, formal languages. Lecture-02-Introduction to finite automaton. Lecture-03-Finite automata continued, deterministic finite automata(DFAs), language accepted by a DFA. Lecture-04-Regular languages, their closure properties. Lecture-05-DFAs solve set membership problems in linear time, pumping lemma. Lecture-06-More examples of nonregular languages, proof of pumping lemma, pumping lemma as a game, converse of pumping lemma does not hold. Lecture-07-A generalization of pumping lemma, nondeterministic finite automata (NFAs), computation trees for NFAs. Lecture-08-Formal description of NFA, language accepted by NFA, such languages are also regular. Lecture-09-'Guess and verify' paradigm for nondeterminism. Lecture-10-NFA's with epsilon transitions. Lecture-11-Regular expressions, they denote regular languages. Lecture-12-Construction of a regular expression for a language given a DFA accepting it. Algebraic closure properies of regular languages. Lecture-13-Closure properties continued. Lecture-14-Closure under reversal, use of closure properties. Lecture-15-Decision problems for regular languages. Lecture-16-About minimization of states of DFAs. Myhill-Nerode theorem. Lecture-17-Continuation of proof of Myhill-Nerode theorem. Lecture-18-Application of Myhill-Nerode theorem. DFA minimization. Lecture-19-DFA minimization continued. Lecture-20-Introduction to context free languages (cfls) and context free grammars (cfgs). Derivation of strings by cfgs. Lecture-21-Languages generated by a cfg, leftmost derivation, more examples of cfgs and cfls. Lecture-22-Parse trees, inductive proof that L is L(G). All regular languages are context free. Lecture-23-Towards Chomsky normal forms: elimination of useless symbols, analysis of reachable symbols, generating nonterminals, order of substeps matter. Lecture-24-Simplification of cfgs continued, Removal of epsilon productions: algorithm and its correctness. Lecture-25-Elimination of unit productions. Converting a cfg into Chomsky normal form. Towards pumping lemma for cfls Lecture-26-Pumping lemma for cfls. Adversarial paradigm. Lecture-27-Completion of pumping lemma proof. Examples of use of pumping lemma. Converse of lemma does not hold. Closure properties of cfls. Lecture-28-Closure properties continued. cfls not closed under complementation. Lecture-29-Another example of a cfl whose complement is not a cfl. Decision problems for cfls. Lecture-30-More decision problems. CYK algorithm for membership decision. Lecture-31-Introduction to pushdown automata (pda). Lecture-32-pda configurations, acceptance notions for pdas. Transition diagrams for pdas Lecture-33-Equivalence of acceptance by empty stack and acceptance by final state. Lecture-34-Turing machines (TM): motivation, informal definition, example, transition diagram. Lecture-35-Execution trace, another example (unary to binary conversion). Lecture-36-Example continued. Finiteness of TM description, TM configuration, language acceptance, definition of recursively enumerable (r.e.) languages. Lecture-37-Notion of non-acceptance or rejection of a string by a TM. Multitrack TM, its equivalence to standard TM. Multitape TMs. Lecture-38-Simulation of multitape TMs by basic model. Nondeterministic TM (NDTM). Equivalence of NDTMs with deterministic TMs. Lecture-39-Counter machines and their equivalence to basic TM model. Lecture-40-TMs can simulate computers, diagonalization proof. Lecture-41-Existence of non-r.e. languages, recursive languages, notion of decidability. Lecture-42-Separation of recursive and r.e. classes, halting problem and its undecidability. Theory of Computation, IIT Kanpur Prof. Somenath Biswas Lecture-01 What is theory of computation? Set membership problem, basic notions like alphabet, strings, formal languages. Concepts Covered: What is theory of computation? Set membership problem, basic notions like alphabet, strings, formal languages.
188312	https://www.ck12.org/flexi/precalculus/double-angle-identities/	Double Angle Identities \| Flexi Homework help & answers \| CK-12 Foundation What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up All Subjects Precalculus Double Angle Identities Double Angle Identities Sum and Difference IdentitiesTrigonometric Equations Concept Summary: Double Angle Identities: sin⁡2 x=2 sin⁡x cos⁡x cos⁡2 x=cos 2⁡x−sin 2⁡x tan⁡2 x=2 tan⁡x 1−tan 2⁡x Half Angle Identities: sin⁡x 2=±1−cos⁡x 2 cos⁡x 2=±1+cos⁡x 2 tan⁡x 2=±1−cos⁡x 1+cos⁡x Power Reducing Identities: sin 2⁡x=1−cos⁡2 x 2 cos 2⁡x=1+cos⁡2 x 2 tan 2⁡x=1−cos⁡2 x 1+cos⁡2 x Ask your own question In Fig. 9.23, BD \|\| CA, E is mid-point of C A and B D=1 2 C A . Prove that ar⁡(A B C)=2 ar⁡(D B C) Using suitable identities, evaluate the following. (132)2−(68)2 Using suitable identities, evaluate the following. 10.1×10.2 Using suitable identities, evaluate the following. 105×95 Using suitable identities, evaluate the following. (49)2 Using suitable identities, evaluate the following. (52)2 Expand the following, using suitable identities. (0.9 p−0.5 q)2 Expand the following, using suitable identities. (a 2+b 2)2 Expand the following, using suitable identities. (x 2+y 2)(x 2−y 2) Expand the following, using suitable identities. (2 x−5 y)(2 x−5 y) Expand the following, using suitable identities. (4 5 p+5 3 q)2 Expand the following, using suitable identities. (4 5 a+5 4 b)2 Expand the following, using suitable identities. (x 2 y−x y 2)2 Expand the following, using suitable identities. (x y+y z)2 Evaluate using suitable identities. 2.07×1.93 In Fig. 6.53, ∠1=∠2 and ∠3=∠4 . (ii) Show that A D=A B and C D=C B . Can we have two obtuse angles whose sum is\ &(a) a reflex angle? Why or why not?\ &(b) a complete angle? Why or why not?-a-reflex-angle-why-or-why-notless-by-pgreater-lesspgreaterandx26(b)-a-complete-angle-why-or-why-notless-by-pgreater/ "Can we have two obtuse angles whose sum is\ &(a) a reflex angle? Why or why not?\ &(b) a complete angle? Why or why not? ") Can we have two acute angles whose sum is\ &(a) an acute angle? Why or why not?\ &(b) a right angle? Why or why not?\ &(c) an obtuse angle? Why or why not?\ &(d) a straight angle? Why or why not?\ &(e) a reflex angle? Why or why not?-an-acute-angle-why-or-why-notless-by-pgreater-lesspgreaterandx26(b)-a-right-angle-why-or-why-notless-by-pgreater-lesspgreaterandx2-b6d6776/ "Can we have two acute angles whose sum is\ &(a) an acute angle? Why or why not?\ &(b) a right angle? Why or why not?\ &(c) an obtuse angle? Why or why not?\ &(d) a straight angle? Why or why not?\ &(e) a reflex angle? Why or why not? ") 1 is the identity for multiplication of whole numbers. In Fig. 6.13, BA ‖E D and B C‖E F . Show that ∠A B C+∠D E F=180∘ In the Fig. 5.1, if ∠1=∠3,∠2=∠4 and ∠3=∠4 , write the relation between ∠1 and ∠2 , using an Euclid's axiom. How does extending the arms of an angle affect its measurement? If the sum of two angles is greater than 180∘ , then which of the following is not possible for the two angles? Write the correct answer in the gap. 4 doubled is [ ]. Write the correct answer in the gap. 8 doubled is [ ]. Does 2(nm + xy) = 2nm + 2xy? Find the definite integral from 3pi/4 to pi/4 of (2 + cot theta)csc^2 theta d theta. Consider the integral ∫ from 0 to pi of (1 + cos(7t))^2 sin(7t) dt. The options are 8/21, 8/3, 1/21, and 1/7. What is the correct answer? What are compound angles in trigonmetry? Define double angle identities in trigonometry. What are the formulae for multiples of x in hyperbolic trigonometric functions? What are trigonometric functions of compound angles? What are the formulas for cos 2A? What is the formula for cosh 2x? What are the formulas for tan 2A? What do we mean by occurence of an event? What is a certain event in the context of probability? What is a sample space in probability? What is the formula for tanh 2x? Describe some important results in trigonometric functions of compound angles. What is the formula for sinh 2x? What are inverse trigonometric functions? How to derive double angle formulas? What are double-angle trigonometric identities? How do you solve double-angle identities? How are half-angle identities handled? How do you do half angle identities? How can double angle formulas be derived? What is the half-angle formula in trigonometry? What is the double-angle formula? What is the double angle formula in trigonometry? What is the angle name for half a revolution? What is a double-angle identity? What is a half angle? What are the trigonometric functions and their inverses? What are double angles? What are double angle identities? What are double angle identities in algebra? How do you find the exact value of double angle? How do you find the derivative of a trigonometric function? CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn -75fd10bcd7 © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Ask me anything! Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? 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188313	https://visaldiary.files.wordpress.com/2011/07/secrets-of-mental-math.pdf	This book has been optimized for viewing at a monitor setting of 1024 x 768 pixels. SECRE+S OF MEN+AL MA+H SECRE+S OF MEN+AL MA+H The Mathemagician’s Guide to Lightning Calculation and Amazing Math Tricks Arthur Benjamin and Michael Shermer Copyright © 2006 by Arthur Benjamin and Michael Shermer All rights reserved. Published in the United States by Three Rivers Press, an imprint of the Crown Publishing Group, a division of Random House, Inc., New York. www.crownpublishing.com Originally published in different form as Mathemagics by Lowell House, Los Angeles, in 1993. Three Rivers Press and the Tugboat design are registered trademarks of Random House, Inc. Library of Congress Cataloging-in-Publication Data Benjamin, Arthur. Secrets of mental math : the mathemagician’s guide to lightning calculation and amazing math tricks / Arthur Benjamin and Michael Shermer.— 1st ed. p. cm. Includes bibliographical references and index. 1. Mental arithmetic—Study and teaching. 2. Magic tricks in mathematics education. 3. Mental calculators. I. Shermer, Michael. II. Title. QA111.B44 2006 510—dc22 2005037289 eISBN: 978-0-307-34746-6 v1.0 I dedicate this book to my wife, Deena, and daughters, Laurel and Ariel. —Arthur Benjamin My dedication is to my wife, Kim, for being my most trusted conﬁdante and personal counselor. —Michael Shermer Acknowledgments The authors wish to thank Steve Ross and Katie McHugh at Random House for their support of this book. Special thanks to Natalya St. Clair for typesetting the initial draft, which was partly supported by a grant from the Mellon Foundation. Arthur Benjamin especially wants to acknowledge those who inspired him to become both a mathematician and a magician— cognitive psychologist William G. Chase, magicians Paul Gertner and James Randi, and mathematicians Alan J. Goldman and Edward R. Scheinerman. Finally, thanks to all of my colleagues and students at Harvey Mudd College, and to my wife, Deena, and daughters, Laurel and Ariel, for constant inspiration. Contents Foreword by Bill Nye (the Science Guy®) xi Foreword by James Randi xvii Prologue by Michael Shermer xix Introduction by Arthur Benjamin xxiii Chapter 0 Quick Tricks: Easy (and Impressive) Calculations 1 Chapter 1 A Little Give and Take: Mental Addition and Subtraction 11 Chapter 2 Products of a Misspent Youth: Basic Multiplication 29 Chapter 3 New and Improved Products: Intermediate Multiplication 53 Chapter 4 Divide and Conquer: Mental Division 80 Chapter 5 Good Enough: The Art of “Guesstimation” 108 Chapter 6 Math for the Board: Pencil-and-Paper Math 131 Chapter 7 A Memorable Chapter: Memorizing Numbers 151 Chapter 8 The T ough Stuff Made Easy: Advanced Multiplication 163 Chapter 9 Presto-digit-ation: The Art of Mathematical Magic 199 Chapter Epilogue by Michael Shermer: How Math Helps Us Think About Weird Things 222 Answers 233 Bibliography 271 Index 273 Contents x Foreword by Bill Nye (the Science Guy®) I like to think about the ﬁrst humans, the people who came up with the idea to count things. They must have noticed right away that ﬁguring on your ﬁngertips works great. Perhaps Og (a typical ancient cave guy) or one of his pals or associates said, “There are one, two, three, four, ﬁve of us here, so we need ﬁve pieces of fruit.” Later, “Hey, look,” someone must have said (or grunted), “you can count the number of people at the campﬁre, the number of birds on a tree, stones in a row, logs for a ﬁre, or grapes in a bunch, just with your ﬁngers.” It was a great start. It’s probably also how you came to ﬁrst know numbers. You’ve probably heard that math is the language of science, or the language of Nature is mathematics. Well, it’s true. The more we understand the universe, the more we discover its mathematical connections. Flowers have spirals that line up with a special sequence of numbers (called Fibonacci numbers) that you can understand and generate yourself. Seashells form in perfect mathematical curves (logarithmic spirals) that come from a chemical balance. Star clusters tug on one another in a mathematical dance that we can observe and understand from millions and even billions of kilometers away. We have spent centuries discovering the mathematical nature of Nature. With each discovery, someone had to go through the math and make sure the numbers were right. Well, Secrets of Mental Math can help you handle all kinds of numbers. You’ll get comfortable with calculations in a way that will let you know some of Nature’s numerical secrets, and who knows where that might take you? As you get to know numbers, the answer really is at your ﬁngertips. That’s not a joke, because that’s where it all begins. Almost everyone has ten ﬁngers, so our system of mathematics started with 1 and went to 10. In fact, we call both our num-bers and our ﬁngers “digits.” Coincidence? Hardly. Pretty soon, though, our ancestors ran out of ﬁngers. The same thing has probably happened to you. But we can’t just ignore those big numbers and (this is a joke) throw up our hands. We need numbers—they’re part of our lives every day, and in ways we typically don’t even notice. Think about a conversation you had with a friend. To call, you needed a phone number, and the time you spent on the phone was measured in numbers of hours and minutes. Every date in history, including an impor-tant one like your birthday, is reckoned with numbers. We even use numbers to represent ideas that have nothing to do with counting. What’s your 20? (I.e., Where are you? From the old police “10” codes, like 10-4 for “yes.”) What’s the 411 on that gal? (I.e., What’s her background; is she dating anyone? From the number for telephone information.) People describe one another in numbers representing height and weight. And, of course, we all like to know how much money we have or how much something costs in numbers: dollars, pesos, yuan, rupees, krona, euros, or yen. Additionally (another joke), this book has a time-saving section on remembering numbers—and large num-bers of numbers. If, for some reason, you’re not crazy about math, read a little further. Of course I, as the Science Guy, hope you do like math. Foreword xii Well, actually, I hope you love math. But no matter how you feel about math, hatred or love, I’d bet that you often ﬁnd yourself just wanting to know the answer right away, without having to write down everything carefully and work slowly and dili-gently—or without even having to stop and grab a calculator. You want the answer, as we say, “as if by magic.” It turns out that you can solve or work many, many math problems almost magically. This book will show you how. What makes any kind of magic so intriguing and fun is that the audience seldom knows how the trick is performed. “How did she do that . . . ?” “I don’t know, but it’s cool.” If you have an audience, the tricks and shortcuts in Secrets of Mental Math are a lot like magic. The audience seldom knows how a trick is performed; they just appreciate it. Notice, though, that in magic, it’s hardly worth doing if no one is watching. But with Secrets, knowing how it works doesn’t subtract from the fun (or pun). When arithmetic is easy, you don’t get bogged down in the calculating; you can concentrate on the wonderful nature of numbers. After all, math runs the universe. Dr. Benjamin got into this business of lightning-fast calculat-ing just for fun. We have to ﬁgure he impressed his teachers and classmates. Magicians might make some in their audience think that they have supernatural powers. Mathemagicians, at ﬁrst, give the impression that they’re geniuses. Getting people to notice what you’re doing is an old part of sharing ideas. If they’re impressed, they’ll probably listen to what you have to say. So try some “mathemagics.” You may impress your friends, all right. But you’ll also ﬁnd yourself performing just for your-self. You’ll ﬁnd you’re able to do problems that you didn’t think you could. You’ll be impressed . . . with yourself. Now, counting on your ﬁngers is one thing (one ﬁnger’s worth). But have you ever found yourself counting out loud or Foreword xiii whispering or making other sounds while you calculate? It almost always makes math easier. The problem, though, is that other people think you’re a little odd . . . not even (more math humor). Well, in Secrets of Mental Math, Dr. Benjamin helps you learn to use that “out-loud” feature of the way your brain works to do math problems more easily, faster, and more accu-rately (which is surprising), all while your brain is thinking away—almost as if you’re thinking out loud. You’ll learn to move through math problems the same way we read in English, left to right. You’ll learn to handle big prob-lems fast with good guesses, actually great guesses, within a per-cent or so. You will learn to do arithmetic fast; that way you can spend your time thinking about what the numbers mean. Og wondered, “Do we have enough fruit for each person sitting around the ﬁre? If not, there might be trouble.” Now you might wonder, “Is there enough space on this computer to keep track of my music ﬁles . . . or my bank account? If not, there might be trouble.” There’s more to Secrets than just ﬁguring. You can learn to take a day, month, and year, then compute what day of the week it was or will be. It’s fantastic, almost magical, to be able to tell someone what day of the week she or he was born. But, it’s really something to be able to ﬁgure that the United States had its ﬁrst big Fourth of July on a Thursday in 1776. April 15, 1912, the day the Titanic sank, was a Monday. The ﬁrst human to walk on the moon set foot there on July 20, 1969, a Sunday. You’ll probably never forget that the United States was attacked by terrorists on September 11, 2001. With Secrets of Mental Math, you’ll always be able to show it was a Tuesday. There are relationships in Nature that numbers describe bet-ter than any other way we know. There are simple numbers that you can count on your hands: one, two, three, and on up. But Foreword xiv there are also an inﬁnite number of numbers in between. There are fractions. There are numbers that never end. They get as big as you want and so small that they’re hard to imagine. You can know them. With Secrets of Mental Math, you can have even these in-between numbers come so quickly to your mind that you’ll have a bit more space in your brain to think about why our world works this way. One way or another, this book will help you see that in Nature, it all adds up. Foreword xv Foreword by James Randi Mathematics is a wonderful, elegant, and exceedingly useful lan-guage. It has its own vocabulary and syntax, its own verbs, nouns, and modiﬁers, and its own dialects and patois. It is used brilliantly by some, poorly by others. Some of us fear to pursue its more eso-teric uses, while a few of us wield it like a sword to attack and conquer income tax forms or masses of data that resist the less courageous. This book does not guarantee to turn you into a Leib-niz, or put you on stage as a Professor Algebra, but it will, I hope, bring you a new, exciting, and even entertaining view of what can be done with that wonderful invention—numbers. We all think we know enough about arithmetic to get by, and we certainly feel no guilt about resorting to the handy pocket calculator that has become so much a part of our lives. But, just as photography may blind us to the beauty of a Vermeer paint-ing, or an electronic keyboard may make us forget the magniﬁ-cence of a Horowitz sonata, too much reliance on technology can deny us the pleasures that you will ﬁnd in these pages. I remember the delight I experienced as a child when I was shown that I could multiply by 25 merely by adding two 0s to my number and dividing by 4. Casting out 9s to check multiplication came next, and when I found out about cross-multiplying I was hooked and became, for a short while, a generally unbearable math nut. Immunizations against such afﬂictions are not avail-able. You have to recover all by yourself. Beware! This is a fun book. You wouldn’t have it in your hands right now if you didn’t have some interest either in improving your math skills or in satisfying a curiosity about this fascinating sub-ject. As with all such instruction books, you may retain and use only a certain percentage of the varied tricks and methods described here, but that alone will make it worth the investment of your time. I know both the authors rather well. Art Benjamin is not only one of those whiz kids we used to groan about in school but also has been known to tread the boards at the Magic Castle in Hollywood, performing demonstrations of his skill, and on one occasion he traveled to Tokyo, Japan, to pit his math skills against a lady savant on live television. Michael Shermer, with his specialized knowledge of science, has an excellent overview of practical applications of math as it is used in the real world. If this is your ﬁrst exposure to this kind of good math stuff, I envy you. You’ll discover, as you come upon each delicious new way to attack numbers, that you missed something in school. Mathematics, particularly arithmetic, is a powerful and depend-able tool for day-to-day use that enables us to handle our compli-cated lives with more assurance and accuracy. Let Art and Michael show you how to round a few of the corners and cut through some of the trafﬁc. Remember these words of Dr. Samuel Johnson, an eminently practical soul in all respects: “Arithemeti-cal inquiries give entertainment in solitude by the practice, and reputation in public by the effect.” Above all, enjoy the book. Let it entertain you, and have fun with it. That, with the occasional good deed, a slice of pizza (no anchovies!), and a selection of good friends is about all you can ask of life. Well, almost all. Maybe a Ferrari . . . Foreword xviii Prologue by Michael Shermer My good friend Dr. Arthur Benjamin, mathematics professor at Harvey Mudd College in Claremont, California, takes the stage to a round of applause at the Magic Castle, a celebrated magic club in Hollywood, where he is about to perform “mathemag-ics,” or what he calls the art of rapid mental calculation. Art appears nothing like a mathematics professor from a prestigious college. Astonishingly quick-witted, he looks at home with the rest of the young magicians playing at the Castle—which he is. What makes Art so special is that he can play in front of any group, including professional mathematicians and magicians, because he can do something that almost no one else can. Art Benjamin can add, subtract, multiply, and divide numbers in his head faster than most people can with a calculator. He can square two-digit, three-digit, and four-digit numbers, as well as ﬁnd square roots and cube roots, without writing anything down on paper. And he can teach you how to perform your own mathematical magic. Traditionally, magicians refuse to disclose how they perform their tricks. If they did, everyone would know how they are done and the mystery and fascination of magic would be lost. But Art wants to get people excited about math. And he knows that one of the best ways to do so is to let you and other readers in on his secrets of “math genius.” With these skills, almost any-one can do what Art Benjamin does every time he gets on stage to perform his magic. This particular night at the Magic Castle, Art begins by ask-ing if anyone in the audience has a calculator. A group of engi-neers raise their hands and join Art on the stage. Offering to test their calculators to make sure they work, Art asks a member of the audience to call out a two-digit number. “Fifty-seven,” shouts one. Art points to another who yells out, “Twenty-three.” Directing his attention to those on stage, Art tells them: “Multiply 57 by 23 on the calculator and make sure you get 1311 or the calculators are not working correctly.” Art waits patiently while the volunteers ﬁnish inputting the numbers. As each participant indicates his calculator reads 1311, the audi-ence lets out a collective gasp. The amazing Art has beaten the calculators at their own game! Art next informs the audience that he will square four two-digit numbers faster than his button-pushers on stage can square them on their calculators. The audience asks him to square the numbers 24, 38, 67, and 97. Then, in large, bold writing for everyone to see, Art writes: 576, 1444, 4489, 9409. Art turns to his engineer volunteers, each of whom is computing a two-digit square, and asks them to call out their answers. Their response triggers gasps and then applause from the audi-ence: “576, 1444, 4489, 9409.” The woman next to me sits with her mouth open in amazement. Art then offers to square three-digit numbers without even writing down the answer. “Five hundred and seventy-two,” a gentleman calls out. Art’s reply comes less than a second later: “572 squared is 327,184.” He immediately points to another member of the audience, who yells, “389,” followed by Art’s unblinking response: “389 squared will give you 151,321.” Prologue xx Someone else blurts out, “262.” “That’ll give you 68,644.” Sens-ing he delayed just an instant on that last one, he promises to make up for it on the next number. The challenge comes—991. With no pause, Art squares the number, “982,081.” Several more three-digit numbers are given and Art responds perfectly. Members of the audience shake their heads in disbelief. With the audience in the palm of his hand, Art now declares that he will attempt to square a four-digit number. A woman calls out, “1,036,” and Art instantly responds, “That’s 1,073,296.” The audience laughs and Art explains, “No, no, that’s much too easy a number. I’m not supposed to beat the calculators on these. Let’s try another one.” A man offers a challenging 2,843. Pausing brieﬂy between digits, Art responds: “Let’s see, the square of that should be 8 million . . . 82 thousand . . . 649.” He is right, of course, and the audience roars their approval, as loudly as they did for the previous magician who sawed a woman in half and made a dog disappear. It is the same everywhere Art Benjamin goes, whether it is a high school auditorium, a college classroom, a professional con-ference, the Magic Castle, or a television studio. Professor Ben-jamin has performed his special brand of magic live all over the country and on numerous television talk shows. He has been the subject of investigation by a cognitive psychologist at Carnegie Mellon University and is featured in a scholarly book by Steven Smith called The Great Mental Calculators: The Psychology, Methods, and Lives of Calculating Prodigies, Past and Present. Art was born in Cleveland on March 19, 1961 (which he calcu-lates was a Sunday, a skill he will teach you in Chapter 9). A hyperactive child, Art drove his teachers mad with his class-room antics, which included correcting the mathematical mis-takes they occasionally made. Throughout this book when teaching you his mathematical secrets, Art recalls when and Prologue xxi where he learned these skills, so I will save the fascinating sto-ries for him to tell you. Art Benjamin is an extraordinary individual with an extraor-dinary program to teach you rapid mental calculation. I offer these claims without hesitation, and ask only that you remem-ber this does not come from a couple of guys promising miracles if you will only call our 800 hotline. Art and I are both creden-tialed in the most conservative of academic professions—Art in mathematics and I, myself, in the history of science—and we would never risk career embarrassment (or worse) by making such powerful claims if they were not true. To put it simply, this stuff works, and virtually everyone can do it because this art of “math genius” is a learned skill. So you can look forward to improving your math skills, impressing your friends, enhancing your memory, and, most of all, having fun! Prologue xxii Introduction Ever since I was a child, I have loved playing with numbers, and in this book I hope to share my passion with you. I have always found numbers to have a certain magical appeal and spent countless hours entertaining myself and others with their beautiful properties. As a teenager, I performed as a magician, and subsequently combined my loves of math and magic into a full-length show, called Mathemagics, where I would demon-strate and explain the secrets of rapid mental calculation to audiences of all ages. Since earning my PhD, I have taught mathematics at Harvey Mudd College, and I still enjoy sharing the joy of numbers with children and adults throughout the world. In this book, I will share all of my secrets for doing math in your head, quickly and easily. (I realize that magicians are not supposed to reveal their secrets, but mathemagicians have a different code of ethics. Mathematics should be awe inspiring, not mysterious.) What will you learn from this book? You will learn to do math in your head faster than you ever thought possible. After a little practice, you will dramatically improve your memory for numbers. You will learn feats of mind that will impress your friends, colleagues, and teachers. But you will also learn to view math as an activity that can actually be fun. Too often, math is taught as a set of rigid rules, leaving little room for creative thinking. But as you will learn from Secrets, there are often several ways to solve the same problem. Large problems can be broken down into smaller, more manageable components. We look for special features to make our problems easier to solve. These strike me as being valuable life lessons that we can use in approaching all kinds of problems, mathematical and otherwise. “But isn’t math talent something that you are born with?” I get this question all the time. Many people are convinced that lightning calculators are prodigiously gifted. Maybe I was born with some curiosity about how things work, whether it be a math problem or a magic trick. But I am convinced, based on many years of teaching experience, that rapid math is a skill that anyone can learn. And like any worthwhile skill, it takes prac-tice and dedication if you wish to become an expert. But to achieve these results efﬁciently, it is important that you practice the right way. Let me show you the way! Mathemagically, Dr. Arthur Benjamin Claremont, California Introduction xxiv Chapter 0 Quick Tricks: Easy (and Impressive) Calculations In the pages that follow, you will learn to do math in your head faster than you ever thought possible. After practicing the meth-ods in this book for just a little while, your ability to work with numbers will increase dramatically. With even more practice, you will be able to perform many calculations faster than some-one using a calculator. But in this chapter, my goal is to teach you some easy yet impressive calculations you can learn to do immediately. We’ll save some of the more serious stuff for later. INSTANT MULTIPLICATION Let’s begin with one of my favorite feats of mental math—how to multiply, in your head, any two-digit number by eleven. It’s very easy once you know the secret. Consider the problem: 32 11 To solve this problem, simply add the digits, 3 2 5 , put the 5 between the 3 and the 2, and there is your answer: 35 2 What could be easier? Now you try: 53 11 Since 5 3 8, your answer is simply 583 One more. Without looking at the answer or writing any-thing down, what is 81 11? Did you get 891? Congratulations! Now before you get too excited, I have shown you only half of what you need to know. Suppose the problem is 85 11 Although 8 5 1 3 , the answer is NOT 81 3 5! As before, the 3 goes in between the numbers, but the 1 needs to be added to the 8 to get the correct answer: 93 5 Think of the problem this way: 1 835 935 Secrets of Mental Math 2 Here is another example. Try 57 11. Since 5 7 12, the answer is Okay, now it’s your turn. As fast as you can, what is 77 11? If you got the answer 847, then give yourself a pat on the back. You are on your way to becoming a mathemagician. Now, I know from experience that if you tell a friend or teacher that you can multiply, in your head, any two-digit num-ber by eleven, it won’t be long before they ask you to do 99 11. Let’s do that one now, so we are ready for it. Since 9 9 18, the answer is: Okay, take a moment to practice your new skill a few times, then start showing off. You will be amazed at the reaction you get. (Whether or not you decide to reveal the secret is up to you!) Welcome back. At this point, you probably have a few ques-tions, such as: “Can we use this method for multiplying three-digit numbers (or larger) by eleven?” 1 989 1089 1 527 627 Quick Tricks: Easy (and Impressive) Calculations 3 Absolutely. For instance, for the problem 314 11, the answer still begins with 3 and ends with 4. Since 3 1 4 , and 1 4 5 , the answer is 34 5 4. But we’ll save larger problems like this for later. More practically, you are probably saying to yourself, “Well, this is ﬁne for multiplying by elevens, but what about larger numbers? How do I multiply numbers by twelve, or thirteen, or thirty-six?” My answer to that is, Patience! That’s what the rest of the book is all about. In Chapters 2, 3, 6, and 8, you will learn meth-ods for multiplying together just about any two numbers. Better still, you don’t have to memorize special rules for every number. Just a handful of techniques is all that it takes to multiply num-bers in your head, quickly and easily. SQUARING AND MORE Here is another quick trick. As you probably know, the square of a number is a number multiplied by itself. For example, the square of 7 is 7 7 49. Later, I will teach you a simple method that will enable you to easily calculate the square of any two-digit or three-digit (or higher) number. That method is especially simple when the number ends in 5, so let’s do that trick now. To square a two-digit number that ends in 5, you need to remember only two things. 1. The answer begins by multiplying the ﬁrst digit by the next higher digit. 2. The answer ends in 25. Secrets of Mental Math 4 For example, to square the number 35, we simply multiply the ﬁrst digit (3) by the next higher digit (4), then attach 25. Since 3 4 12, the answer is 1225. Therefore, 35 35 1225. Our steps can be illustrated this way: How about the square of 85? Since 8 9 72, we immedi-ately get 85 85 7225. We can use a similar trick when multiplying two-digit num-bers with the same ﬁrst digit, and second digits that sum to 10. The answer begins the same way that it did before (the ﬁrst digit multiplied by the next higher digit), followed by the prod-uct of the second digits. For example, let’s try 83 87. (Both numbers begin with 8, and the last digits sum to 3 7 10.) Since 8 9 72, and 3 7 21, the answer is 7221. 83 87 8 9 72 3 7 21 Answer: 7221 85 85 8 9 72 5 5 25 Answer: 7225 35 35 3 4 12 5 5 25 Answer:1225 Quick Tricks: Easy (and Impressive) Calculations 5 Similarly, 84 86 7224. Now it’s your turn. Try 26 24 How does the answer begin? With 2 3 6. How does it end? With 6 4 24. Thus 26 24 624. Remember that to use this method, the ﬁrst digits have to be the same, and the last digits must sum to 10. Thus, we can use this method to instantly determine that 31 39 1209 32 38 1216 33 37 1221 34 36 1224 35 35 1225 You may ask, “What if the last digits do not sum to ten? Can we use this method to multiply twenty-two and twenty-three?” Well, not yet. But in Chapter 8, I will show you an easy way to do problems like this using the close-together method. (For 22 23, you would do 20 25 plus 2 3, to get 500 6 506, but I’m getting ahead of myself!) Not only will you learn how to use these methods, but you will understand why these methods work, too. “Are there any tricks for doing mental addition and subtraction?” Deﬁnitely, and that is what the next chapter is all about. If I were forced to summarize my method in three words, I would say, “Left to right.” Here is a sneak preview. Secrets of Mental Math 6 Consider the subtraction problem Most people would not like to do this problem in their head (or even on paper!), but let’s simplify it. Instead of subtracting 587, subtract 600. Since 1200 600 600, we have that But we have subtracted 13 too much. (We will explain how to quickly determine the 13 in Chapter 1.) Thus, our painful-looking subtraction problem becomes the easy addition problem which is not too hard to calculate in your head (especially from left to right). Thus, 1241 587 654. Using a little bit of mathematical magic, described in Chapter 9, you will be able to instantly compute the sum of the ten num-bers below. 9 5 14 19 33 52 85 137 222 359 935 641 13 654 1241 600 641 1241 587 Quick Tricks: Easy (and Impressive) Calculations 7 Although I won’t reveal the magical secret right now, here is a hint. The answer, 935, has appeared elsewhere in this chapter. More tricks for doing math on paper will be found in Chapter 6. Furthermore, you will be able to quickly give the quotient of the last two numbers: 359 222 1.61 (ﬁrst three digits) We will have much more to say about division (including dec-imals and fractions) in Chapter 4. MORE PRACTICAL TIPS Here’s a quick tip for calculating tips. Suppose your bill at a restaurant came to $42, and you wanted to leave a 15% tip. First we calculate 10% of $42, which is $4.20. If we cut that number in half, we get $2.10, which is 5% of the bill. Adding these numbers together gives us $6.30, which is exactly 15% of the bill. We will discuss strategies for calculating sales tax, dis-counts, compound interest, and other practical items in Chapter 5, along with strategies that you can use for quick mental esti-mation when an exact answer is not required. IMPROVE YOUR MEMORY In Chapter 7, you will learn a useful technique for memorizing numbers. This will be handy in and out of the classroom. Using an easy-to-learn system for turning numbers into words, you will be able to quickly and easily memorize any numbers: dates, phone numbers, whatever you want. Speaking of dates, how would you like to be able to ﬁgure out the day of the week of any date? You can use this to ﬁgure Secrets of Mental Math 8 out birth dates, historical dates, future appointments, and so on. I will show you this in more detail later, but here is a simple way to ﬁgure out the day of January 1 for any year in the twenty-ﬁrst century. First familiarize yourself with the following table. Monday Tuesday Wednesday Thursday Friday Saturday Sunday 1 2 3 4 5 6 7 or 0 For instance, let’s determine the day of the week of January 1, 2030. Take the last two digits of the year, and consider it to be your bill at a restaurant. (In this case, your bill would be $30.) Now add a 25% tip, but keep the change. (You can compute this by cutting the bill in half twice, and ignoring any change. Half of $30 is $15. Then half of $15 is $7.50. Keeping the change results in a $7 tip.) Hence your bill plus tip amounts to $37. To ﬁgure out the day of the week, subtract the biggest mul-tiple of 7 (0, 7, 14, 21, 28, 35, 42, 49, . . .) from your total, and that will tell you the day of the week. In this case, 37 35 2, and so January 1, 2030, will occur on 2’s day, namely Tuesday: Bill: 30 Tip: 7 37 subtract 7s: 3 5 2 Tuesday How about January 1, 2043: Bill: 43 Tip: 1 0 53 subtract 7s: 4 9 4 Thursday Quick Tricks: Easy (and Impressive) Calculations 9 Exception: If the year is a leap year, remove $1 from your tip, then proceed as before. For example, for January 1, 2032, a 25% tip of $32 would be $8. Removing one dollar gives a total of 32 7 39. Subtracting the largest multiple of 7 gives us 39 35 4. So January 1, 2032, will be on 4’s day, namely Thursday. For more details that will allow you to compute the day of the week of any date in history, see Chapter 9. (In fact, it’s perfectly okay to read that chapter ﬁrst!) I know what you are wondering now: “Why didn’t they teach this to us in school?” I’m afraid that there are some questions that even I cannot answer. Are you ready to learn more magical math? Well, what are we waiting for? Let’s go! Secrets of Mental Math 10 Chapter 1 A Little Give and Take: Mental Addition and Subtraction For as long as I can remember, I have always found it easier to add and subtract numbers from left to right instead of from right to left. By adding and subtracting numbers this way, I found that I could call out the answers to math problems in class well before my classmates put down their pencils. And I didn’t even need a pencil! In this chapter you will learn the left-to-right method of doing mental addition and subtraction for most numbers that you encounter on a daily basis. These mental skills are not only important for doing the tricks in this book but are also indis-pensable in school, at work, or any time you use numbers. Soon you will be able to retire your calculator and use the full capac-ity of your mind as you add and subtract two-digit, three-digit, and even four-digit numbers with lightning speed. LEFT -TO-RIGHT ADDITION Most of us are taught to do math on paper from right to left. And that’s ﬁne for doing math on paper. But if you want to do math in your head (even faster than you can on paper) there are many good reasons why it is better to work from left to right. After all, you read numbers from left to right, you pronounce numbers from left to right, and so it’s just more natural to think about (and calculate) numbers from left to right. When you compute the answer from right to left (as you probably do on paper), you generate the answer backward. That’s what makes it so hard to do math in your head. Also, if you want to esti-mate your answer, it’s more important to know that your answer is “a little over 1200” than to know that your answer “ends in 8.” Thus, by working from left to right, you begin with the most signiﬁcant digits of your problem. If you are used to working from right to left on paper, it may seem unnatural to work with numbers from left to right. But with practice you will ﬁnd that it is the most natural and efﬁcient way to do mental calculations. With the ﬁrst set of problems—two-digit addition—the left-to-right method may not seem so advantageous. But be patient. If you stick with me, you will see that the only easy way to solve three-digit and larger addition problems, all subtraction prob-lems, and most deﬁnitely all multiplication and division prob-lems is from left to right. The sooner you get accustomed to computing this way, the better. T wo-Digit Addition Our assumption in this chapter is that you know how to add and subtract one-digit numbers. We will begin with two-digit addition, something I suspect you can already do fairly well in your head. The following exercises are good practice, however, because the two-digit addition skills that you acquire here will be needed for larger addition problems, as well as virtually all Secrets of Mental Math 12 multiplication problems in later chapters. It also illustrates a fundamental principle of mental arithmetic—namely, to sim-plify your problem by breaking it into smaller, more manage-able parts. This is the key to virtually every method you will learn in this book. To paraphrase an old saying, there are three components to success—simplify, simplify, simplify. The easiest two-digit addition problems are those that do not require you to carry any numbers, when the ﬁrst digits sum to 9 or below and the last digits sum to 9 or below. For example: (30 2) To solve 47 32, ﬁrst add 30, then add 2. After adding 30, you have the simpler problem 77 2, which equals 79. We illustrate this as follows: 47 32 77 2 79 (ﬁrst add 30) (then add 2) The above diagram is simply a way of representing the men-tal processes involved in arriving at an answer using our method. While you need to be able to read and understand such diagrams as you work your way through this book, our method does not require you to write down anything yourself. Now let’s try a calculation that requires you to carry a number: (20 8) Adding from left to right, you can simplify the problem by adding 67 20 87; then 87 8 95. 67 28 47 32 A Little Give and Take: Mental Addition and Subtraction 13 67 28 87 8 95 (ﬁrst add 20) (then add 8) Now try one on your own, mentally calculating from left to right, and then check below to see how we did it: (50 7) How was that? You added 84 50 134 and added 134 7 141. 84 57 134 7 141 (ﬁrst add 50) (then add 7) If carrying numbers trips you up a bit, don’t worry about it. This is probably the ﬁrst time you have ever made a systematic attempt at mental calculation, and if you’re like most people, it will take you time to get used to it. With practice, however, you will begin to see and hear these numbers in your mind, and car-rying numbers when you add will come automatically. Try another problem for practice, again computing it in your mind ﬁrst, then checking how we did it: (40 5) You should have added 68 40 108, and then 108 5 113, the ﬁnal answer. Was that easier? If you would like to try your hand at more two-digit addition problems, check out the set of exercises below. (The answers and computations are at the end of the book.) 68 45 84 57 Secrets of Mental Math 14 EXERCISE:TWO-DIGIT ADDITION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Three-Digit Addition The strategy for adding three-digit numbers is the same as for adding two-digit numbers: you add from left to right. After each step, you arrive at a new (and simpler) addition problem. Let’s try the following: (300 20 7) Starting with 538, we add 300, then add 20, then add 7. After adding 300 (538 300 838), the problem becomes 838 27. After adding 20 (838 20 858), the problem simpliﬁes to 858 7 865. This thought process can be diagrammed as follows: 538 327 838 27 858 7 865 300 20 7 All mental addition problems can be done by this method. The goal is to keep simplifying the problem until you are just adding a one-digit number. Notice that 538 327 requires you to hold on to six digits in your head, whereas 838 27 and 538 327 39 38 55 49 19 17 47 36 73 58 89 78 34 26 95 32 64 43 23 16 A Little Give and Take: Mental Addition and Subtraction 15 858 7 require only ﬁve and four digits, respectively. As you simplify the problem, the problem gets easier! Try the following addition problem in your mind before looking to see how we did it: (100 50 9) Did you reduce and simplify the problem by adding left to right? After adding the hundreds (623 100 723), you were left with 723 59. Next you should have added the tens (723 50 773), simplifying the problem to 773 9, which you then summed to get 782. Diagrammed, the problem looks like this: 623 159 723 59 773 9 782 100 50 9 When I do these problems mentally, I do not try to see the numbers in my mind—I try to hear them. I hear the problem 623 159 as six hundred twenty-three plus one hundred ﬁfty-nine; by emphasizing the word hundred to myself, I know where to begin adding. Six plus one equals seven, so my next problem is seven hundred and twenty-three plus ﬁfty-nine, and so on. When ﬁrst doing these problems, practice them out loud. Rein-forcing yourself verbally will help you learn the mental method much more quickly. Three-digit addition problems really do not get much harder than the following: 858 634 623 159 Secrets of Mental Math 16 Now look to see how we did it: 858 634 1458 34 1488 4 1492 600 30 4 At each step I hear (not see) a “new” addition problem. In my mind the problem sounds like this: 858 plus 634 is 1458 plus 34 is 1488 plus 4 is 1492. Your mind-talk may not sound exactly like mine (indeed, you might “see” the numbers instead of “hear” them), but whatever it is you say or visualize to yourself, the point is to reinforce the numbers along the way so that you don’t forget where you are and have to start the addition problem over again. Let’s try another one for practice: (400 90 6) Do it in your mind ﬁrst, then check our computation below: 759 496 1159 96 1249 6 1255 400 90 6 This addition problem is a little more difﬁcult than the last one since it requires you to carry numbers in all three steps. However, with this particular problem you have the option of using an alternative method. I am sure you will agree that it is a 759 496 A Little Give and Take: Mental Addition and Subtraction 17 lot easier to add 500 to 759 than it is to add 496, so try adding 500 and then subtracting the difference: (500 4) 759 496 1259 4 1255 (ﬁrst add 500) (then subtract 4) So far, you have consistently broken up the second number in any problem to add to the ﬁrst. It really does not matter which number you choose to break up, but it is good to be consistent. That way, your mind will never have to waste time deciding which way to go. If the second number happens to be a lot sim-pler than the ﬁrst, I sometimes switch them around, as in the fol-lowing example: 207 528 528 207 728 7 735 (switch) 200 7 Let’s ﬁnish up by adding three-digit to four-digit numbers. Since most human memory can hold only about seven or eight digits at a time, this is about as large a problem as you can handle without resorting to artiﬁcial memory devices, like ﬁngers, calcu-lators, or the mnemonics taught in Chapter 7. In many addition problems that arise in practice, especially within multiplication problems, one or both of the numbers will end in 0, so we shall emphasize those types of problems. We begin with an easy one: 2700 567 207 528 759 496 Secrets of Mental Math 18 Since 27 hundred 5 hundred is 32 hundred, we simply attach the 67 to get 32 hundred and 67, or 3267. The process is the same for the following problems: Because 40 18 58, the ﬁrst answer is 3258. For the sec-ond problem, since 40 72 exceeds 100, you know the answer will be 33 hundred and something. Because 40 72 112, the answer is 3312. These problems are easy because the (nonzero) digits overlap in only one place, and hence can be solved in a single step. Where digits overlap in two places, you require two steps. For instance: (100 71) This problem requires two steps, as diagrammed the follow-ing way: 4560 171 4660 71 4731 100 71 Practice the following three-digit addition exercises, and then add some (pun intended!) of your own if you like until you are comfortable doing them mentally without having to look down at the page. (Answers can be found in the back of the book.) 4560 171 3240 72 3240 18 A Little Give and Take: Mental Addition and Subtraction 19 EXERCISE: THREE-DIGIT ADDITION 1. 2. 3. 4. 5. 912 475 457 241 635 814 312 256 242 137 Secrets of Mental Math 20 Carl Friedrich Gauss: Mathematical Prodigy A prodigy is a highly talented child, usually called precocious or gifted, and almost always ahead of his peers.The German math-ematician Carl Friedrich Gauss (1777–1855) was one such child. He often boasted that he could calculate before he could speak. By the ripe old age of three, before he had been taught any arithmetic, he corrected his father’s payroll by declaring “the reckoning is wrong.” A further check of the numbers proved young Carl correct. As a ten-year-old student,Gauss was presented the following math-ematical problem:What is the sum of numbers from 1 to 100? While his fellow students were frantically calculating with paper and pencil, Gauss immediately envisioned that if he spread out the numbers 1 through 50 from left to right,and the numbers 51 to 100 from right to left directly below the 1–50 numbers,each combination would add up to 101 (1 100,2 99,3 98, . . .).Since there were ﬁfty sums,the answer would be 101 50 5050.T o the astonishment of everyone, including the teacher, young Carl got the answer not only ahead of everyone else, but computed it entirely in his mind. He wrote out the answer on his slate, and ﬂung it on the teacher’s desk with a deﬁant “There it lies.” The teacher was so impressed that he invested his own money to purchase the best available textbook on arithmetic and gave it to Gauss, stating,“He is beyond me, I can teach him nothing more.” Indeed, Gauss became the mathematics teacher of others, and eventually went on to become one of the greatest mathematicians in history, his theories still used today in the service of science. Gauss’s desire to better understand Nature through the language of mathe-matics was summed up in his motto, taken from Shakespeare’s King Lear (substituting “laws” for “law”):“Thou, nature, art my goddess; to thy laws/My services are bound.” 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. LEFT -TO-RIGHT SUBTRACTION For most of us, it is easier to add than to subtract. But if you con-tinue to compute from left to right and to break down problems into simpler components, subtraction can become almost as easy as addition. T wo-Digit Subtraction When subtracting two-digit numbers, your goal is to simplify the problem so that you are reduced to subtracting (or adding) a one-digit number. Let’s begin with a very simple subtraction problem: (20 5) After each step, you arrive at a new and easier subtraction problem. Here, we ﬁrst subtract 20 (86 20 66) then we subtract 5 to reach the simpler subtraction problem 66 5 for your ﬁnal answer of 61. The problem can be diagrammed this way: 86 25 66 5 61 (ﬁrst subtract 20) (then subtract 5) 86 25 4240 371 7830 348 6120 136 1800 855 5400 252 877 539 276 689 878 797 457 269 852 378 A Little Give and Take: Mental Addition and Subtraction 21 Of course, subtraction problems are considerably easier when there is no borrowing (which occurs when a larger digit is being subtracted from a smaller one). But the good news is that “hard” subtraction problems can usually be turned into “easy” addition problems. For example: (20 9) or (30 1) There are two different ways to solve this problem mentally: 1. First subtract 20, then subtract 9: 86 29 66 9 57 (ﬁrst subtract 20) (then subtract 9) But for this problem, I would prefer the following strategy: 2. First subtract 30, then add back 1: 86 29 56 1 57 (ﬁrst subtract 30) (then add 1) Here is the rule for deciding which method to use: If a two-digit subtraction problem would require borrowing, then round the second number up (to a multiple of ten). Subtract the rounded number, then add back the difference. For example, the problem 54 28 would require borrowing (since 8 is greater than 4), so round 28 up to 30, compute 54 30 24, then add back 2 to get 26 as your ﬁnal answer: (30 2) 54 28 24 2 26 30 2 54 28 86 29 Secrets of Mental Math 22 Now try your hand (or head) at 81 37. Since 7 is greater than 1, we round 37 up to 40, subtract it from 81 (81 40 41), then add back the difference of 3 to arrive at the ﬁnal answer: 81 37 41 3 44 40 3 With just a little bit of practice, you will become comfortable working subtraction problems both ways. Just use the rule above to decide which method will work best. EXERCISE: TWO-DIGIT SUBTRACTION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Three-Digit Subtraction Now let’s try a three-digit subtraction problem: (400 10 7) This particular problem does not require you to borrow any numbers (since every digit of the second number is less than the digit above it), so you should not ﬁnd it too hard. Simply sub-tract one digit at a time, simplifying as you go. 958 417 148 86 125 79 89 48 51 27 63 46 79 29 67 48 92 34 84 59 38 23 A Little Give and Take: Mental Addition and Subtraction 23 958 417 558 17 548 7 541 400 10 7 Now let’s look at a three-digit subtraction problem that requires you to borrow a number: (600 2) At ﬁrst glance this probably looks like a pretty tough prob-lem, but if you ﬁrst subtract 747 600 147, then add back 2, you reach your ﬁnal answer of 147 2 149. 747 598 147 2 149 600 2 Now try one yourself: Did you ﬁrst subtract 700 from 853? If so, did you get 853 700 153? Since you subtracted by 8 too much, did you add back 8 to reach 161, the ﬁnal answer? 853 692 153 8 161 700 8 Now, I admit we have been making life easier for you by sub-tracting numbers that were close to a multiple of 100. (Did you notice?) But what about other problems, like: (400 60 8) or (500 ??) 725 468 853 692 747 598 Secrets of Mental Math 24 If you subtract one digit at a time, simplifying as you go, your sequence will look like this: 725 468 325 68 265 8 257 (ﬁrst subtract 400) (then subtract 60) (then subtract 8) What happens if you round up to 500? 725 468 225 ?? ?? (ﬁrst subtract 500) (then add ??) Subtracting 500 is easy: 725 500 225. But you have sub-tracted too much. The trick is to ﬁgure out exactly how much too much. At ﬁrst glance, the answer is far from obvious. To ﬁnd it, you need to know how far 468 is from 500. The answer can be found by using “complements,” a nifty technique that will make many three-digit subtraction problems a lot easier to do. Using Complements (You’re Welcome!) Quick, how far from 100 are each of these numbers? 57 68 49 21 79 Here are the answers: Notice that for each pair of numbers that add to 100, the ﬁrst digits (on the left) add to 9 and the last digits (on the right) add 79 21 100 21 79 100 49 51 100 68 32 100 57 43 100 A Little Give and Take: Mental Addition and Subtraction 25 to 10. We say that 43 is the complement of 57, 32 is the com-plement of 68, and so on. Now you ﬁnd the complements of these two-digit numbers: 37 59 93 44 08 To ﬁnd the complement of 37, ﬁrst ﬁgure out what you need to add to 3 in order to get 9. (The answer is 6.) Then ﬁgure out what you need to add to 7 to get 10. (The answer is 3.) Hence, 63 is the complement of 37. The other complements are 41, 7, 56, 92. Notice that, like everything else you do as a mathemagician, the complements are determined from left to right. As we have seen, the ﬁrst dig-its add to 9, and the second digits add to 10. (An exception occurs in numbers ending in 0—e.g., 30 70 100—but those complements are simple!) What do complements have to do with mental subtraction? Well, they allow you to convert difﬁcult subtraction problems into straightforward addition problems. Let’s consider the last subtraction problem that gave us some trouble: (500 32) To begin, you subtracted 500 instead of 468 to arrive at 225 (725 500 225). But then, having subtracted too much, you needed to ﬁgure out how much to add back. Using complements gives you the answer in a ﬂash. How far is 468 from 500? The same distance as 68 is from 100. If you ﬁnd the complement of 68 the way we have shown you, you will arrive at 32. Add 32 to 225, and you will arrive at 257, your ﬁnal answer. 725 468 Secrets of Mental Math 26 725 468 225 32 257 (ﬁrst subtract 500) (then add 32) Try another three-digit subtraction problem: (300 41) To compute this mentally, subtract 300 from 821 to arrive at 521, then add back the complement of 59, which is 41, to arrive at 562, our ﬁnal answer. The procedure looks like this: 821 259 521 41 562 300 41 Here is another problem for you to try: (400 28) Check your answer and the procedure for solving the prob-lem below: 645 372 245 28 265 8 273 400 20 8 Subtracting a three-digit number from a four-digit number is not much harder, as the next example illustrates: (600 21) 1246 579 645 372 821 259 A Little Give and Take: Mental Addition and Subtraction 27 By rounding up, you subtract 600 from 1246, leaving 646, then add back the complement of 79, which is 21. Your ﬁnal answer is 646 21 667. 1246 579 646 21 667 600 21 Try the three-digit subtraction exercises below, and then cre-ate more of your own for additional (or should that be subtrac-tional?) practice. EXERCISE: THREE-DIGIT SUBTRACTION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 1776 987 2345 678 1428 571 564 228 873 357 772 596 455 319 978 784 219 176 793 402 204 185 763 486 587 298 936 725 583 271 Secrets of Mental Math 28 Chapter 2 Products of a Misspent Youth: Basic Multiplication I probably spent too much time of my childhood thinking about faster and faster ways to perform mental multiplication; I was diagnosed as hyperactive and my parents were told that I had a short attention span and probably would not be successful in school. (Fortunately, my parents ignored that advice. I was also lucky to have some incredibly patient teachers in my ﬁrst few years of school.) It might have been my short attention span that motivated me to develop quick ways to do arithmetic. I don’t think I had the patience to carry out math problems with pencil and paper. Once you have mastered the techniques described in this chapter, you won’t want to rely on pencil and paper again, either. In this chapter you will learn how to multiply in your head one-digit numbers by two-digit numbers and three-digit num-bers. You will also learn a phenomenally fast way to square two-digit numbers. Even friends with calculators won’t be able to keep up with you. Believe me, virtually everyone will be dumb-founded by the fact that such problems can not only be done mentally, but can be computed so quickly. I sometimes wonder whether we were not cheated in school; these methods are so simple once you learn them. There is one small prerequisite for mastering the skills in this chapter—you need to know the multiplication tables through ten. In fact, to really make headway, you need to know your multiplication tables backward and forward. For those of you who need to shake the cobwebs loose, consult the multiplication chart below. Once you’ve got your tables down, you are ready to begin. Multiplication Table of Numbers 1–10 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100 2-BY -1 MULTIPLICATION PROBLEMS If you worked your way through Chapter 1, you got into the habit of adding and subtracting from left to right. You will do vir-tually all the calculations in this chapter from left to right as well. This is undoubtedly the opposite of what you learned in school. Secrets of Mental Math 30 But you’ll soon see how much easier it is to think from left to right than from right to left. (For one thing, you can start to say your answer aloud before you have ﬁnished the calculation. That way you seem to be calculating even faster than you are!) Let’s tackle our ﬁrst problem: First, multiply 40 7 280. (Note that 40 7 is just like 4 7, with a friendly zero attached.) Next, multiply 2 7 14. Then add 280 plus 14 (left to right, of course) to arrive at 294, the correct answer. We illustrate this procedure below. (40 2) We have omitted diagramming the mental addition of 280 14, since you learned how to do that computation in the last chapter. At ﬁrst you will need to look down at the problem while doing the calculation. With practice you will be able to forgo this step and compute the whole thing in your mind. Let’s try another example: Your ﬁrst step is to break down the problem into small multi-plication tasks that you can perform mentally with ease. Since 48 (40 8) 4 42 7 40 7 280 2 7 14 294 42 7 Products of a Misspent Youth: Basic Multiplication 31 48 40 8, multiply 40 4 160, then add 8 4 32. The answer is 192. (Note: If you are wondering why this process works, see the Why These Tricks Work section at the end of the chapter.) (40 8) Here are two more mental multiplication problems that you should be able to solve fairly quickly. First calculate 62 3. Then do 71 9. Try doing them in your head before looking at how we did it. (60 2) (70 1) These two examples are especially simple because the num-bers being added essentially do not overlap at all. When doing 180 6, you can practically hear the answer: One hundred eighty . . . six! Another especially easy type of mental multipli-cation problem involves numbers that begin with ﬁve. When the ﬁve is multiplied by an even digit, the ﬁrst product will be a mul-tiple of 100, which makes the resulting addition problem a snap. 71 9 70 9 630 1 9 9 639 62 3 60 3 180 2 3 6 186 48 4 40 4 160 8 4 32 192 Secrets of Mental Math 32 (50 8) Try your hand at the following problem: (80 7) Notice how much easier this problem is to do from left to right. It takes far less time to calculate “400 plus 35” mentally than it does to apply the pencil-and-paper method of “putting down the 5 and carrying the 3.” The following two problems are a little harder. (30 8) (60 7) As usual, we break these problems down into easier problems. For the one on the left, multiply 30 9 plus 8 9, giving you 270 72. The addition problem is slightly harder because it involves carrying a number. Here 270 70 2 340 2 342. With practice, you will become more adept at juggling 67 8 60 8 480 7 8 56 536 38 9 30 9 270 8 9 72 342 87 5 80 5 400 7 5 35 435 58 4 50 4 200 8 4 32 232 Products of a Misspent Youth: Basic Multiplication 33 problems like these in your head, and those that require you to carry numbers will be almost as easy as those that don’t. Rounding Up You saw in the last chapter how useful rounding up can be when it comes to subtraction. The same goes for multiplication, especially when you are multiplying numbers that end in eight or nine. Let’s take the problem of 69 6, illustrated below. On the left we have calculated it the usual way, by adding 360 54. On the right, however, we have rounded 69 up to 70, and sub-tracted 420 6, which you might ﬁnd easier to do. (60 9) or (70 1) The following example also shows how much easier round-ing up can be: (70 8) or (80 2) The subtraction method works especially well for numbers that are just one or two digits away from a multiple of 10. It does not work so well when you need to round up more than two dig-78 9 80 9 720 2 9 18 702 78 9 70 9 630 8 9 72 702 69 6 70 6 420 1 6 6 414 69 6 60 6 360 9 6 54 414 Secrets of Mental Math 34 its because the subtraction portion of the problem gets difﬁcult. As it is, you may prefer to stick with the addition method. Per-sonally, for problems of this size, I use only the addition method because in the time spent deciding which method to use, I could have already done the calculation! So that you can perfect your technique, I strongly recommend practicing more 2-by-1 multiplication problems. Below are twenty problems for you to tackle. I have supplied you with the answers in the back, including a breakdown of each component of the multiplication. If, after you’ve worked out these prob-lems, you would like to practice more, make up your own. Cal-culate mentally, then check your answer with a calculator. Once you feel conﬁdent that you can perform these problems rapidly in your head, you are ready to move to the next level of mental calculation. EXERCISE: 2-BY -1 MULTIPLICATION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 64 8 29 3 76 8 46 2 37 6 57 7 75 4 96 9 78 2 97 4 58 6 84 5 53 5 28 4 49 9 93 8 71 3 67 5 43 7 82 9 Products of a Misspent Youth: Basic Multiplication 35 3-BY -1 MULTIPLICATION PROBLEMS Now that you know how to do 2-by-1 multiplication problems in your head, you will ﬁnd that multiplying three digits by a sin-gle digit is not much more difﬁcult. You can get started with the following 3-by-1 problem (which is really just a 2-by-1 problem in disguise): Was that easy for you? (If this problem gave you trouble, you might want to review the addition material in Chapter 1.) Let’s try another 3-by-1 problem similar to the one you just did, except we have replaced the 0 with a 6 so you have another step to perform: In this case, you simply add the product of 6 7, which you already know to be 42, to the ﬁrst sum of 2240. Since you do not need to carry any numbers, it is easy to add 42 to 2240 to arrive at the total of 2282. In solving this and other 3-by-1 multiplication problems, the difﬁcult part may be holding in memory the ﬁrst sum (in this case, 2240) while doing the next multiplication problem (in this 326 (300 20 6) 7 300 7 2100 20 7 1 4 0 2240 6 7 4 2 2282 320 (300 20) 7 300 7 2100 20 7 1 4 0 2240 Secrets of Mental Math 36 case, 6 7). There is no magic secret to remembering that ﬁrst number, but with practice I guarantee you will improve your concentration, and holding on to numbers while performing other functions will get easier. Let’s try another problem: Even if the numbers are large, the process is just as simple. For example: When ﬁrst solving these problems, you may have to glance down at the page as you go along to remind yourself what the original problem is. This is okay at ﬁrst. But try to break the habit so that eventually you are holding the problem entirely in memory. In the last section on 2-by-1 multiplication problems, we saw that problems involving numbers that begin with ﬁve are 987 (900 80 7) 9 900 9 8100 80 9 7 2 0 8820 7 9 6 3 8883 647 (600 40 7) 4 600 4 2400 40 4 1 6 0 2560 7 4 2 8 2588 Products of a Misspent Youth: Basic Multiplication 37 sometimes especially easy to solve. The same is true for 3-by-1 problems: Notice that whenever the ﬁrst product is a multiple of 1000, the resulting addition problem is no problem at all. This is because you do not have to carry any numbers and the thou-sands digit does not change. If you were solving the problem above in front of someone else, you would be able to say your ﬁrst product—“three thousand . . .”—out loud with complete conﬁdence that a carried number would not change it to 4000. (As an added bonus, by quickly saying the ﬁrst digit, it gives the illusion that you computed the entire answer immediately!) Even if you are practicing alone, saying your ﬁrst product out loud frees up some memory space while you work on the remaining 2-by-1 problem, which you can say out loud as well—in this case, “. . . three hundred seventy-eight.” Try the same approach in solving the next problem, where the multiplier is a 5: 663 (600 60 3) 5 600 5 3000 60 5 300 3 5 1 5 3315 563 (500 60 3) 6 500 6 3000 60 6 360 3 6 1 8 3378 Secrets of Mental Math 38 Because the ﬁrst two digits of the three-digit number are even, you can say the answer as you calculate it without having to add anything! Don’t you wish all multiplication problems were this easy? Let’s escalate the challenge by trying a couple of problems that require some carrying. (600 80 4) In the next two problems you need to carry a number at the end of the problem instead of at the beginning: (600 40 8) 648 9 600 9 5400 40 9 360 5760 8 9 72 5832 684 9 600 9 5400 80 9 720 6120 4 9 36 6156 184 (100 80 4) 7 100 7 700 80 7 5 6 0 1260 4 7 2 8 1288 Products of a Misspent Youth: Basic Multiplication 39 The ﬁrst part of each of these problems is easy enough to compute mentally. The difﬁcult part comes in holding the pre-liminary answer in your head while computing the ﬁnal answer. In the case of the ﬁrst problem, it is easy to add 5400 360 5760, but you may have to repeat 5760 to yourself several times while you multiply 8 9 72. Then add 5760 72. Some-times at this stage I will start to say my answer aloud before ﬁn-ishing. Because I know I will have to carry when I add 60 72, I know that 5700 will become 5800, so I say “ﬁfty-eight hun-dred and . . .” Then I pause to compute 60 72 132. Because I have already carried, I say only the last two digits, “. . . thirty-two!” And there is the answer: 5832. The next two problems require you to carry two numbers each, so they may take you longer than those you have already done. But with practice you will get faster: 489 (400 80 9) 7 400 7 2800 80 7 5 6 0 3360 9 7 6 3 3423 376 (300 70 6) 4 300 4 1200 70 4 2 8 0 1480 6 4 2 4 1504 Secrets of Mental Math 40 When you are ﬁrst tackling these problems, repeat the answers to each part out loud as you compute the rest. In the ﬁrst problem, for example, start by saying, “Twenty-eight hun-dred plus ﬁve hundred sixty” a couple of times out loud to reinforce the two numbers in memory while you add them together. Repeat the answer—“thirty-three hundred sixty”— several times while you multiply 9 7 63. Then repeat “thirty-three hundred sixty plus sixty-three” aloud until you compute the ﬁnal answer of 3423. If you are thinking fast enough to recognize that adding 60 63 will require you to carry a 1, you can begin to give the ﬁnal answer a split second before you know it—“thirty-four hundred and . . . twenty-three!” Let’s end this section on 3-by-1 multiplication problems with some special problems you can do in a ﬂash because they require one addition step instead of two: 511 (500 11) 7 500 7 3500 11 7 7 7 3577 224 (200 20 4) 9 200 9 1800 20 9 1 8 0 1980 4 9 3 6 2016 Products of a Misspent Youth: Basic Multiplication 41 (900 25) (800 25) In general, if the product of the last two digits of the ﬁrst number and the multiplier is known to you without having to calculate it (for instance, you may know that 25 8 200 automatically since 8 quarters equals $2.00), you will get to the ﬁnal answer much more quickly. For instance, if you know without calculating that 75 4 300, then it is easy to com-pute 975 4: (900 75) To reinforce what you have just learned, solve the following 3-by-1 multiplication problems in your head; then check your computations and answers with ours (in the back of the book). I can assure you from experience that doing mental calculations is just like riding a bicycle or typing. It might seem impossible at ﬁrst, but once you’ve mastered it, you will never forget how to do it. 975 4 900 4 3600 75 4 300 3900 825 3 800 3 2400 25 3 75 2475 925 8 900 8 7200 25 8 200 7400 Secrets of Mental Math 42 EXERCISE: 3-BY -1 MULTIPLICATION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 691 3 312 9 767 3 457 9 722 9 693 6 488 9 285 6 862 5 429 3 670 4 499 9 968 6 188 6 247 5 578 9 611 3 134 8 339 8 457 7 751 9 297 8 259 7 757 8 214 8 184 7 587 4 807 9 529 9 328 6 728 2 927 7 957 6 862 4 637 5 431 6 Products of a Misspent Youth: Basic Multiplication 43 BE THERE OR B2: SQUARING TWO-DIGIT NUMBERS Squaring numbers in your head (multiplying a number by itself) is one of the easiest yet most impressive feats of mental calcula-tion you can do. I can still recall where I was when I discovered how to do it. I was thirteen, sitting on a bus on the way to visit my father at work in downtown Cleveland. It was a trip I made often, so my mind began to wander. I’m not sure why, but I began thinking about the numbers that add up to 20, and I won-dered, how large could the product of two such numbers get? I started in the middle with 10 10 (or 102), the product of which is 100. Next, I multiplied 9 11 99, 8 12 96, 7 13 91, 6 14 84, 5 15 75, 4 16 64, and so on. I noticed that the products were getting smaller, and their differ-ence from 100 was 1, 4, 9, 16, 25, 36, . . . —or 12, 22, 32, 42, 52, 62, . . . (see table below). Numbers that Distance Their Product’s difference add to 20 from 10 product from 100 10 10 0 100 0 9 11 1 99 1 8 12 2 96 4 7 13 3 91 9 6 14 4 84 16 5 15 5 75 25 4 16 6 64 36 3 17 7 51 49 2 18 8 36 64 1 19 9 19 81 I found this pattern astonishing. Next I tried numbers that add to 26 and got similar results. First I worked out 132 169, then computed 12 14 168, 11 15 165, 10 16 160, Secrets of Mental Math 44 9 17 153, and so on. Just as before, the distances these products were from 169 was 12, 22, 32, 42, and so on (see table below). There is actually a simple algebraic explanation for this phe-nomenon (see Why These Tricks Work, page 50). At the time, I didn’t know my algebra well enough to prove that this pattern would always occur, but I experimented with enough examples to become convinced of it. Then I realized that this pattern could help me square num-bers more easily. Suppose I wanted to square the number 13. Instead of multiplying 13 13, Numbers that Distance Their Product’s difference add to 26 from 13 product from 169 13 13 0 169 0 12 14 1 168 1 11 15 2 165 4 10 16 3 160 9 9 17 4 153 16 8 18 5 144 25 why not get an approximate answer by using two numbers that are easier to multiply but also add to 26? I chose 10 16 160. To get the ﬁnal answer, I just added 32 9 (since 10 and 16 are each 3 away from 13). Thus, 132 160 9 169. Neat! This method is diagrammed as follows: 16 132 3 160 32 169 3 10 Now let’s see how this works for another square: Products of a Misspent Youth: Basic Multiplication 45 42 412 1 1680 12 1681 1 40 To square 41, subtract 1 to obtain 40 and add 1 to obtain 42. Next multiply 40 42. Don’t panic! This is simply a 2-by-1 multiplication problem (speciﬁcally, 4 42) in disguise. Since 4 42 168, 40 42 1680. Almost done! All you have to add is the square of 1 (the number by which you went up and down from 41), giving you 1680 1 1681. Can squaring a two-digit number be this easy? Yes, with this method and a little practice, it can. And it works whether you initially round down or round up. For example, let’s examine 772, working it out both by rounding up and by rounding down: 84 772 7 5880 72 5929 7 70 or 80 772 3 5920 32 5929 3 74 In this instance the advantage of rounding up is that you are virtually done as soon as you have completed the multiplication problem because it is simple to add 9 to a number ending in 0! In fact, for all two-digit squares, I always round up or down to the nearest multiple of 10. So if the number to be squared ends in 6, 7, 8, or 9, round up, and if the number to be squared ends in 1, 2, 3, or 4, round down. (If the number ends in 5, you Secrets of Mental Math 46 do both!) With this strategy you will add only the numbers 1, 4, 9, 16, or 25 to your ﬁrst calculation. Let’s try another problem. Calculate 562 in your head before looking at how we did it, below: 60 562 4 3120 42 3136 4 52 Squaring numbers that end in 5 is even easier. Since you will always round up and down by 5, the numbers to be multiplied will both be multiples of 10. Hence, the multiplication and the addition are especially simple. We have worked out 852 and 352, below: 90 852 5 7200 52 7225 5 80 40 352 5 1200 52 1225 5 30 As you saw in Chapter 0, when you are squaring a number that ends in 5, rounding up and down allows you to blurt out the ﬁrst part of the answer immediately and then ﬁnish it with 25. For example, if you want to compute 752, rounding up to 80 and down to 70 will give you “Fifty-six hundred and . . . twenty-ﬁve!” For numbers ending in 5, you should have no trouble beating someone with a calculator, and with a little practice with the Products of a Misspent Youth: Basic Multiplication 47 other squares, it won’t be long before you can beat the calcula-tor with any two-digit square number. Even large numbers are not to be feared. You can ask someone to give you a really big two-digit number, something in the high 90s, and it will sound as though you’ve chosen an impossible problem to compute. But, in fact, these are even easier because they allow you to round up to 100. Let’s say your audience gives you 962. Try it yourself, and then check how we did it. 100 962 4 9200 42 9216 4 92 Wasn’t that easy? You should have rounded up by 4 to 100 and down by 4 to 92, and then multiplied 100 92 to get 9200. At this point you can say out loud, “Ninety-two hundred,” and then ﬁnish up with “sixteen” and enjoy the applause! EXERCISE: TWO-DIGIT SQUARES Compute the following: 1. 142 2. 272 3. 652 4. 892 5. 982 6. 312 7. 412 8. 592 9. 262 10. 532 11. 212 12. 642 13. 422 14. 552 15. 752 16. 452 17. 842 18. 672 19. 1032 20. 2082 Secrets of Mental Math 48 Products of a Misspent Youth: Basic Multiplication 49 Zerah Colburn: Entertaining Calculations O ne of the ﬁrst lightning calculators to capitalize on his talent was Zerah Colburn (1804–1839), an American farmer’s son from Vermont who learned the multiplication tables to 100 before he could even read or write. By the age of six, young Zerah’s father took him on the road,where his performances generated enough capital to send him to school in Paris and London. By age eight he was interna-tionally famous, performing lightning calculations in England, and was described in the Annual Register as “the most singular phenomenon in the history of the human mind that perhaps ever existed.” No less than Michael Faraday and Samuel Morse admired him. No matter where he went,Colburn met all challengers with speed and precision. He tells us in his autobiography of one set of problems he was given in New Hampshire in June 1811:“How many days and hours since the Christian Era commenced, 1811 years ago? Answered in twenty seconds: 661,015 days, 15,864,360 hours. How many sec-onds in eleven years? Answered in four seconds; 346,896,000.” Col-burn used the same techniques described in this book to compute entirely in his head problems given to him.For example,he would fac-tor large numbers into smaller numbers and then multiply: Colburn once multiplied 21,734 543 by factoring 543 into 181 3.He then multiplied 21,734 181 to arrive at 3,933,854, and ﬁnally multiplied that ﬁgure by 3, for a total of 11,801,562. As is often the case with lightning calculators,interest in Colburn’s amazing skills diminished with time, and by the age of twenty he had returned to America and become a Methodist preacher. He died at a youthful thirty-ﬁve. In summarizing his skills as a lightning calculator, and the advantage such an ability affords,Colburn reﬂected,“True,the method . . . requires a much larger number of ﬁgures than the com-mon Rule, but it will be remembered that pen, ink and paper cost Zerah very little when engaged in a sum.” WHY THESE TRICKS WORK This section is presented for teachers, students, math buffs, and anyone curious as to why our methods work. Some people may ﬁnd the theory as interesting as the application. Fortunately, you need not understand why our methods work in order to under-stand how to apply them. All magic tricks have a rational expla-nation behind them, and mathemagical tricks are no different. It is here that the mathemagician reveals his deepest secrets! In this chapter on multiplication problems, the distributive law is what allows us to break down problems into their com-ponent parts. The distributive law states that for any numbers a, b, and c: (b c) a (b a) (c a) That is, the outside term, a, is distributed, or separately applied, to each of the inside terms, b and c. For example, in our ﬁrst mental multiplication problem of 42 7, we arrived at the answer by treating 42 as 40 2, then distributing the 7 as follows: 42 7 (40 2) 7 (40 7) (2 7) 280 14 294 You may wonder why the distributive law works in the ﬁrst place. To understand it intuitively, imagine having 7 bags, each containing 42 coins, 40 of which are gold and 2 of which are sil-ver. How many coins do you have altogether? There are two ways to arrive at the answer. In the ﬁrst place, by the very def-inition of multiplication, there are 42 7 coins. On the other hand, there are 40 7 gold coins and 2 7 silver coins. Hence, Secrets of Mental Math 50 we have (40 7) (2 7) coins altogether. By answering our question two ways, we have 42 7 (40 7) (2 7). Notice that the numbers 7, 40, and 2 could be replaced by any numbers (a, b, or c) and the same logic would apply. That’s why the distributive law works! Using similar reasoning with gold, silver, and copper coins we can derive: (b c d) a (b a) (c a) (d a) Hence, to do the problem 326 7, we break up 326 as 300 20 6, then distribute the 7, as follows: 326 7 (300 20 6) 7 (300 7) (20 7) (6 7), which we then add up to get our answer. As for squaring, the following algebra justiﬁes my method. For any numbers A and d A2 (A d) (A d) d2 Here, A is the number being squared; d can be any number, but I choose it to be the distance from A to the nearest multiple of 10. Hence, for 772, I set d 3 and our formula tells us that 772 (77 3) (77 3) 32 (80 74) 9 5929. The following algebraic relationship also works to explain my squar-ing method: (z d)2 z2 2zd d2 z(z 2d) d2 Hence, to square 41, we set z 40 and d 1 to get: 412 (40 1)2 40 (40 2) 12 1681 Products of a Misspent Youth: Basic Multiplication 51 Similarly, (z d)2 z(z 2d) d2 To ﬁnd 772 when z 80 and d 3, 772 (80 3)2 80 (80 6) 32 80 74 9 5929 Secrets of Mental Math 52 Chapter 3 New and Improved Products: Intermediate Multiplication Mathemagics really gets exciting when you perform in front of an audience. I experienced my ﬁrst public performance in eighth grade, at the fairly advanced age of thirteen. Many mathemagi-cians begin even earlier. Zerah Colburn (1804–1839), for exam-ple, reportedly could do lightning calculations before he could read or write, and he was entertaining audiences by the age of six! When I was thirteen, my algebra teacher did a problem on the board for which the answer was 1082. Not content to stop there, I blurted out, “108 squared is simply 11,664!” The teacher did the calculation on the board and arrived at the same answer. Looking a bit startled, she said, “Yes, that’s right. How did you do it?” So I told her, “I went down 8 to 100 and up 8 to 116. I then multiplied 116 100, which is 11,600, and just added the square of 8, to get 11,664.” She had never seen that method before. I was thrilled. Thoughts of “Benjamin’s Theorem” popped into my head. I actu-ally believed I had discovered something new. When I ﬁnally ran across this method a few years later in a book by Martin Gardner on recreational math, Mathematical Carnival (1965), it ruined my day! Still, the fact that I had discovered it for myself was very exciting to me. You, too, can impress your friends (or teachers) with some fairly amazing mental multiplication. At the end of the last chapter you learned how to multiply a two-digit number by itself. In this chapter you will learn how to multiply two differ-ent two-digit numbers, a challenging yet more creative task. You will then try your hand—or, more accurately, your brain—at three-digit squares. You do not have to know how to do 2-by-2 multiplication problems to tackle three-digit squares, so you can learn either skill ﬁrst. 2-BY -2 MULTIPLICATION PROBLEMS When squaring two-digit numbers, the method is always the same. When multiplying two-digit numbers, however, you can use lots of different methods to arrive at the same answer. For me, this is where the fun begins. The ﬁrst method you will learn is the “addition method,” which can be used to solve all 2-by-2 multiplication problems. The Addition Method To use the addition method to multiply any two two-digit num-bers, all you need to do is perform two 2-by-1 multiplication problems and add the results together. For example: (40 2) 46 42 40 46 1840 2 46 92 1932 Secrets of Mental Math 54 Here you break up 42 into 40 and 2, two numbers that are easy to multiply. Then you multiply 40 46, which is just 4 46 with a 0 attached, or 1840. Then you multiply 2 46 92. Finally, you add 1840 92 1932, as diagrammed above. Here’s another way to do the same problem: (40 6) The catch here is that multiplying 6 42 is harder to do than multiplying 2 46, as in the ﬁrst problem. Moreover, adding 1680 252 is more difﬁcult than adding 1840 92. So how do you decide which number to break up? I try to choose the number that will produce the easier addition problem. In most cases—but not all—you will want to break up the number with the smaller last digit because it usually produces a smaller sec-ond number for you to add. Now try your hand at the following problems: (70 3) (80 1) The last problem illustrates why numbers that end in 1 are especially attractive to break up. If both numbers end in the 81 59 80 59 4720 1 59 59 4779 48 73 70 48 3360 3 48 144 3504 46 42 40 42 1680 6 42 252 1932 New and Improved Products: Intermediate Multiplication 55 same digit, you should break up the larger number as illustrated below: (80 4) If one number is much larger than the other, it often pays to break up the larger number, even if it has a larger last digit. You will see what I mean when you try the following problem two different ways: (70 4) (10 3) Did you ﬁnd the ﬁrst method to be faster than the second? I did. Here’s another exception to the rule of breaking up the num-ber with the smaller last digit. When you multiply a number in the ﬁfties by an even number, you’ll want to break up the num-ber in the ﬁfties: (50 9) 84 59 50 84 4200 9 84 756 4956 74 13 10 74 740 3 74 222 962 74 13 70 13 910 4 13 52 962 84 34 80 34 2720 4 34 136 2856 Secrets of Mental Math 56 rest of your life. Once all future dates are taken care of, we can look back into the past and determine the days of the week for any date in the 1900s or any other century. Every year is assigned a code number, and for 2006 that year code happens to be 0 (see page 218). Now, to calculate the day of the week, you simply add the month code plus the date code plus the year code. Thus for December 3, 2006, we compute Month Code Date Year Code 4 3 0 7 Hence, this date will be on Day 7, which is Sunday. How about November 18, 2006? Since November has a month code of 2, we have Month Code Date Year Code 2 18 0 20. Now since the week repeats every seven days, we can sub-tract any multiple of 7 from our answer (7, 14, 21, 28, 35, . . .) and this will not change the day of the week. So our ﬁnal step is to subtract the biggest multiple of 7 to get 20 14 6. Hence November 18, 2006, occurs on Saturday. How about 2007? Well, what happens to your birthday as you go from one year to the next? For most years, there are 365 days, and since 364 is a multiple of 7 (7 52 364), then the day of the week of your birthday will shift forward by one day in most years. If there are 366 days between your birthdays, then it will shift forward by two days. Hence, for 2007 we cal-culate the day of the week just as before, but now we use a year code of 1. Next, 2008 is a leap year. (Leap years occur every four years, so 2000, 2004, 2008, 2012, . . . , 2096 are the leap years of the twenty-ﬁrst century.) Hence, for 2008, the year Secrets of Mental Math 216 The last digit of the number 84 is smaller than the last digit of 59, but if you break up 59, your product will be a multiple of 100, just as 4200 is, in the example above. This makes the sub-sequent addition problem much easier. Now try an easy problem of a different sort: (10 1) Though the calculation above is pretty simple, there is an even easier and faster way to multiply any two-digit number by 11. This is mathemagics at its best: you won’t believe your eyes when you see it (unless you remember it from Chapter 0)! Here’s how it works. Suppose you have a two-digit number whose digits add up to 9 or less. To multiply this number by 11, merely add the two digits together and insert the total between the original two digits. For example, to do 42 11, ﬁrst do 4 2 6. If you place the 6 between the 4 and the 2, you get 462, the answer to the problem! 462 Try 54 11 by this method. 594 5 4 9 54 11 4 2 6 42 11 42 11 10 42 420 1 42 42 462 New and Improved Products: Intermediate Multiplication 57 (70 2) If you got the right answer the ﬁrst or second time, pat your-self on the back. The 2-by-2 multiplication problems really do not get any tougher than this. If you did not get the answer right away, don’t worry. In the next two sections, I’ll teach you some much easier strategies for dealing with problems like this. But before you read on, practice the addition method on the follow-ing multiplication problems. EXERCISE: 2-BY -2 ADDITION-METHOD MULTIPLICATION PROBLEMS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. The Subtraction Method The subtraction method really comes in handy when one of the numbers you want to multiply ends in 8 or 9. The following problem illustrates what I mean: 85 11 34 11 92 35 88 76 62 94 23 84 77 43 53 58 59 26 27 18 31 41 89 72 70 89 6230 2 89 178 6408 New and Improved Products: Intermediate Multiplication 59 (60 1) Although most people ﬁnd addition easier than subtraction, it is usually easier to subtract a small number than to add a big number. (If we had done this problem by the addition method, we would have added 850 153 1003.) Now let’s do the challenging problem from the end of the last section: (90 1) Wasn’t that a whole lot easier? Now, here’s a problem where one number ends in 8: (90 2) In this case you should treat 88 as 90 2, then multiply 90 23 2070. But you multiplied by too much. How much? By 2 23, or 46 too much. So subtract 46 from 2070 to arrive at 2024, the ﬁnal answer. 88 23 90 23 2070 2 23 46 2024 89 72 90 72 6480 1 72 72 6408 59 17 60 17 1020 1 17 17 1003 Secrets of Mental Math 60 I want to emphasize here that it is important to work out these problems in your head and not simply look to see how we did it in the diagram. Go through them and say the steps to yourself or even out loud to reinforce your thoughts. Not only do I use the subtraction method with numbers that end in 8 or 9, but also for numbers in the high 90s because 100 is such a convenient number to multiply. For example, if some-one asked me to multiply 96 73, I would immediately round up 96 to 100: (100 4) When the subtraction component of a multiplication prob-lem requires you to borrow a number, using complements (as we learned in Chapter 1) can help you arrive at the answer more quickly. You’ll see what I mean as you work your way through the problems below. For example, subtract 340 78. We know the answer will be in the 200s. The difference between 40 and 78 is 38. Now take the complement of 38 to get 62. And that’s the answer, 262! 78 40 38 Complement of 38 62 Now let’s try another problem: 340 78 262 96 73 100 73 7300 4 73 292 7008 New and Improved Products: Intermediate Multiplication 61 (90 2) There are two ways to perform the subtraction component of this problem. The “long” way subtracts 200 and adds back 48: 6840 152 6640 48 6688 (ﬁrst subtract 200) (then add 48) The short way is to realize that the answer will be 66 hundred and something. To determine something, we subtract 52 40 12 and then ﬁnd the complement of 12, which is 88. Hence the answer is 6688. Try this one. (60 1) Again, you can see that the answer will be 3900 and some-thing. Because 67 20 47, the complement 53 means the answer is 3953. As you may have realized, you can use this method with any subtraction problem that requires you to borrow a number, not just those that are part of a multiplication problem. All of this is further proof, if you need it, that complements are a very pow-erful tool in mathemagics. Master this technique and, pretty soon, people will be complimenting you! 67 59 60 67 4020 1 67 67 3953 88 76 90 76 6840 2 76 152 Secrets of Mental Math 62 EXERCISE: 2-BY -2 SUBTRACTION-METHOD MULTIPLICATION PROBLEMS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. The Factoring Method The factoring method is my favorite method of multiplying two-digit numbers since it involves no addition or subtraction at all. You use it when one of the numbers in a two-digit multiplica-tion problem can be factored into one-digit numbers. To factor a number means to break it down into one-digit numbers that, when multiplied together, give the original num-ber. For example, the number 24 can be factored into 8 3 or 6 4. (It can also be factored into 12 2, but we prefer to use only single-digit factors.) Here are some other examples of factored numbers: 42 7 6 63 9 7 84 7 6 2 or 7 4 3 88 49 57 39 85 38 87 22 37 19 79 54 96 29 68 38 47 59 98 43 29 45 New and Improved Products: Intermediate Multiplication 63 To see how factoring makes multiplication easier, consider the following problem: 7 6 Previously we solved this problem by multiplying 46 40 and 46 2 and adding the products together. To use the factor-ing method, treat 42 as 7 6 and begin by multiplying 46 7, which is 322. Then multiply 322 6 for the ﬁnal answer of 1932. You already know how to do 2-by-1 and 3-by-1 multipli-cation problems, so this should not be too hard: 46 42 46 (7 6) (46 7) 6 322 6 1932 Of course, this problem could also have been solved by revers-ing the factors of 42: 46 42 46 (6 7) (46 6) 7 276 7 1932 In this case, it is easier to multiply 322 6 than it is to multi-ply 276 7. In most cases, I like to use the larger factor in solv-ing the initial 2-by-1 problem and to reserve the smaller factor for the 3-by-1 component of the problem. Factoring results in a 2-by-2 multiplication problem being simpliﬁed to an easier 3-by-1 (or sometimes 2-by-1) multiplica-tion problem. The advantage of the factoring method in mental calculation is you do not have to hold much in memory. Let’s look at another example, 75 63: 75 63 75 (9 7) (75 9) 7 675 7 4725 46 42 Secrets of Mental Math 64 As before, you simplify this 2-by-2 problem by factoring 63 into 9 7 and then multiplying 75 by these factors. (By the way, the reason we can shift parentheses in the second step is the associative law of multiplication.) 63 75 63 (5 5 3) (63 5) 5 3 315 5 3 1575 3 4725 Try the following problem for practice: 57 24 57 8 3 456 3 1368 You could have factored 24 as 6 4 for another easy com-putation: 57 24 57 6 4 342 4 1368 Compare this approach with the addition method: (20 4) or (50 7) With the addition method, you have to perform two 2-by-1 problems and then add. With the factoring method, you have just two multiplication problems: a 2-by-1 and a 3-by-1, and then you are done. The factoring method is usually easier on your memory. 57 24 50 24 1200 7 24 168 1368 57 24 20 57 1140 4 57 228 1368 New and Improved Products: Intermediate Multiplication 65 Remember that challenging multiplication problem earlier in this chapter? Here it is again: We tackled that problem easily enough with the subtraction method, but factoring works even faster: 89 72 89 9 8 801 8 6408 The problem is especially easy because of the 0 in the middle of 801. Our next example illustrates that it sometimes pays to factor the numbers in an order that exploits this situation. Let’s look at two ways of computing 67 42: 67 42 67 7 6 469 6 2814 67 42 67 6 7 402 7 2814 Ordinarily you should factor 42 into 7 6, as in the ﬁrst example, following the rule of using the larger factor ﬁrst. But the problem is easier to solve if you factor 42 into 6 7 because it creates a number with a 0 in the center, which is easier to mul-tiply. I call such numbers friendly products. Look for the friendly product in the problem done two ways below: 43 56 43 8 7 344 7 2408 43 56 43 7 8 301 8 2408 Did you think the second way was easier? When using the factoring method, it pays to ﬁnd friendly 89 72 Secrets of Mental Math 66 products whenever you can. The following list should help. I don’t expect you to memorize it so much as to familiarize your-self with it. With practice you will be able to nose out friendly products more often, and the list will become more meaningful. Numbers with Friendly Products 12: 12 9 108 13: 13 8 104 15: 15 7 105 17: 17 6 102 18: 18 6 108 21: 21 5 105 23: 23 9 207 25: 25 4 100, 25 8 200 26: 26 4 104, 26 8 208 27: 27 4 108 29: 29 7 203 34: 34 3 102, 34 6 204, 34 9 306 35: 35 3 105 36: 36 3 108 38: 38 8 304 41: 41 5 205 43: 43 7 301 44: 44 7 308 45: 45 9 405 51: 51 2 102, 51 4 204, 51 6 306, 51 8 408 52: 52 2 104, 52 4 208 53: 53 2 106 54: 54 2 108 56: 56 9 504 61: 61 5 305 63: 63 8 504 New and Improved Products: Intermediate Multiplication 67 67: 67 3 201, 67 6 402, 67 9 603 68: 68 3 204, 68 6 408 72: 72 7 504 76: 76 4 304, 76 8 608 77: 77 4 308 78: 78 9 702 81: 81 5 405 84: 84 6 504 88: 88 8 704 89: 89 9 801 Previously in this chapter you learned how easy it is to multi-ply numbers by 11. It usually pays to use the factoring method when one of the numbers is a multiple of 11, as in the examples below: 52 33 52 11 3 572 3 1716 83 66 83 11 6 913 6 5478 EXERCISE: 2-BY -2 FACTORING-METHOD MULTIPLICATION PROBLEMS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 48 37 45 36 62 77 33 16 85 42 72 17 83 18 56 29 81 48 57 14 86 28 27 14 Secrets of Mental Math 68 APPROACHING MULTIPLICATION CREATIVELY I mentioned at the beginning of the chapter that multiplication problems are fun because they can be solved any number of ways. Now that you know what I mean, let’s apply all three methods explained in this chapter to a single problem, 73 49. We’ll begin by using the addition method: (70 3) Now try the subtraction method: (50 1) Note that the last two digits of the subtraction could be obtained by adding 50 (complement of 73) 50 27 77 or by simply taking the complement of (73 50) complement of 23 77. Finally, try the factoring method: 73 49 73 7 7 511 7 3577 Congratulations! You have mastered 2-by-2 multiplication and now have all the basic skills you need to be a fast mental calculator. All you need to become a lightning calculator is more practice! 73 49 50 73 3650 1 73 73 3577 73 49 70 49 3430 3 49 147 3577 New and Improved Products: Intermediate Multiplication 69 EXERCISE: 2-BY -2 GENERAL MULTIPLICATION—ANYTHING GOES! Many of the following exercises can be solved by more than one method. Try computing them in as many ways as you can think of, then check your answers and computations at the back of the book. Our answers suggest various ways the problem can be mathemagically solved, starting with what I think is the easiest method. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. The following 2-by-2s occur as subproblems to problems appearing later when we do 3-by-2s, 3-by-3s, and 5-by-5s. You can do these problems now for practice, and refer back to them when they are used in the larger problems. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 41 15 29 26 52 47 91 46 83 58 53 53 54 53 36 41 61 37 74 62 43 75 43 76 38 63 57 73 37 72 59 21 56 37 67 58 87 87 92 53 77 36 89 55 73 18 81 57 53 39 Secrets of Mental Math 70 26. 27. 28. 29. 30. 31. 32. 33. THREE-DIGIT SQUARES Squaring three-digit numbers is an impressive feat of mental prestidigitation. Just as you square two-digit numbers by round-ing up or down to the nearest multiple of 10, to square three-digit numbers, you round up or down to the nearest multiple of 100. Take 193: 7 200 1932 37,200 72 37,249 7 186 By rounding up to 200 and down to 186, you’ve transformed a 3-by-3 multiplication problem into a far simpler 3-by-1 prob-lem. After all, 200 186 is just 2 186 372 with two zeros attached. Almost done! Now all you have to add is 72 49 to arrive at 37,249. Now try squaring 706: 6 712 7062 498,400 62 498,436 6 700 41 93 95 26 65 69 65 47 95 81 69 78 34 27 65 19 New and Improved Products: Intermediate Multiplication 71 Rounding down by 6 to 700 requires you to round up by 6 to 712. Since 712 7 4984 (a simple 3-by-1 problem), 712 700 498,400. After adding 62 36, you arrive at 498,436. These last problems are not terribly hard because there is no real addition involved. Moreover, you know the answers to 62 and 72 by heart. Squaring a number that’s farther away from a multiple of 100 is a tougher proposition. Try your hand at 3142: 14 328 3142 98,400 142 98,596 14 300 4 18 142 180 42 196 4 10 For this three-digit square, go down 14 to 300 and up 14 to 328, then multiply 328 3 984. Tack on two 0s to arrive at 98,400. Then add the square of 14. If 142 196 comes to you in a ﬂash (through memory or calculation), you’re in good shape. Just add 98,400 196 to arrive at 98,596. If you need time to compute 142, repeat the number 98,400 to yourself a few times before you go on. (Otherwise you might compute 142 196 and forget what number to add it to.) The farther away you get from a multiple of 100, the more difﬁcult squaring a three-digit number becomes. Try 5292: 29 558 5292 279,000 292 279,841 29 500 1 30 292 840 12 841 1 28 Secrets of Mental Math 72 If you have an audience you want to impress, you can say 279,000 out loud before you compute 292. But this will not work for every problem. For instance, try squaring 636: 36 672 6362 403,200 362 404,496 36 600 4 40 362 1,280 42 1,296 4 32 Now your brain is really working, right? The key here is to repeat 403,200 to yourself several times. Then square 36 to get 1,296 in the usual way. The hard part comes in adding 1,296 to 403,200. Do it one digit at a time, left to right, to arrive at your answer of 404,496. Take my word that as you become more familiar with two-digit squares, these three-digit problems get easier. Here’s an even tougher problem, 8632: 900 8632 8?? The ﬁrst problem is deciding what numbers to multiply together. Clearly one of the numbers will be 900, and the other number will be in the 800s. But what number? You can com-pute it two ways: 1. The hard way: the difference between 863 and 900 is 37 (the complement of 63). Subtract 37 from 863 to arrive at 826. New and Improved Products: Intermediate Multiplication 73 2. The easy way: double the number 63 to get 126, and take the last two digits to give you 826. Here’s why the easy way works. Because both numbers are the same distance from 863, their sum must be twice 863, or 1726. One of your numbers is 900, so the other must be 826. You then compute the problem like this: 37 900 8632 743,400 372 744,769 37 826 3 40 372 1,360 32 1,369 3 34 If you ﬁnd it impossible to remember 743,400 after squaring 37, fear not. In a later chapter, you will learn a memory system that will make remembering such numbers much easier. Try your hand at squaring 359, the hardest problem yet: 41 400 3592 127,200 412 128,881 41 318 1 42 412 1,680 12 1,681 1 40 To obtain 318, either subtract 41 (the complement of 59) from 359, or multiply 2 59 118 and use the last two digits. Next multiply 400 318 127,200. Adding 412, or 1,681, gives you 128,881. Whew! They don’t get much harder than that! If you got it right the ﬁrst time, take a bow! Let’s ﬁnish this section with a big problem that is easy to do, 9872: Secrets of Mental Math 74 New and Improved Products: Intermediate Multiplication 75 What’s Behind Door Number 1? T he mathematical chestnut of 1991 that got everyone hopping mad was an article in Parade magazine by Marilyn vos Savant, the woman listed by the Guinness Book of World Records as having the world’s highest IQ.The paradox has come to be known as the Monty Hall problem, and it goes like this. You are a contestant on Let’s Make a Deal. Monty Hall allows you to pick one of three doors;behind one of these doors is the big prize, behind the other two are goats.You pick Door Number 2. But before Monty reveals the prize of your choice,he shows you what you didn’t pick behind Door Number 3. It’s a goat. Now, in his tantalizing way, Monty gives you another choice. Do you want to stick with Door Number 2, or do you want to risk a chance to see what’s behind Door Number 1? What should you do? Assuming that Monty is only going to reveal where the big prize is not, he will always open one of the consolation doors.This leaves two doors, one with the big prize and the other with another consolation.The odds are now 50-50 for your choice, right? Wrong! The odds that you chose correctly the ﬁrst time remain 1 in 3. The probability that the big prize is behind the other door increases to 2 in 3 because the probability must add to 1. Thus, by switching doors, you double the odds of winning! (The problem assumes that Monty will always give a player the option to switch, that he will always reveal a nonwinning door, and that when your ﬁrst pick is correct he will choose a nonwinning door at ran-dom.) Think of playing the game with ten doors and after your pick he reveals eight other nonwinning doors. Here, your instincts would probably tell you to switch. People confuse this problem for a variant: if Monty Hall does not know where the grand prize is, and reveals Door Number 3, which happens to contain a goat (though it might have contained the prize), then Door Number 1 has a 50 percent chance of being correct.This result is so counterintuitive that Marilyn vos Savant received piles of letters, many from scientists and even mathematicians, telling her she shouldn’t write about math. They were all wrong. 13 1,000 9872 974,000 132 974,169 13 974 3 16 132 160 32 169 3 10 EXERCISE: THREE-DIGIT SQUARES 1. 4092 2. 8052 3. 2172 4. 8962 5. 3452 6. 3462 7. 2762 8. 6822 9. 4312 10. 7812 11. 9752 CUBING We end this chapter with a new method for cubing two-digit numbers. (Recall that the cube of a number is that number mul-tiplied by itself twice. For example, 5 cubed—denoted 53—is equal to 5 5 5 125.) As you will see, this is not much harder than multiplying two-digit numbers. The method is based on the algebraic observation that A3 (A d)A(A d) d2A where d is any number. Just like with squaring two-digit num-bers, I choose d to be the distance to the nearest multiple of ten. For example, when squaring 13, we let d 3, resulting in: 133 (10 13 16) (32 13) Since 13 16 13 4 4 52 4 208, and 9 13 117, we have Secrets of Mental Math 76 133 2080 117 2197 How about the cube of 35? Letting d 5, we get 353 (30 35 40) (52 35) Since 30 35 40 30 1,400 42,000 and 35 5 5 175 5 875, we get 353 42,000 875 42,875 When cubing 49, we let d 1 in order to round up to 50. Here 493 (48 49 50) (12 49) We can solve 48 49 by the factoring method, but for this sort of problem, I prefer to use the close-together method, which will be described in Chapter 8. (Go ahead and look at it now, if you’d like!) Using that method, we get 48 49 (50 47) (1 2) 2352. Multiplying this by 50, we get 117,600 and therefore 493 117,600 49 117,649 Here’s a larger one. Try the cube of 92. 923 (90 92 94) (22 92) If you are fast at squaring two-digit numbers, then you could solve 92 94 932 1 8648, or you could use the close-together method, resulting in 92 94 (90 96) (2 4) 8648. Multiplying this by 9 (as described in the beginning of New and Improved Products: Intermediate Multiplication 77 Chapter 8), we obtain 9 (8,600 48) 77,400 432 77,832, and therefore 90 92 94 778,320. Since 4 92 368, we get 923 778,320 368 778,688 We note that when the close-together method is used for the multiplication problems that arise when cubing two-digit num-bers, the small product being added will either be (depending on whether d 1, 2, 3, 4, or 5) 1 2 2, 2 4 8, 3 6 18, 4 8 32, or 5 10 50. Let’s ﬁnish with the cube of 96. 963 (92 96 100) (42 96) The product 92 96 8,832 can be done many different ways. To celebrate the end of this chapter, let’s do some of them. I’ll start with what I consider to be the hardest way, and end with what I consider the easiest way. By the addition method, (90 2) 96 8,640 192 8,832; by the subtraction method, 92 (100 4) 9,200 368 8,832; by the factoring method, 92 6 4 4 552 4 4 2,208 4 8,832; by squaring, 942 22 8,836 4 8,832; by the close-together method with a base of 90, (90 98) (2 6) 8,820 12 8,832; and by the close-together method with a base of 100, (100 88) (8 4) 8,800 32 8,832. The product 42 96 1,536 can also be done a few differ-ent ways, such as 96 4 4 384 4 1,536, or as 16 (100 4) 1,600 64 1,536. Finally, since 8,832 100 883,200, we have 963 883,200 1,536 884,736 Secrets of Mental Math 78 EXERCISE: TWO-DIGIT CUBES 1. 123 2. 173 3. 213 4. 283 5. 333 6. 393 7. 403 8. 443 9. 523 10. 563 11. 653 12. 713 13. 783 14. 853 15. 873 16. 993 New and Improved Products: Intermediate Multiplication 79 Chapter 4 Divide and Conquer: Mental Division Mental division is a particularly handy skill to have, both in business and in daily life. How many times a week are you confronted with situations that call on you to evenly divide things, such as a check at a restaurant? This same skill comes in handy when you want to ﬁgure out the cost per unit of a case of dog food on sale, or to split the pot in poker, or to ﬁg-ure out how many gallons of gas you can buy with a $20 bill. The ability to divide in your head can save you the inconve-nience of having to pull out a calculator every time you need to compute something. With mental division, the left-to-right method of calculation comes into its own. This is the same method we all learned in school, so you will be doing what comes naturally. I remember as a kid thinking that this left-to-right method of division is the way all arithmetic should be done. I have often speculated that if the schools could have ﬁgured out a way to teach division right-to-left, they probably would have done so! ONE-DIGIT DIVISION The ﬁrst step when dividing mentally is to ﬁgure out how many digits will be in your answer. To see what I mean, try on the fol-lowing problem for size: 179 7 To solve 179 7, we’re looking for a number, Q, such that 7 times Q is 179. Now, since 179 lies between 7 10 70 and 7 100 700, Q must lie between 10 and 100, which means our answer must be a two-digit number. Knowing that, we ﬁrst determine the largest multiple of 10 that can be multiplied by 7 whose answer is below 179. We know that 7 20 140 and 7 30 210, so our answer must be in the twenties. At this point we can actually say the number 20 since that part of our answer will certainly not change. Next we subtract 179 140 39. Our problem has now been reduced to the division problem 39 7. Since 7 5 35, which is 4 away from 39, we have the rest of our answer, namely 5 with a remainder of 4, or 5 and 4 7 . Altogether, we have our answer, 25 with a remainder of 4, or if you prefer, 25 4 7 . Here’s what the process looks like: Answer: 25 with a remainder of 4, or 25 4 7 25 71 7 9 140 39 35 4 Divide and Conquer: Mental Division 81 remainder Let’s try a similar division problem using the same method of mental computation: 675 8 As before, since 675 falls between 8 10 80 and 8 100 800, your answer must be below 100 and therefore is a two-digit number. To divide 8 into 675, notice that 8 80 640 and 8 90 720. Therefore, your answer is 80 something. But what is that “something”? To ﬁnd out, subtract 640 from 675 for a remainder of 35. After saying the 80, our problem has been reduced to 35 8. Since 8 4 32, the ﬁnal answer is 84 with a remainder of 3, or 84 3 8 . We illustrate this problem as follows: Answer: 84 with a remainder of 3, or 84 Like most mental calculations, division can be thought of as a process of simpliﬁcation. The more you calculate, the simpler the problem becomes. What began as 675 8 was simpliﬁed to a smaller problem, 35 8. Now let’s try a division problem that results in a three-digit answer: 947 4 3 8 84 86 7 5 640 35 32 3 Secrets of Mental Math 82 remainder This time, your answer will have three digits because 947 falls between 4 100 400 and 4 1000 4000. Thus we must ﬁrst ﬁnd the largest multiple of 100 that can be squeezed into 947. Since 4 200 800, our answer is deﬁnitely in the 200s, so go ahead and say it! Subtracting 800 from 947 gives us our new division problem, 147 4. Since 4 30 120, we can now say the 30. After subtracting 120 from 147, we compute 27 4 to obtain the rest of the answer: 6 with a remainder of 3. Altogether, we have 236 with a remainder of 3, or 236 3 4 . Answer: 236 The process is just as easy when dividing a one-digit number into a four-digit number, as in our next example. 2196 5 Here the answer will be in the hundreds because 2196 is between 5 100 500 and 5 1000 5000. After subtract-ing 5 400 2000, we can say the 400, and our problem has reduced to 196 5, which can be solved as in the previous examples. 3 4 236 49 4 7 800 147 120 27 24 3 Divide and Conquer: Mental Division 83 remainder Answer: 439 Actually, there is a much easier way to solve this last problem. We can simplify our problem by doubling both numbers. Since 2196 2 4392, we have 2196 5 4392 10 439.2 or 439 1 2 0 . We’ll see more division shortcuts in the next section. EXERCISE: ONE-DIGIT DIVISION 1. 318 9 2. 726 5 3. 428 7 4. 289 8 5. 1328 3 6. 2782 4 THE RULE OF “THUMB” When dividing in your head instead of on paper, you may ﬁnd it difﬁcult to remember parts of the answer as you continue to cal-culate. One option, as you’ve seen, is to say the answer out loud as you go. But for greater dramatic effect, you may prefer, as I do, to hold the answer on your ﬁngers and say it all together at the end. In that case, you may run into problems remembering digits greater than ﬁve if, like most of us, you have only ﬁve ﬁn-gers on each hand. The solution is to use a special technique, based on sign language, which I call the Rule of “Thumb.” It is most effective for remembering three-digit and greater numbers. 1 5 439 52 1 9 6 2000 196 150 46 45 1 Secrets of Mental Math 84 This technique not only is useful in this chapter but also will come in handy (pardon the pun) in subsequent chapters dealing with larger problems and longer numbers to remember. You already know that to represent numbers 0 through 5, all you have to do is raise the equivalent number of ﬁngers on your hand. When you get your thumb involved, it’s just as easy to represent numbers 6 through 9. Here are the Rules of “Thumb”: • To hold on to 6, place your thumb on top of your pinky. • To hold on to 7, place your thumb on top of your ring ﬁnger. • To hold on to 8, place your thumb on top of your middle ﬁnger. • To hold on to 9, place your thumb on top of your index ﬁnger. With three-digit answers, hold the hundreds digit on your left hand and the tens digit on your right. When you get to the ones digit, you’ve reached the end of the problem (except for a possi-ble remainder). Now say the number on your left hand, the num-ber on your right hand, the one-digit you’ve just computed, and the remainder (in your head). Presto—you’ve said your answer! For practice, try to compute the following four-digit division problem: 4579 6 Answer: 763 1 6 763 64 5 7 9 4200 379 360 19 18 1 Divide and Conquer: Mental Division 85 In using the Rule of “Thumb” to remember the answer, you’ll hold the 7 on your left hand by placing your thumb and ring ﬁnger together and the 6 on your right hand by placing your thumb and little ﬁnger together. Once you’ve calculated the ones digit (which is 3) and the remainder (which is 1), you can “read” the ﬁnal answer off your hands from left to right: “seven . . . six . . . three with a remainder of one, or one-sixth.” Some four-digit division problems yield four-digit answers. In that case, since you have only two hands, you will have to say the thousands digit of the answer out loud and use the rule of thumb to remember the rest of the answer. For example: 8352 3 Answer: 2784 For this problem, you divide 3 into 8 to get your thousands digit of 2, say “two thousand” out loud, then divide 3 into 2352 in the usual way. 2784 38 3 5 2 6000 2352 2100 252 240 12 12 0 Secrets of Mental Math 86 TWO-DIGIT DIVISION This section assumes you already have mastered the art of divid-ing by a one-digit number. Naturally, division problems become harder as the number you divide by gets larger. Fortunately, I have some magic up my sleeve to make your life easier. Let’s start with a relatively easy problem ﬁrst: 597 14 Since 597 lies between 14 10 and 14 100, the answer (also called the quotient) lies between 10 and 100. To determine the answer, your ﬁrst step is to ask how many times 14 goes into 590. Because 14 40 560, you know that the answer is 40 something, and so you can say “forty” out loud. Next, subtract 560 from 597, which is 37 and reduces your problem to dividing 14 into 37. Since 14 2 28, your answer is 42. Subtracting 28 from 37 leaves you a remainder of 9. The process of deriving the solution to this problem may be illus-trated as follows: Answer: 42 The following problem is slightly harder because the two-digit divisor (here 23) is larger. 9 14 42 145 9 7 560 37 28 9 Divide and Conquer: Mental Division 87 682 23 In this problem, the answer is a two-digit number because 682 falls between 23 10 230 and 23 100 2300. To ﬁgure out the tens digit of the two-digit answer, you need to ask how many times 23 goes into 680. If you try 30, you’ll see that it’s slightly too much, as 30 23 690. Now you know that the answer is 20 something, and you can say so. Then subtract 23 20 460 from 682 to obtain 222. Since 23 9 207, the answer is 29 with a remainder of 222 207 15. Answer: 29 Now consider: 491 62 Since 491 is less than 62 10 620, your answer will sim-ply be a one-digit number with a remainder. You might guess 8, but 62 8 496, which is a little high. Since 62 7 434, the answer is 7 with a remainder of 491 434 57, or 7 5 6 7 2 . 7 624 9 1 434 57 15 23 29 236 8 2 460 222 207 15 Secrets of Mental Math 88 Answer: 7 Actually, there’s a nifty trick to make problems like this eas-ier. Remember how you ﬁrst tried multiplying 62 8, but found it came out a little high at 496? Well, that wasn’t a wasted effort. Aside from knowing that the answer is 7, you can also compute the remainder right away. Since 496 is 5 more than 491, the remainder will be 5 less than 62, the divisor. Since 62 5 57, your answer is 7 6 5 2 7 . The reason this trick works is because 491 (62 8) 5 62 (7 1) 5 (62 7 62) 5 (62 7) (62 5) 62 7 57. Now try 380 39 using the shortcut we just learned. So 39 10 390, which is too high by 10. Hence the answer is 9 with a remainder of 39 10 29. Your next challenge is to divide a two-digit number into a four-digit number: 3657 54 Since 54 100 5400, you know your answer will be a two-digit number. To arrive at the ﬁrst digit of the answer, you need to ﬁgure how many times 54 goes into 3657. Since 54 70 3780 is a little high, you know the answer must be 60 something. Next, multiply 54 60 3240 and subtract 3657 3240 417. Once you say the 60, your problem has been simpliﬁed to 417 54. Since 54 8 432 is a little too high, your last digit is 7 with the remainder 54 15 39. 57 62 Divide and Conquer: Mental Division 89 Answer: 67 Now try your hand at a problem with a three-digit answer: 9467 13 Answer: 728 Simplifying Division Problems If by this point you’re suffering from brain strain, relax. As promised, I want to share with you a couple of tricks for simpli-fying certain mental division problems. These tricks are based on the principle of dividing both parts of the problem by a com-mon factor. If both numbers in the problem are even numbers, you can make the problem twice as easy by dividing each num-3 13 728 139 4 6 7 9100 367 260 107 104 3 39 54 67 543 6 5 7 3240 417 378 39 Secrets of Mental Math 90 ber by 2 before you begin. For example, 858 16 has two even numbers, and dividing each by 2 yields the much simpler prob-lem of 429 8: Divide by 2 Answer: 53 Answer: 53 As you can see, the remainders 10 and 5 are not the same; but if you write the remainder in the form of a fraction, you get that 1 1 0 6 is the same as 5 8 . Therefore, when using this method, you must always express the answer in fractional form. We’ve done both sets of calculations for you to see how much easier it is. Now you try one for practice: 3618 54 Divide by 2 Answer: 67 The problem on the right is much easier to calculate men-tally. If you’re really alert, you could divide both sides of the 67 271 8 0 9 1620 189 189 0 67 543 6 1 8 3240 378 378 0 5 8 10 16 53 84 2 9 400 29 24 5 53 168 5 8 800 58 48 10 Divide and Conquer: Mental Division 91 original problem by 18 to arrive at an even simpler problem: 201 3 67. Watch for problems that can be divided by 2 twice, such as 1652 36: Answer: 45 I usually ﬁnd it easier to divide the problem by 2 twice than to divide both numbers by 4. Next, when both numbers end in 0, you can divide each by 10: Answer: 8 But if both numbers end in 5, double them and then divide both by 10 to simplify the problem. For example: 13 79 5 70 25 21 4 475 35 950 70 95 7 2 10 2 7 8 75 8 56 2 580 70 58 7 10 8 9 45 94 1 3 360 53 45 8 1652 36 826 18 413 9 2 2 Secrets of Mental Math 92 Answer: 13 Finally, if the divisor ends in 5 and the number you’re divid-ing into ends in 0, multiply both by 2 and then divide by 10, just as you did above: Answer: 19 EXERCISE: TWO-DIGIT DIVISION Here you’ll ﬁnd a variety of two-digit division problems that will test your mental prowess, using the simpliﬁcation tech-niques explained earlier in this chapter. Check the end of the book for answers and explanations. 1. 738 17 2. 591 24 3. 321 79 4. 4268 28 5. 7214 11 6. 3074 18 MATCHING WITS WITH A CALCULATOR: LEARNING DECIMALIZATION As you might guess, I like to work some magic when I convert fractions to decimals. In the case of one-digit fractions, the best 7 9 19 91 7 8 90 88 81 7 890 45 1780 90 178 9 2 10 4 7 Divide and Conquer: Mental Division 93 way is to commit the following fractions—from halves through elevenths—to memory. This isn’t as hard as it sounds. As you’ll see below, most one-digit fractions have special properties that make them hard to forget. Anytime you can reduce a fraction to one you already know, you’ll speed up the process. Chances are you already know the decimal equivalent of the following fractions: .50 1 3 .333 . . . 2 3 .666 . . . Likewise: 1 4 .25 2 4 1 2 .50 3 4 .75 The ﬁfths are easy to remember: 1 5 .20 2 5 .40 3 5 .60 4 5 .80 The sixths require memorizing only two new answers: 1 6 .1666 . . . 2 6 1 3 .333 . . . 3 6 1 2 .50 4 6 2 3 .666 . . . 5 6 .8333 . . . I’ll return to the sevenths in a moment. The eighths are a breeze: 1 8 .125 2 8 1 4 .25 3 8 .375 (3 1 8 3 .125 .375) 4 8 1 2 .50 1 2 Secrets of Mental Math 94 5 8 .625 (5 1 8 5 .125 .625) 6 8 3 4 .75 7 8 .875 (7 1 8 7 .125 .875) The ninths have a magic all their own: 1 9 .1 2 9 .2 3 9 .3 4 9 .4 5 9 .5 6 9 .6 7 9 .7 8 9 .8 where the bar indicates that the decimal repeats. For instance, 4 9 .4 .444. . . . The tenths you already know: 1 1 0 .10 1 2 0 .20 1 3 0 .30 1 4 0 .40 1 5 0 .50 1 6 0 .60 1 7 0 .70 1 8 0 .80 1 9 0 .90 For the elevenths, if you remember that 1 1 1 .0909, the rest is easy: 1 1 1 .0 9 .0909 . . . 1 2 1 .1 8 (2 .0909) 1 3 1 .2 7 (3 .0909) 1 4 1 .3 6 1 5 1 .4 5 Divide and Conquer: Mental Division 95 1 6 1 .5 4 1 7 1 .6 3 1 8 1 .7 2 1 9 1 .8 1 1 1 0 1 .9 0 The sevenths are truly remarkable. Once you memorize 1 7 .1 4 2 8 5 7 , you can get all the other sevenths without having to compute them: 1 7 .1 4 2 8 5 7 2 7 .2 8 5 7 1 4 3 7 .4 2 8 5 7 1 4 7 .5 7 1 4 2 8 5 7 .7 1 4 2 8 5 6 7 .8 5 7 1 4 2 Note that the same pattern of numbers repeats itself in each fraction. Only the starting point varies. You can ﬁgure out the starting point in a ﬂash by multiplying .14 by the numerator. In the case of 2 7 , 2 .14 .28, so use the sequence that begins with the 2, namely .2 8 5 7 1 4 . Likewise with 3 7 , since 3 .14 .42, use the sequence that begins with 4, namely .4 2 8 5 7 1 . The rest follow in a similar way. You will have to calculate fractions higher than 1 1 0 1 as you would any other division problem. However, keep your eyes peeled for ways of simplifying such problems. For example, you can simplify the fraction 1 3 8 4 by dividing both numbers by 2, to reduce it to 1 9 7 , which is easier to compute. If the denominator of the fraction is an even number, you can simplify the fraction by reducing it in half, even if the numerator is odd. For example: 4.5 7 9 14 Secrets of Mental Math 96 Dividing the numerator and denominator in half reduces it to a sevenths fraction. Although the sevenths sequence previously shown doesn’t provide the decimal for 4 7 .5 , once you begin the calculation, the number you memorized will pop up: As you can see, you needn’t work out the entire problem. Once you’ve reduced it to dividing 3 by 7, you can make a great impression on an audience by rattling off this long string of numbers almost instantly! When the divisor ends in 5, it almost always pays to double the problem, then divide by 10. For example, 2 4 9 5 5 9 8 0 5 9 .8 .64 4 2 10 Numbers that end in 25 or 75 should be multiplied by 4 before dividing by 100. 3 2 1 5 1 1 2 0 4 0 1.24 4 100 6 7 2 5 2 3 4 0 8 0 2. 3 48 .826 6 4 100 .64 2 8 5 7 1 74 .5 0 0 0 0 0 0 4.2 3 Divide and Conquer: Mental Division 97 You can even put this trick to use in the middle of a problem. If your fraction is 1 3 6 , look what happens: Once the problem is reduced to 1 1 4 6 , you can further reduce it to 7 8 , which you know to be .875. Thus 1 3 6 .1875. EXERCISE: DECIMALIZATION To solve the following problems, don’t forget to employ the var-ious one-digit fractions you already know as decimals. Wher-ever appropriate, simplify the fraction before converting it to a decimal. 1. 2 5 2. 4 7 3. 3 8 4. 1 9 2 5. 1 5 2 6. 1 6 1 7. 1 2 4 4 8. 1 2 3 7 9. 1 4 8 8 10. 1 1 0 4 11. 3 6 2 12. 1 4 9 5 TESTING FOR DIVISIBILITY In the last section, we saw how division problems could be sim-pliﬁed when both numbers were divisible by a common factor. We end this chapter with a brief discussion of how to determine whether one number is a factor of another number. Being able to ﬁnd the factors of a number helps us simplify division prob-lems and can speed up many multiplication problems. This will also be a very useful tool when we get to advanced multiplica-tion, as you will often be looking for ways to factor a two-, .1 163 .0 0 0 16 14 Secrets of Mental Math 98 three-, or even a ﬁve-digit number in the middle of a multiplica-tion problem. Being able to factor these numbers quickly is very handy. And besides, I think some of the rules are just beautiful. It’s easy to test whether a number is divisible by 2. All you need to do is to check if the last digit is even. If the last digit is 2, 4, 6, 8, or 0, the entire number is divisible by 2. To test whether a number is divisible by 4, check if the two-digit number at the end is divisible by 4. The number 57,852 is a multiple of 4 because 52 13 4. The number 69,346 is not a multiple of 4 because 46 is not a multiple of 4. The reason this works is because 4 divides evenly into 100 and thus into any multiple of 100. Thus, since 4 divides evenly into 57,800, and 4 divides into 52, we know that 4 divides evenly into their sum, 57,852. Likewise, since 8 divides into 1000, to test for divisibility by 8, check the last three digits of the number. For the number 14,918, divide 8 into 918. Since this leaves you with a remain-der (918 8 114 6 8 ), the number is not divisible by 8. You could also have observed this by noticing that 18 (the last two digits of 14,918) is not divisible by 4, and since 14,918 is not divisible by 4, it can’t be divisible by 8 either. When it comes to divisibility by 3, here’s a cool rule that’s easy to remember: A number is divisible by 3 if and only if the sum of its digits are divisible by 3—no matter how many digits are in the number. To test whether 57,852 is divisible by 3, sim-ply add 5 7 8 5 2 27. Since 27 is a multiple of 3, we know 57,852 is a multiple of 3. The same amazing rule holds true for divisibility by 9. A number is divisible by 9 if and only if its digits sum to a multiple of 9. Hence, 57,852 is a multiple of 9, whereas 31,416, which sums to 15, is not. The reason this works is based on the fact that the numbers 1, 10, 100, 1,000, 10,000, and so on, are all 1 greater than a multiple of 9. Divide and Conquer: Mental Division 99 A number is divisible by 6 if and only if it is even and divisi-ble by 3, so it is easy to test divisibility by 6. To establish whether a number is divisible by 5 is even easier. Any number, no matter how large, is a multiple of 5 if and only if it ends in 5 or 0. Establishing divisibility by 11 is almost as easy as determin-ing divisibility by 3 or 9. A number is divisible by 11 if and only if you arrive at either 0 or a multiple of 11 when you alternately subtract and add the digits of a number. For instance, 73,958 is not divisible by 11 since 7 3 9 5 8 16. However, the numbers 8,492 and 73,194 are multiples of 11, since 8 4 9 2 11 and 7 3 1 9 4 0. The reason this works is based, like the rule for 3s and 9s, on the fact that the numbers 1, 100, 10,000, and 1,000,000 are 1 more than a multiple of 11, whereas the numbers 10, 1,000, 100,000, and so on are 1 less than a multiple of 11. Testing for divisibility by 7 is a bit trickier. If you add or sub-tract a number that is a multiple of 7 to the number you are test-ing, and the resulting number is a multiple of 7, then the test is positive. I always choose to add or subtract a multiple of 7 so that the resulting sum or difference ends in 0. For example, to test the number 5292, I subtract 42 (a multiple of 7) to obtain 5250. Next, I get rid of the 0 at the end (since dividing by ten does not affect divisibility by seven), leaving me with 525. Then I repeat the process by adding 35 (a multiple of 7), which gives me 560. When I delete the 0, I’m left with 56, which I know to be a multiple of 7. Therefore, the original number 5292 is divis-ible by 7. This method works not only for 7s, but also for any odd number that doesn’t end in 5. For example, to test whether 8792 is divisible by 13, subtract 4 13 52 from 8792, to Secrets of Mental Math 100 arrive at 8740. Dropping the 0 results in 874. Then add 2 13 26 to arrive at 900. Dropping the two 0s leaves you with 9, which is clearly not a multiple of 13. Therefore, 8792 is not a multiple of 13. EXERCISE: TESTING FOR DIVISIBILITY In this ﬁnal set of exercises, be especially careful when you test for divisibility by 7 and 17. The rest should be easy for you. Divisibility by 2 1. 53,428 2. 293 3. 7241 4. 9846 Divisibility by 4 5. 3932 6. 67,348 7. 358 8. 57,929 Divisibility by 8 9. 59,366 10. 73,488 11. 248 12. 6111 Divisibility by 3 13. 83,671 14. 94,737 15. 7359 16. 3,267,486 Divisibility by 6 17. 5334 18. 67,386 19. 248 20. 5991 Divisibility by 9 21. 1234 22. 8469 23. 4,425,575 24. 314,159,265 Divisibility by 5 25. 47,830 26. 43,762 27. 56,785 28. 37,210 Divide and Conquer: Mental Division 101 Divisibility by 11 29. 53,867 30. 4969 31. 3828 32. 941,369 Divisibility by 7 33. 5784 34. 7336 35. 875 36. 1183 Divisibility by 17 37. 694 38. 629 39. 8273 40. 13,855 FRACTIONS If you can manipulate whole numbers, then doing arithmetic with fractions is almost as easy. In this section, we review the basic methods for adding, subtracting, multiplying, dividing, and simplifying fractions. Those already familiar with fractions can skip this section with no loss of continuity. Multiplying Fractions To multiply two fractions, simply multiply the top numbers (called the numerators), then multiply the bottom numbers (called the denominators). For example, What could be simpler! Try these exercises before going further. EXERCISE: MULTIPLYING FRACTIONS 1. 2. 3. 4. 7 8 9 10 3 4 6 7 11 7 4 9 2 7 3 5 5 18 5 9 1 2 8 15 4 5 2 3 Secrets of Mental Math 102 Dividing Fractions Dividing fractions is just as easy as multiplying fractions. There’s just one extra step. First, turn the second fraction upside down (this is called the reciprocal) then multiply. For instance, the reciprocal of 4 5 is 5 4 . Therefore, EXERCISE: DIVIDING FRACTIONS Now it’s your turn. Divide these fractions. 1. 2. 3. Simplifying Fractions Fractions can be thought of as little division problems. For instance, 6 3 is the same as 6 3 2. The fraction 1 4 is the same as 1 4 (which is .25 in decimal form). Now we know that when we multiply any number by 1, the number stays the same. For example, 3 5 3 5 1. But if we replace 1 with 2 2 , we get 3 5 3 5 1 3 5 2 2 1 6 0 . Hence, 3 5 1 6 0 . Likewise, if we replace 1 with 3 3 , we get 3 5 3 5 3 3 1 9 5 . In other words, if we multiply the numerator and denominator by the same number, we get a frac-tion that is equal to the ﬁrst fraction. For another example, 10 15 5 5 2 3 2 3 3 5 2 5 6 5 1 3 1 2 2 5 9 10 9 5 1 2 5 9 1 2 10 12 5 4 2 3 4 5 2 3 Divide and Conquer: Mental Division 103 It is also true that if we divide the numerator and denomina-tor by the same number, then we get a fraction that is equal to the ﬁrst one. For instance, This is called simplifying the fraction. EXERCISE: SIMPLIFYING FRACTIONS For the fractions below, can you ﬁnd an equal fraction whose denominator is 12? 1. 2. 3. 4. Simplify these fractions. 5. 6. 7. 8. Adding Fractions The easy case is that of equal denominators. If the denominators are equal, then we add the numerators and keep the same denominators. For instance, 6 7 2 7 4 7 4 5 1 5 3 5 20 36 24 36 6 15 8 10 5 2 3 4 5 6 1 3 5 7 5 5 25 35 25 35 2 3 2 2 4 6 4 6 Secrets of Mental Math 104 Sometimes we can simplify our answer. For instance, EXERCISE: ADDING FRACTIONS (EQUAL DENOMINATORS) 1. 2. 3. 4. The trickier case: unequal denominators. When the denomi-nators are not equal, then we replace our fractions with frac-tions where the denominators are equal. For instance, to add we notice that Therefore, To add we notice that 4 8 1 2 7 8 1 2 7 15 2 15 5 15 2 15 1 3 5 15 1 3 2 15 1 3 3 10 3 10 6 18 5 18 4 12 5 12 5 9 2 9 3 4 6 8 5 8 1 8 Divide and Conquer: Mental Division 105 So, To add we see that and So, EXERCISE: ADDING FRACTIONS (UNEQUAL DENOMINATORS) 1. 2. 3. 4. 5. 6. 7. Subtracting Fractions Subtracting fractions works very much like adding them. We have illustrated with examples and provided exercises for you to do. 1 2 4 8 1 8 5 8 2 7 2 7 4 7 2 5 1 5 3 5 5 9 2 11 3 5 3 7 3 4 2 3 5 21 2 7 1 5 1 3 5 18 1 6 1 10 1 5 11 15 6 15 5 15 2 5 1 3 6 15 2 5 5 15 1 3 2 5 1 3 11 8 7 8 4 8 7 8 1 2 Secrets of Mental Math 106 EXERCISE: SUBTRACTING FRACTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 2 8 9 2 5 4 7 1 16 7 8 2 3 3 4 3 5 9 10 1 15 4 5 5 18 13 18 8 7 12 7 3 11 8 11 1 24 15 24 16 24 5 8 2 3 1 28 7 28 8 28 1 4 2 7 3 8 7 8 4 8 7 8 1 2 3 8 4 8 7 8 1 2 7 8 1 5 3 15 2 15 5 15 2 15 1 3 Divide and Conquer: Mental Division 107 Chapter 5 Good Enough:The Art of “Guesstimation” So far you’ve been perfecting the mental techniques necessary to figure out the exact answers to math problems. Often, however, all you’ll want is a ballpark estimate. Say you’re getting quotes from different lenders on reﬁnancing your home. All you really need at this information-gathering stage is a ballpark estimate of what your monthly payments will be. Or say you’re settling a restaurant bill with a group of friends and you don’t want to ﬁg-ure each person’s bill to the penny. The guesstimation methods described in this chapter will make both these tasks—and many more just like them—much easier. Addition, subtraction, divi-sion, and multiplication all lend themselves to guesstimation. As usual, you’ll do your computations from left to right. ADDITION GUESSTIMATION Guesstimation is a good way to make your life easier when the numbers of a problem are too long to remember. The trick is to round the original numbers up or down: ª (ª means approximately) 8,000 6,000 14,000 8,367 5,819 14,186 Good Enough: The Art of “Guesstimation” 109 George Parker Bidder: The Calculating Engineer T he British have had their share of lightning calculators, and the mental performances of George Parker Bidder (1806–1878),born in Devonshire, were as impressive as any. Like most lightning calcula-tors, Bidder began to try his hand (and mind) at mental arithmetic as a young lad. Learning to count, add, subtract, multiply, and divide by playing with marbles, Bidder went on tour with his father at age nine. Almost no question was too difﬁcult for him to handle. “If the moon is 123,256 miles from the earth and sound travels four miles a minute, how long would it take for sound to travel from the earth to the moon?” The young Bidder, his face wrinkled in thought for nearly a minute, replied,“Twenty-one days, nine hours, thirty-four minutes.” (We know now that the distance is closer to 240,000 miles and sound cannot travel through the vacuum of space.) At age ten, Bidder mentally computed the square root of 119,550,669,121 as 345,761 in a mere thirty seconds. In 1818, Bidder and the American lightning cal-culator Zerah Colburn were paired in a mental calculating duel in which Bidder, apparently,“outnumbered” Colburn. Riding on his fame, George Bidder entered the University of Edin-burgh and went on to become one of the more respected engineers in England. In parliamentary debates over railroad conﬂicts, Bidder was frequently called as a witness, which made the opposition shud-der; as one said,“Nature had endowed him with particular qualities that did not place his opponents on a fair footing.” Unlike Colburn, who retired as a lightning calculator at age twenty, Bidder kept it up for his entire life. As late as 1878, in fact, just before his death, Bidder calculated the number of vibrations of light striking the eye in one second, based on the fact that there are 36,918 waves of red light per inch, and light travels at approximately 190,000 miles per second. Notice that we rounded the ﬁrst number down to the nearest thousand and the second number up. Since the exact answer is 14,186, our relative error is small. If you want to be more exact, instead of rounding off to the nearest thousand, round off to the nearest hundred: ª The answer is only 14 off from the exact answer, an error of less than .1%. This is what I call a good guesstimation! Try a ﬁve-digit addition problem, rounding to the nearest hundred: ª By rounding to the nearest hundred, our answer will always be off by less than 100. If the answer is larger than 10,000, your guesstimate will be within 1% of the exact answer. Now let’s try something wild: ª or If you round to the nearest million, you get an answer of 31 million, off by roughly 285,000. Not bad, but you can do better by rounding to the nearest hundred thousand, as we’ve shown in the right-hand column. Again your guesstimate will be within 23.9 million 7.4 million 31.3 million 24,000,000 7,000,000 31,000,000 23,859,379 7,426,087 31,285,466 46,200 19,400 65,600 46,187 19,378 65,565 8,400 5,800 14,200 8,367 5,819 14,186 Secrets of Mental Math 110 1% of the precise answer. If you can compute these smaller problems exactly, you can guesstimate the answer to any addi-tion problem. Guesstimating at the Supermarket Let’s try a real-world example. Have you ever gone to the store and wondered what the total is going to be before the cashier rings it up? For estimating the total, my technique is to round the prices to the nearest 50¢. For example, while the cashier is adding the numbers shown below on the left, I mentally add the numbers shown on the right: My ﬁnal ﬁgure is usually within a dollar of the exact total. SUBTRACTION GUESSTIMATION The way to guesstimate the answers to subtraction problems is the same—you round to the nearest thousand or hundreds digit, preferably the latter: $ 1.50 1.00 2.50 0.50 3.50 3.00 0.00 1.00 0.50 3.00 3.00 $19.50 $ 1.39 0.87 2.46 0.61 3.29 2.99 0.20 1.17 0.65 2.93 3.19 $19.75 Good Enough: The Art of “Guesstimation” 111 ª or You can see that rounding to the nearest thousand leaves you with an answer quite a bit off the mark. By rounding to the second digit (hundreds, in the example), your answer will usu-ally be within 3% of the exact answer. For this problem, your answer is off by only 52, a relative error of 2%. If you round to the third digit, the relative error will usually be below 1%. For instance: ª or By rounding the numbers to the third digit rather than to the second digit, you improve the accuracy of the estimate by a sig-niﬁcant amount. DIVISION GUESSTIMATION The ﬁrst, and most important, step in guesstimating the answer to a division problem is to determine the magnitude of the answer: ª Answer ª 9 2 3 thousand 9,667 9 65 8 ,0 0 0 54 4 9,644.5 65 7 ,8 6 7 .0 439,000 25,000 414,000 440,000 20,000 420,000 439,412 24,926 414,486 8,400 5,800 2,600 8,000 6,000 2,000 8,367 5,819 2,548 Secrets of Mental Math 112 The next step is to round off the larger numbers to the near-est thousand and change the 57,867 to 58,000. Dividing 6 into 58 gives you 9 with a remainder. But the most important com-ponent in this problem is where to place the 9. For example, multiplying 6 90 yields 540, while multiply-ing 6 900 yields 5,400, both of which are too small. But 6 9,000 54,000, which is pretty close to the answer. This tells you the answer is 9,000 and something. You can estimate just what that something is by ﬁrst subtracting 58 54 4. At this point you could bring down the 0 and divide 6 into 40, and so forth. But if you’re on your toes, you’ll realize that dividing 6 into 4 gives you 4 6 2 3 ª .667. Since you know the answer is 9,000 something, you’re now in a position to guess 9,667. In fact, the actual answer is 9,645—darn close! Division on this level is simple. But what about large division problems? Let’s say we want to compute, just for fun, the amount of money a professional athlete earns a day if he makes $5,000,000 a year: 365 days$ 5 ,0 0 0 ,0 0 0 First you must determine the magnitude of the answer. Does this player earn thousands every day? Well, 365 1000 365,000, which is too low. Does he earn tens of thousands every day? Well, 365 10,000 3,650,000, and that’s more like it. To guesstimate your answer, divide the ﬁrst two digits (or 36 into 50) and ﬁgure that’s 1 1 3 4 6 , or 1 1 7 8 . Since 18 goes into 70 about 4 times, your guess is that the athlete earns about $14,000. The exact answer is $13,698.63 per day. Not a bad estimate (and not a bad salary!). Here’s an astronomical calculation for you. How many sec-Good Enough: The Art of “Guesstimation” 113 onds does it take light to get from the sun to the earth? Well, light travels at 186,282 miles per second, and the sun is (on aver-age) 92,960,130 miles away. I doubt you’re particularly eager to attempt this problem by hand. Fortunately, it’s relatively simple to guesstimate an answer. First, simplify the problem: 186,2829 2 ,9 6 0 ,1 3 0 ª 1869 3 ,0 0 0 Now divide 186 into 930, which yields 5 with no remainder. Then append the two 0s you removed from 93,000 and you get 500 seconds. The exact answer is 499.02 seconds, so this is a very respectable guesstimate. MULTIPLICATION GUESSTIMATION You can use much the same techniques to guesstimate your answers to multiplication problems. For example, ª Rounding up to the nearest multiple of 10 simpliﬁes the problem considerably, but you’re still off by 252, or about 5%. You can do better if you round both numbers by the same amount, but in opposite directions. That is, if you round 88 by increasing 2, you should also decrease 54 by 2: ª 90 52 4680 88 54 4752 90 50 4500 88 54 4752 Secrets of Mental Math 114 Instead of a 1-by-1 multiplication problem, you now have a 2-by-1 problem, which should be easy enough for you to do. Your guesstimation is off by only 1.5%. When you guesstimate the answer to multiplication problems by rounding the larger number up and the smaller number down, your guesstimate will be a little low. If you round the larger number down and the smaller number up so that the numbers are closer together, your guesstimate will be a little high. The larger the amount by which you round up or down, the greater your guesstimate will be off from the exact answer. For example: ª Since the numbers are closer together after you round them off, your guesstimate is a little high. ª Since the numbers are farther apart, the estimated answer is too low, though again, not by much. You can see that this multi-plication guesstimation method works quite well. Also notice that this problem is just 672 and that our approximation is just the ﬁrst step of the squaring techniques. Let’s look at one more example: ª 85 50 4250 83 52 4316 70 64 4480 67 67 4489 70 68 4760 73 65 4745 Good Enough: The Art of “Guesstimation” 115 We observe that the approximation is most accurate when the original numbers are close together. Try estimating a 3-by-2 multiplication problem: ª By rounding 63 down to 60 and 728 up to 731, you create a 3-by-1 multiplication problem, which puts your guesstimate within 2004 of the exact answer, an error of 4.3%. Now try guesstimating the following 3-by-3 problem: ª You will notice that although you rounded both numbers up and down by 8, your guesstimate is off by over 1000. That’s because the multiplication problem is larger and the size of the rounding number is larger, so the resulting estimate will be off by a greater amount. But the relative error is still under 1%. How high can you go with this system of guesstimating multi-plication problems? As high as you want. You just need to know the names of large numbers. A thousand thousand is a million, and a thousand million is a billion. Knowing these names and numbers, try this one on for size: ª 29 million 14 thousand 28,657,493 13,864 359 500 179,500 367 492 180,564 731 60 43,860 728 63 45,864 Secrets of Mental Math 116 As before, the objective is to round the numbers to simpler numbers such as 29,000,000 and 14,000. Dropping the 0s for now, this is just a 2-by-2 multiplication problem: 29 14 406 (29 14 29 7 2 203 2 406). Hence the answer is roughly 406 billion, since a thousand million is a billion. SQUARE ROOT ESTIMATION: DIVIDE AND AVERAGE The quantity n , the square root of a number n, is the number which, when multiplied by itself, will give you n. For example, the square root of 9 is 3 because 3 3 9. The square root is used in many science and engineering problems and is almost always solved with a calculator. The following method provides an accurate estimate of the answer. In square root estimation your goal is to come up with a num-ber that when multiplied by itself approximates the original num-ber. Since the square root of most numbers is not a whole number, your estimate is likely to contain a fraction or decimal point. Let’s start by guesstimating the square root of 19. Your ﬁrst step is to think of the number that when multiplied by itself comes closest to 19. Well, 4 4 16 and 5 5 25. Since 25 is too high, the answer must be 4 point something. Your next step is to divide 4 into 19, giving you 4.75. Now, since 4 4 is less than 4 4.75 19, which in turn is less than 4.75 4.75, we know that 19 (or 4 4.75) lies between 42 and 4.752. Hence, the square root of 19 lies between 4 and 4.75. I’d guess the square root of 19 to be about halfway between, at 4.375. In fact, the square root of 19 (rounded to three deci-mal places) is 4.359, so our guesstimate is pretty close. We illus-trate this procedure as follows: Good Enough: The Art of “Guesstimation” 117 Divide: Average: 4.375 Actually, we can obtain this answer another way, which you might ﬁnd easier. We know 4 squared is 16, which is shy of 19 by 3. To improve our guess, we “add the error divided by twice our guess.” Here, we add 3 divided by 8 to get 4 3 8 4.375. We note that this method will always produce an answer that is a little higher than the exact answer. Now you try a slightly harder one. What’s the square root of 87? Divide: Average: 9.3 3 First come up with your ballpark ﬁgure, which you can get fairly quickly by noting that 9 9 81 and 10 10 100, which means the answer is 9 point something. Carrying out the division of 9 into 87 to two decimal places, you get 9.66. To improve your guesstimate, take the average of 9 and 9.66, which is 9.33—exactly the square root of 87 rounded to the sec-ond decimal place! Alternatively, our guesstimate is 9 (error)/18 9 1 6 8 9.3 3 . 9 9.66 2 9.6 6 98 7 .0 4 4.75 2 4.75 41 9 .0 16 30 28 20 20 0 Secrets of Mental Math 118 Using this technique, it’s pretty easy to guesstimate the square root of two-digit numbers. But what about three-digit numbers? Actually, they are not much harder. I can tell you right off the bat that all three-digit and four-digit numbers have two-digit square roots before the decimal point. And the procedure for comput-ing square roots is the same, no matter how large the number. For instance, to compute the square root of 679, ﬁrst ﬁnd your ballpark ﬁgure. Because 20 squared is 400 and 30 squared is 900, the square root of 679 must lie between 20 and 30. When you divide 20 into 679, you get approximately 34. Averaging 20 and 34 gives you a guesstimate of 27, but here’s a better estimate. If you know that 25 squared is 625, then your error is 679 625 54. Dividing that by 50, we have 5 5 4 0 1 1 0 0 8 0 1.08. Hence our improved guesstimate is 25 1.08 26.08. (For an even better estimate, if you know that 26 squared is 676, your error is 3, so add 5 3 2 ª .06 to get 26.06.) The exact answer is 26.06, rounded to two decimal places. To guesstimate the square root of four-digit numbers, look at the ﬁrst two digits of the number to determine the ﬁrst digit of the square root. For example, to ﬁnd the square root of 7369, consider the square root of 73. Since 8 8 64 and 9 9 81, 8 must be the ﬁrst digit of the square root. So the answer is 80 something. Now proceed the usual way. Dividing 80 into 7369 gives 92 plus a fraction, so a good guesstimate is 86. If you squared 86 to get 7396, you would be high by 27, so you should subtract 1 2 7 7 2 ª .16 for a better guesstimate of 85.84, which is right on the money. To guesstimate the square root of a six-digit number like 593,472 would seem like an impossible task for the uninitiated, but for you it’s no sweat. Since 7002 490,000, and 8002 640,000, the square of 593,472 must lie between 700 and 800. In fact, all ﬁve-digit and six-digit numbers have three-digit square Good Enough: The Art of “Guesstimation” 119 roots. In practice, you only need to look at the square root of the ﬁrst two digits of six-digit numbers (or the ﬁrst digit of ﬁve numbers). Once you ﬁgure out that the square root of 59 lies between 7 and 8, you know your answer is in the 700s. Now proceed in the usual manner: Divide: Average: ª 773.5 The exact square root of 593,472 is 770.37 (to ﬁve places), so you’re pretty close. But you could have been closer, as the fol-lowing trick demonstrates. Note that the ﬁrst two digits, 59, are closer to 64 (8 8) than they are to 49 (7 7). Because of this you can start your guesstimation with the number 8 and pro-ceed from there: Divide: Average: ª 770.5 Just for fun, let’s do a real whopper—the square root of 28,674,529. This isn’t as hard as it might seem. Your ﬁrst step is to round to the nearest large number—in this case, just ﬁnd the square root of 29. Divide: Average: 5.4 5 5.8 2 5.8 52 9 .0 25 40 800 741 2 741 85 9 3 4 741 8005 9 3 4 7 2 700 847 2 847 75 9 3 4 847 7005 9 3 4 7 2 Secrets of Mental Math 120 All seven-digit and eight-digit numbers have four-digit square roots, so 5.4 becomes 5400, your estimate. The exact answer is slightly greater than 5354.8. Not bad! This wraps up the chapter on guesstimation math. After doing the exercises below, turn to the next chapter on pencil-and-paper math, where you will learn to write down answers to problems, but in a much quicker way than you’ve done on paper before. Good Enough: The Art of “Guesstimation” 121 The Mathematical Duel of Évariste Galois T he tragic story of the French mathematician Évariste Galois (1811–1832) killed at the age of twenty in a duel over “an infa-mous coquette” is legendary in the annals of the history of mathe-matics. A precociously brilliant student, Galois laid the foundation for a branch of mathematics known as group theory. Legend has it that he penned his theory the night before the duel, anticipating his demise and wanting to leave his legacy to the mathematics commu-nity. Hours before his death, on May 30, 1832, Galois wrote to Auguste Chevalier: “I have made some new discoveries in analysis. The ﬁrst concerns the theory of equations, the others integral func-tions.” After describing these, he asked his friend: “Make a public request of Jacobi or Gauss to give their opinions not as to the truth but as to the importance of these theorems. After that, I hope some men will ﬁnd it proﬁtable to sort out this mess.” Romantic legend and historical truth, however, do not always match.What Galois penned the night before his death were correc-tions and editorial changes to papers that had been accepted by the Academy of Sciences long before. Further, Galois’s initial papers had been submitted three years prior to the duel, when he was all of sev-enteen! It was after this that Galois became embroiled in political controversy, was arrested, spent time in a prison dungeon, and, ulti-mately, got himself mixed up in a dispute over a woman and killed. Aware of his own precocity, Galois noted, “I have carried out researches which will halt many savants in theirs.” For over a century that proved to be the case. MORE TIPS ON TIPS As we indicated in Chapter 0, it is easy to ﬁgure out tips for most situations. For example, to ﬁgure out a 10% tip, we merely multiply the bill by 0.1 (or divide the bill by 10). For example, if a bill came to $42, then a 10% tip would be $4.20. To determine a 20% tip, you simply multiply the bill by 0.2, or double the amount of a 10% tip. Thus a 20% tip on a $42 bill would be $8.40. To determine a 15% tip, we have a few options. If you have mastered the techniques of Chapter 2, and are comfortable with multiplying by 15 5 3, you can simply multiply the bill by 15, then divide by 100. For example, with a $42 bill, we have 42 15 42 5 3 210 3 630, which when divided by 100, gives us a tip of $6.30. Another method is to take the aver-age of a 10% tip and a 20% tip. From our earlier calculation, this would be $6.30 Perhaps the most popular approach to taking a 15% tip is to take 10% of the bill, cut that amount in half (which would be a 5% tip), then add those two numbers together. So, for instance, with a $42 bill, you would add $4.20 plus half that amount, $2.10, to get $4.20 $2.10 $6.30 Let’s use all three methods to compute 15% on a bill that is $67. By the direct method, 67 3 5 201 5 1005, which when divided by 100 gives us $10.05. By the averaging $12.60 2 $4.20 $8.40 2 Secrets of Mental Math 122 method, we average the 10% tip of $6.70 with the 20% tip of $13.40, to get $10.05 Using the last method, we add $6.70 to half that amount, $3.35 to get $6.70 $3.35 $10.05 Finally, to calculate a 25% tip, we offer two methods. Either multiply the amount by 25, then divide by 100, or divide the amount by 4 (perhaps by cutting the amount in half twice). For example, with a $42 bill, you can compute 42 25 42 5 5 210 5 1050, which when divided by 100 produces a tip of $10.50. Or you can divide the original amount directly by 4, or cut in half twice: half of $42 is $21, and half of that is $10.50. With a $67 bill, I would probably divide by 4 directly: since 67 4 16 3 4 , we get a 25% tip of $16.75. NOT -TOO-TAXING CALCULATIONS In this section, I will show you my method for mentally estimat-ing sales tax. For some tax rates, like 5% or 6% or 10%, the calculation is straightforward. For example, to calculate 6% tax, just multiply by 6, then divide by 100. For instance, if the sale came to $58, then 58 6 348, which when divided by 100 gives the exact sales tax of $3.48. (So your total amount would be $61.48.) But how would you calculate a sales tax of 6 1 2 % on $58? $20.10 2 $6.70 $13.40 2 Good Enough: The Art of “Guesstimation” 123 I will show you several ways to do this calculation, and you choose the one that seems easiest for you. Perhaps the easiest way to add half of a percent to any dollar amount is to simply cut the dollar amount in half, then turn it into cents. For exam-ple, with $58, since half of 58 is 29, simply add 29 cents to the 6% tax (already computed as $3.48) to get a sales tax of $3.77. Another way to calculate the answer (or a good mental esti-mate) is to take the 6% tax, divide it by 12, then add those two numbers. For example, since 6% of $58 is $3.48, and 12 goes into 348 almost 30 times, then add 30 cents for an estimate of $3.78, which is only off by a penny. If you would rather divide by 10 instead of 12, go ahead and do it. You will be calculating 6.6% instead of 6.5% (since 1 6 0 0.6) but that should still be a very good estimate. Here, you would take $3.48 and add 34¢ to get $3.82. Let’s try some other sales tax percentages. How can we calcu-late 7 1 4 % sales tax on $124? Begin by computing 7% of 124. From the methods shown in Chapter 2, you know that 124 7 868, so 7% of 124 is $8.68. To add a quarter percent, you can divide the original dollar amount by 4 (or cut it in half, twice) and turn the dollars into cents. Here, 124 4 31, so add 31¢ to $8.68 to get an exact sales tax of $8.99. Another way to arrive at 31¢ is to take your 7% sales tax, $8.68, and divide it by 28. The reason this works is because 2 7 8 1 4 . For a quick mental estimate, I would probably divide $8.68 by 30, to get about 29¢, for an approximate total sales tax of $8.97. When you divide by 30, then you are actually computing a tax of 7 3 7 0 %, which is approximately 7.23% instead of 7.25%. How would you calculate a sales tax of 7.75%? Probably for most approximations, it is sufﬁcient to just say that it is a little less than 8% sales tax. But to get a better approximation, here Secrets of Mental Math 124 are some suggestions. As you saw in the last example, if you can easily calculate a 1 4 % adjustment, then simply triple that amount for the 3 4 % adjustment. For example, to calculate 7.75% of $124, you ﬁrst calculate 7% to obtain $8.68. If you calculated that 1 4 % is 31¢, then 3 4 % would be 93¢, for a grand total of $8.68 0.93 $9.61. For a quick approximation, you can exploit the fact that 7 9 .777 is approximately .75, so you can divide the 7% sales tax by 9 to get slightly more than the .75% adjustment. In this example, since $8.68 divided by 9 is about 96¢, simply add $8.68 0.96 $9.64 for a slight overestimate. We can use this approximation procedure for any sales tax. For a general formula, to estimate the sales tax of $A.B%, ﬁrst multiply the amount of the sale by A%. Then divide this amount by the number D, where A/D is equal to 0.B. (Thus D is equal to A times the reciprocal of B.) Adding these numbers together gives you the total sales tax (or an approximate one, if you rounded D to an easier nearby number). For instance, with 7.75%, the magic divisor D would be 7 4 3 2 3 8 9 1 3 , which we rounded down to 9. For sales tax of 6 3 8 %, ﬁrst compute the sales tax on 6%, then divide that number by 16, since 1 6 6 3 8 . (To divide a number by 16, divide the number by 4 twice, or divide the number by 8, then by 2.) Try to come up with meth-ods for ﬁnding the sales tax in the state that you live in. You will ﬁnd that the problem is not as taxing as it seems! SOME “INTEREST -ING” CALCULATIONS Finally, we’ll brieﬂy mention some practical problems pertain-ing to interest, from the standpoint of watching your invest-ments grow, and paying off money that you owe. We begin with the famous Rule of 70, which tells you approx-imately how long it takes your money to double: To ﬁnd the Good Enough: The Art of “Guesstimation” 125 number of years that it will take for your money to double, divide the number 70 by the rate of interest. Suppose that you ﬁnd an investment that promises to pay you 5% interest per year. Since 70 5 14, then it will take about 14 years for your money to double. For example, if you invested $1000 in a savings account that paid that interest, then after 14 years, it will have $1000(1.05)14 $1979.93. With an inter-est rate of 7%, the Rule of 70 indicates that it will take about 10 years for your money to double. Indeed, if you invest $1000 at that annual interest rate, you will have after ten years $1000(1.07)10 $1967.15. At a rate of 2%, the Rule of 70 says that it should take about 35 years to double, as shown below: $1000(1.02)35 $1999.88 A similar method is called the Rule of 110, which indicates how long it takes for your money to triple. For example, at 5% interest, since 110 5 22, it should take about 22 years to turn $1000 into $3000. This is veriﬁed by the calculation $1000(1.05)22 $2925.26. The Rule of 70 and the Rule of 110 are based on properties of the number e 2.71828 . . . and “natural logarithms” (studied in precalculus), but fortunately we do not need to utilize this higher mathematics to apply the rules. Now suppose that you borrow money, and you have to pay the money back. For example, suppose that you borrow $360,000 at an annual interest rate of 6% per year (which we shall inter-pret to mean that interest will accumulate at a rate of 0.5% per month) and suppose that you have thirty years to pay off the loan. About how much will you need to pay each month? First of all, you will need to pay $360,000 times 0.5% $1,800 each month just to cover what you owe in interest. (Although, actu-ally, what you owe in interest will go down gradually over time.) Secrets of Mental Math 126 Since you will make 30 12 360 monthly payments, then paying an extra $1,000 each month would cover the rest of the loan, so an upper bound on your monthly payment is $1,800 $1,000 $2,800. But fortunately you do not need to pay that much extra. Here is my rule of thumb for estimating your monthly payment. Let i be your monthly interest rate. (This is your annual interest rate divided by 12.) Then to pay back a loan of $P in N months, your monthly payment M is about M In our last example, P $360,000, and i 0.005, so our for-mula indicates that our monthly payment should be about M Notice that the ﬁrst two numbers in the numerator multiply to $1,800. Using a calculator (for a change) to compute (1.005)360 6.02, we have that your monthly payment should be about $1,800(6.02)/5.02, which is about $2,160 per month. Here’s one more example. Suppose you wish to pay for a car, and after your down payment, you owe $18,000 to be paid off in ﬁve years, with an annual interest rate of 4%. If there was no interest, you would have to pay $18,000/60 $300 per month. Since the ﬁrst year’s interest is $18,000(.04) $720, you know that you will need to pay no more than $300 $60 $360. Using our formula, since the monthly interest rate is i .04/12 0.00333, we have M $18,000(0.0333)(1.00333)60 (1.00333)50 1 $360,000(.005)(1.005)360 (1.005)360 1 P(1 i)N (1 i)N 1 Good Enough: The Art of “Guesstimation” 127 and since (1.00333)60 1.22, we have a monthly payment of about $60(1.22)/(0.22) $333. We conclude with some exercises that we hope will hold your interest. GUESSTIMATION EXERCISES Go through the following exercises for guesstimation math; then check your answers and computations at the back of the book. EXERCISE: ADDITION GUESSTIMATION Round these numbers up or down and see how close you come to the exact answer: 1. 2. 3. 4. Mentally estimate the total for the following column of prices by rounding to the nearest 50¢: $ 2.67 1.95 7.35 9.21 0.49 11.21 0.12 6.14 8.31 8,971,011 4,016,367 312,025 79,419 57,293 37,421 1479 1105 Secrets of Mental Math 128 EXERCISE: SUBTRACTION GUESSTIMATION Estimate the following subtraction problems by rounding to the second or third digit: 1. 2. 3. 4. EXERCISE: DIVISION GUESSTIMATION Adjust the numbers in a way that allows you to guesstimate the following division problems: 1. 74 3 7 9 2. 52 3 ,9 5 8 3. 135 4 9 ,2 1 3 4. 2895 ,1 0 2 ,3 5 7 5. 203,6378 ,3 2 9 ,4 8 3 EXERCISE: MULTIPLICATION GUESSTIMATION Adjust the numbers in a way that allows you to guesstimate the following multiplication problems: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 5,462,741 203,413 104,972 11,201 51,276 489 428 313 639 107 312 98 539 17 88 88 76 42 98 27 8,349,241 6,103,839 526,978 42,009 67,221 9,874 4,926 1,659 Good Enough: The Art of “Guesstimation” 129 EXERCISE: SQUARE ROOT GUESSTIMATION Estimate the square roots of the following numbers using the divide and average method: 1. 17 2. 35 3. 163 4. 4279 5. 8039 EXERCISE: EVERYDAY MATH 1. Compute 15% of $88. 2. Compute 15% of $53. 3. Compute 25% of $74. 4. How long does it take an annual interest rate of 10% to double your money? 5. How long does it take an annual interest rate of 6% to double your money? 6. How long does it take an annual interest rate of 7% to triple your money? 7. How long does it take an annual interest rate of 7% to quadruple your money? 8. Estimate the monthly payment to repay a loan of $100,000 at an interest rate of 9% over a ten-year period. 9. Estimate the monthly payment to repay a loan of $30,000 at an interest rate of 5% over a four-year period. Secrets of Mental Math 130 Chapter 6 Math for the Board: Pencil-and-Paper Math In the Introduction to this book I discussed the many beneﬁts you will get from being able to do mental calculations. In this chapter I present some methods for speeding up pencil-and-paper calculations as well. Since calculators have replaced much of the need for pencil-and-paper arithmetic in most practical sit-uations, I’ve chosen to concentrate on the lost art of calculating square roots and the ﬂashy criss-cross method for multiplying large numbers. Since these are, admittedly, mostly for mental gymnastics and not for some practical application, I will ﬁrst touch on addition and subtraction and show you just a couple of little tricks for speeding up the process and for checking your answers. These techniques can be used in daily life, as you’ll see. If you are eager to get to the more challenging multiplication problems, you can skip this chapter and go directly to Chapter 7, which is critical for mastering the big problems in Chapter 8. If you need a break and just want to have some fun, then I rec-ommend going through this chapter—you’ll enjoy playing with pencil and paper once again. COLUMNS OF NUMBERS Adding long columns of numbers is just the sort of problem you might run into in business or while ﬁguring out your personal ﬁnances. Add the following column of numbers as you normally would and then check to see how I do it. 4328 884 620 1477 617 7 2 5 8651 When I have pencil and paper at my disposal, I add the num-bers from top to bottom and from right to left, just as we learned to do in school. With practice, you can do these problems in your head as fast or faster than you can with a calculator. As I sum the digits, the only numbers that I “hear” are the partial sums. That is, when I sum the ﬁrst (rightmost) column 8 4 0 7 7 5, I hear 8 . . . 12 . . . 19 . . . 26 . . . 31. Then I put down the 1, carry the 3, and proceed as usual. The next column would then sound like 3 . . . 5 . . . 13 . . . 15 . . . 22 . . . 23 . . . 25. Once I have my ﬁnal answer, I write it down, then check my computation by adding the numbers from bottom to top and, I hope, arrive at the same answer. For instance, the ﬁrst column would be summed, from bot-tom to top, as 5 7 7 0 4 8 (which in my mind sounds like 5 . . . 12 . . . 19 . . . 23 . . . 31). Then I mentally carry the 3, and add 3 2 1 7 2 8 2, and so on. By adding the Secrets of Mental Math 132 numbers in a different order, you are less likely to make the same mistake twice. Of course, if the answers differ, then at least one of the calculations must be wrong. MOD SUMS If I’m not sure about my answer, I sometimes check my solution by a method I call mod sums (because it is based on the elegant mathematics of modular arithmetic). This method also goes by the names of digital roots and casting out nines. I admit this method is not as practical, but it’s easy to use. With the mod sums method, you sum the digits of each num-ber until you are left with a single digit. For example, to com-pute the mod sum of 4328, add 4 3 2 8 17. Then add the digits of 17 to get 1 7 8. Hence the mod sum of 4328 is 8. For the previous problem the mod sums of each number are computed as follows: Math for the Board: Pencil-and-Paper Math 133 4328 17 8 884 20 2 620 8 8 1477 19 10 1 617 14 5 7 2 5 14 5 8651 29 20 11 2 2 As illustrated above, the next step is to add all the mod sums together (8 2 8 1 5 5). This yields 29, which sums to 11, which in turn sums to 2. Note that the mod sum of 8651, your original total of the original digits, is also 2. This is not a coincidence! If you have computed the answer and the mod sums correctly, your ﬁnal mod sums must be the same. If they are different, you have deﬁnitely made a mistake somewhere: there is about a 1 in 9 chance that the mod sums will match accidentally. If there is a mistake, then this method will detect it 8 times out of 9. The mod sum method is more commonly known to mathe-maticians and accountants as casting out nines because the mod sum of a number happens to be equal to the remainder obtained when the number is divided by 9. In the case of the answer above—8651—the mod sum was 2. If you divide 8651 by 9, the answer is 961 with a remainder of 2. In other words, if you cast out 9 from 8651 a total of 961 times, you’ll have a remain-der of 2. There’s one small exception to this. Recall that the sum of the digits of any multiple of 9 is also a multiple of 9. Thus, if a number is a multiple of 9, it will have a mod sum of 9, even though it has a remainder of 0. SUBTRACTING ON PAPER You can’t, of course, subtract columns of numbers the same way you add them. Rather, you subtract them number by number, which means that all subtraction problems involve just two numbers. Once again, with pencil and paper at our disposal, it is easier to subtract from right to left. To check your answer, just add the answer to the second number. If you are correct, then you should get the top number. If you want, you can also use mod sums to check your Secrets of Mental Math 134 answer. The key is to subtract the mod sums you arrive at and then compare that number to the mod sum of your answer: Math for the Board: Pencil-and-Paper Math 135 65,717 8 3 8 , 4 9 1 7 27,226 1 19 10 42,689 2 1 8 , 7 6 4 8 23,925 6 9 3 21 3 There’s one extra twist. If the difference in the mod sums is a negative number or 0, add 9 to it. For instance: PENCIL-AND-PAPER SQUARE ROOTS With the advent of pocket calculators, the pencil-and-paper method of calculating square roots has practically become a lost art. You’ve already seen how to estimate square roots mentally. Now I’ll show you how to do it exactly, using pencil and paper. Remember how in guesstimating square roots you calculated the square root of nineteen? Let’s look at that problem again, this time using a method that will give you the exact square root. I will describe the general method that ﬁts all cases, and illus-trate it with the above example. Step 1. If the number of digits to the left of the decimal point is one, three, ﬁve, seven, or any odd number of digits, the ﬁrst digit of the answer (or quotient) will be the largest number whose square is less than the ﬁrst digit of the original number. If the num-ber of digits to the left of the decimal point is two, four, six, or any even number of digits, the ﬁrst digit of the quotient will be the largest number whose square is less than the ﬁrst two digits of the dividend. In this case, 19 is a two-digit number, so the ﬁrst digit of the quotient is the largest number whose square is less than 19. That number is 4. Write the answer above either the ﬁrst digit of the dividend (if odd) or the second digit of the dividend (if even). Step 2. Subtract the square of the number in step 1, then bring down two more digits. Since 42 16, we subtract 19 16 3. We bring down two 0s, leaving 300 as the current remainder. Step 3. Double the current quotient (ignoring any decimal point), and put a blank space in following it. Here 4 2 8. Put down 8 to the left of the current remainder, in this case 300. Secrets of Mental Math 136 4. 3 5 8 19.000 000 42 1 6 8 < 3 00 83 3 2 4 9 86 < 5100 865 5 4 3 2 5 870 < 77500 8708 8 69664 Step 4. The next digit of the quotient will be the largest number that can be put in both blanks so that the resulting multiplica-tion problem is less than or equal to the current remainder. In this case the number is 3, because 83 3 249, whereas 84 4 336, which is too high. Write this number above the second digit of the next two numbers; in this case the 3 would go above the second 0. We now have a quotient of 4.3. Step 5. If you want more digits, subtract the product from the remainder (i.e., 300 249 51), and bring down the next two digits; in this case 51 turns into 5100, which becomes the cur-rent remainder. Now repeat steps 3 and 4. To get the third digit of the square root, double the quo-tient, again ignoring the decimal point (i.e., 43 2 86). Place 86 to the left of 5100. The number 5 gives us 865 5 4325, the largest product below 5100. The 5 goes above the next two numbers, in this case two more 0s. We now have a quotient of 4.35. For even more digits, repeat the process as we’ve done in the example. Here’s an example of an odd number of digits before the dec-imal point: Math for the Board: Pencil-and-Paper Math 137 2 8. 9 7 839.40 00 22 4 4 < 439 48 8 3 8 4 56 < 55 40 569 9 5 1 2 1 578 < 4 1900 5787 7 4 0509 Next, we’ll calculate the square root of a four-digit number. In this case—as with two-digit numbers—we consider the ﬁrst two digits of the problem to determine the ﬁrst digit of the square root: Secrets of Mental Math 138 8 2. 0 6 6735.0 000 82 6 4 16 < 335 162 2 3 2 4 164 < 11 00 1640 0 0 1640 < 11 0000 16406 6 9 8436 3. 3 10.89 32 9 6 < 1 89 63 3 1 8 9 0 Finally, if the number for which you are calculating the square root is a perfect square, you will know it as soon as you end up with a remainder of 0 and nothing to bring down. For example: PENCIL-AND-PAPER MULTIPLICATION For pencil-and-paper multiplication I use the criss-cross method, which enables me to write down the entire answer on one line without ever writing any partial results! This is one of the most impressive displays of mathemagics when you have pencil and paper at your disposal. Many lightning calculators in the past earned their reputations with this method. They would be given two large numbers and write down the answer almost instanta-neously. The criss-cross method is best learned by example. Step 1. First, multiply 4 7 to yield 28 , write down the 8, and mentally carry the 2 to the next computation, below. Step 2. Following the diagram, add 2 (4 4) (3 7) 39 , write down the 9, and carry the 3 to the ﬁnal computation, below. 4 7 3 4 47 3 4 1598 Math for the Board: Pencil-and-Paper Math 139 4 7 3 4 Step 3. End by adding 3 (3 4) 1 5 and writing down 15 to arrive at your ﬁnal answer. 4 7 3 4 You have now written the answer: 1 5 9 8 . Let’s solve another 2-by-2 multiplication problem using the criss-cross method: The steps and diagrams appear as follows: Step 1. 5 3 15 Step 2. 1 (5 8) (6 3) 59 Step 3. 5 (6 8) 5 3 Answer: 5 3 9 5 The criss-cross method gets slightly more complicated with 3-by-3 problems. We proceed as suggested by our pattern below: Step 1. 2 3 6 853 762 649,986 83 6 5 5395 Secrets of Mental Math 140 8 3 6 5 8 3 6 5 8 3 6 5 8 5 3 7 6 2 Step 2. (2 5) (6 3) 28 Step 3. 2 (2 8) (7 3) (6 5) 69 Step 4. 6 (6 8) (7 5) 89 Step 5. 8 (8 7) 6 4 Answer: 6 4 9 , 9 8 6 Notice that the number of multiplications at each step is 1, 2, 3, 2, and 1 respectively. The mathematics underlying the criss-cross method is nothing more that the distributive law. For instance, 853 762 (800 50 3) (700 60 2) (3 2) [(5 2) (3 6)] 10 [(8 2) (5 6) (3 7)] 100 [(8 6) (5 7)] 1000 (8 7) 10,000, which are precisely the calculations of the criss-cross method. You can check your answer with the mod sum method by multiplying the mod sums of the two numbers and computing the resulting number’s mod sum. Compare this number to the mod sum of the answer. If your answer is correct, they must match. For example: Math for the Board: Pencil-and-Paper Math 141 8 5 3 7 6 2 8 5 3 7 6 2 8 5 3 7 6 2 8 5 3 7 6 2 If the mod sums don’t match, you made a mistake. This method will detect mistakes, on average, about 8 times out of 9. In the case of 3-by-2 multiplication problems, the procedure is the same except you treat the hundreds digit of the second number as a 0: Step 1. 7 6 42 Step 2. 4 (7 4) (3 6) 50 Step 3. 5 (7 8) (0 6) (3 4) 73 Step 4. 7 (3 8) (0 4) 31 846 037 31,302 Secrets of Mental Math 142 42 6 6 7 6 42 853 762 649,986 8 4 6 0 3 7 8 4 6 0 3 7 8 4 6 0 3 7 8 4 6 0 3 7 Step 5. 3 (0 8) 3 Answer: 3 1 , 3 0 2 Of course, you would normally just ignore the multiplication by zero, in practice. You can use the criss-cross to do any size multiplication prob-lem. To answer the 5-by-5 problem below will require nine steps. The number of multiplications in each step is 1, 2, 3, 4, 5, 4, 3, 2, 1, for twenty-ﬁve multiplications altogether. Step 1. 9 7 63 Step 2. 6 (9 6) (4 7) 88 Step 3. 8 (9 8) (0 7) (4 6) 104 42,867 5 2 , 0 4 9 2,231,184,483 Math for the Board: Pencil-and-Paper Math 143 8 4 6 0 3 7 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 Step 4. 10 (9 2) (2 7) (4 8) (0 6) 74 Step 5. 7 (9 4) (5 7) (4 2) (2 6) (0 8) 98 Step 6. 9 (4 4) (5 6) (0 2) (2 8) 71 Step 7. 7 (0 4) (5 8) (2 2) 51 Step 8. 5 (2 4) (5 2) 23 Step 9. (5 4) 2 2 2 Secrets of Mental Math 144 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 4 2 8 6 7 5 2 0 4 9 Answer: 2 , 2 3 1 , 1 8 4 , 4 8 3 You can check your answer by using the mod sums method. CASTING OUT ELEVENS To double-check your answer another way, you can use the method known as casting out elevens. It’s similar to casting out nines, except you reduce the numbers by alternately subtracting and adding the digits from right to left, ignoring any decimal point. If the result is negative, then add eleven to it. (It may be tempting to do the addition and subtraction from left to right, as you do with mod sums, but in this case you must do it from right to left for it to work.) For example: Math for the Board: Pencil-and-Paper Math 145 234.87 7 8 4 3 2 2 2 5 8 . 6 1 1 6 8 5 2 9 293.48 8 4 3 9 2 0 11 0 42,867 9 5 2 , 0 4 9 2 2,231,184,483 18 36 9 9 The same method works for subtraction problems: 65,717 14 3 3 8 , 4 9 1 ( 9 ) 9 27,226 12 1 1 Secrets of Mental Math 146 Shakuntala Devi: That’s Incalculable! I n 1976 the New York Times reported that an Indian woman named Shakuntala Devi (b. 1939) added 25,842 111,201,721 370,247,830 55,511,315, and then multiplied that sum by 9,878, for a correct total of 5,559,369,456,432, all in less than twenty seconds. Hard to believe, though this uneducated daughter of impoverished parents has made a name for herself in the United States and Europe as a lightning calculator. Unfortunately, most of Devi’s truly amazing feats not done by obvious “tricks of the trade” are poorly documented. Her greatest claimed accomplishment—the timed multiplication of two thirteen-digit numbers on paper—has appeared in the Guinness Book of World Records as an example of a “Human Computer.” The time of the calculation, however, is questionable at best. Devi, a master of the criss-cross method, reportedly multiplied 7,686,369,774,870 2,465,099,745,799, numbers randomly generated at the computer department of Imperial College, London, on June 18, 1980. Her cor-rect answer of 18,947,668,177,995,426,773,730 was allegedly gener-ated in an incredible twenty seconds. Guinness offers this disclaimer: “Some eminent mathematical writers have questioned the condi-tions under which this was apparently achieved and predict that it would be impossible for her to replicate such a feat under highly rig-orous surveillance.” Since she had to calculate 169 multiplication problems and 167 addition problems, for a total of 336 operations, she would have had to do each calculation in under a tenth of a sec-ond, with no mistakes, taking the time to write down all 26 digits of the answer. Reaction time alone makes this record truly in the cate-gory of “that’s incalculable!” Despite this, Devi has proven her abilities doing rapid calculation and has even written her own book on the subject. It even works with multiplication: Math for the Board: Pencil-and-Paper Math 147 853 6 7 6 2 3 649,986 18 4 7 7 If the numbers disagree, you made a mistake somewhere. But if they match, it’s still possible for a mistake to exist. On aver-age, this method will detect an error 10 times out of 11. Thus a mistake has a 1 in 11 chance of sneaking past the elevens check, a 1 in 9 chance of sneaking past the nines check, and only a 1 in 99 chance of being undetected if both checks are used. For more information about this and other fascinating mathemagical top-ics, I would encourage you to read any of Martin Gardner’s recreational math books. You are now ready for the ultimate pencil-and-paper multi-plication problem in the book, a 10-by-10! This has no practical value whatsoever except for showing off! (And personally I think multiplying ﬁve-digit numbers is impressive enough since the answer will be at the capacity of most calculators.) We pre-sent it here just to prove that it can be done. The criss-crosses follow the same basic pattern as that in the 5-by-5 problem. There will be nineteen computation steps and at the tenth step there will be ten criss-crosses! Here you go: 2,766,829,451 4 , 4 2 5 , 5 7 5 , 2 1 6 Here’s how we do it: Step 1. 6 1 6 Step 2. (6 5) (1 1) 31 Step 3. 3 (6 4) (2 1) (1 5) 34 Step 4. 3 (6 9) (5 1) (1 4) (2 5) 76 Step 5. 7 (6 2) (7 1) (1 9) (5 5) (2 4) 68 Step 6. 6 (6 8) (5 1) (1 2) (7 5) (2 9) (5 4) 134 Step 7. 13 (6 6) (5 1) (1 8) (5 5) (2 2) (7 4) (5 9) 164 Step 8. 16 (6 6) (2 1) (1 6) (5 5) (2 8) (5 4) (5 2) (7 9) 194 Step 9. 19 (6 7) (4 1) (1 6) (2 5) (2 6) (5 4) (5 8) (5 9) (7 2) 212 Step 10. 21 (6 2) (4 1) (1 7) (4 5) (2 6) (2 4) (5 6) (5 9) (7 8) (5 2) 225 Step 11. 22 (1 2) (4 5) (2 7) (4 4) (5 6) (2 9) (7 6) (5 2) (5 8) 214 Step 12. 21 (2 2) (4 4) (5 7) (4 9) (7 6) (2 2) (5 6) (5 8) 228 Step 13. 22 (5 2) (4 9) (7 7) (4 2) (5 6) (2 8) (5 6) 201 Step 14. 20 (7 2) (4 2) (5 7) (4 8) (5 6) (2 6) 151 Step 15. 15 (5 2) (4 8) (5 7) (4 6) (2 6) 128 Step 16. 12 (5 2) (4 6) (2 7) (4 6) 84 Step 17. 8 (2 2) (4 6) (4 7) 64 Step 18. 6 (4 2) (4 7) 42 Step 19. 4 (4 2) 1 2 Secrets of Mental Math 148 If you were able to negotiate this extremely difﬁcult problem successfully the ﬁrst time through, you are on the verge of grad-uating from apprentice to master mathemagician! Math for the Board: Pencil-and-Paper Math 149 2,766,829,451 5 4 , 4 2 5 , 5 7 5 , 2 1 6 5 12,244,811,845,244,486,416 7 PENCIL-AND-PAPER MATHEMATICS EXERCISES EXERCISE: COLUMNS OF NUMBERS Sum the following column of numbers, checking your answer by reading the column from bottom to top, and then by the mod sums method. If the two mod sums do not match, recheck your addition: 1. Sum this column of dollars and cents: 2. $ 21.56 19.38 211.02 9.16 26.17 1 . 4 3 672 1367 107 7845 358 210 9 1 6 EXERCISE: SUBTRACTING ON PAPER Subtract the following numbers, using mod sums to check your answer and then by adding the bottom two numbers to get the top number: 1. 2. 3. 4. EXERCISE: SQUARE ROOT GUESSTIMATION For the following numbers, compute the exact square root using the doubling and dividing technique. EXERCISE: PENCIL-AND-PAPER MULTIPLICATION To wrap up this exercise session, use the criss-cross method to compute exact multiplication problems of any size. Place the answer below the problems on one line, from right to left. Check your answers using mod sums. 45,394,358 3 6 , 4 7 2 , 6 5 9 3,249,202 2 , 9 0 3 , 4 4 5 876,452 5 9 3 , 8 7 6 75,423 4 6 , 2 9 8 Secrets of Mental Math 150 1. 15 2. 502 3. 439.2 4. 361 1. 2. 3. 4. 5. 6. 3,923,759 2 , 6 7 4 , 0 9 3 52,819 4 7 , 8 2 0 3,309 2 , 8 6 8 725 6 0 9 273 2 1 7 54 3 7 Chapter 7 A Memorable Chapter: Memorizing Numbers One of the most common questions I am asked pertains to my memory. No, I don’t have an extraordinary memory. Rather, I apply a memory system that can be learned by anyone and is described in the following pages. In fact, experiments have shown that almost anyone of average intelligence can be taught to vastly improve their ability to memorize numbers. In this chapter we’ll teach you a way to improve your mem-ory for numbers. Not only does this have obvious practical ben-eﬁts, such as remembering dates or recalling phone numbers, but it also allows the mathemagician to solve much larger prob-lems mentally. In the next chapter we’ll show you how to apply the techniques of this chapter to multiply ﬁve-digit numbers in your head! USING MNEMONICS The method presented here is an example of a mnemonic—that is, a tool to improve memory encoding and retrieval. Mnemonics work by converting incomprehensible data (such as digit sequences) to something more meaningful. For example, take just a moment to memorize the sentence below: “My turtle Pancho will, my love, pick up my new mover, Ginger.” Read it over several times. Look away from the page and say it to yourself over and over until you don’t have to look back at the page, visualizing the turtle, Pancho, picking up your new mover, Ginger, as you do so. Got it? Congratulations! You have just memorized the ﬁrst twenty-four digits of the mathematical expression π (pi). Recall that π is the ratio of the circumference of a circle to its diameter, usually approximated in school as 3.14, or 2 7 2 . In fact, π is an irrational number (one whose digits continue indeﬁnitely with no repeti-tion or pattern), and computers have been used to calculate π to billions of places. THE PHONETIC CODE I’m sure you’re wondering how “My turtle Pancho will, my love, pick up my new mover, Ginger” translates into 24 places of π. To do this, you ﬁrst need to memorize the phonetic code below, where each number between 0 and 9 is assigned a conso-nant sound. 1 is the t or d sound. 2 is the n sound. 3 is the m sound. 4 is the r sound. 5 is the l sound. Secrets of Mental Math 152 6 is the j, ch, or sh sound. 7 is the k or hard g sound. 8 is the f or v sound. 9 is the p or b sound. 0 is the z or s sound. Memorizing this code isn’t as hard as it looks. For one thing, notice that in cases where more than one letter is associated with a number, they have similar pronunciations. For example, the k sound (as it appears in words like kite or cat) is similar to the hard g sound (as it appears in such words as goat). You can also rely on the following tricks to help you memorize the code: 1 A typewritten t or d has just 1 downstroke. 2 A typewritten n has 2 downstrokes. 3 A typewritten m has 3 downstrokes. 4 The number 4 ends in the letter r. 5 Shape your hand with 4 ﬁngers up and the thumb at a 90-degree angle—that’s 5 ﬁngers in an L shape. 6 A J looks sort of like a backward 6. 7 A K can be drawn by laying two 7s back to back. 8 A lowercase f written in cursive looks like an 8. 9 The number 9 looks like a backward p or an upside-down b. 0 The word zero begins with the letter z. Or you can just remember the list in order, by thinking of the name Tony Marloshkovips! Practice remembering this list. In about ten minutes you should have all the one-digit numbers associated with conso-nant sounds. Next, you can convert numbers into words by placing vowel sounds around or between the consonant sounds. For instance, the number 32 can become any of the following A Memorable Chapter: Memorizing Numbers 153 words: man, men, mine, mane, moon, many, money, menu, amen, omen, amino, mini, minnie, and so on. Notice that the word minnie is acceptable since the n sound is only used once. The following words could not represent the number 32 because they use other consonant sounds: mourn, melon, mint. These words would be represented by the numbers 342, 352, and 321, respectively. The consonant sounds of h, w, and y can be added freely since they don’t appear on the list. Hence, 32 can also become human, woman, yeoman, or my honey. The following list gives you a good idea of the types of words you can create using this phonetic code. Don’t feel obligated to memorize it—use it as inspiration to explore the possibilities on your own. Secrets of Mental Math 154 1 tie 2 knee 3 emu 4 ear 5 law 6 shoe 7 cow 8 ivy 9 bee 10 dice 11 tot 12 tin 13 tomb 14 tire 15 towel 16 dish 17 duck 18 dove 37 mug 38 movie 39 map 40 rose 41 rod 42 rain 43 ram 44 rear 45 roll 46 roach 47 rock 48 roof 49 rope 50 lace 51 light 52 lion 53 lamb 54 lure 19 tub 20 nose 21 nut 22 nun 23 name 24 Nero 25 nail 26 notch 27 neck 28 knife 29 knob 30 mouse 31 mat 32 moon 33 mummy 34 mower 35 mule 36 match 55 lily 56 leash 57 log 58 leaf 59 lip 60 cheese 61 sheet 62 chain 63 chum 64 cherry 65 jail 66 judge 67 chalk 68 chef 69 ship 70 kiss 71 kite 72 coin A Memorable Chapter: Memorizing Numbers 155 The Number-Word List For practice, translate the following numbers into words, then check the correct translation below. When translating numbers into words, there are a variety of words that can be formed: 42 74 67 86 93 10 55 826 951 620 367 Here are some words that I came up with: 42: rain, rhino, Reno, ruin, urn 74: car, cry, guru, carry 67: jug, shock, chalk, joke, shake, hijack 86: ﬁsh, fudge 73 comb 74 car 75 coal 76 cage 77 cake 78 cave 79 cap 80 face 81 ﬁght 82 phone 83 foam 84 ﬁre 85 ﬁle 86 ﬁsh 87 fog 88 ﬁfe 89 V.I.P. 90 bus 91 bat 92 bun 93 bomb 94 bear 95 bell 96 beach 97 book 98 puff 99 puppy 100 daisies 93: bum, bomb, beam, palm, pam 10: toss, dice, toes, dizzy, oats, hats 55: lily, lola, hallelujah! 826: ﬁnch, ﬁnish, vanish 951: pilot, plot, belt, bolt, bullet 620: jeans, chains, genius 367: magic! As an exercise, translate each of the following words into its unique number: dog oven cart fossil banana garage pencil Mudd multiplication Cleveland Ohio Answers: dog: 17 oven: 82 cart: 741 fossil: 805 banana: 922 garage: 746 pencil: 9,205 Secrets of Mental Math 156 A Memorable Chapter: Memorizing Numbers 157 Mudd: 31 multiplication: 35,195,762 Cleveland: 758,521 Ohio: (nothing) Although a number can usually be converted into many words, a word can be translated only into a single number. This is an important property for our applications as it enables us to recall speciﬁc numbers. Using this system you can translate any number or series of numbers (e.g., phone numbers, Social Security numbers, driver’s license numbers, the digits of π) into a word or a sentence. Here’s how the code works to translate the ﬁrst twenty-four digits of π: 3 1415 926 5 3 58 97 9 3 2 384 6264 “My turtle Pancho will, my love, pick up my new mover, Ginger.” Remember that, with this phonetic code, g is a hard sound, as in grass, so a soft g (as in Ginger) sounds like j and is repre-sented by a 6. Also, the word will is, phonetically, just L, and is represented by 5, since the consonant sound w can be used freely. Since this sentence can only be translated back to the twenty-four digits above, you have effectively memorized π to twenty-four digits! There’s no limit to the number of numbers this code will allow you to memorize. For example, the following two sen-tences, when added to “My turtle Pancho will, my love, pick up my new mover, Ginger,” will allow you to memorize the ﬁrst sixty digits of π: 3 38 327 950 2 8841 971 “My movie monkey plays in a favorite bucket.” And: 69 3 99 375 1 05820 97494 “Ship my puppy Michael to Sullivan’s backrubber.” What the heck, let’s go for a hundred digits: 45 92 307 81 640 62 8 620 “A really open music video cheers Jenny F. Jones.” 8 99 86 28 0 3482 5 3421 1 7067 “Have a baby ﬁsh knife so Marvin will marinate the goosechick.” You can really feel proud of yourself once these sentences roll trippingly off your tongue, and you’re able to translate them quickly back into numbers. But you’ve got a ways to go for the world record. Hiroyuki Goto of Japan recited π to 42,195 places, from memory, in seventeen hours and twenty-one min-utes in 1995. HOW MNEMONICS MAKES MENTAL CALCULATION EASIER Aside from improving your ability to memorize long sequences of numbers, mnemonics can be used to store partial results in the middle of a difﬁcult mental calculation. For example, here’s how you can use mnemonics to help you square a three-digit number: Secrets of Mental Math 158 A Memorable Chapter: Memorizing Numbers 159 A Piece of Pi for Alexander Craig Aitken P erhaps one of the most impressive feats of mental calculation was performed by a professor of mathematics at the University of Edinburgh, Alexander Craig Aitken (1895–1967), who not only learned the value of π to 1,000 places but, when asked to demon-strate his amazing memory during a lecture, promptly rattled off the ﬁrst 250 digits of π. He was then challenged to skip ahead and begin at the 551st digit and continue for another 150 places.This he did suc-cessfully without a single error. How did he do it? Aitken explained to his audience that “the secret, to my mind, is relaxation, the complete antithesis of concen-tration as usually understood.” Aitken’s technique was more auditory than usual. He arranged the numbers into chunks of ﬁfty digits and memorized them in a sort of rhythm.With undaunted conﬁdence he explained,“It would have been a reprehensibly useless feat had it not been so easy.” That Aitken could memorize π to a thousand places does not qual-ify him as a lightning calculator. That he could easily multiply ﬁve-digit numbers against each other does. A mathematician named Thomas O’Beirne recalled Aitken at a desk calculator demonstration. “The salesman,” O’Beirne wrote, “said something like ‘We’ll now multiply 23,586 by 71,283.’ Aitken said right off, ‘And get . . .’ (whatever it was). The salesman was too intent on selling even to notice, but his man-ager, who was watching, did.When he saw Aitken was right, he nearly threw a ﬁt (and so did I).” Ironically,Aitken noted that when he bought a desk calculator for himself, his own mental skills deteriorated quickly. Anticipating what the future might hold, he lamented,“Mental calculators may, like the Tasmanian or the Maori,be doomed to extinction. Therefore you may be able to feel an almost anthropological interest in surveying a curi-ous specimen, and some of my auditors here may be able to say in the year A.D. 2000, ‘Yes, I knew one such.’ ” Fortunately, history has proved him wrong! 42 384 “Title” 3422 115,200 422 116,964 42 300 2 44 422 1760 22 1764 2 40 As you recall from Chapter 3, to square 342 you ﬁrst multi-ply 300 384, for 115,200, then add 422. But by the time you’ve computed 422 1,764, you may have forgotten the ear-lier number, 115,200. Here’s where our memory system comes to the rescue. To store the number 115,200, put 200 on your hand by raising two ﬁngers, and convert 115 into a word like title. (By the way, I do not consider storing the 200 on ﬁngers to be cheating. After all, what are hands for if not for holding on to digits!) Repeat the word title to yourself once or twice. That’s easier to remember than 115,200, especially after you start cal-culating 422. Once you’ve arrived at 1,764, you can add that to title 2, or 115,200, for a total of 116,964. Here’s another: 27 300 “Gum” 2732 73,800 272 74,529 27 246 3 30 272 720 32 729 3 24 After multiplying 300 246 73,800, convert 73 into gum and hold 800 on your hand by raising eight ﬁngers. Once you’ve computed 272 729, just add that to gum 8, or 73,800, for a total of 74,529. This may seem a bit cumbersome at ﬁrst, but Secrets of Mental Math 160 with practice the conversion from numbers to words and back to numbers becomes almost second nature. You have seen how easily two-digit numbers can be trans-lated into simple words. Not all three-digit numbers can be translated so easily, but if you’re at a loss for a simple word to act as a mnemonic, an unusual word or a nonsense word will do. For example, if no simple word for 286 or 638 comes quickly to mind, use a combination word like no fudge or a nonsense word like jam-off. Even these unusual words should be easier to recall than 286 or 638 during a long calculation. For some of the huge problems in the next chapter, these mem-ory tricks will be indispensable. MEMORY MAGIC Without using mnemonics, the average human memory (includ-ing mine) can hold only about seven or eight digits at a time. But once you’ve mastered the ability to change numbers into words, you can expand your memory capacity considerably. Have someone slowly call out sixteen digits while someone else writes them down on a blackboard or a piece of paper. Once they are written down, you repeat them back in the exact order they were given without looking at the board or the piece of paper! At a recent lecture demonstration, I was given the following series of numbers: 1, 2, 9, 7, 3, 6, 2, 7, 9, 3, 3, 2, 8, 2, 6, 1 As the numbers were called out, I used the phonetic code to turn them into words and then tied them together with a non-sensical story. In this case, 12 becomes tiny, 97 becomes book, A Memorable Chapter: Memorizing Numbers 161 362 becomes machine, 793 becomes kaboom, 32 becomes moon, and 8261 becomes ﬁnished. As the words were being created, I linked them together to form a silly story to help me remember them. I pictured ﬁnding a tiny book, which I placed inside a machine. This caused the machine to go kaboom, and tossed me to the moon, where I was ﬁnished. This story may sound bizarre, but the more ridiculous the story, the easier it is to remember—and besides, it’s more fun. Secrets of Mental Math 162 Chapter 8 The T ough Stuff Made Easy: Advanced Multiplication At this point in the book—if you’ve gone through it chapter by chapter—you have learned to do mental addition, subtraction, multiplication, and division, as well as the art of guesstimation, pencil-and-paper mathemagics, and the phonetic code for num-ber memory. This chapter is for serious, die-hard mathemagi-cians who want to stretch their minds to the limits of mental calculation. The problems in this chapter range from four-digit squares to the largest problem I perform publicly—the multipli-cation of two different ﬁve-digit numbers. In order to do these problems, it is particularly important that you are comfortable and reasonably fast using the phonetic code. And even though, if you glance ahead in this chapter, the prob-lems seem overwhelming, let me restate the two fundamental premises of this book: (1) all mental calculation skills can be learned by almost anyone; and (2) the key is the simpliﬁcation of all problems into easier problems that can be done quickly. There is no problem in this chapter (or anywhere else) that you will encounter (comparable to these) that cannot be mastered and learned through the simpliﬁcation techniques you’ve learned in previous chapters. Because we are assuming that you’ve mastered the techniques needed for this chapter, we will be teaching pri-marily by illustration, rather than walking you through the prob-lems word for word. As an aid, however, we remind you that many of the simpler problems embedded within these larger prob-lems are ones you have already encountered in previous chapters. We begin with four-digit squares. Good luck! FOUR-DIGIT SQUARES As a preliminary skill for mastering four-digit squares, you need to be able to do 4-by-1 problems. We break these down into two 2-by-1 problems, as in the examples below: (4,800 67) (2,700 81) (6,700 18) 6,718 8 8 6,700 53,600 8 18 144 53,744 2,781 4 4 2,700 10,800 4 81 324 11,124 4,867 9 9 4,800 43,200 9 67 603 43,803 Secrets of Mental Math 164 (4,200 69) Once you’ve mastered 4-by-1s you’re ready to tackle four-digit squares. Let’s square 4,267. Using the same method we used with two-digit and three-digit squares, we’ll square the number 4,267 by rounding down by 267 to 4,000, and up by 267 to 4,534, multi-plying 4,534 4,000 (a 4-by-1), and then adding the square of the number you went up and down by, or 2672, as illustrated below: 267 4,534 “Damage” 4,2672 267 4,000 (4,534 4,000) (2672) 33 300 2672 33 234 (300 234) (332) Now, obviously there is a lot going on in this problem. I real-ize it’s one thing to say, “Just add the square of 267” and quite another to actually do it and remember what it was you were supposed to add to it. First of all, once you multiply 4,534 4 to get 18,136, you can actually say the ﬁrst part of the answer out loud: “Eighteen million . . .” You can do so because the orig-inal number is always rounded to the nearest thousand. Thus the largest three-digit number you will ever square in the next step is 500. The square of 500 is 250,000, so as soon as you see that the rest of your answer (in this case, 136,000) is less than 750,000, you know that the millions digit will not change. 70,200 1,089 71,289 18,136,000 71,289 18,207,289 4,269 5 5 4,200 21,000 5 69 345 21,345 The Tough Stuff Made Easy:Advanced Multiplication 165 Once you’ve said “eighteen million . . . ,” you need to hold on to 136,000 before you square 267. Here’s where our mnemon-ics from the last chapter comes to the rescue! Using the phonetic code, you can convert 136 to damage (1 d, 3 m, 6 j). Now you can work on the next part of the problem and just remember damage (and that there are three zeros following— this will always be the case). If at any time in the computations you forget what the original problem is, you can either glance at the original number or, if it isn’t written down, ask the audience to repeat the problem (which gives the illusion you are starting the problem over from scratch when, in fact, you have already done some of the calculations)! You now do the three-digit square just as you learned to do before, to get 71,289. I used to have trouble remembering the hundreds digits of my answer (2, in this case). I cured this by rais-ing my ﬁngers (two ﬁngers here). If you forget the last two digits (89), you can go back to the original number (4,267), square its last two digits (672 4,489), and take the last two digits of that. To compute the ﬁnal answer, you now add 71,289 to damage (which translates back to 136,000) resulting in 207,289, which you may now say out loud. Let’s do one more four-digit square—8,4312: 431 8,862 “Foppish” 8,4312 431 8,000 (8,862 8,000) (4312) 31 462 4312 31 400 (462 400) (312) 184,800 961 185,761 70,896,000 185,761 71,081,761 Secrets of Mental Math 166 Without repeating all the detailed steps as we did in the last problem, I will point out some of the highlights of this problem. After doing the 8 8,862 70,896, note that 896 is above 750, so a carry is possible. In fact, since 4312 is greater than 4002 160,000, there deﬁnitely will be a carry when you do the addi-tion to the number 896,000. Hence we can safely say aloud, “Seventy-one million . . .” at this point. When you square 431, you get 185,761. Add the 185 to the 896 to get 1,081, and say the rest of the answer. But remember that you already anticipated the carry, so you just say, “. . . 81 thousand . . . 761.” You’re done! We illustrate one more ﬁne point with 2,7532: 247 3,000 “Light off” 2,7532 247 2,506 (3,000 2,506) (2472) 47 294 2472 47 200 (294 200) (472) Since you are rounding up to 3,000, you will be multiplying 3,000 by another number in the 2,000s. You could subtract 2,753 247 2,506, but that’s a little messy. To obtain the last three digits, double 753 to get 1,506. The last three digits of this number, 506, are the last three digits of the 2,000 number: 2,506! This works because the two numbers being multiplied must add to twice the original number. Then proceed in the usual manner of multiplying 3,000 2,506 7,518,000, convert 518 to the words light off, and say 58,800 2,209 61,009 7,518,000 61,009 7,579,009 The Tough Stuff Made Easy:Advanced Multiplication 167 the ﬁrst part of the answer out loud as “Seven million . . .” You can say this with conﬁdence since 518 is below 750, so there will not be a carry. Next, you add the square of 247. Don’t forget that you can Secrets of Mental Math 168 Thomas Fuller: Learned Men and Great Fools I t would be hard to top the physical handicap on learning experi-enced by Helen Keller, though the social handicap imposed on Thomas Fuller, born in Africa in 1710, comes close. He was not only illiterate; he was forced to work in the ﬁelds of Virginia as a slave and never received a single day of education.The “property” of Mrs. Eliza-beth Cox,Thomas Fuller taught himself to count to 100, after which he increased his numerical powers by counting such items at hand as the grains in a bushel of wheat, the seeds in a bushel of ﬂax, and the number of hairs in a cow’s tail (2,872). Extrapolating from mere counting, Fuller learned to compute the number of shingles a house would need to cover its roof, the number of posts needed to enclose it, and other numbers relevant to materi-als needed in construction. His prodigious skills grew, and with them his reputation. In his old age, two Pennsylvanians challenged him to compute, in his head, numbers that would challenge the best lightning calculators. For example, they asked: “Suppose a farmer has six sows, and each sow has six female pigs the ﬁrst year, and they all increase in the same proportion, to the end of eight years, how many sows will the farmer then have?” The problem can be written as 78 6, that is, (7 7 7 7 7 7 7 7) 6.Within ten minutes Fuller gave his response of 34,588,806, the correct answer. Upon Fuller’s death in 1790, the Columbian Centinel reported that “he could give the number of poles, yards, feet, inches, and barley-corns in any given distance, say the diameter of the earth’s orbit; and in every calculation he would produce the true answer, in less time than ninety-nine men in a hundred would take with their pens.” When Fuller was asked if he regretted having never gained a traditional edu-cation he responded:“No,Massa—it is best I got no learning:for many learned men be great fools.” derive 247 quickly as the complement of 753. Then proceed to the ﬁnal answer as you did in the previous four-digit problems. EXERCISE: FOUR-DIGIT SQUARES 1. 12342 2. 86392 3. 53122 4. 98632 5. 36182 6. 29712 3-BY -2 MULTIPLICATION We saw in the 2-by-2 multiplication problems that there are several different ways to tackle the same problem. The variety of methods increases when you increase the number of digits in the problem. In 3-by-2 multiplication I ﬁnd that it pays to take a few moments to look at the problem to determine the method of calculation that will put the least amount of strain on the brain. Factoring Methods The easiest 3-by-2 problems to compute are those in which the two-digit number is factorable. For example: (8 7) 637 56 637 8 7 5,096 7 35,672 These are great because you don’t have to add anything. You just factor 56 into 8 7, then do a 3-by-1 (637 8 5,096), and ﬁnally a 4-by-1 (5,096 7 35,672). There are no addi-tion steps and you don’t have to store any partial results. More than half of all two-digit numbers are factorable into 637 56 The Tough Stuff Made Easy:Advanced Multiplication 169 numbers 11 and below, so you’ll be able to use this method for many problems. Here’s an example: (11 4) 853 11 4 9,383 4 37,532 To multiply 853 11, you treat 853 as 850 3 and proceed as follows: Now multiply 9,383 4 by treating 9,383 as 9,300 83, as follows: If the two-digit number is not factorable into small numbers, examine the three-digit number to see if it can be factored: (6 6 4) 76 144 76 6 6 4 456 6 4 2,736 4 10,944 Notice that the sequence of the multiplication problems is a 2-by-1, a 3-by-1, and ﬁnally a 4-by-1. Since these are all prob-144 76 9,300 4 37,200 83 4 332 37,532 850 11 9,350 3 11 33 9,383 853 44 Secrets of Mental Math 170 lems you can now do with considerable ease, this type of 3-by-2 problem should be no problem at all. Here’s another example where the two-digit number is not factorable but the three-digit number is: (11 7 6) 53 11 7 6 583 7 6 4,081 6 24,486 Here the sequence is a 2-by-2, a 3-by-1, and a 4-by-1, though when the three-digit number is factorable by 11, you can use the elevens method for a very simple 2-by-2 (53 11 583). In this case it pays to be able to recognize when a number is divisible by 11, as described in Chapter 4. If the two-digit number is not factorable, and the three-digit number is factorable into only a 2-by-1, the problem can still be handled easily by a 2-by-2, followed by a 4-by-1: (47 9) 83 47 9 3,901 9 35,109 Here you need to see that 423 is divisible by 9, setting up the problem as 83 47 9. This 2-by-2 is not so easy, but by treat-ing 83 as 80 3, you get: (80 3) 83 47 80 47 3,760 3 47 141 3,901 423 83 462 53 The Tough Stuff Made Easy:Advanced Multiplication 171 Then do the 4-by-1 problem of 3,901 9 for your ﬁnal answer of 35,109. The Addition Method If the two-digit number and the three-digit number cannot be conveniently factored in the 3-by-2 problem you’re doing, you can always resort to the addition method: (720 1) (treating 72 as 9 8) The method requires you to add a 2-by-2 (times 10) to a 2-by-1. These problems tend to be more difﬁcult than prob-lems that can be factored since you have to perform a 2-by-1 while holding on to a ﬁve-digit number, and then add the results together. In fact, it is probably easier to solve this problem by factoring 721 into 103 7, then computing 37 103 7 3,811 7 26,677. Here’s another example: (730 2) (treating 73 as 70 3) Though you will usually break up the three-digit number when using the addition method, it occasionally pays to break up the 732 57 730 57 41,610 2 57 114 41,724 721 37 720 37 26,640 1 37 37 26,677 Secrets of Mental Math 172 two-digit number instead, particularly when the last digit of the two-digit number is a 1 or a 2, as in the following example: (50 1) This reduces the 3-by-2 to a 3-by-1, made especially easy since the second multiplication problem involves a 1. Note, too, that we were aided in multiplying a 5 by an even number, which pro-duces an extra 0 in the answer, so there is only one digit of over-lap in the addition problem. Another example of multiplying a 5 by an even number is illustrated in the following problem: (60 2) When you multiply the 6 in 60 by the 5 in 835, it generates an extra 0 in the answer, making the addition problem espe-cially easy. The Subtraction Method As with 2-by-2s, it is sometimes more convenient to solve a 3-by-2 with subtraction instead of addition, as in the following problems: 835 62 60 835 50,100 2 835 1,670 51,770 386 51 50 386 19,300 1 386 386 19,686 The Tough Stuff Made Easy:Advanced Multiplication 173 (630 1) (63 9 7) (760 2) (43 40 3) By contrast, you can compare the last subtraction method with the addition method, below, for this same problem: (750 8) (75 5 5 3) My preference for tackling this problem would be to use the subtraction method because I always try to leave myself with the easiest possible addition or subtraction problem at the end. In this case, I’d rather subtract 86 than add 344, even though the 2-by-2 multiplication problem in the subtraction method above is slightly harder than the one in the addition method. The subtraction method can also be used for three-digit num-bers below a multiple of 100 or close to 1,000, as in the next two examples: 758 43 750 43 32,250 8 43 344 32,594 758 43 760 43 32,680 2 43 86 32,594 629 38 630 38 23,940 1 38 38 23,902 Secrets of Mental Math 174 (300 7) (1000 12) (12 6 2) The last three digits of the answers were obtained by taking the complements of 609 100 509 and 816, respectively. Finally, in the following illustration we break up the two-digit number using the subtraction method. Notice how we subtract 736 by subtracting 1,000 and adding back the complement: (60 1) (complement of 736) 3-BY -2 EXERCISES USING FACTORING, ADDITION, AND SUBTRACTION METHODS Solve the 3-by-2 problems below, using the factoring, addition, or subtraction method. Factoring, when possible, is usually eas-ier. The solutions appear in the back of the book. 44,160 1,000 43,160 264 43,424 736 59 60 736 44,160 1 736 736 43,424 988 68 1,000 68 68,000 12 68 816 67,184 293 87 300 87 26,100 7 87 609 25,491 The Tough Stuff Made Easy:Advanced Multiplication 175 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. The following 3-by-2s will appear in the ﬁve-digit squares and the 5-by-5 multiplication problems that follow. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 822 95 216 78 653 69 545 27 834 34 154 19 423 45 878 24 853 32 286 64 589 87 383 49 988 22 281 44 144 56 499 25 668 63 817 61 538 53 218 68 361 41 841 72 616 37 157 33 126 87 484 75 967 51 411 93 952 26 906 46 773 42 148 62 796 19 858 15 Secrets of Mental Math 176 FIVE-DIGIT SQUARES Mastering 3-by-2 multiplication takes a fair amount of practice, but once you’ve managed that, you can slide right into doing ﬁve-digit squares because they simplify into a 3-by-2 problem plus a two-digit square and a three-digit square. Watch: To square the following number: 46,7922 Treat it as: Using the distributive law we can break this down to: 1. 2. 3. 46,000 46,000 2(46,000)(792) 792 792 This can be stated more simply as: 462 1 million (46)(792)(2000) 7922 But I do not do them in this order. In fact, I start in the mid-dle, because the 3-by-2 problems are harder than the two-digit and three-digit squares. So in keeping with the principle of get-ting the hard stuff out of the way ﬁrst, I do 792 46 2 and attach three zeroes on the end, as follows: (800 8) “Fisher” 2,000 72,864,000 792 46 800 46 36,800 8 46 368 36,432 46,000 792 46,000 792 The Tough Stuff Made Easy:Advanced Multiplication 177 Using the subtraction method, as shown above, compute 792 46 36,432, then double that number to get 72,864. Using the phonetic code from the last chapter on the number 864 allows you to store this number as 72 Fisher. The next step is to do 462 1 million, which is 2,116,000,000. At this point, you can say, “Two billion . . .” Recalling the 72 of 72 Fisher, you add 116 million to get 188 million. Before saying this number aloud, you need to check ahead to see if there is a carry when adding Fisher, or 864, to 7922. Here you don’t actually calculate 7922; rather, you deter-mine that its product will be large enough to make the 864,000 carry. (You can guesstimate this by noting that 8002 is 640,000, which will easily make the 864,000 carry, thus you bump the 188 up a notch and say, “. . . 189 million. . . .”) Still holding on to Fisher, compute the square of 792, using the three-digit square method (rounding up and down by 8, and so on) to get 627,264. Finally, add 627 to Fisher, or 864, to get 1,491. But since you already made the carry, drop the 1 and say, “. . . 491 thousand 264.” Sometimes I forget the last three digits of the answer because my mind has been so preoccupied with the larger computa-tions. So before doing the ﬁnal addition I will store the 2 of 264 on my ﬁngers and try to remember the 64, which I can usually do because we tend to recall the most recent things heard. If this fails, I can come up with the ﬁnal two digits by squaring the ﬁnal two digits of the original number, 922, or 8,464, the last two digits of which are the last two digits I’m looking for: 64. (Alternatively, you can convert 264 into a word like nature.) I know this is quite a mouthful. To reiterate the entire prob-lem in a single illustration, here is how I computed 46,7922: Secrets of Mental Math 178 (800 8) “Fisher” 2,000 72,864,000 Let’s look at another ﬁve-digit square example: 83,5222 As before, we compute the answer in this order: 83 522 2,000, 832 1 million, then 5222 For the ﬁrst problem, notice that 522 is a multiple of 9. In fact, 522 58 9. Treating 83 as 80 3, we get: (58 9) 83 58 9 4,814 9 43,326 Doubling 43,326 results in 86,652, which can be stored as 86 Julian. Since 832 6,889, we can say, “Six billion . . .” 522 83 72,864,000 46,0002 2,116,000,000 2,188,864,000 627,264 2,189,491,264 792 46 800 46 36,800 8 46 368 36,432 The Tough Stuff Made Easy:Advanced Multiplication 179 8 800 7922 8 784 (82) 627,200 64 627,264 7922 Adding 889 86 gives us 975. Before saying 975 million we check to see if Julian (652,000) will cause a carry after squar-ing 522. Guesstimating 5222 as about 270,000 (500 540), you see it will not carry. Thus you can safely say, “. . . 975 million . . .” Finally, square 522 in the usual way to get 272,484 and add that to Julian (652,000) for the rest of the answer: “. . . 924,484.” Illustrated, this problem looks like: 83,5222 83 58 9 4,814 9 43,326 EXERCISE: FIVE-DIGIT SQUARES 1. 45,7952 2. 21,2312 3. 58,3242 4. 62,4572 5. 89,8542 6. 76,9342 “Julian” 43,326 2,000 86,652,000 83,0002 6,889,000,000 6,975,652,000 5222 272,484 6,975,924,484 522 83 Secrets of Mental Math 180 22 544 5222 22 500 (222) 272,000 484 272,484 3-BY -3 MULTIPLICATION In building to the grand ﬁnale of 5-by-5 multiplication, 3-by-3s are the ﬁnal hurdle. As with 3-by-2s, there are a variety of meth-ods you can use to exploit the numbers in the simpliﬁcation process. Factoring Method We’ll begin with the factoring method. Unfortunately, most three-digit numbers are not factorable into one-digit numbers, but when they are, the calculation is not too bad. (9 8 4) 829 9 8 4 7,461 8 4 59,688 4 238,752 Notice the sequence involved. You simplify the 3-by-3 (829 288) to a 3-by-1-by-1-by-1 through the factoring of 288 into 9 8 4. This then turns into a 4-by-1-by-1 (7,461 8 4), and ﬁnally into a 5-by-1 to yield the ﬁnal answer of 238,752. The beauty of this process is that there is nothing to add and nothing to store in memory. When you get to the 5-by-1, you are one step away from completion. The 5-by-1 problem can be solved in two steps by treating 59,688 as 59,000 688, then adding the results of the 2-by-1 (59,000 4) and the 3-by-1 (688 4), as below: (59,000 688) 59,688 4 59,000 4 236,000 688 4 2,752 238,752 829 288 The Tough Stuff Made Easy:Advanced Multiplication 181 If both three-digit numbers are factorable into 2-by-1s, then the 3-by-3 can be simpliﬁed to a 2-by-2-by-1-by-1, as in the fol-lowing problem: (57 9) (41 6) 57 41 9 6 2,337 9 6 21,033 6 126,198 As usual, it is best to get the hard part of the problem over ﬁrst (the 2-by-2). Once you’ve got this, the problem is then reduced to a 4-by-1, then to a 5-by-1. More often than not, only one of the three-digit numbers will be factorable, in which case it is reduced to a 3-by-2-by-1, as in the following problem: (51 9) 526 459 526 51 9 526 (50 1) 9 26,826 9 241,434 The next 3-by-3 is really just a 3-by-2 in disguise: By doubling the 435 and cutting 624 in half, we obtain the equivalent problem: 624 435 459 526 513 246 Secrets of Mental Math 182 87 52 6 10 87 (50 2) 6 10 4,524 6 10 27,144 10 271,440 Close-T ogether Method Are you ready for something easier? The next multiplication shortcut, which we introduced in Chapter 0, is based on the fol-lowing algebraic formula: (z a)(z b) z2 za zb ab Which we rewrite as: (z a)(z b) z(z a b) ab This formula is valid for any values of z, a, and b. We shall take advantage of this whenever the three-digit numbers to be multiplied (z a) and (z b) are both near an easy number z (typically one with lots of zeroes in it). For example, to multiply: We treat this problem as (100 7)(100 11). Using z 100, a 7, b 11, our formula gives us: 100(100 7 11) 7 11 100 118 77 11,877 107 111 312 (52 6) 870 (87 10) The Tough Stuff Made Easy:Advanced Multiplication 183 I diagram the problem this way: (7) (11) The numbers in parentheses denote the difference between the number and our convenient “base number” (here, z 100). The number 118 can be obtained either by adding 107 11 or 111 7. Algebraically these sums are always equal since (z a) b (z b) a. With fewer words this time, here’s another quickie: (9) (4) Neat! Let’s up the ante a little with a higher base number: (8) (9) Although this method is usually used for three-digit multipli-cation, we can use it for 2-by-2s as well: 408 409 400 417 166,800 8 9 72 166,872 109 104 100 113 11,300 9 4 36 11,336 107 111 100 118 11,800 7 11 77 11,877 Secrets of Mental Math 184 (8) (3) Here, the base number is 70, which we multiply by 81 (78 3). Even the addition component is usually very simple. We can also apply this method when the two numbers are both lower than the base number, as in the following problem where both numbers are just under 400: (4) (13) The number 383 can be obtained from 396 13, or from 387 4. I would use this method for 2-by-2 problems like the ones below: (3) (6) (1) (2) 79 78 80 77 6,160 1 2 2 6,162 97 94 100 91 9,100 3 6 18 9,118 396 387 400 383 153,200 4 13 52 153,252 78 73 70 81 5,670 8 3 24 5,694 The Tough Stuff Made Easy:Advanced Multiplication 185 In our next example, the base number falls between the two numbers: (4) (13) The number 409 is obtained from 396 13, or 413 4. Notice that since 4 and 13 are of opposite signs we must sub-tract 52 here. Let’s raise the ante higher still, to where the second step requires a 2-by-2 multiplication: (21) (37) (37 7 3) We note here that step 1 in the multiplication problem (600 658) is itself a reasonable estimate. Our method enables you to go from an estimate to the exact answer. (24) (47) (47 6 4) Also notice that in all these examples, the numbers we multi-ply in the ﬁrst step have the same sum as the original numbers. 876 853 900 829 746,100 24 47 1,128 747,228 621 637 600 658 394,800 21 37 777 395,577 396 413 400 409 163,600 4 13 52 163,548 Secrets of Mental Math 186 For example, in the problem above, 900 829 1729 just as 876 853 1729. This is because: z [(z a) b] (z a) (z b) Thus, to obtain the number to be multiplied by 900 (which you know will be 800 plus something), you need only look at the last two digits of 76 53 129 to determine 829. In the next problem, adding 827 761 1588 tells us that we should simply multiply 800 788, then subtract 27 39 as follows: (27) (39) (39 9 3) This method is so effective that if the 3-by-3 problem you are presented with has numbers that are not close together, you can sometimes modify the problem by dividing one and multiplying the other, both by the same number, to bring them closer to-gether. For instance, 672 157 can be solved by: (36) (14) (36 7 2) When the numbers being multiplied are the same (you can’t get any closer than that!), notice that the close-together method 300 350 105,000 36 14 504 105,504 336 314 2 2 672 157 827 761 800 788 630,400 39 27 1,053 629,347 The Tough Stuff Made Easy:Advanced Multiplication 187 produces the exact same calculations you did in our traditional squaring procedure: (47) (47) Addition Method When none of the previous methods work, I look for an addition-method possibility, particularly when the ﬁrst two dig-its of one of the three-digit numbers is easy to work with. For example, in the problem below, the 64 of 641 is factorable into 8 8, so I would solve the problem as illustrated: (640 1) (373 8 8 10) In a similar way, in the next problem the 42 of 427 is fac-torable into 7 6, so you can use the addition method and treat 427 as 420 7: (420 7) (656 7 6 10) 656 427 420 656 275,520 7 656 4,592 280,112 373 641 640 373 238,720 1 373 373 239,093 347 347 300 394 118,200 472 2,209 120,409 Secrets of Mental Math 188 Often I break the last addition problem into two steps, as follows: Since addition-method problems can be very strenuous, I usu-ally go out of my way to ﬁnd a method that will produce a sim-ple computation at the end. For example, the above problem could have been done using the factoring method. In fact, that is how I would choose to do it: (61 7) 656 61 7 656 (60 1) 7 40,016 7 280,112 The simplest addition-method problems are those in which one number has a 0 in the middle, as below: (300 8) These problems tend to be so much easier than other addition-method problems that it pays to see whether the 3-by-3 can 732 308 300 732 219,600 8 732 5,856 225,456 656 427 275,520 7 600 4,200 279,720 7 56 392 280,112 The Tough Stuff Made Easy:Advanced Multiplication 189 be converted to a problem like this. For instance, 732 308 could have been obtained by either of the “non-zero” prob-lems below: or We mention that another way to do this problem is by 308 366 2, and take advantage of the closeness of 308 and 366. Let’s do one more toughie: (440 3) (739 11 4 10) Subtraction Method The subtraction method is one that I sometimes use when one of the three-digit numbers can be rounded up to a convenient two-digit number with a 0 at the end, as in the next problem: (720 1) (247 9 8 10) 719 247 720 247 177,840 1 247 247 177,593 739 443 440 739 325,160 3 700 2,100 327,260 3 39 117 327,377 732 308 2 2 366 616 732 308 3 3 244 924 Secrets of Mental Math 190 Likewise in the following example: (540 2) (346 6 9 10) When-All-Else-Fails Method When all else fails, I use the following method, which is fool-proof when you can ﬁnd no other method to exploit the num-bers. In the when-all-else-fails method, the 3-by-3 problem is broken down into three parts: a 3-by-1, a 2-by-1, and a 2-by-2. As you do each computation, you sum the totals as you go. These problems are difﬁcult, especially if you cannot see the original number. In my presentation of 3-by-3s and 5-by-5s, I have the problems written down, but I do all the calculations mentally. Here’s an example: In practice, the calculation actually proceeds as shown below. Sometimes I use the phonetic code to store the thousands digits 851 527 500 851 425,500 27 800 21,600 447,100 27 51 1,377 448,477 538 346 540 346 186,840 2 346 692 186,148 The Tough Stuff Made Easy:Advanced Multiplication 191 (e.g., 447 our rug) and put the hundreds digit (1) on my ﬁngers: “Our rug” Let’s do another example, but this time I’ll break up the ﬁrst number. (I usually break up the larger one so that the addition problem is easier.) “Shut up” EXERCISE: 3-BY -3 MULTIPLICATION 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 273 138 658 468 446 176 692 644 942 879 809 527 343 226 853 325 596 167 644 286 100 619,500 73 23 1,679 621,179 923 673 9 673 6,057 6 23 138 6,195 100 447,100 51 27 1,377 448,477 851 527 5 851 4,255 8 27 216 4,471 Secrets of Mental Math 192 11. 12. 13. 14. 15. 16. 17. 18. 19. The following problems are embedded in the 5-by-5 multipli-cation problems in the next section: 20. 21. 22. 23. 5-BY -5 MULTIPLICATION The largest problem we will attempt is the mental multiplication of two ﬁve-digit numbers. To do a 5-by-5, you need to have mastered 2-by-2s, 2-by-3s, and 3-by-3s, as well as the phonetic code. Now it is just a matter of putting it all together. As you did in the ﬁve-digit square problems, you will use the distributive law to break down the numbers. For example: Based on this you can break down the problem into four eas-ier multiplication problems, which I illustrate below in criss-cross fashion as a 2-by-2, two 3-by-2s, and ﬁnally a 3-by-3, summing the results for a grand total. That is, (27 52) million [(27 196) (52 639)] thousand (639 196) 27,639 (27,000 639) 52,196 (52,000 196) 393 822 216 653 545 834 154 423 765 350 976 878 557 756 417 298 341 715 871 926 783 589 642 249 824 206 The Tough Stuff Made Easy:Advanced Multiplication 193 As with the ﬁve-digit squares, I begin in the middle with the 3-by-2s, starting with the harder 3-by-2: “Mom, no knife” 1. 52 639 52 71 9 3,692 9 33,228 Committing 33,228 to memory with the mnemonic Mom, no knife, you then turn to the second 3-by-2: 2. 27 196 27 (200 4) 5,400 108 5,292 and add it to the number that you are storing: 3. (“Mom, no knife”) for a new total, which we store as: “Movie lines” (38 million, 520 thousand) Holding on to movie lines, compute the 2-by-2: 4. 52 27 52 9 3 1,404 At this point, you can give part of the answer. Since this 2-by-2 represents 52 27 million, 1,404 represents 1 billion, 404 mil-lion. Since 404 million will not cause a carry, you can safely say, “One billion . . .” 5. 404 “Movie” (38) 442 33,228 5,292 38,520 Secrets of Mental Math 194 In this step you add 404 to movie (38) to get 442, at which point you can say, “. . . 442 million. . . .” You can say this because you know 442 will not carry—you’ve peeked ahead at the 3-by-3 to see whether it will cause 442 to carry to a higher number. If you found that it would carry, you would say, “443 million.” But since lines is 520,000 and the 3-by-3 (639 196) will not exceed 500,000 (a rough guesstimate of 600 200 120,000 shows this), you are safe. 6. 639 196 639 7 7 4 4,473 7 4 31,311 4 125,244 While still holding on to lines, you now compute the 3-by-3 using the factoring method, to get 125,244. You might convert 244 into a word like nearer. The ﬁnal step is a simple addition of: 7. 125,244 “Lines” (520,000) This allows you to say the rest of the answer: “. . . 645,244.” Since a picture is worth a thousand calculations, here’s our picture of how this would look: “Mom, no knife” “Movie lines” 38,520 1,000 38,520,000 52 27 1 million 1,404,000,000 1,442,520,000 639 196 125,244 1,442,645,244 639 52 33,228 196 27 5,292 27,639 52,196 The Tough Stuff Made Easy:Advanced Multiplication 195 I should make a parenthetical note here that I am assuming in doing 5-by-5s that you can write down the problem on a black-board or piece of paper. If you can’t, you will have to create a mnemonic for each of the four numbers. For example, in the last problem, you could store the problem as: —“Neck jump” —“Lion dopish” Then you would multiply lion jump, dopish neck, lion neck, and ﬁnally dopish jump. Obviously this would slow you down a bit, but if you want the extra challenge of not being able to see the numbers, you can still solve the problem. We conclude with one more 5-by-5 multiplication: The steps are the same as those in the last problem. You start with the harder 3-by-2 and store the answer with a mnemonic: 1. 547 79 547 (80 1) 43,760 547 43,213—“Rome anatomy” Then you compute the other 3-by-2. 2. 838 45 838 5 9 4,190 9 37,710 Summing the 3-by-2s you commit the new total to memory. 79,838 45,547 27,639 52,196 Secrets of Mental Math 196 3. —“Rome anatomy” —“Face Panama” 4. 79 45 79 9 5 711 5 3,555 The 2-by-2 gives you the ﬁrst digit of the ﬁnal answer, which you can say out loud with conﬁdence: “Three billion . . .” 5. 555 “Face” (80) 635 The millions digits of the answer involve a carry from 635 to 636, because Panama (923) needs only 77,000 to cause it to carry, and the 3-by-3 (838 547) will easily exceed that ﬁgure. So you say: “. . . 636 million . . .” The 3-by-3 is computed using the addition method: 6. (540 7) (838 9 6 10) And in the next step you add this total to Panama (923,000): 7. 923,000 458,386 1,381,386 838 547 540 838 452,520 7 800 5,600 458,120 7 38 266 458,386 43,213 37,710 80,923 The Tough Stuff Made Easy:Advanced Multiplication 197 Since you’ve already used the 1 in the carry to 636 million, you just say the thousands: “. . . 381 thousand . . . 386,” and take a bow! This problem may be illustrated in the following way: “Rome anatomy” “Face Panama” EXERCISE: 5-BY -5 MULTIPLICATION 1. 2. 3. 4. 95,393 81,822 69,216 78,653 34,545 27,834 65,154 19,423 80,923 1,000 80,923,000 79 45 1 million 3,555,000,000 3,635,923,000 838 547 458,386 3,636,381,386 547 79 43,213 838 45 37,710 79,838 45,547 Secrets of Mental Math 198 Chapter 9 Presto-digit-ation: The Art of Mathematical Magic Playing with numbers has brought me great joy in life. I ﬁnd that arithmetic can be just as entertaining as magic. But under-standing the magical secrets of arithmetic requires algebra. Of course, there are other reasons to learn algebra (SATs, modeling real-world problems, computer programming, and understand-ing higher mathematics, to name just a few), but what ﬁrst got me interested in algebra was the desire to understand some mathematical magic tricks, which I now present to you! PSYCHIC MATH Say to a volunteer in the audience, “Think of a number, any number.” And you should also say, “But to make it easy on yourself, think of a one-digit or two-digit number.” After you’ve reminded your volunteer that there’s no way you could know her number, ask her to: 1. double the number, 2. add 12, 3. divide the total by 2, 4. subtract the original number. Then say, “Are you now thinking of the number six?” Try this one on yourself ﬁrst and you will see that the sequence always produces the number 6 no matter what number is origi-nally selected. Why This Trick Works This trick is based on simple algebra. In fact, I sometimes use this as a way to introduce algebra to students. The secret num-ber that your volunteer chose can be represented by the letter x. Here are the functions you performed in the order you per-formed them: 1. 2x (double the number) 2. 2x 12 (then add 12) 3. (2x 12) 2 x 6 (then divide by 2) 4. x 6 x 6 (then subtract original number) So no matter what number your volunteer chooses, the ﬁnal answer will always be 6. If you repeat this trick, have the volun-teer add a different number at step 2 (say 18). The ﬁnal answer will be half that number (namely 9). THE MAGIC 1089! Here is a trick that has been around for centuries. Have your audience member take out paper and pencil and: 1. secretly write down a three-digit number where the digits are decreasing (like 851 or 973), Secrets of Mental Math 200 2. reverse that number and subtract it from the ﬁrst number, 3. take that answer and add it to the reverse of itself. At the end of this sequence, the answer 1089 will magically appear, no matter what number your volunteer originally chose. For example: Why This Trick Works No matter what three-digit number you or anyone else chooses in this game, the ﬁnal result will always be 1089. Why? Let abc denote the unknown three-digit number. Algebraically, this is equal to: 100a 10b c When you reverse the number and subtract it from the origi-nal number, you get the number cba, algebraically equal to: 100c 10b a Upon subtracting cba from abc, you get: 100a 10b c (100c 10b a) 100(a c) (c a) 99(a c) 851 158 693 396 1089 Presto-digit-ation: The Art of Mathematical Magic 201 Hence, after subtracting in step 2, we must have one of the following multiples of 99: 198, 297, 396, 495, 594, 693, 792, or 891, each one of which will produce 1089 after adding it to the reverse of itself, as we did in step 3. MISSING-DIGIT TRICKS Using the number 1,089 from the last effect, hand a volunteer a calculator and ask her to multiply 1089 by any three-digit num-ber she likes, but not to tell you the three-digit number. (Say she secretly multiplies 1,089 256 278,784.) Ask her how many digits are in her answer. She’ll reply, “Six.” Next you say: “Call out ﬁve of your six digits to me in any order you like. I shall try to determine the missing digit.” Suppose she calls out, “Two . . . four . . . seven . . . eight . . . eight.” You correctly tell her that she left out the number 7. The secret is based on the fact that a number is a multiple of 9 if and only if its digits sum to a multiple of 9. Since 1 0 8 9 18 is a multiple of 9, then so is 1089. Thus 1089 times any whole number will also be a multiple of 9. Since the digits called out add up to 29, and the next multiple of 9 greater than 29 is 36, our volunteer must have left out the number 7 (since 29 7 36). There are more subtle ways to force the volunteer to end up with a multiple of 9. Here are some of my favorites: 1. Have the volunteer randomly choose a six-digit number, scramble its digits, then subtract the smaller six-digit number from the larger one. Since we’re subtracting two numbers with the same mod sum (indeed, the identical sum of digits), the resulting difference will have a mod sum of 0, and hence be a multiple of 9. Then continue as before to ﬁnd the missing digit. Secrets of Mental Math 202 2. Have the volunteer secretly choose a four-digit number, reverse the digits, then subtract the smaller number from the larger. (This will be a multiple of 9.) Then multiply this by any three-digit number, and continue as before. 3. Ask the volunteer to multiply one-digit numbers randomly until the product is seven digits long. This is not “guaranteed” to produce a multiple of 9, but in practice it will do so at least 90% of the time (the chances are high that the one-digit numbers being multiplied include a 9 or two 3s or two 6s, or a 3 and a 6). I often use this method in front of mathematically advanced audiences who might see through other methods. There is one problem to watch out for. Suppose the numbers called out add up to a multiple of 9 (say 18). Then you have no way of determining whether the missing digit is 0 or 9. How do you remedy that? Simple—you cheat! You merely say, “You didn’t leave out a zero, did you?” If she did leave out a 0, you have completed the trick successfully. If she did not leave out the 0, you say: “Oh, it seemed as though you were thinking of noth-ing! You didn’t leave out a one, two, three, or four, did you?” She’ll either shake her head, or say no. Then you follow with, “Nor did you leave out a ﬁve, six, seven, or eight, either. You left out the number nine, didn’t you?” She’ll respond in the afﬁr-mative, and you will receive your well-deserved applause! LEAPFROG ADDITION This trick combines a quick mental calculation with an aston-ishing prediction. Handing the spectator a card with ten lines, numbered 1 through 10, have the spectator think of two posi-tive numbers between 1 and 20, and enter them on lines 1 and 2 of the card. Next have the spectator write the sum of lines 1 and Presto-digit-ation: The Art of Mathematical Magic 203 2 on line 3, then the sum of lines 2 and 3 on line 4, and so on as illustrated below. 1 9 2 2 3 11 4 13 5 24 6 37 7 61 8 98 9 159 10 257 Finally, have the spectator show you the card. At a glance, you can tell him the sum of all the numbers on the card. For instance, in our example, you could instantly announce that the numbers sum up to 671 faster than the spectator could do using a calculator! As a kicker, hand the spectator a calculator, and ask him to divide the number on line 10 by the number on line 9. In our example, the quotient 2 1 5 5 7 9 1.616. . . . Have the spec-tator announce the ﬁrst three digits of the quotient, then turn the card over (where you have already written a prediction). He’ll be surprised to see that you’ve already written the number 1.61! Why This Trick Works To perform the quick calculation, you simply multiply the num-ber on line 7 by 11. Here 61 11 671. The reason this works is illustrated in the table below. If we denote the numbers on lines 1 and 2 by x and y, respectively, then the sum of lines 1 Secrets of Mental Math 204 through 10 must be 55x 88y, which equals 11 times (5x 8y), that is, eleven times the number on line 7. 1 x 2 y 3 x y 4 x 2y 5 2x 3y 6 3x 5y 7 5x 8y 8 8x 13y 9 13x 21y 10 21x 34y Total: 55x 88y As for the prediction, we exploit the fact that for any positive numbers, a, b, c, d, if a/b < c/d, then it can be shown that the fraction you get by “adding fractions badly” (i.e., adding the numerators and adding the denominators) must lie in between the original two fractions. That is, < < Thus the quotient of line 10 divided by line 9 (21x 34y)/ (13x 21y), must lie between 1.615 . . . < < 1.619 . . . Hence, the ratio must begin with the digits 1.61, as predicted. In fact, if you continue the leapfrog process indeﬁnitely, the ratio of consecutive terms gets closer and closer to 34y 21y 21x 34y 13x 21y 21x 13x c d a c b d a b Presto-digit-ation: The Art of Mathematical Magic 205 ≈1.6180339887 . . . a number with so many amazingly beautiful and mysterious properties that it is often called the golden ratio. MAGIC SQUARES Are you ready for a challenge of a different sort? Below you will ﬁnd what is called a magic square. There has been much written on magic squares and how to construct them, going back as far as ancient China. Here we describe a way to present magic squares in an entertaining fashion. This is a routine I’ve been doing for years. I bring out a business card with the following written on the back: 8 11 14 1 13 2 7 12 3 16 9 6 10 5 4 15 34 I say, “This is called a magic square. In fact, it’s the smallest magic square you can create, using the numbers one through sixteen. You’ll notice that every row and every column adds to the same number—thirty-four. Now I’ve done such an extensive study on magic squares that I propose to create one for you right before your very eyes.” I then ask someone from the audience to give me any number larger than 34. Let’s suppose she says 67. I then bring out another business card and draw a blank 4-by-4 grid, and place the number 67 to the right of it. Next I 1 5 2 Secrets of Mental Math 206 ask her to point to the squares, one at a time, in any order. As she points to an empty box, I immediately write a number inside it. The end result looks like this: 16 19 23 9 22 10 15 20 11 25 17 14 18 13 12 24 67 I continue: “Now with the ﬁrst magic square, every row and column added to thirty-four. [I usually put the thirty-four card away at this point.] Let’s see how we did with your square.” After checking that each row and column adds up to 67, I say: “But I did not stop there. For you, I decided to go one step fur-ther. Notice that both diagonals add up to sixty-seven!” Then I point out that the four squares in the upper left corner sum to 67 (16 19 22 10 67), as do the other three four-square corners, the four squares in the middle, and the four corner squares! “They all sum to sixty-seven. But don’t take my word for any of this. Please keep this magic square as a souvenir from me—and check it out for yourself!” HOW TO CONSTRUCT A MAGIC SQUARE You can create a magic square that sums to any number by tak-ing advantage of the original magic square that sums to 34. Keep that square within eyeshot while you construct the volun-teer’s magic square. As you draw the 4-by-4 grid, mentally per-form the calculations of steps 1 and 2: 1. Subtract 34 from the given number (e.g., 67 34 33). 2. Divide this number by 4 (e.g., 33 4 8 with a remainder of 1). Presto-digit-ation: The Art of Mathematical Magic 207 The quotient is the ﬁrst “magic” number. The quotient plus the remainder is the second “magic” number. (Here our magic numbers are 8 and 9.) 3. When the volunteer points out a square, inconspicuously look at the 34-square and see what is in the corresponding square. If it is a 13, 14, 15, or 16, add the second number to it (e.g., 9). If not, add the ﬁrst magic number (e.g., 8). 4. Insert the appropriate number until the magic square is completed. Note that when the given number is even, but not a multiple of 4, then your ﬁrst and second magic numbers will be the same, so you’ll have just one magic number to add to the numbers in your 34-square. Why This Trick Works The reason this method works is based on the fact that every row, column, diagonal (and more) from the originally displayed magic square sums to 34. Suppose the given number had been 82. Since 82 34 48 (and 48 4 12), we would add 12 to each square. Then every group of four that had previously summed to 34 would add to 34 48 82. See the magic square below. 20 23 26 13 25 14 19 24 15 28 21 18 22 17 16 27 82 On the other hand, if the given number were 85, our magic numbers would be 12 and 15, so we would be adding 3 more to the squares showing 13, 14, 15, and 16. Since each row, col-Secrets of Mental Math 208 umn, and group of four contains exactly one of these numbers, each group of four would now add to 34 48 3 85 in the following magic square. 20 23 29 13 28 14 19 24 15 31 21 18 22 17 16 30 85 As an interesting piece of mathemagical trivia, let me point out another astonishing property of the famous 3-by-3 magic square below. 4 9 2 3 5 7 8 1 6 15 Not only do the rows, columns, and diagonals add up to 15, but if you treat the rows of the magic square as three-digit numbers, you can verify on your calculator that 4922 3572 8162 2942 7532 6182. Also, 4382 9512 2762 8342 1592 6722. If you are curious about why this property hap-pens, you might want to explore my paper Magic “Squares” Indeed! (included in the bibliography). QUICK CUBE ROOTS Ask someone to select a two-digit number and keep it secret. Then have him cube the number; that is, multiply it by itself twice (using a calculator). For instance, if the secret number is 68, have the volunteer compute 68 68 68 314,432. Then Presto-digit-ation: The Art of Mathematical Magic 209 ask the volunteer to tell you his answer. Once he tells you the cube, 314,432, you can instantly reveal the original secret num-ber, the cube root, 68. How? To calculate cube roots, you need to learn the cubes from 1 to 10: 13 1 23 8 33 27 43 64 53 125 63 216 73 343 83 512 93 729 103 1000 Once you have learned these, calculating the cube roots is as easy as π. For instance, with this example problem: What is the cube root of 314,432? Seems like a pretty tough one to begin with, but don’t panic, it’s actually quite simple. As usual, we’ll take it one step at a time: 1. Look at the magnitude of the thousands number (the numbers to the left of the comma), 314 in this example. 2. Since 314 lies between 63 216 and 73 343, the cube root lies in the 60s (since 603 216,000 and 703 343,000). Hence the ﬁrst digit of the cube root is 6. 3. To determine the last digit of the cube root, note that only the number 8 has a cube that ends in 2 (83 512), so the last digit ends in 8. Secrets of Mental Math 210 Therefore, the cube root of 314,432 is 68. Three simple steps and you’re there. Notice that every digit, 0 through 9, appears once among the last digits of the cubes. (In fact, the last digit of the cube root is equal to the last digit of the cube of the last digit of the cube. Go ﬁgure out that one!) Now you try one for practice: What is the cube root of 19,683? 1. 19 lies between 8 and 27 (23 and 33). 2. Therefore the cube root is 20-something. 3. The last digit of the problem is 3, which corresponds to 343 73, so 7 is the last digit. The answer is 27. Notice that our derivation of the last digit will only work if the original number is the cube root of a whole number. For instance, the cube root of 19,684 is 27.0004572 . . . deﬁnitely not 24. That’s why we included this in our mathemagical magic section and not in an earlier chapter. (Besides, the calculation goes so fast, it seems like magic!) SIMPLIFIED SQUARE ROOTS Square roots can also be calculated easily if you are given a per-fect square. For instance, if someone told you that the square of a two-digit number was 7569, you could immediately tell her that the original number (the square root) is 87. Here’s how. 1. Look at the magnitude of the “hundreds number” (the numbers preceding the last two digits), 75 in this example. 2. Since 75 lies between 82 (8 8 64) and 92 (9 9 81), then we know that the square root lies in the 80s. Hence the ﬁrst digit of Presto-digit-ation: The Art of Mathematical Magic 211 the square root is 8. Now there are two numbers whose square ends in 9: 32 9 and 72 49. So the last digit must be 3 or 7. Hence the square root is either 83 or 87. Which one? 3. Compare the original number with the square of 85 (which we can easily compute as 80 90 25 7225. Since 7569 is larger than 7225, the square root is the larger number, 87. Let’s do one more example. What is the square root of 4761? Since 47 lies between 62 36 and 72 49, the answer must be in the 60s. Since the last digit of the square is 1, the last digit of the square root must be 1 or 9. Since 4761 is greater than 652 4225, the square root must be 69. As with the previous cube root trick, this method can be applied only when the origi-nal number given is a perfect square. AN “AMAZING” SUM The following trick was ﬁrst shown to me by James “the Amaz-ing” Randi, who has used it effectively in his magic. Here, the magician is able to predict the total of four randomly chosen three-digit numbers. To prepare this trick you will need three sets of nine cards each, and a piece of paper with the number 2247 written down on it and then sealed in an envelope. Next, on each of the three sets of cards do the following: On Set A write the following numbers, one number on each card: 4286 5771 9083 6518 2396 6860 2909 5546 8174 Secrets of Mental Math 212 On Set B write the following numbers: 5792 6881 7547 3299 7187 6557 7097 5288 6548 On Set C write the following numbers: 2708 5435 6812 7343 1286 5237 6470 8234 5129 Select three people in the audience and give each one a set of cards. Have each of your volunteers randomly pick one of the nine cards they hold. Let’s say they choose the numbers 4286, 5792, and 5435. Now, in the sequence, have each one call out one digit from the four-digit number, ﬁrst person A, then per-son B, and ﬁnally person C. Say they call out the numbers 8, 9, and 5. Write down the numbers 8, 9, and 5 (895) and say, “You must admit that this number was picked entirely at random and could not possibly have been predicted in advance.” Next, have the three people call out a different number from their cards. Say they call out 4, 5, and 3. Write 453 below 895. Then repeat this two more times with their remaining two num-bers, resulting in four three-digit numbers, such as: A B C 8 9 5 4 5 3 2 2 4 6 7 5 2 2 4 7 Next have someone add the four numbers and announce the total. Then have someone open the envelope to reveal your pre-diction. Take a bow! Presto-digit-ation: The Art of Mathematical Magic 213 Why This Trick Works Look at the numbers in each set of cards and see if you can ﬁnd anything consistent about them. Each set of numbers sums to the same total. Set A numbers total to 20. Set B numbers total to 23. Set C numbers total to 17. With person C’s numbers in the right column totaling to 17, you will always put down the 7 and carry the 1. With person B’s numbers totaling to 23, plus the 1, you will always put down the 4 and carry the 2. Finally with person A’s numbers totaling to 20, adding the 2 gives you a total of 2247! A DAY FOR ANY DATE We conclude our book with one of the classic feats of mental calculation: how to ﬁgure out the day of the week of anyone’s birthday. This is actually a very practical skill. It’s not every day that someone will ask you to square a three-digit number, but hardly a day goes by without somebody mentioning a date to you in the past or future. With just a little bit of practice, you can quickly and easily determine the day of the week of practi-cally any date in history. First we assign a code number to every day of the week. They are easy to remember: Number Day 1 Monday 2 Tuesday 3 Wednesday 4 Thursday 5 Friday 6 Saturday 7 or 0 Sunday Secrets of Mental Math 214 I have found the above list to be easy to remember, especially since 2’s day is Tuesday. The other days can be given similar mnemonics: “1 day is Monday, 4’s day is Thursday, and 5 day is Friday. For Wednesday, notice that if you hold up three ﬁngers, you get the letter w. For the weekends (at the end of our list), you might think of Saturday as 6-urday and Sunday as 7 day (especially if you pronounce the v softly). Or to remember the zero, you could think of Sunday as none-day or nun-day! Next we need a code for every month of the year. These month codes are used for every year, with two exceptions. In a leap year (like 2000 or 2004 or 2008 or . . .) the month code for January is 5, and the month code for February is 1. To make the month codes easier to remember, we provide a table of mnemonics below. Month Code Mnemonic January 6 W-I-N-T -E-R has 6 letters. February 2 February is the 2nd month of the year. March 2 March 2 the beat of the drum! April 5 A-P-R-I-L and F-O-O-L-S have 5 letters. May 0 May I have a sandwich? Hold the May-0! June 3 June B-U-G has 3 letters. July 5 Watching FIVER-works and FIVER-crackers! August 1 August begins with A, the 1st letter. September 4 September is the beginning of F-A-L-L. October 6 Halloween T -R-I-C-K-S and T -R-E-A-T -S. November 2 I’ll have 2 servings of TUrkey, please! December 4 December is the L-A-S-T month, or X-M-A-S. In a leap year, the code for January is 5 and the code for February is 1. Now let’s calculate the day of the week for any date in 2006. After that, we will describe 2007, then 2008, and so on, for the Presto-digit-ation: The Art of Mathematical Magic 215 rest of your life. Once all future dates are taken care of, we can look back into the past and determine the days of the week for any date in the 1900s or any other century. Every year is assigned a code number, and for 2006 that year code happens to be 0 (see page 218). Now, to calculate the day of the week, you simply add the month code plus the date code plus the year code. Thus for December 3, 2006, we compute Month Code Date Year Code 4 3 0 7 Hence, this date will be on Day 7, which is Sunday. How about November 18, 2006? Since November has a month code of 2, we have Month Code Date Year Code 2 18 0 20. Now since the week repeats every seven days, we can sub-tract any multiple of 7 from our answer (7, 14, 21, 28, 35, . . .) and this will not change the day of the week. So our ﬁnal step is to subtract the biggest multiple of 7 to get 20 14 6. Hence November 18, 2006, occurs on Saturday. How about 2007? Well, what happens to your birthday as you go from one year to the next? For most years, there are 365 days, and since 364 is a multiple of 7 (7 52 364), then the day of the week of your birthday will shift forward by one day in most years. If there are 366 days between your birthdays, then it will shift forward by two days. Hence, for 2007 we cal-culate the day of the week just as before, but now we use a year code of 1. Next, 2008 is a leap year. (Leap years occur every four years, so 2000, 2004, 2008, 2012, . . . , 2096 are the leap years of the twenty-ﬁrst century.) Hence, for 2008, the year Secrets of Mental Math 216 code increases by two, so it will have a year code of 3. The next year, 2009, is not a leap year, so the year code increases by one, to 4. Thus, for example, May 2, 2007, has Month Code Date Year Code 0 2 1 3 so this date is a Wednesday. For September 9, 2008, we have Month Code Date Year Code 4 9 3 16 Subtracting the biggest multiple of 7, we have 16 14 2, so this date is a Tuesday. On the other hand, January 16, 2008, is a leap year January, so its month code will be 5 instead of 6. Thus, we have Month Code Date Year Code 5 16 3 24 and therefore occurs on day 24 21 3, which is Wednesday. For your reference, we have listed all of the year codes for the twenty-ﬁrst century in the ﬁgure on the following page. The good news is that we do not have to memorize this table. We can mentally calculate the year code for any date between 2000 and 2099. For the year code of 2000 x, we simply take the number x/4 (ignoring any remainder) and add that to x. The year code can be reduced by subtracting any multiple of 7. For example, with 2061, we see that 6 4 1 15 (with a remain-der of 1 that we ignore). Thus 2061 has a year code of 61 15 76. And since we can subtract any multiple of 7, we use the sim-pler year code of 76 70 6. Presto-digit-ation: The Art of Mathematical Magic 217 Hence March 19, 2061, has Month Code Date Year Code 2 19 6 27 Secrets of Mental Math 218 Year Code Year Code Year Code Year Code 2000 0 2025 3 2050 6 2075 2 2001 1 2026 4 2051 0 2076 4 2002 2 2027 5 2052 2 2077 5 2003 3 2028 0 2053 3 2078 6 2004 5 2029 1 2054 4 2079 0 2005 6 2030 2 2055 5 2080 2 2006 0 2031 3 2056 0 2081 3 2007 1 2032 5 2057 1 2082 4 2008 3 2033 6 2058 2 2083 5 2009 4 2034 0 2059 3 2084 0 2010 5 2035 1 2060 5 2085 1 2011 6 2036 3 2061 6 2086 2 2012 1 2037 4 2062 0 2087 3 2013 2 2038 5 2063 1 2088 5 2014 3 2039 6 2064 3 2089 6 2015 4 2040 1 2065 4 2090 0 2016 6 2041 2 2066 5 2091 1 2017 0 2042 3 2067 6 2092 3 2018 1 2043 4 2068 1 2093 4 2019 2 2044 6 2069 2 2094 5 2020 4 2045 0 2070 3 2095 6 2021 5 2046 1 2071 4 2096 1 2022 6 2047 2 2072 6 2097 2 2023 0 2048 4 2073 0 2098 3 2024 2 2049 5 2074 1 2099 4 Subtracting 27 21 6 tells us that this date will take place on Saturday. What about birth dates between 1900 and 1999? Do the problem exactly as in the previous calculation, but shift the ﬁnal answer forward by one day (or simply add 1 to the year code). Thus March 19, 1961, occurred on a Sunday. For the date December 3, 1998, we have 9 4 8 24 (with a remainder of 2 that we ignore). Thus 1998 has year code 98 24 1 123, where the plus one is added for dates in the 1900s. Next, subtract the biggest multiple of 7. For handy refer-ence, here are the multiples of 7 that you might need: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, . . . Since 123 119 4, 1998 has a year code of 4. Therefore, December 3, 1998, has Month Code Date Year Code 4 3 4 11 and 11 7 4, so this date occurred on a Thursday. For dates in the 1800s, we add 3 to the year code. For exam-ple, Charles Darwin and Abraham Lincoln were both born on February 12, 1809. Since 2009 has a year code of 4, then 1809 has a year code of 4 3 7, which can be reduced to 0. Thus, February 12, 1809, has Month Code Date Year Code 2 12 0 14 and 14 14 0, so they were both born on a Sunday. Presto-digit-ation: The Art of Mathematical Magic 219 For dates in the 2100s, we add 5 to the year code (or, equiv-alently, subtract 2 from the year code). For example, since 2009 has a year code of 4, then 2109 has a year code of 4 5 9, which, after subtracting 7 is the same as a year code of 2. Dates in the 1700s are treated just like in the 2100s (by adding 5 or subtracting 2) but we need to be careful. The days that we are calculating are based on the Gregorian calendar, established in 1582. But this calendar was not adopted by England (and the American colonies) until 1752, when Wed-nesday, September 2, was followed by Thursday, September 14. Let’s verify that September 14, 1752, was indeed a Thurs-day. Since 2052 has a year code of 2 (from page 218, or by 52 13 63 2), then 1752 has a year code of 0. Thus Sep-tember 14, 1752, has Month Code Date Year Code 4 14 0 18 and 18 14 4, so it was indeed a Thursday. However, our for-mula will not work for earlier dates (which were governed by the Julian calendar). Finally, we remark that under the Gregorian calendar, a leap year occurs every four years, with the exception of years that are divisible by 100, although there is an exception to the exception: years divisible by 400 are leap years. Thus 1600, 2000, 2400, and 2800 are leap years, but 1700, 1800, 1900, 2100, 2200, 2300, and 2500 are not leap years. In fact, the Gregorian calendar repeats every 400 years, so you can con-vert any future date into a date near 2000. For example, March 19, 2361, and March 19, 2761, will have the same day of the week as March 19, 1961, which, as we already deter-mined, is a Sunday. Secrets of Mental Math 220 EXERCISE: A DAY FOR ANY DATE Determine the days of the week for the following dates: 1. January 19, 2007 2. February 14, 2012 3. June 20, 1993 4. September 1, 1983 5. September 8, 1954 6. November 19, 1863 7. July 4, 1776 8. February 22, 2222 9. June 31, 2468 10. January 1, 2358 Presto-digit-ation: The Art of Mathematical Magic 221 Chapter ∞ Epilogue: How Math Helps Us Think About Weird Things by Michael Shermer As the publisher of Skeptic magazine, the executive director of the Skeptics Society, and a Scientiﬁc American editor with a monthly column entitled “Skeptic,” I receive volumes of mail from people who challenge me with stories about their unusual experiences, such as haunted houses, ghosts, near-death and out-of-body experiences, UFO sightings, alien abductions, death-premonition dreams, and much more. The most interesting stories to me are those about highly improbable events. The implication of the letter writer’s tale is that if I cannot offer a satisfactory natural explanation for that particular event, the general principle of supernaturalism is pre-served. A common story is the one about having a dream about the death of a friend or relative, then a phone call comes the next day about the unexpected death of the person in the dream. What are the odds of that? I am asked. Here is where math comes in to play in our thinking and rea-soning. I don’t want to pontiﬁcate about how mathematics in school teaches students to think critically, because that has probably been said by nearly every math teacher in nearly every math class in nearly every school in America, at least once a year. I want to give some speciﬁc examples of how I use very simple math to help me on the job in explaining why weird things happen to people. Although I cannot always explain such speciﬁc occurrences, a principle of probability called the Law of Large Numbers shows that an event with a low probability of occurrence in a small number of trials has a high probability of occurrence in a large number of trials. Or, as I like to say, million-to-one odds happen 295 times a day in America. Let’s begin with death premonitions. Here is a little “back-of-the-envelope” calculation I did. Psychologists tell us that the average person has about ﬁve dreams per day, which equals 1,825 dreams per year. Even if we remember only one out of ten dreams, that still results in 182.5 remembered dreams a year. There are 295 million Americans, so that means there will be 53.8 billion remembered dreams per year. Now, anthropologists and sociologists tell us that each of us knows about 150 people fairly well (that is, the average person has about 150 names in his or her address book about which can be said something sig-niﬁcant). That means there is a network grid of 44.3 billion per-sonal relationships among those 295 million Americans. The annual U.S. death rate from all causes across all ages is .008, or 2.6 million per year. It is inevitable that some of those 53.8 bil-lion remembered dreams will be about some of these 2.6 million deaths among the 295 million Americans and their 44.3 billion relationships. It would be a miracle, in fact, if some “death pre-monition” dreams did not come true. Even if my numbers are off, even way off, the point still stands. What are the odds of a death premonition dream com-ing true? Pretty darn good. Epilogue: How Math Helps Us Think About Weird Things 223 There is an additional psychological factor at work here called the conﬁrmation bias, where we notice the hits and ignore the misses in support of our favorite beliefs. The conﬁrmation bias explains how conspiracy theories work, for example. Peo-ple who adhere to a particular conspiracy theory (9/11 was orchestrated by the Bush administration in order to launch a war in the Middle East), will look for and ﬁnd little factoids here and there that seem to indicate that it might be true (Bush sat in that classroom reading to the children about goats as if he knew he was safe), while ignoring the vast body of evidence that points to another more likely explanation (Osama bin Laden and his band of international terrorists orchestrated 9/11). The conﬁrmation bias also helps explain how astrologers, tarot-card readers, and psychics seem so successful at “reading” people. People who get a reading are likely to remember the handful of hits and forget the countless misses. When such hits and misses are actually counted—which I once did for an ABC television special on psychics—it turns out that there is nothing more than guessing and random chance at work. In the case of the death-premonition dream, if just a couple of these people who have such dreams recount their miraculous tales in a public forum (next on Oprah!), the paranormal seems vindicated. In fact, it is nothing more than the laws of probabil-ity writ large. This mathematical process of thinking about weird things led me to another back-of-the-envelope calculation about miracles. People typically invoke the term miracle to describe really unusual events, events whose odds of occurring are a “million to one.” Okay, let’s take that as our benchmark deﬁnition. A miracle is an event whose odds of occurrence are a million to one. Now, as we go about our day, we see and hear things hap-pen about once per second. That is, data from the world and Secrets of Mental Math 224 events around us are pouring in through our senses at a rate of about one per second. If we are awake and alert and out in the world for, say, eight hours a day, that means there are thirty thousand bits of data per day, or one million events per month that we take in. The vast majority of these data and events are completely meaningless, of course, and our brains are wired to ﬁlter out and forget the vast majority of them because we would be overwhelmed otherwise. But, in the course of a month, we would expect million-to-one odds to happen at least once. Add to that the conﬁrmation bias where we will remember the most unusual events and forget all the rest, and it is inevitable that someone somewhere will report a miracle every month. And the tabloids will be there to record it! This is a short primer on how science works. In our quest to understand how the world works, we need to determine what is real and what is not, what happens by chance and what hap-pens because of some particular predictable cause. The problem we face is that the human brain was designed by evolution to pay attention to the really unusual events and ignore the vast body of data ﬂowing by; as such, thinking statistically and with probabilities does not come naturally. Science, to that extent, does not come naturally. It takes some training and practice. In addition, there are those pesky cognitive biases I men-tioned, such as the conﬁrmation bias. And there are others. The data do not just speak for themselves. Data are ﬁltered through very subjective and biased brains. The self-serving bias, for example, dictates that we tend to see ourselves in a more posi-tive light than others see us: national surveys show that most business people believe they are more moral than other business people, while psychologists who study moral intuition think they are more moral than other such psychologists. In one Col-lege Entrance Examination Board survey of 829,000 high school Epilogue: How Math Helps Us Think About Weird Things 225 seniors, 0 percent rated themselves below average in “ability to get along with others,” while 60 percent put themselves in the top 10 percent (presumably not all were from Lake Woebegone). And according to a 1997 U.S. News & World Report study on who Americans believe are most likely to go to heaven, 52 per-cent said Bill Clinton, 60 percent thought Princess Diana, 65 per-cent chose Michael Jordan, 79 percent selected Mother Teresa, and, at 87 percent, the person most likely to go to heaven was the survey taker! Princeton University psychology professor Emily Pronin and her colleagues tested a bias called blind spot, in which subjects recognized the existence and inﬂuence in others of eight different cognitive biases, but they failed to see those same biases in themselves. In one study on Stanford University students, when asked to compare themselves to their peers on such personal qualities as friendliness and selﬁshness, they pre-dictably rated themselves higher. Even when the subjects were warned about the better-than-average bias and were asked to reevaluate their original assessments, 63 percent claimed that their initial evaluations were objective, and 13 percent even claimed that they were originally too modest! In a second study, Pronin randomly assigned subjects high or low scores on a “social intelligence” test. Unsurprisingly, those given the high marks rated the test fairer and more useful than those receiving low marks. When asked if it was possible that they had been inﬂuenced by the score on the test, subjects responded that other participants had been far more biased than they were. In a third study in which Pronin queried subjects about what method they used to assess their own and others’ biases, she found that people tend to use general theories of behavior when evaluating others, but use introspection when appraising Secrets of Mental Math 226 themselves; however, in what is called the introspection illu-sion, people do not believe that others can be trusted to do the same. Okay for me but not for thee. The University of California at Berkeley psychologist Frank J. Sulloway and I made a similar discovery of an attribution bias in a study we conducted on why people say they believe in God, and why they think other people believe in God. In general, most people attribute their own belief in God to such intellec-tual reasons as the good design and complexity of the world, whereas they attribute others’ belief in God to such emotional reasons as it is comforting, gives meaning, and that they were raised to believe. Political scientists have made a similar discov-ery about political attitudes, where Republicans justify their conservative attitudes with rational arguments but claim that Democrats are “bleeding-heart liberals,” and where Democrats claim that their liberal attitudes are the most rational but claim that Republicans are “heartless.” How does science deal with such subjective biases? How do we know when a claim is bogus or real? We want to be open-minded enough to accept radical new ideas when they occasionally come along, but we don’t want to be so open-minded that our brains fall out. This problem led us at the Skeptics Society to create an educational tool called the Baloney Detection Kit, inspired by Carl Sagan’s discussion of how to detect “baloney” in his mar-velous book The Demon-Haunted World. In this Baloney Detec-tion Kit, we suggest ten questions to ask when encountering any claim that can help us decide if we are being too open-minded in accepting it or too closed-minded in rejecting it. 1. How reliable is the source of the claim? As Daniel Kevles showed so effectively in his 1999 book The Baltimore Affair, in Epilogue: How Math Helps Us Think About Weird Things 227 investigating possible scientiﬁc fraud there is a boundary prob-lem in detecting a fraudulent signal within the background noise of mistakes and sloppiness that is a normal part of the scientiﬁc process. The investigation of research notes in a laboratory afﬁl-iated with Nobel laureate David Baltimore by an independent committee established by Congress to investigate potential fraud revealed a surprising number of mistakes. But science is messier than most people realize. Baltimore was exonerated when it became clear that there was no purposeful data manipulation. 2. Does this source often make similar claims? Pseudoscientists have a habit of going well beyond the facts, so when individuals make numerous extraordinary claims, they may be more than just iconoclasts. This is a matter of quantitative scaling, since some great thinkers often go beyond the data in their creative speculations. Cornell’s Thomas Gold is notorious for his radical ideas, but he has been right often enough that other scientists listen to what he has to say. Gold proposes, for example, that oil is not a fossil fuel at all, but the by-product of a deep hot bio-sphere. Hardly any earth scientists I have spoken with take this thesis seriously, yet they do not consider Gold a crank. What we are looking for here is a pattern of fringe thinking that consis-tently ignores or distorts data. 3. Have the claims been veriﬁed by another source? Typically pseudoscientists will make statements that are unveriﬁed, or veriﬁed by a source within their own belief circle. We must ask who is checking the claims, and even who is checking the check-ers. The biggest problem with the cold fusion debacle, for exam-ple, was not that scientists Stanley Pons and Martin Fleischman were wrong; it was that they announced their spectacular dis-covery before it was veriﬁed by other laboratories (at a press Secrets of Mental Math 228 conference no less), and, worse, when cold fusion was not repli-cated, they continued to cling to their claim. 4. How does the claim ﬁt with what we know about how the world works? An extraordinary claim must be placed into a larger context to see how it ﬁts. When people claim that the pyramids and the Sphinx were built more than ten thousand years ago by an advanced race of humans, they are not present-ing any context for that earlier civilization. Where are the rest of the artifacts of those people? Where are their works of art, their weapons, their clothing, their tools, their trash? This is simply not how archaeology works. 5.Has anyone gone out of the way to disprove the claim,or has only conﬁrmatory evidence been sought? This is the conﬁrma-tion bias, or the tendency to seek conﬁrmatory evidence and reject or ignore disconﬁrmatory evidence. The conﬁrmation bias is powerful and pervasive and is almost impossible for any of us to avoid. It is why the methods of science that emphasize check-ing and rechecking, veriﬁcation and replication, and especially attempts to falsify a claim are so critical. 6. Does the preponderance of evidence converge to the claimant’s conclusion, or a different one? The theory of evolu-tion, for example, is proven through a convergence of evidence from a number of independent lines of inquiry. No one fossil, no one piece of biological or paleontological evidence has “evolu-tion” written on it; instead there is a convergence of evidence from tens of thousands of evidentiary bits that adds up to a story of the evolution of life. Creationists conveniently ignore this convergence, focusing instead on trivial anomalies or cur-rently unexplained phenomena in the history of life. Epilogue: How Math Helps Us Think About Weird Things 229 7. Is the claimant employing the accepted rules of reason and tools of research, or have these been abandoned in favor of others that lead to the desired conclusion? UFOlogists suffer this fallacy in their continued focus on a handful of unexplained atmospheric anomalies and visual misperceptions by eyewit-nesses, while conveniently ignoring the fact that the vast major-ity (90 to 95 percent) of UFO sightings are fully explicable with prosaic answers. 8. Has the claimant provided a different explanation for the observed phenomena, or is it strictly a process of denying the existing explanation? This is a classic debate strategy—criticize your opponent and never afﬁrm what you believe in order to avoid criticism. But this stratagem is unacceptable in science. Big Bang skeptics, for example, ignore the convergence of evidence of this cosmological model, focus on the few ﬂaws in the accepted model, and have yet to offer a viable cosmological alternative that carries a preponderance of evidence in favor of it. 9. If the claimant has proffered a new explanation, does it account for as many phenomena as the old explanation? The HIV-AIDS skeptics argue that lifestyle, not HIV, causes AIDS. Yet, to make this argument they must ignore the convergence of evidence in support of HIV as the causal vector in AIDS, and simultaneously ignore such blatant evidence as the signiﬁcant correlation between the rise in AIDS among hemophiliacs shortly after HIV was inadvertently introduced into the blood supply. On top of this, their alternative theory does not explain nearly as much of the data as the HIV theory. 10. Do the claimants’ personal beliefs and biases drive the con-clusions, or vice versa? All scientists hold social, political, and Secrets of Mental Math 230 ideological beliefs that could potentially slant their interpreta-tions of the data, but how do those biases and beliefs affect their research? At some point, usually during the peer-review system, such biases and beliefs are rooted out, or the paper or book is rejected for publication. This is why one should not work in an intellectual vacuum. If you don’t catch the biases in your research, someone else will. There is no deﬁnitive set of criteria we can apply in determin-ing how open-minded we should be when encountering new claims and ideas, but with mathematical calculations on the odds of weird things happening and with an analysis of the sorts of questions we should ask when we encounter weird things, we have made a start toward coming to grips with our weird and wonderful world. Epilogue: How Math Helps Us Think About Weird Things 231 Answers CHAPTER 1: A LITTLE GIVE AND TAKE T wo-Digit Addition (page 15) 1. 23 16 23 10 6 33 6 39 2. 64 43 64 40 3 104 3 107 3. 95 32 95 30 2 125 2 127 4. 34 26 34 20 6 54 6 60 5. 89 78 89 70 8 159 8 167 6. 73 58 73 50 8 123 8 131 7. 47 36 47 30 6 77 6 83 8. 19 17 19 10 7 29 7 36 9. 55 49 55 40 9 95 9 104 10. 39 38 39 30 8 69 8 77 Three-Digit Addition (page 20) 1. 242 137 242 100 30 7 342 30 7 372 7 379 2. 312 256 312 200 50 6 512 50 6 562 6 568 3. 635 814 635 800 10 4 1435 10 4 1445 4 1449 4. 457 241 457 200 40 1 657 40 1 697 1 698 5. 912 475 912 400 70 5 1312 70 5 1382 5 1387 6. 852 378 852 300 70 8 1152 70 8 1222 8 1230 7. 457 269 457 200 60 9 657 60 9 717 9 726 8. 878 797 878 700 90 7 1578 90 7 1668 7 1675 or 878 797 878 800 3 1678 3 1675 9. 276 689 276 600 80 9 876 80 9 956 9 965 10. 877 539 877 500 30 9 1377 30 9 1407 9 1416 11. 5400 252 5400 200 52 5600 52 5652 12. 1800 855 1800 800 55 2600 55 2655 13. 6120 136 6120 100 30 6 6220 30 6 6250 6 6256 14. 7830 348 7830 300 40 8 8130 40 8 8170 8 8178 15. 4240 371 4240 300 70 1 4540 70 1 4610 1 4611 T wo-Digit Subtraction (page 23) 1. 38 23 38 20 3 18 3 15 2. 84 59 84 60 1 24 1 25 3. 92 34 92 40 6 52 6 58 4. 67 48 67 50 2 17 2 19 5. 79 29 79 20 9 59 9 50 or 79 29 79 30 1 49 1 50 6. 63 46 63 50 4 13 4 17 7. 51 27 51 30 3 21 3 24 8. 89 48 89 40 8 49 8 41 9. 125 79 125 80 1 45 1 46 10. 148 86 148 90 4 58 4 62 Answers 234 Three-Digit Subtraction (page 28) 1. 583 271 583 200 70 1 383 70 1 313 1 312 2. 936 725 936 700 20 5 236 20 5 216 5 211 3. 587 298 587 300 2 287 2 289 4. 763 486 763 500 14 263 14 277 5. 204 185 204 200 15 04 15 19 6. 793 402 793 400 2 393 2 391 7. 219 176 219 200 24 19 24 43 8. 978 784 978 800 16 178 16 194 9. 455 319 455 400 81 55 81 136 10. 772 596 772 600 4 172 4 176 11. 873 357 873 400 43 473 43 516 12. 564 228 564 300 72 264 72 336 13. 1428 571 1428 600 29 828 29 857 14. 2345 678 2345 700 22 1645 22 1667 15. 1776 987 1776 1000 13 776 13 789 CHAPTER 2: PRODUCTS OF A MISSPENT YOUTH 2-by-1 Multiplication (page 35) 1. 2. 3. 4. 5. 6. or 7. 8. 9. 10. 58 6 300 48 348 84 5 400 20 420 53 5 250 15 265 28 4 80 32 112 49 9 450 9 441 49 9 360 81 441 93 8 720 24 744 71 3 210 3 213 67 5 300 35 335 43 7 280 21 301 82 9 720 18 738 Answers 235 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 3-by-1 Multiplication (page 43) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 214 8 1600 80 1680 32 1712 184 7 700 560 1260 28 1288 587 4 2000 320 2320 28 2348 807 9 7200 63 7263 529 9 4500 180 4680 81 4761 328 6 1800 120 1920 48 1968 728 2 1400 40 1440 16 1456 927 7 6300 140 6440 49 6489 957 6 5400 300 5700 42 5742 862 4 3200 240 3440 8 3448 637 5 3000 150 3150 35 3185 431 6 2400 180 2580 6 2586 64 8 480 32 512 29 3 60 27 87 76 8 560 48 608 46 2 80 12 92 37 6 180 42 222 57 7 350 49 399 75 4 280 20 300 96 9 810 54 864 78 2 140 6 156 97 4 360 28 388 Answers 236 With this kind of problem you can easily say the answer out loud as you go. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 862 5 4000 300 4300 10 4310 429 3 1200 60 1260 27 1287 670 4 2400 280 2680 499 9 4500 9 4491 or 500 9 1 9 499 9 3600 810 4410 81 4491 968 6 5400 360 5760 48 5808 188 6 600 480 1080 48 1128 247 5 1000 200 1200 35 1235 578 9 4500 630 5130 72 5202 611 3 1800 33 1833 134 8 800 240 1040 32 1072 339 8 2400 240 2640 72 2712 457 7 2800 350 3150 49 3199 751 9 6300 450 6750 9 6759 297 8 2400 24 2376 or 300 8 3 8 297 8 1600 720 2320 56 2376 259 7 1400 350 1750 63 1813 757 8 5600 400 6000 56 6056 Answers 237 With this kind of problem you can easily say the answer out loud as you go. 29. 30. 31. 32. 33. 34. 35. 36. T wo-Digit Squares (page 48) 1. 142 180 42 196 2. 272 720 32 729 3. 652 4200 52 4225 4. 892 7920 12 7921 90 88 1 1 70 60 5 5 30 24 3 3 18 10 4 4 691 3 1800 270 2070 3 2073 312 9 2700 90 2790 18 2808 767 3 2100 180 2280 21 2301 457 9 3600 450 4050 63 4113 722 9 6300 180 6480 18 6498 693 6 3600 540 4140 18 4158 488 9 3600 720 4320 72 4392 285 6 1200 480 1680 30 1710 Answers 238 5. 982 9600 22 9604 6. 312 960 12 961 7. 412 1680 12 1681 8. 592 3480 12 3481 9. 262 660 42 676 10. 532 2800 32 2809 11. 212 440 12 441 12. 642 4080 42 4096 13. 422 1760 22 1764 44 40 2 2 68 60 4 4 22 20 1 1 56 50 3 3 30 22 4 4 60 58 1 1 42 40 1 1 32 30 1 1 100 96 2 2 Answers 239 14. 552 3000 52 3025 15. 752 5600 52 5625 16. 452 2000 52 2025 17. 842 7040 42 7056 18. 672 4480 32 5625 19. 1032 10,600 32 10,609 20. 2082 43,200 82 43,264 CHAPTER 3: NEW AND IMPROVED PRODUCTS Multiplying by 11 (page 58) 1. 385 2. 528 3. 1034 9 4 13 94 11 4 8 12 48 11 3 5 8 35 11 216 200 8 8 106 100 3 3 70 64 3 3 88 80 4 4 50 40 5 5 80 70 5 5 60 50 5 5 Answers 240 2-by-2 Addition-Method Multiplication Problems (page 59) 1. or 2. 3. 4. 5. 6. or 7. 8. 9. 10. 374 or 11. 935 or 850 85 935 8 5 13 85 11 340 34 374 3 4 7 34 11 92 (90 2) 35 90 35 1230 2 35 70 3220 88 (80 8) 76 80 76 6080 8 76 608 6688 62 (60 2) 94 60 94 5640 2 94 188 5828 23 (80 4) 84 80 23 1840 4 23 92 1932 23 (20 3) 84 20 84 1680 3 84 252 1932 77 (40 3) 43 40 77 3080 1 41 231 3311 53 (50 3) 58 50 58 2900 3 58 174 3074 59 (50 9) 26 50 26 1300 9 26 234 1534 27 (20 7) 18 20 18 360 7 18 126 486 31 (40 1) 41 40 31 1240 1 31 31 1271 31 (30 1) 41 30 41 1230 1 41 41 1271 Answers 241 2-by-2 Subtraction-Method Multiplication Problems (page 63) 1. 2. 3. 4. 5. or 6. 7. 8. 9. 10. 11. 88 (50 1) 49 50 88 4400 1 88 88 4312 57 (40 1) 39 40 57 2280 1 57 57 2223 85 (40 2) 38 40 85 3400 2 85 170 3230 87 (90 3) 22 90 22 1980 3 22 66 1914 37 (20 1) 19 20 37 740 1 37 37 703 79 (80 1) 54 80 54 4320 1 54 54 4266 96 (30 1) 29 30 96 2880 1 96 96 2784 96 (100 4) 29 100 29 2900 4 29 116 2784 68 (70 2) 43 70 38 2660 2 38 76 2584 47 (60 1) 59 60 47 2820 1 47 47 2773 98 (100 2) 43 100 43 4300 2 43 86 4214 29 (30 1) 45 30 45 1350 1 45 45 1305 Answers 242 2-by-2 Factoring-Method Multiplication Problems (page 68) 1. 27 14 27 7 2 189 2 378 or 14 27 14 9 3 126 3 378 2. 86 28 86 7 4 602 4 2408 3. 57 14 57 7 2 399 2 798 4. 81 48 81 8 6 648 6 3888 or 48 81 48 9 9 432 9 3888 5. 56 29 29 7 8 203 8 1624 6. 83 18 83 6 3 498 3 1494 7. 72 17 17 9 8 153 8 1224 8. 85 42 85 6 7 510 7 3570 9. 33 16 33 8 2 264 2 528 or 16 33 16 11 3 176 3 528 10. 62 77 62 11 7 682 7 4774 11. 45 36 45 6 6 270 6 1620 or 45 36 45 9 4 405 4 1620 or 36 45 36 9 5 324 5 1620 or 36 45 36 5 9 180 9 1620 12. 37 48 37 8 6 296 6 1776 2-by-2 General Multiplication—Anything Goes! (page 70) 1. or 2. or 57 81 57 9 9 513 9 4617 3. 73 18 73 9 2 657 2 1314 or 73 18 73 6 3 438 3 1314 73 18 (9 2) 81 (80 1) 57 80 57 4560 1 57 57 4617 53 (50 3) 39 50 39 1950 3 39 117 2067 53 (40 1) 39 40 53 2120 1 53 53 2067 Answers 243 4. or 89 55 89 11 5 979 5 4895 5. 77 36 77 4 9 308 9 2772 or 77 36 77 9 4 693 4 2772 6. 7. 872 7560 32 7569 8. 9. 37 56 37 8 7 296 7 2072 or 37 56 37 7 8 259 8 2072 10. or or 59 21 59 7 3 413 3 1239 11. (9 8) 12. 37 9 8 333 8 2664 57 73 (70 3) 70 57 3990 3 57 171 4161 37 72 59 (60 1) 21 60 21 1260 1 21 21 1239 59 (20 1) 21 20 59 1180 1 59 59 1239 56 37 (8 7) 67 (60 2) 58 60 67 4020 2 67 154 3886 90 84 3 3 92 (50 3) 53 50 92 4600 3 92 276 4876 77 36 (4 9) 89 (90 1) 55 90 55 4950 1 55 55 4895 Answers 244 13. (9 7) 14. 38 63 38 9 7 342 7 2394 15. (5 5 3) 16. 43 75 43 5 5 3 215 5 3 1075 3 3225 17. 18. 41 36 41 6 6 246 6 1476 19. 20. 532 2800 32 2809 54 53 53 9 6 477 6 2862 21. 22. 23. 24. 29 (30 1) 26 30 26 780 1 26 26 754 52 (50 2) 47 50 47 2350 2 47 94 2444 91 (90 1) 46 90 46 4140 1 46 46 4186 83 (80 3) 58 80 58 4640 3 58 174 4814 56 50 3 3 54 (9 6) 53 36 (6 6) 41 61 (60 1) 37 60 37 2220 1 37 37 2257 74 62 (60 2) 60 74 4440 2 74 148 4588 43 75 43 (40 3) 76 40 76 3040 3 76 228 3268 38 63 Answers 245 25. 26. 41 15 41 5 3 205 3 615 27. (9 3) 28. 34 27 34 9 3 306 3 918 29. (9 9) 30. 95 81 95 9 9 855 9 7695 31. 32. 33. Three-Digit Squares (page 76) 1. 4092 167,200 92 167,281 2. 8052 648,000 52 648,025 810 800 5 5 418 400 9 9 41 (40 1) 93 40 93 3720 1 93 93 3813 95 26 (20 6) 20 95 1900 6 95 570 2470 65 69 (70 1) 70 65 4550 1 65 65 4485 65 (60 5) 47 60 47 2820 5 47 235 3055 95 81 69 (70 1) 78 70 78 5460 1 78 78 5382 34 27 65 19 (20 1) 20 65 1300 1 65 65 1235 41 15 (5 3) Answers 246 3. 2172 46,800 172 47,089 172 280 32 289 4. 8962 802,800 42 802,816 5. 3452 117,000 452 119,025 452 2,000 52 2025 6. 3462 117,600 462 119,716 462 2,100 42 2116 7. 2762 75,600 242 76,176 242 560 42 576 8. 6822 464,800 182 465,124 182 320 22 324 20 16 2 2 700 664 18 18 28 20 4 4 300 252 24 24 50 42 4 4 392 300 46 46 50 40 5 5 390 300 45 45 900 892 4 4 20 14 3 3 234 200 17 17 Answers 247 Answers 248 9. 4312 184,800 312 185,761 312 960 12 961 10. 7812 609,600 192 609,961 192 360 12 361 11. 9752 950,000 252 950,625 252 600 52 625 T wo-Digit Cubes (page 79) 1. 123 (10 12 14) (22 12) 1680 48 1728 2. 173 (14 17 20) (32 17) 4760 153 4913 3. 213 (20 21 22) (12 21) 9240 21 9261 4. 283 (26 28 30) (22 28) 21,840 112 21,952 5. 333 (30 33 36) (32 33) 35,640 297 35,937 6. 393 (38 39 40) (12 39) 59,280 39 59,319 7. 403 40 40 40 64,000 8. 443 (40 44 48) (42 44) 84,480 704 85,184 9. 523 (50 52 54) (22 52) 140,400 208 140,608 10. 563 (52 56 60) (42 56) 174,720 896 175,616 11. 653 (60 65 70) (52 65) 273,000 1,625 274,625 12. 713 (70 71 72) (12 71) 357,840 71 357,911 13. 783 (76 78 80) (22 78) 474,240 312 474,552 14. 853 (80 85 90) (52 85) 612,000 2,125 614,125 15. 873 (84 87 90) (32 87) 657,720 783 658,503 16. 993 (98 99 100) (12 99) 970,200 99 970,299 30 20 5 5 1,000 950 25 25 20 18 1 1 800 762 19 19 32 30 1 1 462 400 31 31 CHAPTER 4: DIVIDE AND CONQUER One-Digit Division (page 84) 1. 2. 3. 4. 5. 6. T wo-Digit Division (page 93) 1. 2. 3. 4. 5. 6. 170 1 1 4 8 183 0 7 4 18 127 126 14 655 1 9 1 117 2 1 4 66 61 55 64 55 9 152 1 2 2 8 284 2 6 8 28 146 140 68 56 12 4 7 5 9 793 2 1 316 5 24 1 2 5 4 245 9 1 48 111 96 15 43 1 7 7 177 3 8 68 58 51 7 695 2 4 42 7 8 2 24 38 36 22 20 2 442 2 3 31 3 2 8 12 12 08 6 2 36 1 6 82 8 9 24 49 48 1 61 1 7 74 2 8 42 08 7 1 145 1 5 57 2 6 5 22 20 26 25 1 35 3 9 93 1 8 27 48 45 3 Answers 249 Decimalization (page 98) 1. 2 5 .40 2. 4 7 .571428 3. 3 8 .375 4. 1 9 2 .75 5. 1 5 2 .4166 6. 1 6 1 .5454 7. 1 2 4 4 .5833 8. 1 2 3 7 .481 9. 1 4 8 8 .375 10. 1 1 0 4 .714285 11. 3 6 2 .1875 12. 1 4 9 5 .422 T esting for Divisibility (page 101) Divisibility by 2 1. 2. 3. 4. Divisibility by 4 5. 6. 7. 8. Divisibility by 8 9. 10. 11. 12. Divisibility by 3 13. 14. 15. 16. Divisibility by 6 17. 18. 19. 20. Divisibility by 9 21. 22. 8469 Yes: 8 4 6 9 27 1234 No: 1 2 3 4 10 5991 No: odd 248 No: 2 4 8 14 67,386 Yes: 6 7 3 8 6 30 5334 Yes: 5 3 3 4 15 3,267,486 Yes: 3 2 6 7 4 8 6 36 7359 Yes: 7 3 5 9 24 94,737 Yes: 9 4 7 3 7 30 83,671 No: 8 3 6 7 1 25 6111 No 248 Yes 73,488 Yes 59,366 No 57,929 No 358 No 67,348 Yes 3932 Yes 9846 Yes 7241 No 293 No 53,428 Yes Answers 250 23. 24. Divisibility by 5 25. 26. 27. 28. Divisibility by 11 29. 30. 31. 32. Divisibility by 7 33. 34. 35. 36. Divisibility by 17 37. 6 38. 39. 40. 13,855 Yes: 13,855 85 13,940 1394 34 1360 136 34 170 17 8273 No: 8273 17 8290 829 51 880 88 629 Yes: 629 51 680 68 94 No: 694 34 660 66 1183 Yes: 1183 7 1190 119 21 140 14 875 Yes: 875 35 840 84 14 70 7 7336 Yes: 7336 14 7350 735 35 700 7 5784 No: 5784 7 5777 577 7 570 57 941,369 Yes: 9 4 1 3 6 9 0 3828 Yes: 3 8 2 8 11 4969 No: 4 9 6 9 8 53,867 Yes: 5 3 8 6 7 11 37,210 Yes 56,785 Yes 43,762 No 47,830 Yes 314,159,265 Yes: 3 1 4 1 5 9 2 6 5 36 4,425,575 No: 4 4 2 5 5 7 5 32 Answers 251 Multiplying Fractions (page 102) 1. 2. 3. 4. Dividing Fractions (page 103) 1. 2. 3. Simplifying Fractions (page 104) 1. 1 4 2 2. 5 6 1 1 0 2 3. 3 4 1 9 2 4. 5 2 3 1 0 2 5. 1 8 0 4 5 6. 1 6 5 2 5 7. 2 3 4 6 2 3 8. 2 3 0 6 5 9 Adding Fractions (Equal Denominators) (page 105) 1. 2 9 5 9 7 9 2. 1 5 2 1 4 2 1 9 2 3 4 3. 1 5 8 1 6 8 1 1 1 8 4. 1 3 0 1 3 0 1 6 0 3 5 Adding Fractions (Unequal Denominators) (page 106) 1. 1 5 1 1 0 1 2 0 1 1 0 1 3 0 2. 1 6 1 5 8 1 3 8 1 5 8 1 8 8 4 9 3. 1 3 1 5 1 5 5 1 3 5 1 8 5 4. 2 7 2 5 1 2 6 1 2 5 1 1 2 1 1 5. 2 3 3 4 1 8 2 1 9 2 1 1 7 2 6. 3 7 3 5 1 3 5 5 2 3 1 5 3 3 6 5 7. 1 2 1 5 9 1 9 8 9 5 9 5 9 7 9 3 9 Subtracting Fractions (page 107) 1. 1 8 1 1 3 1 1 5 1 2. 1 7 2 8 7 4 7 3. 1 1 3 8 1 5 8 1 8 8 4 9 4. 4 5 1 1 5 1 1 2 5 1 1 5 1 1 1 5 5. 1 9 0 3 5 1 9 0 1 6 0 1 3 0 6. 3 4 2 3 1 9 2 1 8 2 1 1 2 1 3 2 3 10 15 5 18 4 5 63 80 9 14 18 28 44 63 6 35 Answers 252 7. 7 8 1 1 6 1 1 4 6 1 1 6 1 1 3 6 8. 4 7 2 5 2 3 0 5 1 3 4 5 3 6 5 9. 8 9 1 2 1 1 6 8 1 9 8 1 7 8 CHAPTER 5: GOOD ENOUGH Addition Guesstimation (page 128) Exact 1. 2. 3. 4. Guesstimates 1. or 2. or 3. or 4. or or Exact Guesstimates $ 2.50 2.00 7.50 9.00 0.50 11.00 0.00 6.00 8.50 $47.00 $ 2.67 1.95 7.35 9.21 0.49 11.21 0.12 6.14 8.31 $47.35 8.97 million 4.02 million 12.99 million 8.9 million 4.0 million 12.9 million 9 million 4 million 13 million 312,000 79,000 391,000 310,000 80,000 390,000 57,300 37,400 94,700 57,000 37,000 94,000 1480 1100 2580 1500 1100 2600 8,971,011 4,016,367 12,987,378 312,025 79,419 391,444 57,293 37,421 94,714 1479 1105 2584 Answers 253 Subtraction Guesstimation (page 129) Exact 1. 2. 3. 4. Guesstimates 1. 2. or 3. or 4. or Division Guesstimation (page 129) Exact 1. 2. 3. 4. 5. Guesstimates 1. 2. 3. 4. 5. Multiplication Guesstimation (page 129) Exact 1. 2. 3. 4. 5. 6. 7. 8. 51,276 489 25,073,964 428 313 133,964 639 107 68,373 312 98 30,576 539 17 9163 88 88 7744 76 42 3192 98 27 2646 40 ≈200,0008 ,0 0 0 ,0 0 0 2008 0 0 0 17,000 ≈3005 ,1 0 0 ,0 0 0 35 1 ,0 0 0 42,000 135 5 0 ,0 0 0 4,800 52 4 ,0 0 0 630 74 4 0 0 40.90 203,6378 ,3 2 9 ,4 8 3 17,655.21 2895 ,1 0 2 ,3 5 7 42,247.15 135 4 9 ,2 1 3 4,791.6 52 3 ,9 5 8 625.57 74 3 7 9 8.35 million 6.10 million 2.25 million 8.3 million 6.1 million 2.2 million 527,000 42,000 485,000 530,000 40,000 490,000 67,200 9,900 57,300 67,000 10,000 57,000 4900 1700 3200 8,349,241 6,103,839 2,245,402 526,978 42,009 484,969 67,221 9,874 57,347 4926 1659 3267 Answers 254 9. 10. Guesstimates 1. 2. 3. 4. 5. 6. or 7. 8. 9. 10. Square Root Guesstimation (page 130) Exact (to two decimal places) 1. 2. 3. 4. 5. Divide and Average 1. 4.1 2. 5.9 3. 13.15 4. 65.5 5. 89.5 90 89 2 89 908 0 3 9 60 71 2 71 604 2 7 9 10 16.3 2 16.3 101 6 3 6 5.8 2 5.8 63 5 4 4.2 2 4.2 41 7 89.66 8039 65.41 4279 12.76 163 5.91 35 4.12 17 5,500,000 200,00 1100 billion 1.1 trillion 105,000 11,000 1155 million 1.155 billion 51,000 490 24,990,000 430 310 133,300 640 110 70,400 646 100 64,600 310 100 31,000 540 17 9180 90 86 7740 78 40 3120 100 25 2500 5,462,741 203,413 1,111,192,535,033 104,972 11,201 1,175,791,372 Answers 255 Everyday Math (page 130) 1. $8.80 $4.40 $13.20 2. $5.30 $2.65 $7.95 3. $74 2 2 $37 2 $18.50 4. Since 70 10 7, seven years 5. Since 70 6 11.67, it will take twelve years to double 6. Since 110 7 15.714, it will take sixteen years to triple 7. Since 70 7 10, it will take ten years to double, then another ten years to double again.Thus it will take twenty years to quadruple. 8. M $1267 9. M $693 CHAPTER 6: MATH FOR THE BOARD Columns of Numbers (page 149) 1. 2. Subtracting on Paper (page 150) 1. 2. 3. 4. 45,394,358 5 36,472,659 6 8,921,699 8 3,249,202 4 2,903,445 9 345,757 4 876,452 5 593,876 2 282,576 3 75,423 3 46,298 2 29,125 1 $ 21.56 5 19.38 3 211.02 6 9.16 7 26.17 7 1.43 3 $288.72 9 672 6 1,367 8 107 8 7,845 6 358 7 210 3 916 7 11,475 9 $125(1.22) 0.22 $30,000(0.004167)(1.004167)42 (1.004167)42 1 $750(2.451) 1.451 $100,000(0.0075)(1.0075)120 (1.00333)120 1 Answers 256 Square Root Guesstimation (page 150) 1. 2. 3. 4. Pencil-and-Paper Multiplication (page 150) 1. 2. 3. 4. 5. 6. 3,923,759 3 2,674,093 4 10,492,496,475,587 3 52,819 7 47,820 3 2,525,804,580 3 3,309 6 2,868 6 9,490,212 9 725 5 609 6 441,525 3 273 3 217 1 59,241 3 54 9 37 1 1998 9 1 9 exactly 361 12 1 261 29 9 261 0 2 0. 9 5 439.20 00 22 4 039 40 0 0 3920 409 9 3681 23900 4480 0 20925 2 2. 4 0 502.00 00 22 4 102 42 2 84 1800 444 4 1776 2400 4480 0 0 3. 8 7 15.000 0 32 9 600 68 8 544 5600 767 7 5369 Answers 257 CHAPTER 8: THE TOUGH STUFF MADE EASY Four-Digit Squares (page 169) 1. 1,2342 2342 2. 8,6392 3612 3. 5,3122 3122 4. 9,8632 1372 5. 3,6182 3822 6. 2,9712 (No mnemonic needed) 8,826,000 841 (292) 8,826,841 3,000 2,942 29 29 145,600 324 (182) 145,924 400 364 18 18 “Prayer” 12,944,000 145,924 (3822) 13,089,924 4,000 3,236 382 382 17,400 1,369 (372) 18,769 174 100 37 37 “Nachos” 97,260,000 18,769 (1372) 97,278,769 10,000 9,726 137 137 97,200 144 (122) 97,344 324 300 12 12 “Tons” 28,120,000 97,344 (3122) 28,217,344 5,624 5,000 312 312 128,800 1,521 (392) 130,321 400 322 39 39 “Lesson” 74,502,000 130,321 (3612) 74,632,321 9,000 8,278 361 361 53,600 1,156 (342) 54,756 268 200 34 34 “Reach off” 1,468,000 54,756 (2342) 1,522,756 1,468 1,000 234 234 Answers 258 3-by-2 Exercises Using Factoring, Addition, and Subtraction Methods (page 175) 1. 2. 3. or 4. 5. 6. 7. 8. 9. 484 75 (5 5 3) 484 75 484 5 5 3 2,420 5 3 12,100 3 36,300 967 (50 1) 51 50 967 48,350 1 967 967 49,317 411 (410 1) 93 410 93 38,130 1 93 93 38,223 952 (950 2) 26 950 26 24,700 2 26 52 24,752 906 (900 6) 46 900 46 41,400 6 46 276 41,676 773 42 (7 6) 773 42 773 7 6 5,411 6 32,466 148 (74 2) 62 (60 2) 62 148 62 74 2 4,588 2 9,176 148 (60 2) 62 148 60 8,880 148 2 296 9,176 796 (800 4) 19 800 19 15,200 4 19 76 15,124 858 15 (5 3) 858 15 858 5 3 4,290 3 12,870 Answers 259 10. 11. 12. 13. 14. 15. 16. or 17. 18. 19. 20. 144 56 (7 8) 144 56 144 7 8 1008 8 8064 499 (500 1) 25 500 25 12,500 1 25 25 12,475 668 63 (9 7) 668 63 668 9 7 6,012 7 42,084 817 (60 1) 61 60 817 49,020 1 817 817 49,837 538 (530 8) 53 530 53 28,090 8 53 424 28,514 538 (540 2) 53 540 53 28,620 2 53 106 28,514 218 (70 2) 68 70 218 15,260 2 218 436 14,824 361 (360 1) 41 360 41 14,760 1 41 41 14,801 841 72 (9 8) 841 72 841 9 8 7,569 8 60,552 616 (610 6) 37 610 37 22,570 6 37 222 22,792 157 33 (11 3) 157 33 157 11 3 1727 3 5181 126 87 (9 7 2) 126 87 87 9 7 2 783 7 2 5,481 2 10,962 Answers 260 21. or 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 545 27 (9 3) 545 27 545 9 3 4,905 3 14,715 834 (800 34) 34 800 34 27,200 34 34 1,156 28,356 154 19 (11 14) 154 19 19 11 14 209 7 2 1463 2 2926 423 65 (47 9) 423 65 65 47 9 3,055 9 27,495 878 24 (8 3) 878 24 878 8 3 7,024 3 21,072 853 32 (8 4) 853 32 853 8 4 6,824 4 27,296 286 64 (8 8) 286 64 286 8 8 2,288 8 18,304 589 (600 11) 87 600 87 52,200 11 87 957 51,243 383 49 (7 7) 383 49 383 7 7 2,681 7 18,767 988 (1000 12) 22 1000 22 22,000 12 22 264 21,736 281 (280 1) 44 280 44 12,320 1 44 44 12,364 281 44 (11 4) 281 44 281 11 4 3,091 4 12,364 Answers 261 32. 33. 34. Five-Digit Squares (page 180) 1. 45,7952 7952 2. 21,2312 “Cousin” 2312 52,400 961 (312) 53,361 262 200 31 31 9,702,000 21,0002 441,000,000 450,702,000 2312 53,361 450,755,361 231 (7 3) 21 231 7 3 1,617 3 4,851 632,000 25 (52) 632,025 800 790 5 5 71,550,000 45,0002 2,025,000,000 2,096,550,000 7952 632,025 2,097,182,025 795 (800 5) 45 800 45 36,000 5 45 225 “Lilies” 35,775 2,000 71,550,000 822 (100 5) 95 100 822 82,200 5 822 4,110 78,090 216 78 (6 6 6) 216 78 78 6 6 6 468 6 6 2,808 6 16,848 653 (650 3) 69 650 69 44,850 3 69 207 45,057 Answers 262 4,851 2,000 3. 58,3242 “Liver” 3242 4. 62,4572 4572 207,000 (500 414) 1,849 (432) 208,849 500 414 43 43 56,668,000 62,0002 3,844,000,000 3,900,668,000 4572 208,849 3,900,876,849 457 (60 2) 62 60 457 27,420 2 457 914 “Judge off” 28,334 2,000 56,668,000 104,400 (348 300) 576 (242) 104,976 348 300 24 24 18,792 2,000 37,584,000 58,0002 3,364,000,000 3,401,584,000 3242 104,976 3,401,688,976 324 (9 6 6) 58 324 58 58 9 6 6 522 6 6 3,132 6 18,792 Answers 263 5. 89,8542 8542 6. 76,9342 9342 871,200 (968 900) 1,156 (342) 872,356 968 900 34 34 141,968,000 76,0002 5,776,000,000 5,917,968,000 9342 872,356 5,918,840,356 934 (930 4) 76 930 76 70,680 4 76 304 “Pie Chief” 76,984 2,000 141,968,000 727,200 (900 808) 2,116 (462) 729,316 900 808 46 46 152,012,000 89,0002 7,921,012,000 8,073,012,000 8542 729,316 8,073,741,316 854 (90 1) 89 90 854 76,860 1 854 854 “Stone” 76,006 2,000 152,012,000 Answers 264 3-by-3 Multiplication (page 192) 1. or 2. 3. 4. 5. 6. 942 (42) 879 (21) 900 921 828,900 21 42 882 828,018 809 527 (800 9) 800 527 421,600 9 527 4,743 426,343 343 226 (7 7 7) 343 226 226 7 7 7 1,582 7 7 11,074 7 77,518 853 325 (320 5) 320 853 272,960 5 800 4,000 276,960 5 53 265 277,225 596 167 (600 4) 600 167 100,200 4 167 668 99,532 644 286 (7 92) 684 286 286 7 92 2,002 92 184,184 644 286 (640 4) 640 286 183,040 (8 8 10) 4 200 800 183,840 4 86 344 184,184 Answers 265 7. 8. 9. 10. 11. 12. 642 249 107 83 18 8,881 9 2 79,929 2 159,858 642 (107 6) 249 (83 3) 824 206 (4122) 400 424 169,600 12 12 144 169,744 273 (91 3) 138 (46 3) 273 138 91 46 9 4,186 9 37,674 658 (47 7 2) 468 (52 9) 658 468 52 47 9 7 2 2,444 9 7 2 21,996 7 2 153,972 2 307,944 446 176 (11 8 2) 446 176 446 11 8 2 4,906 8 2 39,248 2 78,496 692 (8) 644 (56) 700 636 445,200 (8) (56) 448 445,648 Answers 266 13. 783 589 589 87 9 51,243 9 461,187 14. 15. 16. 17. 557 756 557 9 84 5,013 7 6 2 35,091 6 2 210,546 2 421,092 18. 976 (1000 24) 878 878 1,000 878,000 878 24 21,072 856,928 557 756 (9 84) 417 298 (300 2) 300 417 125,100 2 417 834 124,266 341 715 7 341 2,387 3 15 45 2,432 100 243,200 41 15 615 243,815 871 (29) 926 (26) 900 897 807,300 29 26 754 806,546 783 (87 9) 589 Answers 267 19. 765 350 765 7 5 10 5,355 5 10 26,775 10 267,750 20. 154 423 47 11 9 14 517 9 7 2 4,653 2 7 9,306 7 65,142 21. 22. 216 653 653 6 6 6 3,918 6 6 23,508 6 141,048 23. 393 (400 7) 822 400 822 328,800 7 822 5,754 323,046 216 (6 6 6) 653 545 (109 5) 834 100 834 83,400 9 834 7,506 90,906 5 454,530 154 (11 14) 423 (47 9) 765 350 (7 5 10) Answers 268 5-by-5 Multiplication (page 198) 1. “Neck ripple” “Mouse round” 2. “Knife mulch” “Room scout” 3. “Roll silk” “Shoot busily” 61,905 1,000 61,905,000 69 78 1 million 5,382,000,000 5,443,905,000 216 653 141,048 5,444,046,048 653 69 45,057 216 78 16,848 69,216 78,653 43,071 1,000 43,071,000 34 27 1 million 918,000,000 961,071,000 834 545 454,530 961,525,530 834 34 28,356 545 27 14,715 34,545 27,834 30,421 1,000 30,421,000 65 19 1 million 1,235,000,000 1,265,421,000 154 423 65,142 1,265,486,142 423 65 27,495 154 19 2,926 65,154 19,423 Answers 269 4. “Cave soups” “Toss-up Panama” A Day for Any Date (page 221) 1. January 19, 2007, is Friday: 6 19 1 26; 26 21 5 2. February 14, 2012, is Tuesday: 1 14 1 16; 16 14 2 3. June 20, 1993, is Sunday: 3 5 20 28; 28 28 0 4. September 1, 1983, is Thursday: 4 1 6 11; 11 7 4 5. September 8, 1954, is Wednesday: 4 8 5 17; 17 14 3 6. November 19, 1863, is Thursday: 2 19 4 25; 25 21 4 7. July 4, 1776, is Thursday: 5 4 2 11; 11 7 4 8. February 22, 2222, is Friday: 2 22 2 26; 26 21 5 9. June 31, 2468, doesn’t exist (only 30 days in June)! But June 30, 2468, is Saturday, so the next day would be Sunday. 10. January 1, 2358, is Wednesday: 6 1 3 10; 10 7 3 109,923 1,000 109,923,000 95 81 1 million 7,804,000,000 7,804,923,000 393 822 323,046 7,805,246,046 822 95 78,090 393 81 31,833 95,393 81,822 Answers 270 Bibliography RAPID CALCULATION Cutler, Ann, and Rudolph McShane. The Trachtenberg Speed System of Basic Mathematics. New York: Doubleday, 1960. Devi, Shakuntala. Figuring: The Joys of Numbers. New York: Basic Books, 1964. Doerﬂer, Ronald W. Dead Reckoning: Calculating Without Instruments. Houston: Gulf Publishing Company, 1993. Flansburg, Scott, and Victoria Hay. Math Magic. New York: William Mor-row and Co., 1993. Handley, Bill. Speed Mathematics: Secrets of Lightning Mental Calcula-tion. Queensland, Australia: Wrightbooks, 2003. Julius, Edward H. Rapid Math Tricks and Tips: 30 Days to Number Power. New York: John Wiley & Sons, 1992. Lucas, Jerry. Becoming a Mental Math Wizard. Crozet, Virginia: Shoe Tree Press, 1991. Menninger, K. Calculator’s Cunning. New York: Basic Books, 1964. Smith, Steven B. The Great Mental Calculators: The Psychology, Methods, and Lives of Calculating Prodigies, Past and Present. New York: Columbia University Press, 1983. Sticker, Henry. How to Calculate Quickly. New York: Dover, 1955. Stoddard, Edward. Speed Mathematics Simpliﬁed. New York: Dover, 1994. Tirtha, Jagadguru Swami Bharati Krishna, Shankaracharya of Govard-hana Pitha. Vedic Mathematics or “Sixteen Simple Mathematical For-mulae from the Vedas.” Banaras, India: Hindu University Press, 1965. MEMORY Lorayne, Harry, and Jerry Lucas. The Memory Book. New York: Ballan-tine Books, 1974. Sanstrom, Robert. The Ultimate Memory Book. Los Angeles: Stepping Stone Books, 1990. RECREATIONAL MATHEMATICS Gardner, Martin. Magic and Mystery. New York: Random House, 1956. ———. Mathematical Carnival. Washington, D.C.: Mathematical Associa-tion of America, 1965. ———. Mathematical Magic Show. New York: Random House, 1977. ———. The Unexpected Hanging and Other Mathematical Diversions. New York: Simon & Schuster, 1969. Huff, Darrell. How to Lie with Statistics. New York: Norton, 1954. Paulos, John Allen. Innumeracy: Mathematical Illiteracy and Its Conse-quences. New York: Hill and Wang, 1988. Stewart, Ian. Game, Set, and Math: Enigmas and Conundrums. New York: Penguin Books, 1989. ADVANCED MATHEMATICS (BY ARTHUR BENJAMIN) Benjamin, Arthur T., and Jennifer J. Quinn. Proofs That Really Count: The Art of Combinatorial Proof. Washington: Mathematical Association of America, 2003. Benjamin, Arthur T., and Kan Yasuda. “Magic ‘Squares’ Indeed!,” The American Mathematical Monthly 106, no. 2 (February 1999): 152–56. Bibliography 272 Index addition: an “amazing” sum (Randi), 212–14 carrying a number in, 13–14 columns of numbers (on paper), 132–33; exercises, 149; answers, 256 fractions with equal denominators, 104–5; exercises, 105; answers, 252 fractions with unequal denominators, 105–6; exercises, 106; answers, 252 guesstimation, 108–11; at the supermarket, 111; exercises, 128; answers, 253 leapfrog, 203–6 left to right, 6–8, 11–21 in multiplying three-by-three numbers, 188–90; exercises, 192–93; answers, 265–68 in multiplying three-by-two numbers, 172–73; exercises, 175–76; answers, 259–62 in multiplying two-digit numbers, 54–59; exercises, 59; answers, 241 with overlapping digits, 19 three-digit numbers, 15–21; exercises, 19–21; answers, 233–34 three-digit to four-digit numbers, 18–19 two-digit numbers, 12–15; exercises, 15; answers, 233 Aitken, Alexander Craig, 153 “amazing” sum (Randi), 212–14 associative law of multiplication, 65 astrology, 224 attribution bias, 227 Baloney Detection Kit, 227–31 Baltimore, David, 228 Baltimore Affair, The (Kevles), 227–28 belief in God, 227 better-than-average bias, 226 Bidder, George Parker, 109 Big Bang theory, 230 blind spots, 226 casting out elevens, 145, 147–49 casting out nines, 133, 134 chance, random, 224, 225 checking solutions: by casting out elevens, 145, 147–49 by casting out nines, 133, 134 digital roots method, 133 mod sums method, 133–34, 135 Chevalier, Auguste, 121 close-together method: of cubing, 77, 78 in multiplying three-by-three numbers, 183–88; exercises, 192–93; answers, 265–68 Colburn, Zerah, 49, 53, 109 cold fusion debacle, 228–29 columns of numbers, adding (on paper), 132–33; exercises, 149; answers, 256 complements, 25–28 conﬁrmation bias, 224, 225, 229 conspiracy theories, 224 creationism, 229 criss-cross method of multiplication on paper, 138–45; exercises, 150; answers, 257 cube roots, quick, 209–11 cubing: close-together method, 77, 78 two-digit numbers, 76–79; exercises, 79; answers, 248 data ﬁlters, 225–26 day of the week, 8–10, 214–21 day codes, 214–15 exercises, 221; answers, 270 Julian vs. Gregorian calendar, 220 leap years, 220 month codes, 215 year codes, 216–18 death premonitions, 223, 224 decimalization, 93–98; exercises, 98; answers, 250 Demon-Haunted World, The (Sagan), 227 Devi, Shakuntala, 146 digital roots, checking solutions via, 133 digits (ﬁngers), xii distributive law, 50–51 divisibility, testing for, 98–102 by 2, 99; exercises, 101; answers, 250 by 3, 99; exercises, 101; answers, 250 by 4, 99; exercises, 101; answers, 250 by 5, 100; exercises, 101; answers, 251 by 6, 100; exercises, 101; answers, 250 by 7, 100; exercises, 102; answers, 251 by 8, 99; exercises, 101; answers, 250 by 9, 99; exercises, 101; answers, 250–51 by 11, 100; exercises, 102; answers, 251 any odd number not ending in 5, 100–101; exercises (by 17), 102; answers, 251 division, 80–107 decimalization, 93–98; exercises, 98; answers, 250 fractions, 103; exercises, 103; answers, 252 guesstimation, 112–14; exercises, 129; answers, 254–55 left to right, 80 numbers ending in 25 or 75, 97 one-digit numbers, 81–84; exercises, 84; answers, 249 quotient in, 87 rule of “thumb” in, 84–86 as simpliﬁcation process, 82 simplifying problems of, 90–93, 98 square root estimation, 117–21; exercises, 130; answers, 255 square roots with pencil and paper, 135–38; exercises, 150; answers, 257 two-digit number into four-digit number, 89–90 two-digit numbers, 87–93; exercises, 93; answers, 249 when divisor ends in 5, 97 dreams, frequency of, 223 estimations, 12; see also guesstimation evolution, theory of, 229 Index 274 factoring a number: in three-by-three multiplication, 181–83; exercises, 192–93; answers, 265–68 in three-by-two multiplication, 169–72; exercises, 175–76; answers, 259–62 in two-digit multiplication, 63–68; exercises, 68; answers, 243 Fibonacci numbers, xi Fleischman, Martin, 228–29 fractions, 102–7 adding with equal denominators, 104–5; exercises, 105; answers, 252 adding with unequal denominators, 105–6; exercises, 106; answers, 252 dividing, 103; exercises, 103; answers, 252 multiplying, 102; exercises, 102; answers, 252 one-digit, 93–96 simplifying, 96–97, 103–4; exercises, 104; answers, 252 subtracting, 106–7; exercises, 107; answers, 252–53 friendly products, 66–68 Fuller, Thomas, 168 Gardner, Martin, 53–54 Gauss, Carl Friedrich, 20 Glois, Évariste, 121 Gold, Thomas, 228 Great Mental Calculators, The (Smith), xxi group theory, 121 guesstimation, 108–30 addition, 108–11; at the supermarket, 111; exercises, 128; answers, 253 division, 112–14; exercises, 129; answers, 254–55 multiplication, 114–17; exercises, 129; answers, 254–55 sign for “approximately,” 109 square roots, 117–21; exercises, 130; answers, 255 subtraction, 111–12; exercises, 129; answers, 254 hearing numbers, 16 HIV-AIDS skeptics, 230 instant multiplication, 1–4 interest: calculating, 125–28; exercises, 130; answers, 256 repaying a loan, 126–27 Rule of 70, 125–26 Rule of 110, 126 introspection illusion, 227 Kevles, Daniel, 227–28 Law of Large Numbers, 223 laws of probability, 224 leapfrog addition, 203–6 leap years, day of the week, 10, 220 left-to-right computations: addition, 6–8, 11–21 division, 80 multiplication, 30–31, 33 subtraction, 6–7, 21–28 loan, repaying with interest, 126–27; exercises, 130; answers, 256 logarithmic spirals, xi magic squares, 206–9 how to construct, 207–8 why this trick works, 208–9 magic 1089, 200–202 math: creative thinking in, xxiv ﬁguring out loud, xiv as language of science, xi useful language of, xvii Mathemagics, xiii, xix, xxiii “amazing” sum (Randi), 212–14 leapfrog addition, 203–6 magic squares, 206–9 magic 1089, 200–202 Index 275 Mathemagics (cont.): missing-digit tricks, 202–3 psychic math, 199–200 quick cube roots, 209–11 simpliﬁed square roots, 211–12 why these tricks work, 50–52 Mathematical Carnival (Gardner), 54 memorizing numbers, 151–62 forming stories for, 161–62 mnemonics in, 151–52 number-word list for, 156–57; exercises, 157–58 phonetic code in, 152, 153–58 memory, remembering sixteen digits, 161–62 million-to-one odds, 224–25 miracles, 224–25 missing-digit tricks, 202–3 mnemonics: in memorizing numbers, 151–52 number-word list, 155–56; exercise, 156–57 phonetic code for, 152, 153–58 squaring three-digit numbers with, 158–61 for storing partial results, 158–61 mod sums, checking solutions via, 133–34, 135 Monty Hall problem, 75 multiplication: advanced, 163–98 associative law of, 65 basic, 29–52 breaking down the problem in, 31–32 calculating restaurant tips, 8, 122–23; exercises, 130; answers, 256 calculating sales tax, 123–25 complements in, 61–62 creative approach to, 69–71; exercises, 70–71; answers, 243–46 criss-cross method (on paper), 138–45; exercises, 150; answers, 257 cubing two-digit numbers, 76–79; exercises, 79; answers, 248 distributive law in, 50–51 ﬁve-by-ﬁve problems, 193–98; exercises, 198; answers, 269–70 ﬁve-digit squares, 177–80; exercises, 180; answers, 262–64 four-digit squares, 164–69; exercises, 169; answers, 258 fractions, 102; exercises, 102; answers, 252 friendly products in, 66–68 guesstimation, 114–17; exercises, 129; answers, 254–55 instant, 1–4 intermediate, 53–79 left to right, 30–31, 33 multiplier is 5 in, 38–39 no overlap in, 32 number by itself, see squaring numbers that begin with 5, 32–33, 37–38 with pencil and paper, 138–45; exercises, 150; answers, 257 problems requiring some carrying, 39–40 rounding up, 34–35, 71 squaring a number, see squaring ten-by-ten digit numbers (on paper), 147–49 three-by-one problems, 36–43; exercises, 43; answers, 236–38 three-by-three problems, 181–93; addition method, 188–90; close-together method, 183–88; factoring method, 181–83; subtraction method, 190–91; when-all-else-fails method, 191–92; exercises, 192–93; answers, 265–68 Index 276 three-by-two problems, 169–76; addition method, 172–73; factoring method, 169–72; subtraction method, 173–75; exercises, 175–76; subproblems, 176; answers, 259–62 three-digit numbers by 11, 3–4 two-by-one problems, 30–34; exercises, 35; answers, 235–36 two-by-twos as subproblems, 70–71; answers, 243–46 two-digit numbers, 54–68; addition method, 54–59; exercises, 59; answers, 241; factoring method, 63–68; exercises, 68; answers, 243; subtraction method, 59–63; exercises, 63; answers, 242 two-digit numbers by 11, 1–3, 57–58; factoring method, 68; exercises, 58; answers, 240 two-digit numbers with same ﬁrst digit, and second digits that add to 10, 5–6 multiplication tables, 30 Nature, mathematical nature of, xi–xii, xiv number-word list, 155–56; exercises, 156–57 Nye, Bill, xi–xv O’Beirne, Thomas, 159 odds of a million to one, 224–25 partial results, mnemonics for storage of, 158–61 peer-review system, 231 pencil-and-paper math, 131–50 adding columns of numbers, 132–33; exercises, 149; answers, 256 casting out elevens, 145, 147–49 mod sums, 133–34, 135 multiplication by criss-cross method, 138–45; exercises, 150; answers, 257 square roots, 135–38; exercises, 150; answers, 257 subtraction, 134–35; exercises, 150; answers, 256 percentages: interest, 125–28; exercises, 130; answers, 256 restaurant tips, 8, 122–23; exercises, 130; answers, 256 sales tax, 123–25 personal beliefs and biases, 227, 230–31 phonetic code, 152, 153–58 Pons, Stanley, 228–29 predictability, 225 premonitions of death, 223, 224 preponderance of evidence, 229 probability, laws of, 224 products, friendly, 66–68 Pronin, Emily, 226 psychic math, 199–200 psychic readings, 224 pyramids, construction of, 229 quick cube roots, 209–11 quotient, 87 Randi, James “the Amazing,” xvii–xiii; “amazing” sum, 212–14 random chance, 224, 225 reality, 225 remembering sixteen digits, 161–62 rounding down, 72 rounding up, 34–35, 71 Rule of 70, 125–26 Rule of 110, 126 rule of “thumb,” 84–86 Sagan, Carl, 227 sales tax, calculating, 123–25 Savant, Marilyn vos, 75 science, math as language of, xi self-serving bias, 225–26 Shermer, Michael, xix–xxii Index 277 simpliﬁcation, 13 division, 82, 90–93, 98 fractions, 96–97, 103–4; exercises, 104; answers, 252 square roots, 211–12 Skeptics Society, 222 Smith, Steven, xxi Sphinx, construction of, 229 square roots: of four-digit numbers, 119 guesstimation, 117–21; exercises, 130; answers, 255 with pencil and paper, 135–38; exercises, 150; answers, 257 simpliﬁed, 211–12 of six-digit numbers, 119–20 of three-digit numbers, 119 of two-digit numbers, 119 squaring, 4–8 ﬁve-digit numbers, 177–80; exercises, 180; answers, 262–64 four-digit numbers, 164–69; exercises, 169; answers, 258 numbers that end in 5, 4–5, 47–48 rounding down in, 72 rounding up in, 71 three-digit numbers, 71–76; exercises, 76; answers, 246–48 three-digit numbers using mnemonics, 158–61 two-digit numbers, 44–48; exercises, 48; answers, 238–40 why these tricks work, 51–52 subtraction: borrowing in, 22–23, 24 complements, 25–28 fractions, 106–7; exercises, 107; answers, 252–53 guesstimation, 111–12; exercises, 129; answers, 254 left to right, 6–7, 21–28 in multiplying three-by-three numbers, 190–91; exercises, 192–93; answers, 265–68 in multiplying three-by-two numbers, 173–75; exercises, 175–76; answers, 259–62 in multiplying two-digit numbers, 59–63; exercises, 63; answers, 242 no borrowing in, 22, 23–24 with paper and pencil, 134–35; exercises, 150; answers, 256 three-digit numbers, 23–25; exercises, 28; answers, 235 two-digit numbers, 21–23; exercises, 23; answers, 234 Sulloway, Frank J., 227 supermarket, guesstimation of total charge, 111; exercise, 128; answer, 253 tarot-card readings, 224 1089 magic, 200–202 tips (restaurant), calculating, 8, 122–23; exercises, 130; answers, 256 UFO sightings, 230 when-all-else-fails method, 191–92 Index 278 About the Authors DR. ARTHUR BENJAMIN is a professor of mathematics at Harvey Mudd College in Claremont, California, having received his PhD in mathematical sciences from Johns Hopkins University in 1989. In 2000, the Mathematical Association of America awarded him the Haimo Prize for Distinguished College Teach-ing. He is also a professional magician and frequently performs at the Magic Castle in Hollywood. He has demonstrated and explained his calculating talents to audiences all over the world. In 2005, Reader’s Digest called him “America’s Best Math Whiz.” DR. MICHAEL SHERMER is a contributing editor to and monthly columnist for Scientiﬁc American, the publisher of Skeptic magazine (www.skeptic.com), the executive director of the Skeptics Society, and the host of the Caltech public science lecture series. He is the author of numerous science books, including Why People Believe Weird Things, How We Believe, The Science of Good and Evil, The Borderlands of Science, and Science Friction.
188314	https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Support_Course_for_Elementary_Statistics/Decimals_Fractions_and_Percents/Decimals%3A__Rounding_and_Scientific_Notation	Decimals: Rounding and Scientific Notation - Statistics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode Decimals Fractions and Percents Support Course for Elementary Statistics { } { Comparing_Fractions_Decimals_and_Percents : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Converting_Between_Fractions_Decimals_and_Percents : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Decimals:__Rounding_and_Scientific_Notation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Using_Fractions_Decimals_and_Percents_to_Describe_Charts : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Decimals_Fractions_and_Percents : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Expressions_Equations_and_Inequalities : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Graphing_Points_and_Lines_in_Two_Dimensions : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Operations_on_Numbers : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Sets : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", The_Number_Line : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 09 Apr 2022 19:10:33 GMT Decimals: Rounding and Scientific Notation 4732 4732 admin { } Anonymous Anonymous 2 false false [ "article:topic", "rounding", "authorname:green", "showtoc:no", "license:ccby", "licenseversion:40" ] [ "article:topic", "rounding", "authorname:green", "showtoc:no", "license:ccby", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Introductory Statistics 4. Support Course for Elementary Statistics 5. Decimals Fractions and Percents 6. Decimals: Rounding and Scientific Notation Expand/collapse global location Support Course for Elementary Statistics Front Matter Decimals Fractions and Percents Comparing Fractions, Decimals, and Percents Converting Between Fractions, Decimals and Percents Decimals: Rounding and Scientific Notation Using Fractions, Decimals and Percents to Describe Charts The Number Line Operations on Numbers Sets Expressions, Equations and Inequalities Graphing Points and Lines in Two Dimensions Back Matter Decimals: Rounding and Scientific Notation Last updated Apr 9, 2022 Save as PDF Converting Between Fractions, Decimals and Percents Using Fractions, Decimals and Percents to Describe Charts picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review HomeworkView on CommonsDonate Page ID 4732 Larry Green Lake Tahoe Community College ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Brief Review of Decimal Language 2. Rules of Rounding 3. Applications 4. Rounding and Arithmetic 5. Scientific Notation Learning Outcomes Understand what it means to have a number rounded to a certain number of decimal places. Round a number to a fixed number of digits. Convert from scientific notation to decimal notation and back. In this section, we will go over how to round decimals to the nearest whole number, nearest tenth, nearest hundredth, etc. In most statistics applications that you will encounter, the numbers will not come out evenly, and you will need to round the decimal. We will also look at how to read scientific notation. A very common error that statistics students make is not noticing that the calculator is giving an answer in scientific notation. For example, suppose that you used a calculator to find the probability that a randomly selected day in July will have a high temperature of over 90 degrees. Your calculator gives the answer: 0.4987230156. This is far too many digits for practical use, so it makes sense to round to just a few digits. By the end of this section you will be able to perform the rounding that is necessary to make unmanageable numbers manageable. Brief Review of Decimal Language Consider the decimal number: 62.5739. There is a defined way to refer to each of the digits. The digit 6 is in the "Tens Place" The digit 2 is in the "Ones Place" The digit 5 is in the "Tenths Place" The digit 7 is in the "Hundredths Place" The digit 3 is in the "Thousandths Place" The digit 9 is in the "Ten-thousandths Place" We also say that 62 is the "Whole Number" part. Keeping this example in mind will help you when you are asked to round to a specific place value. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.1 It is reported that the mean number of classes that college students take each semester is 3.2541. Then the digit in the hundredths place is 5. Rules of Rounding Now that we have reviewed place values of numbers, we are ready to go over the process of rounding to a specified place value. When asked to round to a specified place value, the answer will erase all the digits after the specified digit. The process to deal with the other digits is best shown by examples. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.2: Case 1 - The Test Digit is Less Than 5 Round 3.741 to the nearest tenth. Solution Since the test digit (4) is less than 5, we just erase everything to the right of the tenths digit, 7. The answer is: 3.7. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.3: Case 2 - The Test Digit is 5 or Greater Round 8.53792 to the nearest hundredth. Solution Since the test digit (6) is 5 or greater, we add one to the hundredths digit and erase everything to the right of the hundredths digit, 3. Thus the 3 becomes a 4. The answer is: 8.54. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.4: Case 3 - The Test Digit is 5 or Greater and the rounding position digit is a 9 Round 0.014952 to four decimal places. Solution The test digit is 5, so we must round up. The rounding position is a 9 and adding 1 gives 10, which is not a single digit number. Instead look at the two digits to the left of the test digit: 49. If we add 1 to 49, we get 50. Thus the answer is 0.0150. Applications Rounding is used in most areas of statistics, since the calculator or computer will produce numerical answers with far more digits than are useful. If you are not told how many decimal places to round to, then you often want to think about the smallest number of decimals to keep so that no important information is lost. For example suppose you conducted a sample to find the proportion of college students who receive financial aid and the calculator presented 0.568429314. You could turn this into a percent at 56.8429314%. There are no applications where keeping this many decimal places is useful. If, for example, you wanted to present this finding to the student government, you might want to round to the nearest whole number. In this case the ones digit is 6 and the test digit is 8. Since 8 > 5, you add 1 to the ones digit. You can tell the student government that 57% of all college students receive financial aid. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.5 Suppose that you found out that the probability that a randomly selected person with who has misused prescription opioids will transition to heroin is 0.04998713. Round this number to four decimal places. Solution The first four decimal places are 0.0499 and the test digit is 8. Since 8 > 5, we would like to add 1 to the fourth digit. Since this is a 9, we go to the next digit to the left. This is also a 9, so we go to the next one which is a 4. We can think of adding 0499 + 1 = 0500. Thus the answer is 0.0500. Note that we keep the last two 0's after the 5 to emphasize that this is accurate to the fourth decimal place. Rounding and Arithmetic Many times, we have to do arithmetic on numbers with several decimal places and want the answer rounded to a smaller number of decimal places. One question you might ask is should you round before you perform the arithmetic or after. For the most accurate result, you should always round after you preform the arithmetic if possible. When asked to do arithmetic and present you answer rounded to a fixed number of decimal places, only round after performing the arithmetic. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.6 Suppose you pick three cards from a 52 card deck with replacement and want to find the probability of the event, A, that none of the three cards will be a 2 through 7 of hearts. This probability is: P⁡(A)=(0.8846)3 Round the answer to 2 decimal places. Solution Note that we have to first perform the arithmetic. With a computer or calculator we get: 0.8846 3=0.69221467973 Now we round to two decimal places. Notice that the hundredths digit is a 9 and the test digit is a 2. Thus the 9 remains unchanged and everything to the right of the 9 goes away. the result is P⁡(A)≈0.69 If we mistakenly rounded 0.8846 to two decimal places (0.88) and then cubed the answer we would have gotten 0.68 which is not the correct answer. Scientific Notation When a calculator presents a number in scientific notation, we must pay attention to what this represents. The standard way of writing a number in scientific notation is writing the number as a product of a number greater than or equal 1 but less than 10 followed by a power of 10. For example: 602,000,000,000,000,000,000,000=6.02×10 23 The main purpose of scientific notation is to allow us to write very large numbers or numbers very close to 0 without having to use so many digits. Most calculators and computers use a different notation for scientific notation, most likely because the superscript is difficult to render on a screen. For example, with a calculator: 0.00000032=3.2⁢E−7 Notice that to arrive at 3.2, the decimal needed to be moved 7 places to the right. Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.7 A calculator displays: 2.0541⁢E⁢6 Write this number in decimal form. Solution Notice that the number following E is 6. This means move the decimal over 6 places to the right. The first 4 moves is natural, but for the last 2 moves, there are no numbers to move the decimal place past. We can always add extra zeros after the last number to the right of the decimal place: 2.0541⁢E⁢6=2.054100⁢E⁢6 Now we can move the decimal place to the right 6 places to get 2.0541⁢E⁢6=2.054100⁢E⁢6=2,054,100 Example D⁢e⁢c⁢i⁢m⁢a⁢l⁢s.8 If you use a calculator or computer to find the probability of flipping a coin 27 times and getting all heads, then it will display: 7.45⁢E−9 Write this number in decimal form. Solution Many students will forget to look for the "E" and just write that the probability is 7.45, but probabilities can never be bigger than 1. You can not have a 745% chance of it occurring. Notice that the number following E is −9. Since the power is negative, this means move the decimal to the left, and in particular 9 places to the left. There is only one digit to the left of the decimal place, so we need to insert 8 zeros: 7.45⁢E−9=000000007.45⁢E−9 Now we can move the decimal place to the right 9 places to the left to get 7.45⁢E−9=000000007.45⁢E−9=0.00000000745 Application of Rounding Decimal Numbers Here is a video that explains rounding. This page titled Decimals: Rounding and Scientific Notation is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Larry Green. Back to top Converting Between Fractions, Decimals and Percents Using Fractions, Decimals and Percents to Describe Charts Was this article helpful? Yes No Recommended articles 1.4: Levels of MeasurementSome calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that... 1.4: Frequency, Frequency Tables, and Levels of MeasurementSome calculations generate numbers that are artificially precise. 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188315	https://www.ck12.org/flexi/cbse-math/venn-diagrams-and-operations-of-sets/what-is-the-complement-of-a-set/	What is the complement of a set? - Examples, Symbol, & Definition \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Venn Diagrams and Operations of Sets Question What is the complement of a set? Flexi Says: The complement of a set refers to the elements not in the set, but which belong to a universal set that the original set is a subset of. The universal set is usually established by the context of the problem. i For example, if we have a universal set U={1,2,3,4,5,6,7,8,9,10} and a set A={1,2,3}, the complement of set A (denoted as A′ or A c) would be A′={4,5,6,7,8,9,10}. So, the complement of a set A with respect to a universal set U is the set of elements in U that are not in A. Analogy / Example Try Asking: Define the Property of Ⲫ/ Identity Law in set theory.What is the idempotent property of sets?What are the properties of the complement of a set? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
188316	https://www.teachoo.com/8703/2767/Properties-of-Square-Root/category/Square-root/	Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Maths Class 11 Computer Science (Python) Class 11 English Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Remove Ads Login Square root Square root Square Root of Numbers from 1 to 100 List of perfect square roots Is zero a perfect square? Properties of Square Root You are here Ex 5.3, 1 (i) Ex 5.3, 2 (i) Important Finding Square root through repeated subtraction → Chapter 5 Class 8 Squares and Square Roots Concept wise Square numbers Properties of square numbers Sum of consecutive odd numbers Numbers between square numbers Pattern Solving Finding square of large numbers Pythagorean triplets Square root Finding Square root through repeated subtraction Finding Square root through prime factorisation Checking if perfect square by prime factorisation Smallest number multiplied to get perfect square Smallest number divided to get perfect square Smallest square number divisible by numbers Finding number of digits in square root (without calculation) Finding square root by division method - Integers Least number subtracted to get a perfect square Least number added to get a perfect square Finding square root by division method - Decimals Statement Questions Properties of Square Root Last updated at Dec. 16, 2024 by Teachoo Let’s look at the square of numbers from 1 to 50 \| \| \| --- \| \| Number \| Square root \| \| 1 \| 1 \| \| 4 \| 2 \| \| 9 \| 3 \| \| 16 \| 4 \| \| 25 \| 5 \| \| 36 \| 6 \| \| 49 \| 7 \| \| 64 \| 8 \| \| 81 \| 9 \| \| 100 \| 10 \| \| 121 \| 11 \| \| 144 \| 12 \| \| 169 \| 13 \| \| 196 \| 14 \| \| 225 \| 15 \| \| 256 \| 16 \| \| 289 \| 17 \| \| 324 \| 18 \| \| 361 \| 19 \| \| 400 \| 20 \| \| 441 \| 21 \| \| 484 \| 22 \| \| 529 \| 23 \| \| 576 \| 24 \| \| 625 \| 25 \| \| 900 \| 30 \| \| 1225 \| 35 \| \| 1600 \| 40 \| \| 2025 \| 45 \| \| 2500 \| 50 \| Let’s see some pattern in it, and find properties of square root If a number ends with 1, its square root will end with 1 or 9 Example: As 1 2 = 1 & 9 2 = 81 ∴ √81 = 9 √1 = 1 If a number ends with 6, its square root will end with 4 or 6 Example : As 4 2 = 16 & 6 2 = 36 ∴ √16 = 4 √36 = 6 If a number ends with 5, its square root will always end with 5 Example : As 5 2 = 25 & 15 2 = 225 ∴ √25 = 5 √225 = 1 5 If number has even number of zeroes at the end, its square root will have half of it Example : 10 2 = 100 200 2 =40000 So, If a number ends with 2, 3, 7, 8, it is not a perfect square Example : 2422, 373, 918, 27 are not perfect square. So, their square root will be in decimals If a number ends with odd number of zeroes, it is not a perfect square Example : 24000, 10, 2500000 do not have a square root So, their square root will be in decimals Square root of even number is even, Square root of odd number is odd Example : Odd √1=1 √9= 3 √81=9 Even √4=2 √16= 4 √100= 10 To summarize Unit digit of square roots have this property \| \| \| --- \| \| One’s digit of Number \| One’s digit of Square Root \| \| 1 \| 1 or 9 \| \| 6 \| 4 or 6 \| \| 5 \| 5 \| \| Even number of zeroes \| Half of it \| Square root of even number is even, odd number is odd A number is not a perfect square, if it ends with 2, 3, 7, 8 or if it has odd number of zeroes Square root of Negative number is not possible √(-9) is not possible Next: Ex 5.3, 1 (i) → Remove Ads Made by Davneet Singh Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo Hi, it looks like you're using AdBlock :( Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription. Join Teachoo Black at ₹39 only Please login to view more pages. It's free :) Teachoo gives you a better experience when you're logged in. Please login :) Login
188317	https://dspace.mit.edu/bitstream/handle/1721.1/123321/18-312-spring-2009/contents/readings-and-lecture-notes/MIT18_312S09_Lecture32-36.pdf	Course 18.312: Algebraic Combinatorics Lecture Notes # 32-36 Addendum by Gregg Musiker May 1st - 11th, 2009 1 Perfect Matchings and Domino Tilings Deﬁniton. A Perfect Matching of a graph G = (V, E) is a set M ⊂ E of distringuished edges such that each vertex v ∈V is incident to exactly one edge of M. Notice that in particular that this implies that M = V /2 and only graphs with \| \| \| \| an even number of vertices can contain a perfect matching. Perfect matching enu meration can in general be a diﬃcult problem but we consider the case of bipartite planar graphs where their number is easier to compute. For such graphs, it is also possible to view a perfect matching of a graph as a tiling by dominoes. In particular, let G∗ be the dual graph to G, which means that a vertex of G∗ corresponds to a face (bounded or unbounded) region of G and vice-versa. Edges also correspond to one another. In the case that G embeds inside the Z × Z lattice, a domino covers two faces of G∗ and perfect matchings of G exactly correspond to a domino tiling of G∗ . See Kuo’s paper and the reference on Aztec diamonds for typical examples of such dual graphs. These references can be found at the bottom of these lecture notes. 2 Pfaﬃans and Matching Enumeration We enumerate perfect matchings by computing Pfaﬃans, which turn out to be determinants in the case of certain planar graphs. We warn the reader that today’s lecture notes are more of a sketch than previous lecture notes. We begin with a few deﬁnitions. Given an edge (vi, vj) in graph G and an orientation σ on G, we let ai,j = +1 if vi vj, ai,j = −1 if vj vi and ai,j = 0 otherwise. Given a perfect matching → → 1 M = {(i1, j1), . . . , (in/2, jn/2)} (where ir < jr by convention), we let the sign of the matching be sgn(M) = (−1)# crossings in the nesting pattern of M where we draw 1, 2 through n in a line, and the nesting pattern connects ir and jr for each edge of M. Deﬁnition. An orientation σ is a Pfaﬃan orientation of G if and only if for all perfect matchings M of G, we have that each expression Qn/ r=1 2 air ,jr sgn (M) has the same sign, regardless of the choice of M. Theorem. Let G be a planar simple ﬁnite graph with no bridges (i.e. no edges whose removal would disconnect the set of vertices of G). Let σ be an orientation such that the number of clockwise edges about any bounded face is odd. Then σ is Pfaﬃan. To sketch the proof of this Theorem, we will use a sign-preserving map between pairs of matchings of the line {1, 2, . . ., n} and permutations of Sn: φ : (M1, M2) − →σ such that sgn(M1)aM1 · sgn(M2)aM2 = sgn(σ)aσ, where σ denotes the match ing {1, σ(1)}, {2, σ(2)}, . . ., {n, σ(n)}. Recall that the sign of a permutation is (−1)# cycles of even length. In particular, we draw the line from 1 to n and then connect vertices on the top according to matching M1 and connect vertices on the bottom using matching M2. We then make cycles starting with the smallest element. Example. Let M1 = {1, 2}, {3, 6}, {4, 5}, M2 = {1, 4}, {2, 5}, {3, 6}. These are both matchings of the 2-by-3 grid graph with vertices 1, 2, 3 on top and 4, 5, 6 in the second row. The corresponding permutation is σ = (1254)(36). Claim. Any σ constructed in this way only has cycles of even length. In fact, we obtain a bijection since the process is reversible by alternating assign ments of edges in the cycles to M1 and M2. For example, for the permutation σ = (1, 3)(2, 6, 5, 8)(4, 10, 9, 7), we obtain M1 = {1, 3}, {2, 6}, {5, 8}, {4, 10}, {9, 7} and M2 = {3, 1}, {6, 5}, {8, 2}, {10, 9}, {7, 4}. ̸ ̸ P Furthermore, we can pull apart cycles by eliminating two crossings. This is equiva lent to rearranging numbers on the line so that if i, j are in diﬀerent cycles, then their placements do not aﬀect the sign. The sign is only aﬀected by the ordering within individual cycles, and a self-crossing can be eliminating by applying a transposition. Getting back to the main proof, if we assume that an orientation of G is not Pfaf ﬁan, then there exists a pair of matchings (M1, M2) so that the signs of M1 and M2 diﬀer: sgn(M1)aM1 = sgn(M2)aM2 which would imply by the theorem that aσ sgn(σ) = +1. Thus it suﬃces to show that this sign is −1 instead. Given an orientation with an odd number of clockwise edges around each bounded face, we build a permutation σ corresponding to a pair of matchings M1 and M2. Then σ breaks into a product of cycles, each of which must have even length. It fol lows that sgn(σ) = (−1)# cycles of even length which decomposes as σ = σ1 σt. How · · · ever, cycles of σ decmposes V into edges and even cycles (cycles not nec. contained on a single face) so aσi = # edges oriented clockwise on face induced by cycle σi. So if each face has an odd number of clockise edges, then we have the result. Lemma. Let G = (V, E, F) be a simple connected planar graph without bridges with an odd number of clockwise edges on each face. Let C be a directed cycle. Then the number of edges oriented clockwise in C has the opposite parity as the number of vertices in the interior of C. Proof omitted, but follows by application of Euler’s formula applied to the subgraph of G consisting of the interior of C together with cycle C. Thus we have shown the Theorem that an orientation of G with each face having an odd number of clockwise edges is Pfaﬃan. We will not show it in these notes, but it is possible by induction to show that such an orientation exists. Main Theorem. If σ is a Pfaﬃan orientation (which we know exists in certain cases and we can inductively construct such an orientation) then we have the formula det A(G) = (# perfect matchings of G)2 , where det A(G) = σ∈Sn sgn(σ)aσ. The matrix A(G) is the signed adjacency matrix for the oriented graph. Application. Consider the m-by-n grid graph Lm,n with m rows and n columns, for m n vertices in all. We give this graph an orientation so that all vertical edges are · oriented downwards, the horizontal of the odd (i.e. ﬁrst, third, etc.) labeled rows are oriented rightwards, while the even labeled rows are oriented leftwards. Let B be the n-by-n matrix with a superdiagonal of +1’s, a subdiagonal of −1’s and zeros everywhere else. The signed adjancency matrix for Lm,n is the (mn)-by-(mn) block matrix obtained by taking an m-by-m matrix whose superdiagonal are the n-by-n identity matrices In, subdiagonal entries are −In and diagonal entries are the B’s deﬁned above. Then all other entries are zero. By the Theorem, M(Lm,n), the number of perfect matchings in Lm,n is √ det A. Corollary. For n even, ⌊m/2⌋ n/2 Y Y kπ ℓπ M(Lm,n) = 4⌊m/2⌋n/2 cos 2( ) + cos 2( ) . m + 1 n + 1 k=1 ℓ=1 Other applications of Pﬀaﬁans provide formulas for Aztec Diamonds and vari ants. 3 References for Height functions, Domino Tilings, and Aztec Diamonds “Alternating sign matrices and domino tilings” by Noam Elkies, Greg Kuperberg, Michael Larsen, and James Propp and “Applications of graphical condensation for enumerating matchings and tilings” by Eric Kuo MIT OpenCourseWare 18.312 Algebraic Combinatorics Spring 2009 For information about citing these materials or our Terms of Use, visit:
188318	https://www.doubtnut.com/qna/234805995	[Bengali] The parametric equations of the ellipse (x^(2))/(a^(2))+( Download Allen App Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Ncert Solutions Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Exam IIT-JEE NEET CBSE UP Board Bihar Board Results Study with ALLEN JEE NEET Class 6-10 My Profile Logout Ncert Solutions English Medium Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Exam IIT-JEE NEET CBSE UP Board Bihar Board Study with ALLEN JEE NEET Class 6-10 Books Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Online Class Blog Select Theme Class 12 MATHS The parametric equations of the ellips... The parametric equations of the ellipse x 2 a 2+y 2 b 2=1 are _ Solution in Bengali A x=a cos ϕ,y=b sin ϕ B x=a cos ϕ,y=a sin ϕ C x=a tan ϕ,y=b sec ϕ D x=a sec ϕ,y=b tan ϕ Video Solution To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video Video Player is loading. Play Video Play Skip Backward Mute Current Time 0:00 / Duration-:- Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. 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View Solution Find (i) the lengths of axes (ii) the length of latus rectum (iii) coordinates of vertices (iv) eccentricity (v) coordinates of foci and (iv) equations of directrices of each of the following ellipses : 16x^(2) + 25y^(2) = 400 View Solution Find the length of mejor and minor axis of the ellipse 25(x+1)2+9(y+2)2=225 . View Solution Consider the points A(0, 0, 0), B(2, 0, 0), C(1,√3,0) and D(1,1√3,2√2√3) Statement - I : ABCD is a square. Statement - II : \|A B\|=\|B C\|=\|C D\|=\|D A\|. View Solution Statement-I: General solution of the equation tan 4 x−tan 2 x 1+tan 4 x tan 2 x=1 is x=n π 2+π 8,n∫Z Statement-II: tan α=1, therefore general solution is α=n π+π 4,n∫Z View Solution Find the parametric equation of the parabola (x-1)^(2)=-16(y-2) View Solution दीर्घवृत्त x 2 a 2+y 2 b 2=1 का अवकल समीकरण ज्ञात कीजिये जबकि a तथा b स्वेच्छ अचर है। View Solution the equation of a diameter conjugate to a diameter y=b a x of the ellipse x 2 a 2+y 2 b 2=1 is View Solution The parametric equations of the hyperbola x 2 a 2−y 2 b 2=1 are- View Solution Obtain the parametric equations of the circle x 2+y 2=1 View Solution Define ellipse and derive its equation in the form (x^(2))/(a^(2))+(y^(2))/(b^(2)) = 1(a gt b) . View Solution x 2 a 2−y 2 b 2=1 एक दीर्घवृत्त का समीकरण है। View Solution Find the length of mejor and minor axis of the ellipse 25(x+1)2+9(y+2)2=225 . View Solution If the distance between two latus rectum of a ellipse is 10 unit and length of major axis 12 unit then find its eccentricity. View Solution Find the length of the latus rectum of the ellipse 5 x 2+3 y 2=15 . View Solution CHHAYA PUBLICATION-ELLIPSE-M . C . Q If e be the eccentricity of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2... 01:36 If the length of the minor axis of an ellipse is equal to the distan... 01:51 The parametric equations of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^... 01:20 The eqation of auxiliary circle of the ellipse (x^(2))/(a^(2))+(y^(2... 01:28 If a point moves on a plane is such a way that the sum of its dista... 02:55 The length of latus rectum of the ellipse 9x^(2) + 25y^(2) = 225 is 02:55 The coordinates of the vertices of the ellipse 4x^(2) + y^(2) = 16 ... 02:22 The eccentricity of the ellipse 4x^(2) + 25y^(2) = 100 is 02:33 The length of latus rectum of the ellipse 25x^(2) + 9y ^(2) = 225 is 03:11 The coordinates of the vertices of the ellipse x^(2) + 4y^(2) = 16 ... 02:09 The length of major axis of the ellipse 4x^(2) + 9y^(2) = 36 is 02:22 The eccentricity of the ellipse 25 x^(2) + 4y^(2) = 100 is 02:51 The length of minor axis of the ellipse 9x^(2) + 4y ^(2) = 36 is 02:22 The coordinates of the point on the ellipse 9x^(2) + 16y^(2) = 144 ... 04:04 If the equation (x^(2))/(4-m)+(y^(2))/(m-7)+ 1 = 0 represents an e... 02:12 The sum of the focal distances of any point on the ellipse 4x^(2) +... 06:51 If the distance between the foci of an ellipse is equal to the length ... 03:35 The eccentricity of the ellipse 5x^(2) + 9y^(2) = 1 is 01:33 If e and a be the eccentricity and length of semi - major axis of an e... 02:41 The coordinates of the centre of the ellipse 4x^(2) + 9y^(2) - 16 x +... 02:45 Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics Sunil Batra Solutions for Physics Pradeep Solutions for Physics Errorless Solutions for Physics Narendra Awasthi Solutions for Chemistry MS Chouhan Solutions for Chemistry Errorless Solutions for Biology Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium NCERT Solutions for Class 10 English Medium NCERT Solutions for Class 9 English Medium NCERT Solutions for Class 8 English Medium NCERT Solutions for Class 7 English Medium NCERT Solutions for Class 6 English Medium Free Ncert Solutions Hindi Medium NCERT Solutions NCERT Solutions for Class 12 Hindi Medium NCERT Solutions for Class 11 Hindi Medium NCERT Solutions for Class 10 Hindi Medium NCERT Solutions for Class 9 Hindi Medium NCERT Solutions for Class 8 Hindi Medium NCERT Solutions for Class 7 Hindi Medium NCERT Solutions for Class 6 Hindi Medium Boards CBSE UP Board Bihar Board UP Board Resources Unit Converters Calculators Books Store Seminar Results Blogs Class 12 All Books Class 11 All Books Class 10 All Books Class 9 All Books Class 8 All Books Class 7 All Books Class 6 All Books Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. 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188319	https://www.youtube.com/watch?v=mkyZ45KQYi4	Coin flipping probability \| Probability and Statistics \| Khan Academy Khan Academy 9040000 subscribers 3004 likes Description 907992 views Posted: 9 Aug 2011 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? Probability and statistics on Khan Academy: We dare you to go through a day in which you never consider or use probability. Did you check the weather forecast? Busted! Did you decide to go through the drive through lane vs walk in? Busted again! We are constantly creating hypotheses, making predictions, testing, and analyzing. Our lives are full of probabilities! Statistics is related to probability because much of the data we use when determining probable outcomes comes from our understanding of statistics. In these tutorials, we will cover a range of topics, some which include: independent events, dependent probability, combinatorics, hypothesis testing, descriptive statistics, random variables, probability distributions, regression, and inferential statistics. So buckle up and hop on for a wild ride. We bet you're going to be challenged AND love it! About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to KhanAcademy’s Probability and Statistics channel: Subscribe to KhanAcademy: 164 comments Transcript: Now let's start to do some more interesting problems. And one of these things that you'll find in probability is that you can always do a more interesting problem. So now I'm going to think about-- I'm going to take a fair coin, and I'm going to flip it three times. And I want to find the probability of at least one head out of the three flips. So the easiest way to think about this is how many equally likely possibilities there are. In the last video, we saw if we flip a coin 3 times, there's 8 possibilities. For the first flip, there's 2 possibilities. Second flip, there's 2 possibilities. And in the third flip, there are 2 possibilities. So 2 times 2 times 2-- there are 8 equally likely possibilities if I'm flipping a coin 3 times. Now how many of those possibilities have at least 1 head? Well, we drew all the possibilities over here. So we just have to count how many of these have at least 1 head. So that's 1, 2, 3, 4, 5, 6, 7. So 7 of these have at least 1 head in them. And this last one does not. So 7 of the 8 have at least 1 head. Now you're probably thinking, OK, Sal. You were able to do it by writing out all of the possibilities. But that would be really hard if I said at least one head out of 20 flips. This had worked well because I only had 3 flips. Let me make it clear, this is in 3 flips. This would have been a lot harder to do or more time consuming to do if I had 20 flips. Is there some shortcut here? Is there some other way to think about it? And you couldn't just do it in some simple way. You can't just say, oh, the probability of heads times the probability of heads, because if you got heads the first time, then now you don't have to get heads anymore. Or you could get heads again-- you don't have to. So it becomes a little bit more complicated. But there is an easy way to think about it where you could use this methodology right over here. You'll actually see this on a lot of exams where they make it seem like a harder problem, but if you just think about in the right way, all of a sudden it becomes simpler. One way to think about it is the probability of at least 1 head in 3 flips is the same thing-- this is the same thing-- as the probability of not getting all tails, right? If we got all tails, then we don't have at least 1 head. So these two things are equivalent. The probability of getting at least 1 head in 3 flips is the same thing as the probability of not getting all tails in 3 flips. So what's the probability of not getting all tails? Well, that's going to be 1 minus the probability of getting all tails. The probability of getting all tails, since it's 3 flips, it's the probability of tails, tails, and tails. Because any of the other situations are going to have at least 1 head in them. And that's all of the other possibilities, and then this is the only other leftover possibility. If you add them together, you're going to get 1. Let me write it this way. Let me write it a new color just so you see where this is coming from. The probability of not all tails plus the probability of all tails-- well, this is essentially exhaustive. This is all of the possible circumstances. So your chances of getting either not all tails or all tails-- and these are mutually exclusive, so we can add them. The probability of not all tails or, just to be clear what we're doing, the probability of not all tails or the probability of all tails is going to be equal to one. These are mutually exclusive. You're either going to have not all tails, which means a head shows up. Or you're going to have all tails. But you can't have both of these things happening. And since they're mutually exclusive and you're saying the probability of this or this happening, you could add their probabilities. And this is essentially all of the possible events. So this is essentially, if you combine these, this is the probability of any of the events happening. And that's going to be a 1 or 100% chance. So another way to think about is the probability of not all tails is going to be 1 minus the probability of all tails. So that's what we did right over here. And the probability of all tails is pretty straightforward. That's the probability of it's going to be 1/2, because you have a 1/2 chance of getting a tails on the first flip, times-- let me write it here, so we can have it a little clearer. So this is going to be 1 minus the probability of getting all tails. You will have a 1/2 chance of getting tails on the first flip, and then you're going to have to get another tails on the second flip, and then you're going to have to get another tails on the third flip. And then 1/2 times 1/2 times 1/2. This is going to be 1/8. And then 1 minus 1/8 or 8/8 minus 1/8 is going to be equal to 7/8. So we can apply that to a problem that is harder to do than writing all of the scenarios like we did in the first problem. Let's say we have 10 flips, the probability of at least one head in 10 flips-- well, we use the same idea. This is going to be equal to the probability of not all tails in 10 flips. So we're just saying the probability of not getting all of the flips going to be tail. All of the flips is tails-- not all tails in 10 flips. And this is going to be 1 minus the probability of flipping tails 10 times. So it's 1 minus 10 tails in a row. And so this is going to be equal to this part right over here. Let me write this. So this is going to be this one. Let me just rewrite it. This is equal to 1 minus-- and this part is going to be, well, one tail, another tail. So it's 1/2 times 1/2. And I'm going to do this 10 times. Let me write this a little neater. 1/2-- so that's 5, 6, 7, 8, 9, and 10. And so we really just have to-- the numerator is going to be 1. So this is going to be 1. This is going to be equal to 1. Let me do it in that same color of green. This is going to be equal to 1 minus-- our numerator, you just have 1 times itself 10 times. So that's 1. And then on the denominator, you have 2 times 2 is 4. 4 times 2 is 8, 16, 32, 64, 128, 256, 512, 1,024-- over 1,024. This is the exact same thing as 1 is 1024 over 1024 minus 1 over 1024, which is equal to 1,023 over 1,024. We have a common denominator here. So 1,000-- I'm doing that same blue-- over 1,024. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. It's 1,023 over 1,024. And you can get a calculator out to figure that out in terms of a percentage. Actually, let me just do that just for fun. So if we have 1,023 divided by 1,024 that gives us-- you have a 99.9% chance that you're going to have at least one heads. So this is if we round. This is equal to 99.9% chance. And I rounded a little bit. It's actually slightly, even slightly, higher than that. And this is a pretty powerful tool or a pretty powerful way to think about it because it would have taken you forever to write all of the scenarios down. In fact, there would have been 1,024 scenarios to write down. So doing this exercise for 10 flips would have taken up all of our time. But when you think about in a slightly different way, when you just say, look the probability of getting at least 1 heads in 10 flips is the same thing as the probably of not getting all tails. And that's 1 minus the probability of getting all tails. And this is actually a pretty easy thing to think about.
188320	https://personal.math.ubc.ca/~CLP/CLP4/clp_4_vc/sec_reparam.html	Reparametrization Skip to main content CLP-4 Vector Calculus Joel Feldman, Andrew Rechnitzer, Elyse Yeager Contents Search Book close Search Results: No results. PrevUpNext Front Matter Colophon Preface Feedback about the text 1 Curves 1.1 Derivatives, Velocity, Etc. 1.1 Exercises 1.2 Reparametrization 1.2 Exercises 1.3 Curvature 1.3 Exercises 1.4 Curves in Three Dimensions 1.4 Exercises 1.5 A Compendium of Curve Formula 1.6 Integrating Along a Curve 1.6 Exercises 1.7 Sliding on a Curve 1.7.1 The Sliding Bead 1.7.2 The Skier 1.7.3 The Skate Boarder 1.7.4 Exercises 1.8 Optional — Polar Coordinates 1.8 Exercises 1.9 Optional — Central Forces 1.9 Exercises 1.10 Optional — Planetary Motion 1.11 Optional — The Astroid 1.12 Optional — Parametrizing Circles 2 Vector Fields 2.1 Definitions and First Examples 2.1 Exercises 2.2 Optional — Field Lines 2.2.1 More about r′(t)×v(r(t))=0 2.2.2 Exercises 2.3 Conservative Vector Fields 2.3 Exercises 2.4 Line Integrals 2.4.1 Path Independence 2.4.2 Exercises 2.5 Optional — The Pendulum 3 Surface Integrals 3.1 Parametrized Surfaces 3.1 Exercises 3.2 Tangent Planes 3.2 Exercises 3.3 Surface Integrals 3.3.1 Parametrized Surfaces 3.3.2 Graphs 3.3.3 Surfaces Given by Implicit Equations 3.3.4 Examples of ∬S ρ d S 3.3.5 Optional — Dropping Higher Order Terms in d u,d v 3.3.6 Exercises 3.4 Interpretation of Flux Integrals 3.4.1 Examples of Flux Integrals 3.5 Orientation of Surfaces 4 Integral Theorems 4.1 Gradient, Divergence and Curl 4.1.1 Vector Identities 4.1.2 Vector Potentials 4.1.3 Interpretation of the Gradient 4.1.4 Interpretation of the Divergence 4.1.5 Interpretation of the Curl 4.1.6 Exercises 4.2 The Divergence Theorem 4.2.1 Optional — An Application of the Divergence Theorem — the Heat Equation 4.2.1.1 Derivation of the Heat Equation 4.2.1.2 An Application of the Heat Equation 4.2.2 Variations of the Divergence Theorem 4.2.3 An Application of the Divergence Theorem — Buoyancy 4.2.4 Optional — Torque 4.2.5 Optional — Solving Poisson’s Equation 4.2.6 Exercises 4.3 Green’s Theorem 4.3 Exercises 4.4 Stokes’ Theorem 4.4.1 The Interpretation of Div and Curl Revisited 4.4.1.1 Divergence 4.4.1.2 Curl 4.4.2 Optional — An Application of Stokes’ Theorem — Faraday’s Law 4.4.3 Exercises 4.5 Optional — Which Vector Fields Obey ∇∇×F=0 4.6 Really Optional — More Interpretation of Div and Curl 4.7 Optional — A Generalized Stokes’ Theorem 5 True/False and Other Short Questions 5.1 True/False and Other Short Questions 5.2 Exercises Appendices A Appendices A.1 Trigonometry A.1.1 Trigonometry — Graphs A.1.2 Trigonometry — Special Triangles A.1.3 Trigonometry — Simple Identities A.1.4 Trigonometry — Add and Subtract Angles A.1.5 Inverse Trigonometric Functions A.2 Powers and Logarithms A.2.1 Powers A.2.2 Logarithms A.3 Table of Derivatives A.4 Table of Integrals A.5 Table of Taylor Expansions A.6 3d Coordinate Systems A.6.1 Cartesian Coordinates A.6.2 Cylindrical Coordinates A.6.3 Spherical Coordinates A.7 ISO Coordinate System Notation A.7.1 Polar Coordinates A.7.2 Cylindrical Coordinates A.7.3 Spherical Coordinates A.8 Conic Sections and Quadric Surfaces A.9 Review of Linear Ordinary Differential Equations B Hints for Exercises C Answers to Exercises D Solutions to Exercises Section 1.2 Reparametrization There are invariably many ways to parametrize a given curve. Kind of trivially, one can always replace t by, for example, .3 u. But there are also more substantial ways to reparametrize curves. It often pays to tailor the parametrization used to the application of interest. For example, we shall see in the next couple of sections that many curve formulae simplify a lot when arc length is used as the parameter. Example 1.2.1. Here are three different parametrizations of the semi-circle ,x 2+y 2=r 2,.y≥0. The first uses the polar angle θ as the parameter. We have already seen, in Example 1.0.1, the parametrization r 1(θ)=(r cos⁡θ,r sin⁡θ)0≤θ≤π The second uses x as the parameter. Just solving ,x 2+y 2=r 2,y≥0 for y as a function of ,x, gives y(x)=r 2−x 2 and so gives the parametrization r 2(x)=(x,r 2−x 2)−r≤x≤r The third uses arc length from (r,0) as the parameter. We have seen, in Example 1.1.6, that the arc length from (r,0) to r 1(θ) is just .s=r θ. So the point on the semicircle that is arc length s away from (r,0) is r 3(s)=r 1(s r)=(r cos⁡s r,r sin⁡s r)0≤s≤π r We shall see that, for some purposes, it is convenient to use parametrization by arc length. Here is a messier example in which we reparametrize a curve so as to use the arc length as the parameter. Example 1.2.2. We saw in Example 1.1.9, that, as t runs from 0 to ,π 2,r(t)=a cos 3⁡t ı ı^+a sin 3⁡t ȷ ȷ^ runs from (a,0) to (0,a) along the astroid .x 2/3+y 2/3=a 2/3. Suppose that we want a new parametrization R(s) chosen so that, as s runs from 0 to some appropriate value, R(s) runs from (a,0) to (0,a) along ,x 2/3+y 2/3=a 2/3, with s being the arc length from (a,0) to R(s) along .x 2/3+y 2/3=a 2/3. We saw, in Example 1.1.9, that, for ,0≤t≤π 2,d s d t=3 a 2 sin⁡(2 t) so that the arclength from (a,0)=r(0) to r(t) is s(t)=∫0 t 3 a 2 sin⁡(2 t′)d t′=3 a 4[1−cos⁡(2 t)] which runs from ,0, at ,t=0, to ,3 a 2, at .t=π 2. This is relatively clean and we can invert s(t) to find t as a function of .s. The value, ,T(s), of t that corresponds to any given 0≤s≤3 a 2 is determined by s=3 a 4[1−cos⁡(2 T(s))]⟺T(s)=1 2 arccos⁡(1−4 s 3 a) and R(s)=r(T(s))=a cos 3⁡(T(s))ı ı^+a sin 3⁡(T(s))ȷ ȷ^ We can simplify cos 3⁡(T(s)) and sin 3⁡(T(s)) by just using trig identities to convert the cos⁡(2 T(s)) in s=3 a 4[1−cos⁡(2 T(s))] into cos⁡(T(s))’s and sin⁡(T(s))’s. s=3 a 4[1−cos⁡(2 T(s))]=3 a 4[1−{2 cos 2⁡(T(s))−1}]⟺cos 2⁡(T(s))=1−2 s 3 a s=3 a 4[1−cos⁡(2 T(s))]=3 a 4[1−{1−2 sin 2⁡(T(s))}]⟺sin 2⁡(T(s))=2 s 3 a Consequently the desired parametrization is R(s)=a[1−2 s 3 a]3/2 ı ı^+a[2 s 3 a]3/2 ȷ ȷ^0≤s≤3 a 2 which is remarkably simple. Exercises Exercises Exercise Group. Exercises — Stage 1 1. A curve r(s) is parametrized in terms of arclength. What is ∫1 t\|r′(s)\|d s when ?t≥1? 2. The function r(s)=sin⁡(s+1 2)ı ı^+cos⁡(s+1 2)ȷ ȷ^+3 2(s+1)k^ is parametrized in terms of arclength, starting from the point .P. What is ?P? 3. A curve R=a(t) is reparametrized in terms of arclength as .R=b(s)=a(t(s)). Of the following options, which best describes the relationship between the vectors a′(t 0) and ,b′(s 0), where ?t(s 0)=t 0? You may assume a′(t) and b′(s) exist and are nonzero for all .t,s≥0. they are parallel and point in the same direction they are parallel and point in opposite directions they are perpendicular they have the same magnitude they are equal Exercise Group. Exercises — Stage 2 4.(✳). Let r(t)=(2 sin 3⁡t,2 cos 3⁡t,3 sin⁡t cos⁡t) Find the unit tangent vector to this parametrized curve at ,t=π/3, pointing in the direction of increasing .t. Reparametrize the vector function r(t) from (a) with respect to arc length measured from the point t=0 in the direction of increasing .t. 5.(✳). This problem is about the logarithmic spiral in the plane r(t)=e t(cos⁡t,sin⁡t),t∈R Find the arc length of the piece of this spiral which is contained in the unit circle. Reparametrize the logarithmic spiral with respect to arc length, measured from .t=−∞. Exercise Group. Exercises — Stage 3 6. Define r(t)=(1 1+t 2,arctan⁡t 1+t−2,arctan⁡t) for .0≤t. Reparametrize the function using ,z=arctan⁡t, and describe the curve it defines. What is the geometric interpretation of the new parameter ?z? 7. Reparametrize the function r(t)=(1 2 t 2,1 3 t 3) in terms of arclength from .t=−1. PrevTopNext Feedback
188321	https://study.com/skill/learn/using-mass-concentration-to-find-solute-mass-explanation.html	Using Mass Concentration to Find Solute Mass \| Chemistry \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Using Mass Concentration to Find Solute Mass High School Chemistry Skills Practice Click for sound 4:10 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 Using mass… 02:36 Using mass… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Jiwon Park, Kirsten Wordeman Instructors Jiwon Park Jiwon has a B.S. degree in the mathematics/ science field and over 4 years of tutoring experience. She fell in love with math when she discovered geometry proofs and that calculus can help her describe the world around her like never before. View bio Kirsten Wordeman Kirsten has taught high school biology, chemistry, physics, and genetics/biotechnology for three years. She has a Bachelor's in Biochemistry from The University of Mount Union and a Master's in Biochemistry from The Ohio State University. She holds teaching certificates in biology and chemistry. View bio Example SolutionsPractice Questions Using Mass Concentration to Find Solute Mass Step 1: Identify the total volume of the solution. Step 2: Identify the mass concentration. Step 3: To find the mass of the solute, multiply the volume and the mass concentration of the solution. Using Mass Concentration to Find Solute Mass Vocabulary Solution: When compounds mix completely to become a uniform mixture, the mixture is called a solution. Solute: A solute is the substance dissolved in a solution. Solvent: A solvent is the substance that dissolves a solute in a solution. Mass Concentration: The measure of a solute in grams per 1 liter or 1 milliliter of solution. Concentration (g/L)=mass of solute (g)volume of total solution (L) Let's practice using mass concentration to find solute mass with the following two examples. Using Mass Concentration to Find Solute Mass Example Sodium chloride was dissolved in water and created a sodium chloride solution of 0.500L with a mass concentration of 14.5 g/L. What mass of sodium chloride, in grams, was dissolved in water to create the solution with the given concentration? Step 1: Identify the total volume of the solution. The total volume of the solution is 0.500 L Step 2: Identify the mass concentration. The mass concentration of the solution is 14.5 g/L. Step 3: To find the mass of the solute, multiply the volume and the mass concentration of the solution. The mass of the solute can be found by: mass of solute=volume of solution×concentration of solution=0.500 L×14.5 g L=7.25 gThe mass of the sodium chloride dissolved in water was 7.25 g. Using Mass Concentration to Find Solute Mass Example Sugar was dissolved in water and created a sugar solution of 125 mL with a mass concentration of 24.4 g/L. What mass of sugar, in grams, was dissolved in water to create the solution with the given concentration? Step 1: Identify the total volume of the solution. The total volume of the solution is 125 mL. Since the mass concentration is in unit g/L, we need to convert 125 mL to L. 125 m L=125 m L×1 L 1000 m L=0.125 L Step 2: Identify the mass concentration. The mass concentration of the solution is 24.4 g/L. Step 3: To find the mass of the solute, multiply the volume and the mass concentration of the solution. The mass of the solute can be found by: mass of solute=volume of solution×concentration of solution=0.125 L×24.4 g L=3.05 gThe mass of the sugar dissolved in water was 3.05 g. Get access to thousands of practice questions and explanations! 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188322	https://en.wikipedia.org/wiki/Graph_traversal	Jump to content Search Contents (Top) 1 Redundancy 2 Graph traversal algorithms 2.1 Depth-first search 2.1.1 Pseudocode 2.2 Breadth-first search 2.2.1 Pseudocode 3 Applications 4 Graph exploration 5 Universal traversal sequences 6 See also 7 References Graph traversal Deutsch فارسی Français 한국어 Magyar Polski Српски / srpski Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia "Graph search" redirects here; not to be confused with Facebook Graph Search. Computer science algorithm \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Graph traversal" – news · newspapers · books · scholar · JSTOR (October 2014) (Learn how and when to remove this message) \| In computer science, graph traversal (also known as graph search) refers to the process of visiting (checking and/or updating) each vertex in a graph. Such traversals are classified by the order in which the vertices are visited. Tree traversal is a special case of graph traversal. Redundancy [edit] Unlike tree traversal, graph traversal may require that some vertices be visited more than once, since it is not necessarily known before transitioning to a vertex that it has already been explored. As graphs become more dense, this redundancy becomes more prevalent, causing computation time to increase; as graphs become more sparse, the opposite holds true. Thus, it is usually necessary to remember which vertices have already been explored by the algorithm, so that vertices are revisited as infrequently as possible (or in the worst case, to prevent the traversal from continuing indefinitely). This may be accomplished by associating each vertex of the graph with a "color" or "visitation" state during the traversal, which is then checked and updated as the algorithm visits each vertex. If the vertex has already been visited, it is ignored and the path is pursued no further; otherwise, the algorithm checks/updates the vertex and continues down its current path. Several special cases of graphs imply the visitation of other vertices in their structure, and thus do not require that visitation be explicitly recorded during the traversal. An important example of this is a tree: during a traversal it may be assumed that all "ancestor" vertices of the current vertex (and others depending on the algorithm) have already been visited. Both the depth-first and breadth-first graph searches are adaptations of tree-based algorithms, distinguished primarily by the lack of a structurally determined "root" vertex and the addition of a data structure to record the traversal's visitation state. Graph traversal algorithms [edit] Note. — If each vertex in a graph is to be traversed by a tree-based algorithm (such as DFS or BFS), then the algorithm must be called at least once for each connected component of the graph. This is easily accomplished by iterating through all the vertices of the graph, performing the algorithm on each vertex that is still unvisited when examined. Depth-first search [edit] Main article: Depth-first search A depth-first search (DFS) is an algorithm for traversing a finite graph. DFS visits the child vertices before visiting the sibling vertices; that is, it traverses the depth of any particular path before exploring its breadth. A stack (often the program's call stack via recursion) is generally used when implementing the algorithm. The algorithm begins with a chosen "root" vertex; it then iteratively transitions from the current vertex to an adjacent, unvisited vertex, until it can no longer find an unexplored vertex to transition to from its current location. The algorithm then backtracks along previously visited vertices, until it finds a vertex connected to yet more uncharted territory. It will then proceed down the new path as it had before, backtracking as it encounters dead-ends, and ending only when the algorithm has backtracked past the original "root" vertex from the very first step. DFS is the basis for many graph-related algorithms, including topological sorts and planarity testing. Pseudocode [edit] Input: A graph G and a vertex v of G. Output: A labeling of the edges in the connected component of v as discovery edges and back edges. procedure DFS(G, v) is label v as explored for all edges e in G.incidentEdges(v) do if edge e is unexplored then w ← G.adjacentVertex(v, e) if vertex w is unexplored then label e as a discovered edge recursively call DFS(G, w) else label e as a back edge Breadth-first search [edit] Main article: Breadth-first search \| \| \| This section needs expansion. You can help by adding to it. (October 2012) \| A breadth-first search (BFS) is another technique for traversing a finite graph. BFS visits the sibling vertices before visiting the child vertices, and a queue is used in the search process. This algorithm is often used to find the shortest path from one vertex to another. Pseudocode [edit] Input: A graph G and a vertex v of G. Output: The closest vertex to v satisfying some conditions, or null if no such vertex exists. procedure BFS(G, v) is create a queue Q enqueue v onto Q mark v while Q is not empty do w ← Q.dequeue() if w is what we are looking for then return w for all edges e in G.adjacentEdges(w) do x ← G.adjacentVertex(w, e) if x is not marked then mark x enqueue x onto Q return null Applications [edit] Breadth-first search can be used to solve many problems in graph theory, for example: finding all vertices within one connected component; Cheney's algorithm; finding the shortest path between two vertices; testing a graph for bipartiteness; Cuthill–McKee algorithm mesh numbering; Ford–Fulkerson algorithm for computing the maximum flow in a flow network; serialization/deserialization of a binary tree vs serialization in sorted order (allows the tree to be re-constructed in an efficient manner); maze generation algorithms; flood fill algorithm for marking contiguous regions of a two dimensional image or n-dimensional array; analysis of networks and relationships. Graph exploration [edit] The problem of graph exploration can be seen as a variant of graph traversal. It is an online problem, meaning that the information about the graph is only revealed during the runtime of the algorithm. A common model is as follows: given a connected graph G = (V, E) with non-negative edge weights. The algorithm starts at some vertex, and knows all incident outgoing edges and the vertices at the end of these edges—but not more. When a new vertex is visited, then again all incident outgoing edges and the vertices at the end are known. The goal is to visit all n vertices and return to the starting vertex, but the sum of the weights of the tour should be as small as possible. The problem can also be understood as a specific version of the travelling salesman problem, where the salesman has to discover the graph on the go. For general graphs, the best known algorithms for both undirected and directed graphs is a simple greedy algorithm: In the undirected case, the greedy tour is at most O(ln n)-times longer than an optimal tour. The best lower bound known for any deterministic online algorithm is 10/3. Unit weight undirected graphs can be explored with a competitive ration of 2 − ε, which is already a tight bound on Tadpole graphs. In the directed case, the greedy tour is at most (n − 1)-times longer than an optimal tour. This matches the lower bound of n − 1. An analogous competitive lower bound of Ω(n) also holds for randomized algorithms that know the coordinates of each node in a geometric embedding. If instead of visiting all nodes just a single "treasure" node has to be found, the competitive bounds are Θ(n2) on unit weight directed graphs, for both deterministic and randomized algorithms. Universal traversal sequences [edit] \| \| \| This section needs expansion. You can help by adding to it. (December 2016) \| A universal traversal sequence is a sequence of instructions comprising a graph traversal for any regular graph with a set number of vertices and for any starting vertex. A probabilistic proof was used by Aleliunas et al. to show that there exists a universal traversal sequence with number of instructions proportional to O(n5) for any regular graph with n vertices. The steps specified in the sequence are relative to the current node, not absolute. For example, if the current node is vj, and vj has d neighbors, then the traversal sequence will specify the next node to visit, vj+1, as the ith neighbor of vj, where 1 ≤ i ≤ d. See also [edit] External memory graph traversal References [edit] ^ Rosenkrantz, Daniel J.; Stearns, Richard E.; Lewis, II, Philip M. (1977). "An Analysis of Several Heuristics for the Traveling Salesman Problem". SIAM Journal on Computing. 6 (3): 563–581. doi:10.1137/0206041. S2CID 14764079. ^ Birx, Alexander; Disser, Yann; Hopp, Alexander V.; Karousatou, Christina (May 2021). "An improved lower bound for competitive graph exploration". Theoretical Computer Science. 868: 65–86. arXiv:2002.10958. doi:10.1016/j.tcs.2021.04.003. S2CID 211296296. ^ Miyazaki, Shuichi; Morimoto, Naoyuki; Okabe, Yasuo (2009). "The Online Graph Exploration Problem on Restricted Graphs". IEICE Transactions on Information and Systems. E92-D (9): 1620–1627. Bibcode:2009IEITI..92.1620M. doi:10.1587/transinf.E92.D.1620. hdl:2433/226939. S2CID 8355092. ^ Brandt, Sebastian; Foerster, Klaus-Tycho; Maurer, Jonathan; Wattenhofer, Roger (November 2020). "Online graph exploration on a restricted graph class: Optimal solutions for tadpole graphs". Theoretical Computer Science. 839: 176–185. arXiv:1903.00581. doi:10.1016/j.tcs.2020.06.007. S2CID 67856035. ^ Foerster, Klaus-Tycho; Wattenhofer, Roger (December 2016). "Lower and upper competitive bounds for online directed graph exploration". Theoretical Computer Science. 655: 15–29. doi:10.1016/j.tcs.2015.11.017. ^ Aleliunas, R.; Karp, R.; Lipton, R.; Lovász, L.; Rackoff, C. (1979). "Random walks, universal traversal sequences, and the complexity of maze problems". 20th Annual Symposium on Foundations of Computer Science (SFCS 1979). pp. 218–223. doi:10.1109/SFCS.1979.34. S2CID 18719861. \| v t e Graph and tree traversal algorithms \| \| Search \| α–β pruning A + IDA + LPA + SMA Best-first search Beam search Bidirectional search Breadth-first search + Lexicographic + Parallel B Depth-first search + Iterative deepening D Fringe search Jump point search Monte Carlo tree search SSS \| \| Shortest path \| Bellman–Ford Dijkstra's Floyd–Warshall Johnson's Shortest path faster Yen's \| \| Minimum spanning tree \| Borůvka's Kruskal's Prim's Reverse-delete \| \| List of graph search algorithms \| \| v t e Data structures and algorithms \| \| Data structures \| Array Associative array Binary search tree Fenwick tree Graph Hash table Heap Linked list Queue Segment tree Stack String Tree Trie \| \| Algorithms and algorithmic paradigms \| Backtracking Binary search Breadth-first search Brute-force search Depth-first search Divide and conquer Dynamic programming Graph traversal Fold Greedy Hash function Minimax Online Randomized Recursion Root-finding Sorting Streaming Sweep line String-searching Topological sorting \| \| List of data structures List of algorithms \| Retrieved from " Category: Graph algorithms Hidden categories: CS1: long volume value Articles with short description Short description is different from Wikidata Articles needing additional references from October 2014 All articles needing additional references Articles to be expanded from October 2012 All articles to be expanded Articles to be expanded from December 2016 Articles with example pseudocode Graph traversal Add topic
188323	https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/10-6-torque/	10 Fixed-Axis Rotation 10.6 Torque Learning Objectives By the end of this section, you will be able to: Describe how the magnitude of a torque depends on the magnitude of the lever arm and the angle the force vector makes with the lever arm Determine the sign (positive or negative) of a torque using the right-hand rule Calculate individual torques about a common axis and sum them to find the net torque An important quantity for describing the dynamics of a rotating rigid body is torque. We see the application of torque in many ways in our world. We all have an intuition about torque, as when we use a large wrench to unscrew a stubborn bolt. Torque is at work in unseen ways, as when we press on the accelerator in a car, causing the engine to put additional torque on the drive train. Or every time we move our bodies from a standing position, we apply a torque to our limbs. In this section, we define torque and make an argument for the equation for calculating torque for a rigid body with fixed-axis rotation. Defining Torque So far we have defined many variables that are rotational equivalents to their translational counterparts. Let’s consider what the counterpart to force must be. Since forces change the translational motion of objects, the rotational counterpart must be related to changing the rotational motion of an object about an axis. We call this rotational counterpart torque. In everyday life, we rotate objects about an axis all the time, so intuitively we already know much about torque. Consider, for example, how we rotate a door to open it. First, we know that a door opens slowly if we push too close to its hinges; it is more efficient to rotate a door open if we push far from the hinges. Second, we know that we should push perpendicular to the plane of the door; if we push parallel to the plane of the door, we are not able to rotate it. Third, the larger the force, the more effective it is in opening the door; the harder you push, the more rapidly the door opens. The first point implies that the farther the force is applied from the axis of rotation, the greater the angular acceleration; the second implies that the effectiveness depends on the angle at which the force is applied; the third implies that the magnitude of the force must also be part of the equation. Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point. (Figure) shows counterclockwise rotations. Figure 10.31 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) A counterclockwise torque is produced by a force $$ \overset{\to }{F} $$ acting at a distance r from the hinges (the pivot point). (b) A smaller counterclockwise torque is produced when a smaller force $$ {\overset{\to }{F}}^{\text{′}} $$ acts at the same distance r from the hinges. (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) A smaller counterclockwise torque is produced by the same magnitude force as (a) acting at the same distance as (a) but at an angle $$ \theta $$ that is less than $$ 90\text{°}$$. Now let’s consider how to define torques in the general three-dimensional case. When a force $$ \overset{\to }{F} $$ is applied to a point P whose position is $$ \overset{\to }{r} $$ relative to O ((Figure)), the torque $$ \overset{\to }{\tau } $$ around O is $$\overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F}.$$ Figure 10.32 The torque is perpendicular to the plane defined by $$ \overset{\to }{r}\,\text{and}\,\overset{\to }{F} $$ and its direction is determined by the right-hand rule. From the definition of the cross product, the torque $$ \overset{\to }{\tau } $$ is perpendicular to the plane containing $$ \overset{\to }{r}\,\text{and}\,\overset{\to }{F} $$ and has magnitude $$\|\overset{\to }{\tau }\|=\|\overset{\to }{r}\,×\,\overset{\to }{F}\|=rF\text{sin}\,\theta ,$$ where $$ \theta $$ is the angle between the vectors $$ \overset{\to }{r} $$ and $$ \overset{\to }{F}$$. The SI unit of torque is newtons times meters, usually written as $$ \text{N}·\text{m}$$. The quantity $$ {r}_{\perp }=r\text{sin}\,\theta $$ is the perpendicular distance from O to the line determined by the vector $$ \overset{\to }{F} $$ and is called the lever arm. Note that the greater the lever arm, the greater the magnitude of the torque. In terms of the lever arm, the magnitude of the torque is $$\|\overset{\to }{\tau }\|={r}_{\perp }F.$$ The cross product $$ \overset{\to }{r}\,×\,\overset{\to }{F} $$ also tells us the sign of the torque. In (Figure), the cross product $$ \overset{\to }{r}\,×\,\overset{\to }{F} $$ is along the positive z-axis, which by convention is a positive torque. If $$ \overset{\to }{r}\,×\,\overset{\to }{F} $$ is along the negative z-axis, this produces a negative torque. If we consider a disk that is free to rotate about an axis through the center, as shown in (Figure), we can see how the angle between the radius $$ \overset{\to }{r} $$ and the force $$ \overset{\to }{F} $$ affects the magnitude of the torque. If the angle is zero, the torque is zero; if the angle is $$ 90\text{°}$$, the torque is maximum. The torque in (Figure) is positive because the direction of the torque by the right-hand rule is out of the page along the positive z-axis. The disk rotates counterclockwise due to the torque, in the same direction as a positive angular acceleration. Figure 10.33 A disk is free to rotate about its axis through the center. The magnitude of the torque on the disk is $$ rF\text{sin}\,\theta $$.When $$ \theta =0\text{°}$$, the torque is zero and the disk does not rotate. When $$ \theta =90\text{°}$$, the torque is maximum and the disk rotates with maximum angular acceleration. Any number of torques can be calculated about a given axis. The individual torques add to produce a net torque about the axis. When the appropriate sign (positive or negative) is assigned to the magnitudes of individual torques about a specified axis, the net torque about the axis is the sum of the individual torques: $${\overset{\to }{\tau }}_{\text{net}}=\sum _{i}\|{\overset{\to }{\tau }}_{i}\|.$$ Calculating Net Torque for Rigid Bodies on a Fixed Axis In the following examples, we calculate the torque both abstractly and as applied to a rigid body. We first introduce a problem-solving strategy. Problem-Solving Strategy: Finding Net Torque Choose a coordinate system with the pivot point or axis of rotation as the origin of the selected coordinate system. Determine the angle between the lever arm $$ \overset{\to }{r} $$ and the force vector. Take the cross product of $$ \overset{\to }{r}\,\text{and}\,\overset{\to }{F} $$ to determine if the torque is positive or negative about the pivot point or axis. Evaluate the magnitude of the torque using $$ {r}_{\perp }F$$. Assign the appropriate sign, positive or negative, to the magnitude. Sum the torques to find the net torque. Example Calculating Torque Four forces are shown in (Figure) at particular locations and orientations with respect to a given xy-coordinate system. Find the torque due to each force about the origin, then use your results to find the net torque about the origin. Figure 10.34 Four forces producing torques. Strategy This problem requires calculating torque. All known quantities––forces with directions and lever arms––are given in the figure. The goal is to find each individual torque and the net torque by summing the individual torques. Be careful to assign the correct sign to each torque by using the cross product of $$ \overset{\to }{r} $$ and the force vector $$ \overset{\to }{F}$$. Solution Use $$ \|\overset{\to }{\tau }\|={r}_{\perp }F=rF\text{sin}\,\theta $$ to find the magnitude and $$ \overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F} $$ to determine the sign of the torque. The torque from force 40 N in the first quadrant is given by $$ (4)(40)\text{sin}\,90\text{°}=160\,\text{N}·\text{m}$$. The cross product of $$ \overset{\to }{r} $$ and $$ \overset{\to }{F} $$ is out of the page, positive. The torque from force 20 N in the third quadrant is given by$$\text{−}(3)(20)\text{sin}\,90\text{°}=-60\,\text{N}·\text{m}$$. The cross product of $$ \overset{\to }{r} $$ and $$ \overset{\to }{F} $$ is into the page, so it is negative. The torque from force 30 N in the third quadrant is given by $$ (5)(30)\text{sin}\,53\text{°}=120\,\text{N}·\text{m}$$. The cross product of $$ \overset{\to }{r} $$ and $$ \overset{\to }{F} $$ is out of the page, positive. The torque from force 20 N in the second quadrant is given by $$ (1)(20)\text{sin}\,30\text{°}=10\,\text{N}·\text{m}$$. The cross product of $$ \overset{\to }{r} $$ and $$ \overset{\to }{F} $$ is out of the page. The net torque is therefore $$ {\tau }_{\text{net}}=\sum _{i}\|{\tau }_{i}\|=160-60+120+10=230\,\text{N}·\text{m}\text{.}$$ Significance Note that each force that acts in the counterclockwise direction has a positive torque, whereas each force that acts in the clockwise direction has a negative torque. The torque is greater when the distance, force, or perpendicular components are greater. Example Calculating Torque on a rigid body(Figure) shows several forces acting at different locations and angles on a flywheel. We have $$ \|{\overset{\to }{F}}_{1}\|=20\,\text{N}, $$ $$\|{\overset{\to }{F}}_{2}\|=30\,\text{N}$$, $$ \|{\overset{\to }{F}}_{3}\|=30\,\text{N}$$, and $$ r=0.5\,\text{m}$$. Find the net torque on the flywheel about an axis through the center. Figure 10.35 Three forces acting on a flywheel. Strategy We calculate each torque individually, using the cross product, and determine the sign of the torque. Then we sum the torques to find the net torque. Solution We start with $$ {\overset{\to }{F}}_{1}$$. If we look at (Figure), we see that $$ {\overset{\to }{F}}_{1} $$ makes an angle of $$ 90\text{°}+60\text{°} $$ with the radius vector $$ \overset{\to }{r}$$. Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude: $$\|{\overset{\to }{\tau }}_{1}\|=r{F}_{1}\text{sin}\,150\text{°}=0.5\,\text{m}(20\,\text{N})(0.5)=5.0\,\text{N}·\text{m}.$$ Next we look at $$ {\overset{\to }{F}}_{2}$$. The angle between $$ {\overset{\to }{F}}_{2} $$ and $$ \overset{\to }{r} $$ is $$ 90\text{°} $$ and the cross product is into the page so the torque is negative. Its value is $$\|{\overset{\to }{\tau }}_{2}\|=\text{−}r{F}_{2}\text{sin}\,90\text{°}=-0.5\,\text{m}(30\,\text{N})=-15.0\,\text{N}·\text{m}.$$ When we evaluate the torque due to $$ {\overset{\to }{F}}_{3}$$, we see that the angle it makes with $$ \overset{\to }{r} $$ is zero so $$ \overset{\to }{r}\,×\,{\overset{\to }{F}}_{3}=0. $$ Therefore, $$ {\overset{\to }{F}}_{3} $$ does not produce any torque on the flywheel. We evaluate the sum of the torques: $${\tau }_{\text{net}}=\sum _{i}\|{\tau }_{i}\|=5-15=-10\,\text{N}·\text{m}.$$ Significance The axis of rotation is at the center of mass of the flywheel. Since the flywheel is on a fixed axis, it is not free to translate. If it were on a frictionless surface and not fixed in place, $$ {\overset{\to }{F}}_{3} $$ would cause the flywheel to translate, as well as $$ {\overset{\to }{F}}_{1}$$. Its motion would be a combination of translation and rotation. Check Your Understanding A large ocean-going ship runs aground near the coastline, similar to the fate of the Costa Concordia, and lies at an angle as shown below. Salvage crews must apply a torque to right the ship in order to float the vessel for transport. A force of $$ 5.0\,×\,{10}^{5}\,\text{N} $$ acting at point A must be applied to right the ship. What is the torque about the point of contact of the ship with the ground ((Figure))? Figure 10.36 A ship runs aground and tilts, requiring torque to be applied to return the vessel to an upright position. Show Answer The angle between the lever arm and the force vector is $$ 80\text{°}; $$ therefore, $$ {r}_{\perp }=100\text{m(sin80}\text{°})=98.5\,\text{m}$$. The cross product $$ \overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F} $$ gives a negative or clockwise torque. The torque is then $$ \tau =\text{−}{r}_{\perp }F=-98.5\,\text{m}(5.0\,×\,{10}^{5}\text{N})=-4.9\,×\,{10}^{7}\text{N}·\text{m}$$. Summary The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation $$ \|\overset{\to }{\tau }\|={r}_{\perp }F$$, where $$ {r}_{\perp } $$ is the perpendicular distance from the axis to the line upon which the force vector lies. The sign of the torque is found using the right hand rule. If the page is the plane containing $$ \overset{\to }{r} $$ and $$ \overset{\to }{F}$$, then $$ \overset{\to }{r}\,×\,\overset{\to }{F} $$ is out of the page for positive torques and into the page for negative torques. The net torque can be found from summing the individual torques about a given axis. Conceptual Questions What three factors affect the torque created by a force relative to a specific pivot point? Show Solution magnitude of the force, length of the lever arm, and angle of the lever arm and force vector Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque. When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame? Show Solution The moment of inertia of the wheels is reduced, so a smaller torque is needed to accelerate them. Can a single force produce a zero torque? Can a set of forces have a net torque that is zero and a net force that is not zero? Show Solution yes Can a set of forces have a net force that is zero and a net torque that is not zero? In the expression $$ \overset{\to }{r}\,×\,\overset{\to }{F} $$ can $$ \|\overset{\to }{r}\| $$ ever be less than the lever arm? Can it be equal to the lever arm? Show Solution $$\|\overset{\to }{r}\| $$ can be equal to the lever arm but never less than the lever arm Problems Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller flywheel of radius 30 cm has a cord that has a pulling force of 50 N on it. What pulling force needs to be applied to the cord connecting the larger flywheel of radius 50 cm such that the combination does not rotate? Show Answer $$F=30\,\text{N}$$ The cylindrical head bolts on a car are to be tightened with a torque of 62.0 N$$·\text{m}$$. If a mechanic uses a wrench of length 20 cm, what perpendicular force must he exert on the end of the wrench to tighten a bolt correctly? (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? There is only one pair of hinges. Show Solution a. $$ 0.85\,\text{m}(55.0\,\text{N})=46.75\,\text{N}·\text{m}$$; b. It does not matter at what height you push. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. How much torque are you exerting in newton-meters (relative to the center of the bolt)? What hanging mass must be placed on the cord to keep the pulley from rotating (see the following figure)? The mass on the frictionless plane is 5.0 kg. The inner radius of the pulley is 20 cm and the outer radius is 30 cm. Show Answer $${m}_{2}=\frac{4.9\,\text{N}·\text{m}}{9.8(0.3\,\text{m})}=1.67\,\text{kg}$$ A simple pendulum consists of a massless tether 50 cm in length connected to a pivot and a small mass of 1.0 kg attached at the other end. What is the torque about the pivot when the pendulum makes an angle of $$ 40\text{°} $$ with respect to the vertical? Calculate the torque about the z-axis that is out of the page at the origin in the following figure, given that $$ {F}_{1}=3\,\text{N},\enspace{F}_{2}=2\,\text{N},\enspace{F}_{3}=3\,\text{N},\enspace{F}_{4}=1.8\,\text{N}$$. Show Answer $${\tau }_{net}=-9.0\,\text{N}·\text{m}+3.46\,\text{N}·\text{m}+0-3.28\,\text{N}·\text{m}=-8.82\,\text{N}·\text{m}$$ A seesaw has length 10.0 m and uniform mass 10.0 kg and is resting at an angle of $$ 30\text{°} $$ with respect to the ground (see the following figure). The pivot is located at 6.0 m. What magnitude of force needs to be applied perpendicular to the seesaw at the raised end so as to allow the seesaw to barely start to rotate? A pendulum consists of a rod of mass 1 kg and length 1 m connected to a pivot with a solid sphere attached at the other end with mass 0.5 kg and radius 30 cm. What is the torque about the pivot when the pendulum makes an angle of $$ 30\text{°} $$ with respect to the vertical? Show Solution $$\tau =5.66\,\text{N}·\text{m}$$ A torque of $$ 5.00\,×\,{10}^{3}\text{N}·\text{m}\, $$ is required to raise a drawbridge (see the following figure). What is the tension necessary to produce this torque? Would it be easier to raise the drawbridge if the angle $$ \theta $$ were larger or smaller? A horizontal beam of length 3 m and mass 2.0 kg has a mass of 1.0 kg and width 0.2 m sitting at the end of the beam (see the following figure). What is the torque of the system about the support at the wall? Show Answer $$\sum \tau =57.82\,\text{N}·\text{m}$$ What force must be applied to end of a rod along the x-axis of length 2.0 m in order to produce a torque on the rod about the origin of $$ 8.0\hat{k}\,\text{N}·\text{m}$$? What is the torque about the origin of the force $$ (5.0\hat{i}-2.0\hat{j}+1.0\hat{k})\,\text{N} $$ if it is applied at the point whose position is: $$ \overset{\to }{r}=(-2.0\hat{i}+4.0\hat{j})\,\text{m?}$$ Show Solution $$\overset{\to }{r}\,×\,\overset{\to }{F}=4.0\hat{i}+2.0\hat{j}-16.0\hat{k}\text{N}·\text{m}$$ Glossary lever arm : perpendicular distance from the line that the force vector lies on to a given axis torque : cross product of a force and a lever arm to a given axis Candela Citations CC licensed content, Shared previously OpenStax University Physics. Authored by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at Licenses and Attributions CC licensed content, Shared previously OpenStax University Physics. 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188324	https://www.frontiersin.org/journals/neuroscience/articles/10.3389/fnins.2020.599812/full	Your new experience awaits. Try the new design now and help us make it even better ORIGINAL RESEARCH article Front. Neurosci., 18 November 2020 Sec. Neuropharmacology Volume 14 - 2020 \| This article is part of the Research TopicPharmacological Aspects of Ligand-gated Ion Channels as Targets of Natural and Synthetic AgentsView all 18 articles The Z-Drugs Zolpidem, Zaleplon, and Eszopiclone Have Varying Actions on Human GABAA Receptors Containing γ1, γ2, and γ3 Subunits Grant RichterVivian W. Y. LiaoPhilip K. AhringMary Chebib Brain and Mind Centre, Sydney Pharmacy School, The University of Sydney, Sydney, NSW, Australia γ-Aminobutyric-acid type A (GABAA) receptors expressing the γ1 or γ3 subunit are only found within a few regions of the brain, some of which are involved in sleep. No known compounds have been reported to selectively target γ1- or γ3-containing GABAA receptors. Pharmacological assessments of this are conflicting, possibly due to differences in experimental models, conditions, and exact protocols when reporting efficacies and potencies. In this study, we evaluated the modulatory properties of five non-benzodiazepine Z-drugs (zaleplon, indiplon, eszopiclone, zolpidem, and alpidem) used in sleep management and the benzodiazepine, diazepam on human α1β2γ receptors using all three γ subtypes. This was accomplished using concatenated GABAA pentamers expressed in Xenopus laevis oocytes and measured via two-electrode voltage clamp. This approach removes the potential for single subunits to form erroneous receptors that could contribute to the pharmacological assessment of these compounds. No compound tested had significant effects on γ1-containing receptors below 10 μM. Interestingly, zaleplon and indiplon were found to modulate γ3-containing receptors equally as efficacious as γ2-containing receptors. Furthermore, zaleplon had a higher potency for γ3- than for γ2-containing receptors, indicating certain therapeutic effects could occur via these γ3-containing receptors. Eszopiclone modulated γ3-containing receptors with reduced efficacy but no reduction in potency. These data demonstrate that the imidazopyridines zaleplon and indiplon are well suited to further investigate potential γ3 effects on sleep in vivo. Introduction γ-Aminobutyric-acid type A (GABAA) receptors are ligand-gated ion channels that mediate most inhibitory responses in the brain. These receptors are made up of five building block subunits, and in mammals, there are nineteen identified subunits, α1-6, β1-3, γ1-3, δ, ε, θ, π, and ρ1-3 (Sieghart and Savic, 2018). Receptors typically form from two α, two β, and one of either γ or δ with the most widely expressed combination made from two α1, two β2/3, and one γ2, denoted as α1β2/3γ2 (Olsen and Sieghart, 2008). Distinctive GABAA receptor subtypes are found based on their cellular and anatomical locations and behave differently in response to agonists and modulating compounds. Each GABAA subunit contains a principal side (+) and a complimentary side (−). GABA binding within the β(+) and α(−) interface induces a conformational change in the receptor channel allowing Cl– ions to pass into the cell to hyperpolarize neurons and make action potentials less likely (Figure 1). Benzodiazepines and Z-drugs allosterically modulate GABAA receptors making the frequency of Cl– channel opening more likely. These drugs bind to the interface within the α(+) and γ(−) (Sigel and Buhr, 1997; Zhu et al., 2018) to reduce the brain’s excitability and thus are primarily prescribed for their effects as anxiolytics, hypnotics, anti-epileptics, and muscle relaxants. Z-drugs are the most commonly prescribed treatment for insomnia and compared with benzodiazepines they more closely induce normal physiological sleep (Klimm et al., 1987; Fleming et al., 1988). However, it is still not precisely characterized which regions Z-drugs act on to induce sleep. FIGURE 1 Figure 1. A schematic diagram of a concatenated pentamer GABAA receptor construct. Linkers concatenating subunits are shown as arrows. The GABA binding site (orange arrowhead) is shown between the β2(+) and α(–) interfaces and the benzodiazepine binding site (gray arrowhead) is between the α1(+) and γ(–) interfaces. While three γ subunits exist, the actions of benzodiazepines and Z-drugs have typically been associated with the γ2 subunit with little information available for γ1 and γ3 subunits. The γ1 or γ3-subunits are found in at most 10 or 15% of GABAA receptors, respectively (Quirk et al., 1994; Benke et al., 1996; Sieghart and Sperk, 2002). Temporally, the γ2 subunit is expressed throughout all stages of development, while γ1 subunit expression peaks around birth and γ3 subunit expression peaks in 2-week old animals (Laurie et al., 1992; Allen Institute for Brain Science, 2008) GABAA receptors with a γ1 subunit have been detected mainly in the amygdala, basal ganglia, hypothalamus, thalamus, and in astrocytes, while receptors with γ3 subunits show some expression in the basal ganglia, thalamus, and midbrain (Bovolin et al., 1992; Quirk et al., 1994; Pirker et al., 2000; Sieghart and Sperk, 2002; Hertz and Chen, 2010). The thalamus and hypothalamus regions are intricately involved in the maintenance of the sleep-wake cycle (Gent et al., 2018) and Z-drugs have been shown to affect clusters of nuclei in these regions (Jia et al., 2009; Kumar et al., 2011; Uygun et al., 2016). Hence, it is a genuine possibility that γ1- or γ3-containing receptors could also play a role in the hypnotic effects of Z-drugs. Indeed the interface between α(+)/γ(−) is believed to be sensitive to benzodiazepine binding in both γ1 and γ3 containing receptors, though some ligands might have lower potencies and/or efficacies because of amino acid sequence differences (Knoflach et al., 1991; Sieghart, 1995; Khom et al., 2006). Although some Z-drugs have been evaluated on γ1 or γ3-containing GABAA receptors, it is difficult to conclude any clear effects mediated from these subunits as there is conflicting literary evidence of the modulative ability of Z-drugs. This may be due to differences in experimental models, conditions, and exact protocols reported for efficacy and potencies of these compounds. Furthermore, studies that utilize Xenopus laevis oocytes to investigate the pharmacology of Z-drugs have conflicting results potentially due to using single subunit cRNAs to express recombinant receptors. Using single subunit cRNAs in a heterologous expression system can potentially result in a mix of receptor populations. For example, if unlinked cRNAs for α1, β2, and γ subunits are injected into a cell, there is potential for GABAA receptors to assemble from only α1 and β2 with two different stoichiometries [i.e., (α1)2(β2)3 or (α1)3(β2)2], potentially confounding results. Therefore, our group has recently optimized receptor concatenation technology to ensure a single receptor subtype population with assembly in the correct orientation (Liao et al., 2019). In the present study, we systematically evaluated the pharmacology of five Z-drugs including the pyrazolopyrimidines (zaleplon and indiplon), cyclopyrrolones (zopiclone and its isolated S-enantiomer eszopiclone), and imidazopyridines (zolpidem and alpidem), along with diazepam on γ1, γ2, and γ3 concatenated pentameric GABAA receptors (Figure 2). We found that zaleplon, indiplon, and eszopiclone show comparable efficacy and potency on γ3 as γ2-containing receptors. Furthermore, zolpidem and alpidem modulate γ2 receptors with exclusive selectivity at concentrations below 10 μM. These data clarify conflicting observations and provide further insight into the receptor subtype populations targeted by Z-drugs, and identifies zaleplon, indiplon, and possibly eszopiclone as useful tools for further studies that understand the role γ3-containing receptors in sleep. FIGURE 2 Figure 2. Chemical structures and classes of the drugs used in this study. Materials and Methods Materials GABA, diazepam, alpidem, and all salts and chemicals not specifically mentioned were purchased from Sigma-Aldrich. Zolpidem was purchased from Chemieliva (Yubei District, Chongqing, China), zaleplon was purchased from Alomone Labs (Jerusalem, Israel), eszopiclone was purchased from Clearsynth (NJ, United States), and indiplon was purchased from Tocris (VIC, Australia). Human cDNA for α1 β2, γ1,2,3 GABAA receptor subunits were gifts from Saniona A/S. Oligonucleotides were purchased from Sigma-Aldrich. Restriction enzymes, Q5 polymerase, T4 DNA ligase, and 10-beta competent Escherichia coli were from New England Biolabs (Ipswich, MA, United States). Collagenase A was purchased from Roche (Basel, Switzerland). DNA purification kits were from Qiagen (Hilden, Germany). The QuickChange II Site-Directed Mutagenesis Kit was from Agilent Technologies (Santa Clara, CA, United States). The mMessage mMachine T7 transcription kit–were purchased from Thermo Fisher Scientific (Waltham, MA, United States). Molecular Biology To ensure homogenous receptor populations and subunit orientation, we used concatenated receptors expressed in X. laevis oocytes. Concatenated pentameric constructs were created using the subunits γx-β2-α1-β2-α1 (where x = 1, 2, or 3). A detailed description of the creation of concatenated receptor constructs has been previously described (Liao et al., 2019). Briefly, natural restriction sites BamHI, HindIII, and KpnI restriction sites in the γ1, 2, 3, β2, or α1 subunits were removed through silent mutations using site-directed mutagenesis. Linker sequences of 13 amino acids inserted between the natural C-terminal in the transmembrane segment 4 of the γ subunit and the N-terminal leucine anchor of the β subunit through standard PCR reactions and subunit cDNA ligated together (corresponding to a total linker length of 28 total amino acids between subunits) This was found to be an optimal length for relatively pure receptor expression and orientation without compromising function (Liao et al., 2019). Inserted linker lengths are as follows γx-13a-β2-27a-α1-18a-β2-27a-α1. E. coli bacteria were hosts for plasmid amplification and plasmid purification was performed using standard kits. RNA was produced from DNA using the mMessage mMachine T7 Transcription kit (Thermo Fisher Scientific, Waltham, MA, United States), but due to the size of the pentameric constructs (>10 kb), guanosine triphosphate concentration was increased to give a final cap analog to guanosine triphosphate ratio of 2:1. Expression of GABAA Receptors in X. laevis Oocytes The collection and preparation of oocytes were done as previously described (Ahring et al., 2016). Briefly, ovarian lobes were removed from anesthetized adult X. laevis following protocol approval by the Animal Ethics Committee of The University of Sydney (AEC No. 2016/970) in accordance with the National Health and Medical Research Council of Australia code for the care and use of animals. Oocytes were prepared by slicing lobes into small pieces and defolliculated through collagenase A treatment. Stage V and VI oocytes were injected with around 50 nL of 0.5 ng/nL RNA for each concatenated construct or α1/β2 subunits in a 1:1 ratio and incubated for 3–4 days at 18°C in modified Barth’s solution (96 mM NaCl, 2.0 mM KCl, 1 mM MgCl2, 1.8 mM CaCl2, 5 mM HEPES, 2.5 mM sodium pyruvate, 0.5 mM theophylline, and 100 μg/mL gentamicin; pH 7.4). Electrophysiological Recordings Using Two-Electrode Voltage Clamp This technique was performed as previously described (Ahring et al., 2016, 2018; Liao et al., 2019). Briefly, oocytes sit in a custom-built chamber and continuously perfused with a saline solution, ‘ND96’ (96 mM NaCl, 2 mM KCl, 1 mM MgCl2, 1.8 mM CaCl2, and 10 mM HEPES; pH 7.4). Glass electrode pipettes were filled with 3 M KCl, with resistances ranging from 0.4 to 2 MΩ. Oocytes were clamped to −60 mV using an Axon GeneClamp 500B amplifier (Molecular Devices). Currents were filtered at 20 Hz with a four-pole low pass Bessel filter (Axon GeneClamp 500 B) and digitized by a Digidata 1440A (Molecular Devices). Sampling was taken at 200 Hz and analyzed using pClamp 10.2 suite (Molecular Devices). Stock solutions of 3.16 M GABA in ultrapure water and drug solutions of 100 mM in DMSO were stored at −20°C and aliquoted to avoid repeated freeze-thaw cycles. Each recording day, a fresh stock was used to prepare dilutions. The maximal concentration of DMSO in final drug ND96 solutions was <0.1%. Experimental Design GABA concentration-response curves were determined for each construct as follows. To ensure RNA expression and reproducibility, a set of control applications were first applied consisting of three applications of 40 μM GABA, one 316 μM application, and three more 40 μM applications. After this, ten solutions of GABA each increasing in concentration by a factor of 3.16 were used starting with 100 nM and ending with 3.16 mM. Applications lasted for 30 s and were followed by 2–5 min of washout. EC50 and EC10 were calculated from this curve. The drug modulation experiments were done as follows. Like the GABA dose-response curves, first, a set of three control applications were run consisting of GABA EC10, then a maximal response of GABA 3.16 mM, followed by three more GABA EC10. Before the application of modulators, EC10 was confirmed by comparing the ratio of the current of the last control application to the maximal response current. For each drug, 6 concentrations increasing by a factor of 10, ranging from 0.1 nM to 10 μM, were co-applied with GABA EC10 for 30 s followed by 2–5 min of washout. Data and Statistical Analysis The final dataset was from a minimum of four experiments and a minimum of two different X. laevis donors. Raw traces were analyzed using pClamp 10.2. Episodic traces for each application were overlaid and the baseline was subtracted. Peak current amplitude was quantified by measuring maximum inward current for each response. Peak current amplitudes (I) were fitted to the Hill equation and normalized to the maximal fitted response (Imax). The calculated Emax response is expressed as a percentage of the current obtained through GABA EC10 (actual GABA control percentage for each experiment is listed in Table 1). The Emax response and EC50 values were calculated by using non-linear regression to fit the data to the Hill equation in a monophasic model with three variables (top, bottom, EC50) using GraphPad Prism 8. Efficacy at infinitely low compound concentration was set to 0, and the slope was constrained to 1. For GABA concentration-response curves, the slope was unconstrained and listed in Table 1. Means are reported ± one SD. To compare differences in Emax response, EC50 values within drug groups and across γ3 and γ2 receptors, or to compare γ1/3 receptor responses at 10 μM with binary α1β2 receptors, one-way ANOVAs were run with Sidak multiple comparisons test. F tests, respectively, are [F (7, 48) = 32.77, p < 0.0001] and [F (7, 48) = 75.69, p < 0.0001], and [F (14,57) = 48.68, p < 0.0001]. All reported statistically significant comparisons within the results section are p < 0.01. TABLE 1 Table 1. Potency and efficacy of Z-drugs on GABAA receptors with varying γ subunits. Results GABA Response of Concatenated γ1-, γ2-, and γ3-Containing GABAA Receptors To ensure homogenous receptor populations and subunit orientation, we used concatenated receptors expressed in X. laevis oocytes. Concatenated pentameric constructs were created using the subunits γ-β2-α1-β2-α1 (where γ = γ1, γ2, or γ3). Subunits were linked with artificial linker sequences optimized to give relatively pure receptor expression and orientation without compromising function (Liao et al., 2019). We first measured the concentration-response for GABA on each construct (Figure 3A). Upon visual inspection of representative traces (Figure 3B), γ1 and γ2 receptors presented similar current decay profiles at the highest GABA concentrations while the γ3 receptor showed a shorter current decay time. This could indicate that γ3 receptors undergo a higher degree of desensitization upon prolonged GABA exposure than the γ1 and γ2 receptor counterparts. FIGURE 3 Figure 3. (A) Normalized GABA concentration-response curves of α1β2γx receptors (x = 1,2,3) expressed as concatenated pentamers in Xenopus laevis oocytes measured via two-electrode voltage clamp. Oocytes were injected with 50 nL of 0.5 ng/nL cRNA for each concatenated construct. Datapoints are depicted as means ± SD (n = 12–14). Data were fitted by non-linear regression to the Hill equation with an unconstrained Hill slope. Log(EC50) and Hill slope parameters are as follows; γ1 Log(EC50) = –3.82 ± 0.07 Hill slope = 1.22 ± 0.32, γ2 Log(EC50) = –3.96 ± 0.10 Hill slope = 1.19 ± 0.55, γ3 Log(EC50) = –4.38 ± 0.07 Hill slope = 1.36 ± 0.63. (B) Representative traces of each construct with indicated concatenated subunit combination. Application bars designate 30 s application time and concentrations of GABA are indicated at the peak of each trace. The three receptor subtypes presented EC50 values in the range of 42–153 μM with γ3 being the most sensitive and γ1 being the least sensitive to GABA. The value for the γ2-containing concatenated receptor (EC50 of 111 μM) is in good agreement with Liao et al. (2019). Previously reported GABA EC50 values using single subunit injections of GABAA γ1, 2, 3 cRNAs in X. laevis oocytes show substantial variations in obtained GABA potencies ranging from 5–100 μM, but generally, γ3 receptors appear more sensitive to GABA (Knoflach et al., 1991; Wafford et al., 1993; Ebert et al., 1994; Khom et al., 2006; Esmaeili et al., 2009). Comparing the Efficacy of Modulators Between GABAA Receptor Subtypes Positive allosteric modulators work by increasing the open-state probability of a receptor in the presence of an endogenous ligand (GABA). If the receptor is already at its maximal open-state probability, then the modulator will have no additional effect. For all γ-containing GABAA receptors, applications of high concentrations of GABA (>1 mM) are typically able to reach activation levels close to the maximal open-state probability, hence, modulators show no efficacy under conditions with high GABA concentrations. Whereas allosteric modulators, by definition, should not gate the receptor in the absence of GABA, substantial modulatory efficacies can be observed as GABA concentrations are lowered toward zero. Therefore, any efficacy of modulators described in percent will depend entirely on the selected concentration of the endogenous ligand. Low concentrations of GABA co-applied with modulators will yield large modulatory percent changes. Conversely, higher concentrations of GABA co-applied with modulators give small percent changes. For our experiments, we selected to co-apply modulators with a GABAcontrol concentration that yields 10% of the maximum response (EC10) at the given receptor. Modulator efficacy is reported as a percent change of evoked current amplitude relative to the GABAcontrol application alone. To directly compare modulator efficacy across different receptors, it was critical that each experiment is run as close as possible to the EC10 of that receptor subtype. Due to GABA potency variations both between batches of oocytes and between individual oocytes, each experiment began with a full GABA concentration-response to determine EC10. Then for each oocyte, a set of 3 control applications at EC10 followed by a max GABA application, followed by three more EC10 applications were applied to confirm that the chosen GABAcontrol concentration yielded ∼10% of the maximum response. Any oocytes responding outside this narrow range (10% ± 5) were discarded before continuing with modulator experiments. GABAcontrol variation is reported in Table 1. Modulatory Potency and Efficacy of Z-Drugs and Diazepam on GABAA γ1-, γ2-, and γ3-Containing Receptors We examined the modulatory effects of the non-benzodiazepines ‘Z-drugs’ (zaleplon, indiplon, eszopiclone, zolpidem, and alpidem) and the benzodiazepine, diazepam on GABAA receptors with varying γ subunits (Figures 4A–C). Representative traces for each compound and receptor subtype are shown in Figures 4D–F. Concentrations ranging from 10–10 to 10–5 M were co-applied with GABA EC10. Full experimental results with Log(EC50) ± SD are listed in Table 1. Unless stated otherwise, all reported statistically significant comparisons have a p < 0.01. FIGURE 4 Figure 4. Modulatory actions of zaleplon, indiplon, eszopiclone, diazepam, zolpidem, and alpidem, on GABA evoked Cl– currents measured in human (A) α1β2γ1, (B) α1β2γ2, and (C) α1β2γ3 GABAA receptors expressed in Xenopus laevis oocytes measured via two-electrode voltage-clamp. The data are expressed as a percentage potentiation of GABA EC10 and are means ± SD (n = 4–8 from at least 2 separate Xenopus laevis donors). Data points were fitted to the Hill equation with bottom set to 0 and slope constrained to 1. (D–F) Representative traces illustrating modulator concentration-response experiments. Pyrazolopyrimidines The pyrazolopyrimidines, zaleplon and indiplon, showed a reverse potency preference for γ2 and γ3 receptors (Figures 5A,B). Zaleplon had a ∼4-fold greater potency at γ3 receptors compared with γ2 (EC50 of approximately 50 vs. 200 nM), while indiplon had a ∼4-fold greater potency for γ2 vs. γ3 (10 vs. 45 nM). Neither compound had statistically significant different efficacies when γ2 was replaced by γ3 with Emax both in the range of ∼250–300%. On γ1 receptors, neither compound showed sufficient potency to enable fitting to the Hill equation within the concentration range tested. At the highest concentration applied (10 μM), zaleplon elicited a modulatory response of 125% and indiplon, 20%. FIGURE 5 Figure 5. Modulatory actions of (A) zaleplon, (B) indiplon, (C) eszopiclone, (D) diazepam, (E) zolpidem, and (F) alpidem, on GABA evoked Cl– currents measured in human α1β2γ1, α1β2γ2, and α1β2γ3 GABAA receptors expressed in Xenopus laevis oocytes measured via two-electrode voltage-clamp. The data are expressed as a percentage potentiation of GABA EC10 and are means ± SD (n = 4–8 from at least 2 separate Xenopus laevis donors). Data points were fitted to the Hill equation with bottom set to 0 and slope constrained to 1. In support of these findings, previous competitive binding studies using Ro15-4513 have suggested that zaleplon binds to γ3 GABAA receptors with an eightfold higher affinity than when γ2 is present (Dämgen and Lüddens, 1999). However, efficacy and potency have only been studied for the α2β2γ3 receptor which shows 10-fold less potency than what we have seen on α1β2γ3 with an EC50 of ∼500 nM. Eszopiclone and Diazepam The cyclopyrrolone, eszopiclone, and the benzodiazepine, diazepam modulated both γ2 and γ3 containing receptors with varying potency and efficacies and did not significantly modulate γ1 containing receptors (Figures 5C,D). Substituting the γ3 receptor for γ2 had no statistically significant difference on eszopiclone’s potency (in the range of 300–500 nM), but diazepam had a ∼15-fold reduction (1900 vs. 150 nM). Both compounds had ∼1.5-fold reductions in Emax when γ3 replaced γ2 (300 vs. 200%). At 10 μM, eszopiclone modulated γ1 receptors by 20% and diazepam by 40% above GABA EC10. No literary data are available for eszopiclone, however, the racemic mixture zopiclone has been investigated. In a competitive binding study from Dämgen and Lüddens (1999), a marginal reduction in binding affinity was observed for zopiclone when γ3 was replaced by γ2. Yet, in another study zopiclone was observed to modulate α1β2γ3 receptors with comparable efficacy and potency to that of α1β2γ2 (Davies et al., 2000). Hence, eszopiclone and zopiclone seem to behave in a similar fashion at γ2- and γ3 containing receptors A previous study of diazepam on recombinant α1β2γ3 GABAA receptors shows good agreement for the potency (EC50 of 1.95 μM), but they observed no reduction in Emax comparing α1β2γ2 vs. α1β2γ3 receptors (Lippa et al., 2005). Imidazopyridines The imidazopyridines, zolpidem, and alpidem were selective for the γ2 subunit, not showing significant potencies to be able to estimate an EC50 from fitting to the Hill equation for γ1 and γ3 receptors (Figures 5E,F). Zolpidem and alpidem had Emax on γ2 receptors ranging from 475–550%. Zolpidem’s EC50 on γ2 receptors was 230 nM and alpidem’s 500 nM. On γ3 receptors, both compounds had a measured response at concentrations of 10 μM of near 125% of GABA EC10. Neither compound showed robust efficacy on γ1 containing receptors. At 10 μM, zolpidem elicited a response of 15% and alpidem 40% above GABAcontrol. Overall this data indicates that zolpidem’s pharmacological activity is likely to be related only to the γ2 subunit. Zolpidem’s selectivity for the γ2 subunit below 10 μM correlates with previous studies both on the binding for the γ1 (Benke et al., 1996) and γ3 subunit (Herb et al., 1992; Lüddens et al., 1994; Tögel et al., 1994; Hadingham et al., 1995; Sieghart, 1995; Dämgen and Lüddens, 1999), and with measurements in oocytes showing 20% or less efficacy (Wafford et al., 1993; Mckernan et al., 1995; Khom et al., 2006). These observations contrast with studies using HEK293 cells expressing α1βγ1 receptors observing zolpidem potentiating near 50–75% (Puia et al., 1991) and with an EC50 around 200 nM (Esmaeili et al., 2009). α1β2 Binary Receptors To investigate whether the modulation observed at high compound concentrations on γ1 or γ3 receptors was specific to the γ subunit, 10 μM of each compound was applied to α1β2 binary receptors. Potentiation values are depicted along with the respective values at the γ-containing receptors in Figure 6. All 6 tested compounds elicited small responses on α1β2 receptors, with mean values ranging from 10–35%. Zaleplon was the only compound to show significantly higher γ1 receptor modulation above the value seen for α1/β2 receptors (p < 0.01) indicating that potentiation observed is specific to the γ1 subunit. Importantly, all tested compounds showed significantly higher potentiation values at γ3 receptors compared with α1β2 receptors (p < 0.01) indicating that modulation is specific to γ3. FIGURE 6 Figure 6. Modulatory actions of 10 μM (A) zaleplon, (B) indiplon, (C) eszopiclone, (D) diazepam, (E) zolpidem, and (F) alpidem, on GABA evoked Cl– currents measured in human α1β2γ1, α1β2γ2, α1β2γ3, and binary α1β2 GABAA receptors expressed in Xenopus laevis oocytes measured via two-electrode voltage-clamp. Data are expressed as percentage potentiation of GABA EC10 and are means ± SD (n = 3–8 from at least 2 separate Xenopus laevis donors). One-way ANOVA with post hoc Sidak multiple comparisons test were calculated between all compounds on γ1 vs. α1β2 and between zolpidem and alpidem on γ3 vs. α1β2; p < 0.01. Discussion In this study, we examined the effectiveness of the Z-drugs (zaleplon, indiplon, eszopiclone, zolpidem, and alpidem) and the benzodiazepine, diazepam on GABAA receptors containing γ1, γ2, or γ3 subunits under highly controlled experimental conditions. We used concatenated pentamers expressed in Xenopus laevis oocytes to reduce the potential of confounding mixed receptor populations arising when single subunits are injected (Boileau et al., 2002; Sigel et al., 2006; Ahring et al., 2016; Liao et al., 2019). Furthermore, all experiments were performed identically for each oocyte. Modulators were co-applied with a GABAcontrol concentration eliciting 10% of the maximum response. α1β2γ2 Receptors All the tested drugs are efficient and potent modulators of γ2 receptors. Maximum efficacies ranged from 250–500% with the most and least efficacious being alpidem and indiplon, respectively. Potencies ranged from 10–500 nM with the most and least potent being indiplon and alpidem, respectively. In general, our results are within the range of variation from previous studies (Puia et al., 1991; Davies et al., 2000; Sanna et al., 2002; Petroski et al., 2006). α1β2γ1 Receptors None of the Z-drugs exhibited sufficient potency within the tested concentration range to allow reliable fitting of the data to the Hill equation at γ1 receptors. At the highest tested concentration (10 μM) zaleplon had an efficacy of 125%. This contrasts the structurally similar compound, indiplon which at the same concentration did not affect α1β2γ1 receptors. Notably, zaleplon’s modulation was likely specific to the γ1 subunit, as the same concentration applied to α1/β2 receptors only elicited 25% above GABAcontrol. It remains a possibility that even higher concentrations of zaleplon could reveal further robust modulation at γ1-containing receptors. However, we generally chose to limit the concentration range tested to a maximum of 10 μM to avoid issues with compound solubility and potential interfering efficacies from binding to secondary modulatory sites as previously described for diazepam (Walters et al., 2000; Sieghart, 2015; Masiulis et al., 2019). There is some discrepancy regarding γ1-containing receptors and zolpidem in the literature. Several studies observed that zolpidem displays no binding (Benke et al., 1996), or low maximum efficacies below 20% in α1βγ1 (Khom et al., 2006) and α2βγ1 receptors (Wafford et al., 1993; Mckernan et al., 1995) expressed in X. laevis oocytes. In contrast, other studies using HEK293 cells expressing α1βγ1 receptors observe zolpidem potentiations of near 50–75% (Puia et al., 1991) with an EC50 around 200 nM (Esmaeili et al., 2009). Differences in the experimental protocol, expression systems, assembling receptor populations, or chosen GABA control concentration may account for some of these divergences. Nevertheless, our data showing that zolpidem and the structurally similar compound, alpidem have negligible effects on concatenated pentameric α1β2γ1 receptors align with the findings that these compounds do not modulate γ1-containing receptors. While definitive high-resolution crystal or Cryo-EM structures of GABAA receptors with bound diazepam exist (Zhu et al., 2018), they are still lacking for Z-drugs. Mutational studies and molecular modeling have provided insights into the nature of the important amino acids determining Z-drugs’ binding within the α1(+)–γ2(−) interface. The necessary His101 residue on the α1(+) interface is a well-characterized component of the benzodiazepine binding site (Wieland et al., 1992; Benson et al., 1998; Mckernan et al., 2000), but there are also important residues on the γ2(−) side. The amino acids Met130 and Phe77 have interactions with zolpidem, and mutating one or more of these abolishes binding (Buhr and Sigel, 1997; Wingrove et al., 1997). These residues are not present on the γ1 subunit, yet introducing them into the γ1 subunit does not fully restore zolpidem binding (Wingrove et al., 1997). Furthermore, the γ2 Phe77 mutation when expressed in the mouse eliminated zolpidem (but not flurazepam) dependent sedation and decreases motor exploration (Cope et al., 2004). Overall, our data pose the question of whether any of the tested drugs bind efficiently to α1-γ1 interfaces within the tested concentration range. α1β2γ3 Receptors Zaleplon, indiplon, and eszopiclone modulate γ3-containing GABAA receptors at therapeutically relevant doses while diazepam, zolpidem, and alpidem do not. On α1β2γ3 receptors, zaleplon has equal efficacy compared with α1β2γ2, and a four-fold increase in potency. The structurally similar indiplon was also equally as efficacious on γ3- as γ2 containing receptors, but with reduced potency indicating that small differences in pyrazolopyrimidines can alter selectivity preferences between γ3- and γ2-containing GABAA subunits. Eszopiclone potentiates α1β2γ3 receptors with equal potency to α1β2γ2, but with a 1.5-fold reduction in efficacy. Overall these data indicate that even though classes of Z-drugs are quite similar, the arylamide moiety located at C4 of the pyrazolopyrimidines may be important for binding to the γ3 subunit. Interestingly, high concentrations of zolpidem and alpidem potentiated GABA at γ3 receptors. This effect is specific to the γ3 subunit, as the same concentration applied to α1/β2 receptors elicited little response. Previous competitive binding studies using high-affinity benzodiazepine site ligands such as flunitrazepam or Ro-154513 have indicated that zolpidem has no binding to the classical γ3 receptor benzodiazepine site (Herb et al., 1992; Lüddens et al., 1994; Tögel et al., 1994; Hadingham et al., 1995; Sieghart, 1995; Dämgen and Lüddens, 1999). Implications for Z-Drugs Hypnotic Effect While clinical studies observing the pharmacokinetics and pharmacodynamics of Z-drug mediated sleep are extensive, there have been relatively few studies comparing how hypnotic drugs target specific brain areas to induce sleep. Within the thalamus and hypothalamus are clusters of nuclei that relay information from subcortical structures to the cortex and both these regions are important for sleep-wake maintenance. The thalamic reticular nucleus generates characteristic sleep EEG firing rhythms, and the lateral hypothalamus is part of an ascending pathway stimulating cortical activity and wakefulness (Saper et al., 2001; Gent et al., 2018). Interestingly, eszopiclone but not zolpidem modulates GABAergic postsynaptic potentials in the thalamic reticular nucleus (Jia et al., 2009) and suppresses activity in the lateral hypothalamus (Kumar et al., 2011) to bring about sleep. Both of these regions contain a wider variety of GABAA subunits including the γ3 subunit (Pirker et al., 2000) which may, in part, account for the differences. Compared to zolpidem, eszopiclone has a faster sleep onset, more time spent in the restorative non-rapid eye movement stage, and a differing EEG signature (Xi and Chase, 2008). There is a need to understand how hypnotics mediate their effect to aid in future drug development. Z-drugs were designed well before our detailed understanding of GABAA receptor subtypes (Bardone et al., 1978; Arbilla et al., 1985; Beer et al., 1997), and different GABAA subunit preferences contribute to differences in drug action along with pharmacokinetic factors like plasma concentration and drug half-life. In this study, we limited receptors to only contain α1 in combination with γ1, γ2, or γ3. While Z-drugs preferentially modulate α1 receptors at low concentrations, at moderate to high concentrations they also modulate receptors with α2 and α3 subunits (Petroski et al., 2006; Nutt and Stahl, 2010; Ramerstorfer et al., 2010; Sieghart and Savic, 2018), and these subunits may also play a role in sleep generation (Kopp et al., 2004). In addition to α subunit preference variations, we provide evidence here that there are also differences in how Z-drugs modulate GABAA receptors with γ3 subunits, but the significance of this in vivo is still unknown. In addition, future studies should characterize receptors with γ3 subunits in combination with α2/3. The γ2 Phe77 mutation which abolishes zolpidem binding has been used as an in vivo pharmacogenetic model to explore zolpidem’s effects in particular brain regions (Wulff et al., 2007). This approach revealed that zolpidem specifically prolongs postsynaptic potentials within the hypothalamic tuberomammillary nucleus, reducing histamine levels across the brain sufficiently to induce sleep (Uygun et al., 2016). Because γ3-containing receptors are expressed within the same networks controlling sleep, elucidating any potential role they play would be important for the development of better hypnotics. Utilizing the approach of expressing the γ2 Phe77 mutation may reveal residual non-γ2 mediated behavioral effects related to zaleplon, indiplon, or eszopiclone administration. Moreover, because indiplon is efficacious on γ3, but not γ1-containing receptors, it would be well suited to specifically target γ3-containing receptors. In conclusion, the approach taken of using concatenated GABAA receptors has overcome issues of forming unexpected receptor populations when using single subunit cRNAs to express recombinant receptors in X. laevis oocytes. We used this strategy to clarify inconsistencies within the literature on what effects Z-drugs have on γ1- and γ3-containing GABAA receptors. Using this strategy, we have shown that zaleplon, indiplon, and eszopiclone modulate γ3-containing GABAA receptors with no effects on γ1-containing GABAA receptors below 10 μM. Zolpidem and alpidem show no significant modulation on γ1 or γ3 subunits below 10 μM indicating that their pharmacological effects are likely limited to GABAA receptors with γ2 subunits. Gaining a complete picture of the GABAA receptor subtypes targeted by Z-drugs will help in the understanding of hypnotics and aid in developing drugs that more closely replicate physiological sleep with less adverse side effects. Data Availability Statement The raw data supporting the conclusions of this article will be made available by the authors, without undue reservation. Ethics Statement The animal study was reviewed and approved by The University of Sydney Animal Ethics Committee. Author Contributions GR, VL, PA, and MC conceptualized the study and designed the experiments. GR, VL, and PA collected and analyzed the data. GR wrote the manuscript and prepared the figures. VL, PA, and MC reviewed the manuscript. All authors contributed to the article and approved the submitted version. 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Nature 559, 67–72. doi: 10.1038/s41586-018-0255-3 PubMed Abstract \| CrossRef Full Text \| Google Scholar Keywords: GABAA receptors, Z-drugs, modulators, γ1 subunit, γ3 subunit, zolpidem, zaleplon, eszopiclone Citation: Richter G, Liao VWY, Ahring PK and Chebib M (2020) The Z-Drugs Zolpidem, Zaleplon, and Eszopiclone Have Varying Actions on Human GABAA Receptors Containing γ1, γ2, and γ3 Subunits. Front. Neurosci. 14:599812. doi: 10.3389/fnins.2020.599812 Received: 28 August 2020; Accepted: 26 October 2020; Published: 19 November 2020. Edited by: Christian Legros, Université d’Angers, France Reviewed by: Petra Scholze, Medical University of Vienna, Austria Enrico Sanna, University of Cagliari, Italy Copyright © 2020 Richter, Liao, Ahring and Chebib. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Grant Richter, grant.richter@sydney.edu.au; Mary Chebib, mary.collins@sydney.edu.au Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
188325	https://journals.lww.com/jtrauma/fulltext/2011/04000/fallen_lung_sign__on_chest_radiograph_.40.aspx	Journal of Trauma and Acute Care Surgery Log in or Register Subscribe to journal Subscribe Get new issue alerts Get alerts;;) Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation SubscribeRegisterLogin Browsing History Articles & Issues Current Issue Previous Issues Epub Ahead Collections EGS Algorithm Hottest Articles Best of Videos Presidential Addresses NTRAP Manuscripts AAST OIS papers EAST Practice Management WTA Critical Decisions in Trauma Classics of Trauma What You Need to Know The Journal of Trauma Military Supplements View All Collections Multimedia Videos Podcasts Reports CME For Authors Submit a Manuscript Instructions for Authors Language Editing Services Embargo Policy Author Permissions Journal Info About the Journal Editorial Board Ethics Policies Affiliated Society Advertising Open Access Subscription Services Reprints Rights and Permissions DEI Resources Featured Reviewers Image of the Month Articles Advanced Search April 2011 - Volume 70 - Issue 4 Previous Abstract Next Abstract Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection The Image of Trauma Fallen Lung Sign (on Chest Radiograph) Magu, Sarita MD; Agarwal, Kanupriya MD, DNB, FRCR; Lohchab, Shamsher S. MD, Mch; Agarwal, Shalini MD, DNB Author Information From the Department of Radiology and Cardiothoracic Surgery, Pandit Bhagwat Dayal Sharma, Post Graduate Institute of Medical Sciences, Rohtak (Haryana) India. Submitted for publication November 16, 2008. Accepted for publication January 3, 2009. Address for reprints: Sarita Magu, MD, 22/8 FM, Medical campus, Pt. BD Sharma, PGIMS, Rohtak, Haryana 124001, India: email: snkmagu@gmail.com. The Journal of Trauma: Injury, Infection, and Critical Care 70(4):p 1012, April 2011. \| DOI: 10.1097/TA.0b013e31819adb51 Buy © 2011 Lippincott Williams & Wilkins, Inc. 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188326	https://www.lumoslearning.com/llwp/practice-tests-sample-questions-92281/grade-5-aasa-math/multiplication-of-whole-numbers/277401-question-32-5.NBT.B.5.html	Loading [Contrib]/a11y/accessibility-menu.js Multiplication of whole numbers 5.NBT.B.5 Grade Practice Test Questions TOC \| Lumos Learning Toggle navigation What We Do For Schools State Test Prep Blended program State Test Prep Online Program Summer Learning Writing Program Oral Reading Fluency Program Reading Skills Improvement Program After School Program High School Program Professional Development Platform Teacher PD Online Course Smart Communication Platform For Teachers State Assessment Prep Classroom Tools State Test Teacher PD Course For Families State Test Prep Blended Program State Test Prep Online Program Summer Learning Oral Reading Fluency Online Program Oral Reading Fluency Workbooks College Readiness – SAT/ACT High School Math & ELA Curious Reader Series For Libraries For High School Students High School Math, Reading and Writing College Readiness – SAT/ACT For Professionals Online Professional Certification Exams Job Interview Practice Tool For Enterprises and Associations HR Software Online Pre-employment Assessment Platform Online Employee Training & Development Platform Enterprise Video Library Association Software Association Learning Management System Online Course Platform After-event Success Platform Self-service FAQ and Help Center Who Are We Resources Free Resources Success Stories Podcasts Program Evaluation Reports Events Login Multiplication of whole numbers 5.NBT.B.5 Question & Answer Key Resources Grade 5 Mathematics - Skill Builder + LEAP2025 Rehearsal (UPDATED) Grade 5 Mathematics - Skill Builder + LEAP2025 Rehearsal (UPDATED) Multiplication of whole numbers GO BACKReview Selection 1 0 out of 5 stars 5 0 0 4 0 0 3 0 0 2 0 0 1 0 0 Click on the value of 1620 x 944. 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Prothrombin Complex Concentrates (aPCCs) Rh 0(D) Immunoglobulin (RhIG) ►Platelet Concentrates Leukocyte-Reduced Platelet Concentrate Reconstituted Platelet Concentrate Refractoriness to Platelet Transfusions Washed Platelet Concentrate Transfusion Reactions and Procedure Whole Blood ►Infectious Diseases Acquired Immunodeficiency Syndrome (AIDS) ►Antimicrobial Agents Antibacterial Agents Antifungal Agents ►Antiparasitic Agents Agents with Activity Against Ectoparasites Anthelmintic Agents Antiprotozoal Agents Antiviral Agents ►Bacterial Diseases Brucellosis Cholera ►_Clostridium_ spp Infections Botulism Clostridial Myonecrosis (Gas Gangrene) Tetanus Enteric (Typhoid) Fever Leprosy Melioidosis Nocardiosis Scrub Typhus ►Sexually Transmitted Bacterial Diseases Gonorrhea Lymphogranuloma Venereum (LGV) Sexually Transmitted Chlamydial Infections Other Than Lymphogranuloma Venereum (LGV) Syphilis ►Skin and Soft Tissue Infections Nonpurulent Skin and Soft Tissue Infections: Erysipelas and 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Poliomyelitis Vaccines: Rabies Vaccines: Seasonal Influenza Vaccines: Tick-Borne Encephalitis Vaccines: Varicella Zoster (Chickenpox and Shingles) Vaccines: Yellow Fever ►Infection Prevention Among Medical Staff Nonspecific Methods of Infection Prevention Among Medical Staff Occupational Exposures to Blood-Borne Viral Infections Nonspecific Insect and Tick Bite Precautions Intravascular Catheter–Related Infections ►Mycoses ►Superficial Mycoses Dermatophytoses Tinea Versicolor Sporotrichosis Superficial Candidiasis ►Systemic Mycoses Candidiasis Cryptococcosis Histoplasmosis Mucormycosis ►Parasitic Diseases ►Ectoparasitic Infestations Pediculosis Scabies ►Helminthiases Alveolar Echinococcosis Anisakiasis Ascariasis Cystic Echinococcosis Cysticercosis Enterobiasis Filarial Infections Hookworm Infections Tapeworm Infections Toxocariasis Trichinellosis Trichuriasis ►Protozoal Diseases Acanthamoeba Keratitis Amebiasis Balantidiasis Blastocystosis Giardiasis Leishmaniasis Malaria 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Segmental Glomerulosclerosis IgA Nephropathy Lupus Nephritis Membranoproliferative Glomerulonephritis (MPGN) Membranous Nephropathy Minimal Change Disease (MCD) Postinfectious Glomerulonephritis Renal Amyloidosis ►Kidney Cysts Acquired Cystic Kidney Disease Autosomal Dominant Polycystic Kidney Disease (ADPKD) Medullary Sponge Kidney Simple Kidney Cyst Kidney Transplant Nephrolithiasis Obstructive Nephropathy ►Renal Replacement Therapy Hemodialysis Peritoneal Dialysis ►Tubulointerstitial Nephritis Acute Interstitial Nephritis (AIN) Autosomal Dominant Tubulointerstitial Kidney Disease (ADTKD) Chronic Interstitial Nephritis (CIN) ►Tubulopathies Bartter and Gitelman Syndromes Cystinosis Cystinuria Familial Hypophosphatemic Rickets Fanconi Syndrome Nephrogenic Diabetes Insipidus (NDI) Renal Glycosuria ►Renal Tubular Acidosis Distal (Type 1) Renal Tubular Acidosis (dRTA) Hyperkalemic (Type 4) Renal Tubular Acidosis (RTA) Proximal (Type 2) Renal Tubular Acidosis (pRTA) ►Urinary Tract 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Arteritis (GCA) Granulomatosis with Polyangiitis IgA Vasculitis (Henoch-Schönlein Purpura) Microscopic Polyangiitis Polyarteritis Nodosa Takayasu Arteritis ►Toxicology and Addiction Approach to the Poisoned Patient Acetaminophen (Paracetamol) ►Alcohols ►Ethyl Alcohol (Ethanol) Management of Acute Alcohol Withdrawal Ethylene Glycol Isopropyl Alcohol Methyl Alcohol (Methanol) Anticholinergic Syndrome (Anticholinergic Toxicity) Benzodiazepines Carbon Monoxide Cholinergic Syndrome (Cholinergic Toxicity) Digoxin and Other Cardiac Glycosides Lithium Toxicity ►Opioids Acute Opioid Toxicity Acute Opioid Withdrawal in the Emergency Department Opioids and Opioid Use Disorder: General Considerations Course What’s New Interviews Lectures Refreshers Articles Resources Editorial Board Mobile Sponsors Diseases Allergy & Immunology Cardiovascular Diseases Dermatology Endocrinology Fluids, Electrolytes, Acids, Bases Gastroenterology Geriatrics Hematology Infectious Diseases Nephrology Neurology Palliative Care Poisonings & Intoxications Psychiatry Respiratory Diseases Rheumatology Noninvasive Diagnostic Tests Procedures Signs & Symptoms Trauma & Injuries Acute & Critical Medical Care Abbreviations CME GRADE Log in/ McMaster Textbook of Internal Medicine Diseases Endocrinology Thyroid Gland Diseases Thyroiditis Subacute Painful Thyroiditis (de Quervain Thyroiditis) How to Cite This Chapter: Braga M, Brito JP, Lewiński A, Płaczkiewicz-Jankowska E. Subacute Painful Thyroiditis (de Quervain Thyroiditis). McMaster Textbook of Internal Medicine. Kraków: Medycyna Praktyczna. Accessed September 29, 2025. Last Reviewed: January 17, 2021 Last Updated: January 17, 2021 Chapter Information McMaster Textbook of Internal Medicine Editorial Offices Editorial Office (Canada) Section Editors: Ally P. H. Prebtani, Juan P. Brito Authors: Manoela Braga, Juan P. Brito Editorial Office (Poland) Section Editors: Barbara Jarząb, Ewa Płaczkiewicz-Jankowska Authors: Andrzej Lewiński, Ewa Płaczkiewicz-Jankowska Main Documents Taken Into Account: Bekkering GE, Agoritsas T, Lytvyn L, et al. Thyroid hormones treatment for subclinical hypothyroidism: a clinical practice guideline. BMJ. 2019 May 14;365:l2006. doi: 10.1136/bmj.l2006. PubMed PMID: 31088853. Alexander EK, Pearce EN, Brent GA, et al. 2017 Guidelines of the American Thyroid Association for the Diagnosis and Management of Thyroid Disease During Pregnancy and the Postpartum. Thyroid. 2017 Mar;27(3):315-389. doi: 10.1089/thy.2016.0457. Erratum in: Thyroid. 2017 Sep;27(9):1212. PubMed PMID: 28056690. Ross DS, Burch HB, Cooper DS, et al. 2016 American Thyroid Association Guidelines for Diagnosis and Management of Hyperthyroidism and Other Causes of Thyrotoxicosis. Thyroid. 2016 Oct;26(10):1343-1421. Erratum in: Thyroid. 2017 Nov;27(11):1462. PubMed PMID: 27521067. Gharib H, Papini E, Garber JR, et al; AACE/ACE/AME Task Force on Thyroid Nodules. AMERICAN ASSOCIATION OF CLINICAL ENDOCRINOLOGISTS, AMERICAN COLLEGE OF ENDOCRINOLOGY, AND ASSOCIAZIONE MEDICI ENDOCRINOLOGI MEDICAL GUIDELINES FOR CLINICAL PRACTICE FOR THE DIAGNOSIS AND MANAGEMENT OF THYROID NODULES--2016 UPDATE. Endocr Pract. 2016 May;22(5):622-39. doi: 10.4158/EP161208.GL. PubMed PMID: 27167915. Jonklaas J, Bianco AC, Bauer AJ, et al; American Thyroid Association Task Force on Thyroid Hormone Replacement. Guidelines for the treatment of hypothyroidism: prepared by the american thyroid association task force on thyroid hormone replacement. Thyroid. 2014 Dec;24(12):1670-751. doi: 10.1089/thy.2014.0028. PubMed PMID: 25266247; PubMed Central PMCID: PMC4267409. Garber JR, Cobin RH, Gharib H, et al; American Association of Clinical Endocrinologists and American Thyroid Association Taskforce on Hypothyroidism in Adults. Clinical practice guidelines for hypothyroidism in adults: cosponsored by the American Association of Clinical Endocrinologists and the American Thyroid Association. Endocr Pract. 2012 Nov-Dec;18(6):988-1028. Erratum in: Endocr Pract. 2013 Jan-Feb;19(1):175. PubMed PMID: 23246686. American Thyroid Association Guidelines Task Force, Kloos RT, Eng C, Evans DB, et al. Medullary thyroid cancer: management guidelines of the American Thyroid Association. Thyroid. 2009 Jun;19(6):565-612. doi: 10.1089/thy.2008.0403. Review. Erratum in: Thyroid. 2009 Nov;19(11):1295. PubMed PMID: 19469690. Elsheikh TM. Diagnostic Terminology and Criteria for the Cytologic Diagnosis of Thyroid Lesions. Papanicolaou Society of Cytopathology Guidelines. Accessed June 21, 2019. Definition, Etiology, PathogenesisTop Subacute painful thyroiditis (de Quervain thyroiditis, granulocytic thyroiditis, subacute granulomatous thyroiditis, giant cell thyroiditis) is probably of viral origin and typically follows a 4-phase course. There is a strong correlation of subacute thyroiditis with the presence of human leukocyte antigen (HLA)-B35. The disease is usually preceded (2-8 weeks earlier) by an upper respiratory tract infection. Clinical Features and Natural HistoryTop The disease can be divided into 4 distinct phases (see Table 1). Initially the dominant features are painful swelling of the thyroid gland and fever; pain is referred to the ears, mandibular angle, throat, and upper chest, and it is accompanied by exquisite tenderness of the thyroid gland. Thyrotoxicosis (lasting 2-8 weeks) results from destruction of the glandular parenchyma and release of thyroid hormones. Fatigue, malaise, anorexia, and muscle pain are common. The pain and fever subside spontaneously and hormone levels normalize after 8 to 16 weeks. Phase 3(hypothyroidism) is not always present. Permanent hypothyroidism is rare and recovery is almost always complete (phase 4). A rapidly growing nodule, requiring cytologic assessment to exclude malignancy, may actually be an inflammatory infiltrate in the course of subacute thyroiditis. In ~2% of patients the disease may recur after a long remission (up to 20 years). DiagnosisTop Diagnostic Tests Laboratory tests: 1) Extremely elevated erythrocyte sedimentation rate (ESR), usually >50 mm/h, and C-reactive protein (CRP) are common. 2) Thyroid-stimulating hormone (TSH) and thyroid hormones: see Table 1. 3) Antithyroid antibodies are uncommonly present; thyroglobulin (Tg) antibodies are more frequent than thyroperoxidase (TPO) antibodies. Imaging studies: Thyroid ultrasonography reveals diffuse or focal hypoechogenicity of the thyroid gland and it is uncommonly needed for the diagnosis. Radionuclide thyroid imaging shows a very low iodine uptake, usually <1% to 3% (in the early phase of the disease). Cytology (fine-needle aspiration biopsy): Rarely needed. The dominant cells are neutrophils, giant cells (characteristic polynuclear macrophages), and epithelioid cells (mononuclear macrophages). Diagnostic Criteria The key diagnostic criteria are a painful or tender goiter; elevated ESR,CRP, or both; transient thyrotoxicosis; and significantly decreased iodine uptake. Differential Diagnosis Acute infectious thyroiditis with abscess formation. Other rare types of granulomatous thyroiditis: tuberculous thyroiditis, fungal thyroiditis (Aspergillus, Candida, Cryptococcus); Pneumocystis jiroveci thyroiditis in immunodeficient patients. Traumatic thyroiditis. Radiation thyroiditis. Acute hemorrhagic nodule. If pain is not a dominant feature, differential diagnosis should include painless (“silent”) thyroiditis and other forms of thyrotoxicosis (see Thyroiditis; see Thyrotoxicosis and Hyperthyroidism). TreatmentTop The hyperthyroid phase requires no antithyroid treatment, as it is not caused by excess thyroid hormone synthesis. Beta-blockers, such as propranolol, bisoprolol, or metoprolol, may be used in patients with bothersome symptoms of hyperthyroidism. Administer acetylsalicylic acid 2 to 4 g/d or nonsteroidal anti-inflammatory drugs to control pain and inflammation if not contraindicated; if pain is severe, consider prednisone 40 to 60 mg/d for 1 to 2 weeks, then taper the dose down to discontinue treatment in ≤4 weeks. The use of glucocorticoids does not lower the probability of transient hypothyroidism, but it decreases and shortens discomfort. In the hypothyroid phase consider levothyroxine (L-T 4) replacement therapy in those with a TSH level >10 mIU/L, significant symptoms, or both; note that hypothyroidism is usually transient and therapy should not be continued indefinitely. Replacement therapy may be withheld after 6 to 8 weeks with a follow-up of thyroid function tests to be sure that hypothyroidism is not permanent. There is no indication for surgery, as the disease is self-limiting and does not cause permanent thyroid damage. TablesTop Table 6.7-3. Clinical phases of subacute thyroiditis (de Quervain thyroiditis)Hormone levels Iodine uptake Clinical features Phase 1↑ FT 4, ↑ FT 3, ↓ TSH Low Thyrotoxicosis Phase 2 Normal Low Euthyroidism Phase 3↓ FT 4, ↓ FT 3, ↑ TSH High Hypothyroidism Phase 4 Normal Normal Euthyroidism ↑, increase; ↓, decrease; FT 3, free triiodothyronine; FT 4, free thyroxine; TSH, thyroid-stimulating hormone. × Close Welcome to the McMaster Textbook of Internal Medicine In one week we will ask you to register to continue full complimentary access to both our website and app. Continue DEFINITION, ETIOLOGY, PATHOGENESIS CLINICAL FEATURES AND NATURAL HISTORY DIAGNOSIS TREATMENT TABLES We would love to hear from you Comments, mistakes, suggestions? Send feedback Social Media We use cookies to ensure you get the best browsing experience on our website. 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188328	https://brainly.com/question/17158049	[FREE] Use Newton's method to find all solutions of the equation correct to six decimal places. \ln(x) = - brainly.com 2 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +51,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +47,7k Ace exams faster, with practice that adapts to you Practice Worksheets +5,7k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Use Newton's method to find all solutions of the equation correct to six decimal places. ln(x)=x 1−3 (Enter your answers as a comma-separated list.) 2 See answers Explain with Learning Companion NEW Asked by yellomustard • 08/06/2020 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1137178851 people 1137M 4.7 4 Upload your school material for a more relevant answer x ≈ {0.653059729092, 3.75570086464} Explanation A graphing calculator can tell you the roots of ... f(x) = ln(x) -1/(x -3) are near 0.653 and 3.756. These values are sufficiently close that Newton's method iteration can find solutions to full calculator precision in a few iterations. In the attachment, we use g(x) as the iteration function. Since its value is shown even as its argument is being typed, we can start typing with the graphical solution value, then simply copy the digits of the iterated value as they appear. After about 6 or 8 input digits, the output stops changing, so that is our solution. Rounded to 6 decimal places, the solutions are {0.653060, 3.755701}. A similar method can be used on a calculator such as the TI-84. One function can be defined a.s f(x) is above. Another can be defined as g(x) is in the attachment, by making use of the calculator's derivative function. After the first g(0.653) value is found, for example, remaining iterations can be g(Ans) until the result stops changing, Answered by sqdancefan •61.9K answers•1.1B people helped Thanks 4 4.7 (6 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 1137178851 people 1137M 4.5 3 COMPUTATIONAL PHYSICS: A Practical Introduction to Computational Physics and Scientific Computing (using C++) - Steven W. Ellingson Celestial Mechanics - Jeremy Tatum Fundamentals of Calculus - Joel Robbin, Sigurd Angenent Upload your school material for a more relevant answer Using Newton's method on the equation ln(x)=x 1−3, we find two roots. The approximated solutions correct to six decimal places are 0.653060 and 3.755701. Explanation To find all solutions of the equation ln(x)=x 1−3 using Newton's method, we first need to rearrange the equation into a function. Let's define the function: f(x)=ln(x)−x 1+3 Next, we compute the derivative of this function: f′(x)=x 1+x 2 1 This derivative helps us apply Newton's formula: x n+1=x n−f′(x n)f(x n) Now, we need a reasonable initial guess. By inspecting the function defined, we can find that it has roots around 0.6 and 3.8. Using x 0=0.5 and x 0=3.5 as initial guesses, we can apply the iterations. Iterations for x = 0.5: x 1=0.5−f′(0.5)f(0.5)≈0.746 x 2=0.746−f′(0.746)f(0.746)≈0.655 x 3=0.655−f′(0.655)f(0.655)≈0.653 Repeat until the result stabilizes around 0.653060. Iterations for x = 3.5: 5. x 1=3.5−f′(3.5)f(3.5)≈3.76 6. x 2=3.76−f′(3.76)f(3.76)≈3.755 7. Repeat until the result stabilizes around 3.755701. After this process, rounding to six decimal places gives us the solutions: 0.653060,3.755701. Examples & Evidence Examples of Newton's method include solving equations such as x 2−2=0 where one starts with a guess like x 0=1 and iterates to refine the approximation. In practice, this method is widely applicable as it quickly converges to roots for a variety of functions. Newton's method is a well-established numerical technique in calculus, widely documented in mathematical literature and used for obtaining increasingly accurate estimates of roots of real-valued functions. Thanks 3 4.5 (4 votes) Advertisement Community Answer This answer helped 8873585 people 8M 0.0 0 Newton's method involves using the derivative of the function to get increasingly accurate approximations to the function's roots. The given equation is rearranged into the form f(x) = 0, then the iterative procedure of Newton's method is applied. Achieving the precision of six decimal places usually requires multiple iterations and software support. Explanation To use Newton's method to find solutions to an equation, we first need to understand the concept. Newton's method or Newton-Raphson method is an iterative method for finding successively better approximations for the roots (or zeroes) of a real-valued function. It uses the derivative of the function to approximate the steps towards a solution. In the case of the equation ln(x) = 1 /x − 3, we first rearrange it to f(x) = ln(x) - 1/x + 3 = 0. Then we compute the derivative of the function f'(x) = 1/x - 1/x². Choosing an initial approximation x₀, we use the formula xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ) to get increasingly accurate values for x which eventually give us the solution to the equation. Note that this method assumes that the function is reasonably well behaved in the region of the root and that the initial guess is close enough to the true root. Also, getting the solution to six decimal places needs an iterative process using a software or calculator to reach the desired precision. Learn more about Newton's Method here: brainly.com/question/31910767 SPJ3 Answered by BritMarling •16.3K answers•8.9M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Mathematics solutions and answers Community Answer Use Newton's method to find all solutions of the equation correct to six decimal places. (Enter your answers as a comma-separated list.) ln(x)=x−31 x= Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics Which graph represents the function f(x)=(x−3)2? Graph the function f(x)=(x+1)(x−5). 1. Identify the x-intercepts: (−1,0) and (5,0) 2. Find the midpoint between the intercepts (calculate, don't plot). 3. Find the vertex: □ 4. Find the y-intercept: □ 5. Plot another point, then draw the graph.> Identify the $x$-intercepts: $(-1,0)$ and $(5,0)$ Find the midpoint between the intercepts (calculate, don't plot). Find the vertex: $\square$ Find the $y$-intercept: $\square$ Plot another point, then draw the graph.") 4 m 7(3 m 9)2 simplifies to 36 m 25. Explain in detail each step it takes to simplify this expression. You must have at least two separate steps explained in words, but you could have more. The graph of the function f(x)=−(x+1)2 is shown. The vertex is the □ , The function is positive □ The function is decreasing □ The domain of the function is □ The range of the function is □ Mr. Walker gave his class the function f(x)=(x+3)(x+5). Four students made a claim about the function. • Jeremiah: The y-intercept is at (15,0). • Lindsay: The x-intercepts are at (−3,0) and (5,0). • Stephen: The vertex is at (−4,−1). • Alexis: The midpoint between the x-intercepts is at (4,0).Which student's claim about the function is correct?> Jeremiah: The $y$-intercept is at $(15,0)$. Lindsay: The $x$-intercepts are at $(-3,0)$ and $(5,0)$. Stephen: The vertex is at $(-4,-1)$. Alexis: The midpoint between the $x$-intercepts is at $(4,0)$. Which student's claim about the function is correct?") 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188329	https://www.zhihu.com/question/356778994	如果一条直线与两条平行直线都相交，那么这三条直线是否共面？ - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册如果一条直线与两条平行直线都相交，那么这三条直线是否共面？关注问题写回答登录/注册空间几何直线解析几何如果一条直线与两条平行直线都相交，那么这三条直线是否共面？求证，有图显示全部关注者 5 被浏览 743 关注问题写回答邀请回答好问题 1 条评论分享 3 个回答默认排序 chay96 学生关注先说结论:这三条直线一定共面。为了说明方便，我令两条平行线分别为l1,l2,令相交的那条线为l3,令l3与l1,l2的交点分别为P和Q 证明:因为l1,l2可以确定一个平面β,又P和Q在l1,l2上，所以同时过P,Q的直线l3上所有的点都在平面β内(平面的基本性质1)，也就是l3在平面β内，证明成立如果想象不出来的话，你可以画图发布于 2023-11-14 15:44 赞同添加评论分享收藏喜欢知乎用户h8Aau3 关注不一定共面，如果两条平行线在同一平面，那三条线共面发布于 2019-11-21 02:07 赞同1 条评论分享收藏喜欢 a a 关注异面平行发布于 2019-11-20 05:41 赞同添加评论分享收藏喜欢写回答 1 个回答被折叠（为什么？）下载知乎客户端与世界分享知识、经验和见解相关问题如何证明两两相交的直线要么交于一点要么在同一平面？ 1 个回答这两条直线的位置关系为什么是相交(求详解)？ 1 个回答两条直线被第三条直线所截同旁内角为什么不一定互补? 1 个回答广告帮助中心知乎隐私保护指引申请开通机构号联系我们举报中心涉未成年举报网络谣言举报涉企侵权举报更多关于知乎下载知乎知乎招聘知乎指南知乎协议更多京 ICP 证 110745 号 · 京 ICP 备 13052560 号 - 1 · 京公网安备 11010802020088 号 · 互联网新闻信息服务许可证：11220250001 · 京网文2674-081 号 · 药品医疗器械网络信息服务备案（京）网药械信息备字（2022）第00334号 · 广播电视节目制作经营许可证:（京）字第06591号 · 互联网宗教信息服务许可证：京（2022）0000078 · 服务热线：400-919-0001 · Investor Relations · © 2025 知乎北京智者天下科技有限公司版权所有 · 违法和不良信息举报：010-82716601 · 举报邮箱：jubao@zhihu.com 想来知乎工作？请发送邮件到 jobs@zhihu.com 登录知乎，问答干货一键收藏打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
188330	https://theorems.home.blog/2020/06/01/the-julia-set-of-the-feigenbaum-polynomial-has-zero-lebesgue-measure/	The Julia set of the Feigenbaum polynomial has zero Lebesgue measure – Theorems of the 21st century Skip to content Theorems of the 21st century Menu Home Contact The Julia set of the Feigenbaum polynomial has zero Lebesgue measure Bogdan GrechukAnalysis1st Jun 2020 1st Jun 2020 2 Minutes You need to know: Basic complex analysis, set of complex numbers, absolute value of complex number z, boundary of a set, area (that is, Lebesgue measure) of a set , notation for the n-th iterate of function . Background: For polynomial , let be the set of points such that the sequence is bounded, that is, for some . The boundary of is called the Julia set of f. We say that point is periodic if for some positive integer n, and the smallest n such that this holds is called the period of . For each n, let be the (unique) real number such that is a periodic point with period of polynomial . It is known that the limit exists and numerically equal to . The polynomial is called the Feigenbaum polynomial. The Theorem: On 22nd December 2017, Artem Dudko and Scott Sutherland submitted to arxiv a paper in which they proved that the Julia set of the Feigenbaum polynomial has zero Lebesgue measure. Short context: Julia set is a fundamental concept in the theory of complex dynamics, because it consists of values such that an arbitrarily small perturbation can cause significant changes in the sequence of iterated function values. The Julia set is known to have zero area for almost all but not all quadratic polynomials. The Feigenbaum polynomial is important in so-called renormalization theory in dynamics, and the question whether its Julia set has zero area was a long-standing open question. The Theorem resolves it affirmatively. Links:Free arxiv version of the original paper is here, journal version is here. Go to the list of all theorems Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Related The Julia set of typical quadratic map is locally connected and has measure zero10th Dec 2019 In "Analysis" There exist quadratic polynomials that have a Julia set of positive area20th Feb 2020 In "Analysis" Small set of initial points for Newton’s method to find all roots of polynomials14th Nov 2019 In "Analysis" Published by Bogdan Grechuk View all posts by Bogdan Grechuk Published1st Jun 2020 1st Jun 2020 Post navigation Previous Post The Schinzel-Zassenhaus conjecture on integer polynomials is true Next Post Characterisation of polyhedral types inscribed in the hyperboloid or cylinder Leave a comment Cancel reply Δ Create a free website or blog at WordPress.com. Comment Reblog SubscribeSubscribed Theorems of the 21st century Sign me up Already have a WordPress.com account? Log in now. Theorems of the 21st century SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy %d Design a site like this with WordPress.com Get started Advertisement
188331	https://community.typeform.com/build-your-typeform-7/how-to-setup-a-percentage-question-that-adds-up-to-100-2241	How to setup a percentage question that adds up to 100% \| Community Skip to main content Create a post Log in Community Typeform conversations Build your typeform How to setup a percentage question that adds up to 100% Answered How to setup a percentage question that adds up to 100% June 29, 2021 3 replies 2605 views A akb Explorer 5 replies Hi, I can’t figure out how to set this specific up on Typeform. Is there a way, I can setup a Multiple choice question where users can add their inputs to each choice as a % and the survey form makes sure that the total adds up to 100%. How much of your overall time is spent on each of these categories of work? Make sure that the total % adds up to 100% [ %] Supporting ad-hoc requests from business teams [ %] Supporting periodic repetitive requests from business teams [ %] Planning ahead and delivering strategic insights to business teams [ %] Building tools or systems to scale your time [ %] Other activities Best answer by john.desborough @akb @Liz - the only workaround that i could add to this is one that doesn’t yet have the in-Typeform logic processing capability: Matrix Question You could use a matrix question with the Question statements down the side and percentage across the top ie 5,10,15,20,25 Currently you would have to put in a message in the Description field that said something like “We won’t do the validation that this adds up to exactly 100 Percent, but try to give us the best representation” If you could perform logic on the Matrix question you would you be able to add the selections of each of the questions and validate the sum. @akb - one thing you might want to think on - instead of using percent, maybe use the ranking or opinion question to say something like: Out of an 8-hour day, how many hours do you usually spend on this activity? and having 8 options to select, that would give you a 1-8 hour selection for that activity. at the end all the question series, you could then have a yes/no question that would allow you to sum the hours - let’s say 10 for example - and say “hey, your answers of: q1 - a q2 - c etc total up to 10 hours in an 8 hour day, do you want to go back and adjust your answers to try and keep it to 8? no worries if you really do spend this extra time daily” that might give you a more solid data point. just some thoughts des View original Typeform calculator Like Quote Subscribe Share 3 replies +5 Liz Tech Community Advocate 15035 replies June 29, 2021 Hi @akb welcome to the community! Happy you’re here. Our multiple choice question doesn’t allow for custom inputs, so it wouldn’t exactly be possible to add this to your form. Would a ranking question work? It’s not exactly what you’re looking for, but it would save you the troubles of setting up a lot of logic jumps to make this work. I could also be missing a workaround that @john.desborough may know! Life is a series of awkward moments separated by snacks. Like Quote A akb Author Explorer 5 replies June 29, 2021 Hey @Liz Thanks for the prompt response! I resorted to a Ranking question but to be honest, we felt it was a bit awkward to answer with the ranking format. The % format also helps us to carve out a clear data point which would be missing from a ranking styled format. LMK if there is a workaround. Like Quote +6 john.desborough Certified Partner & Champion 5379 replies Answer June 30, 2021 @akb @Liz - the only workaround that i could add to this is one that doesn’t yet have the in-Typeform logic processing capability: Matrix Question You could use a matrix question with the Question statements down the side and percentage across the top ie 5,10,15,20,25 Currently you would have to put in a message in the Description field that said something like “We won’t do the validation that this adds up to exactly 100 Percent, but try to give us the best representation” If you could perform logic on the Matrix question you would you be able to add the selections of each of the questions and validate the sum. @akb - one thing you might want to think on - instead of using percent, maybe use the ranking or opinion question to say something like: Out of an 8-hour day, how many hours do you usually spend on this activity? and having 8 options to select, that would give you a 1-8 hour selection for that activity. at the end all the question series, you could then have a yes/no question that would allow you to sum the hours - let’s say 10 for example - and say “hey, your answers of: q1 - a q2 - c etc total up to 10 hours in an 8 hour day, do you want to go back and adjust your answers to try and keep it to 8? no worries if you really do spend this extra time daily” that might give you a more solid data point. just some thoughts des If your third coffee of the day is not helping you with your Typeform, maybe we can help you find your answers... 1 person likes this Liz Like Quote Reply Rich Text Editor, editor1 Editor toolbarsStylesStylesStyles "Formatting Styles")BoldKeyboard shortcut Ctrl+B "Bold (Ctrl+B)")ItalicKeyboard shortcut Ctrl+I "Italic (Ctrl+I)")UnderlineKeyboard shortcut Ctrl+U "Underline (Ctrl+U)")StrikethroughKeyboard shortcut Ctrl+Shift+X "Strikethrough (Ctrl+Shift+X)")Text Color "Text Color")Background Color "Background Color")Bullet list "Bullet list")Numbered list "Numbered list")Align Left "Align Left")Center "Center")Align Right "Align Right")Emoji "Emoji")Insert linkKeyboard shortcut Ctrl+K "Insert link (Ctrl+K)")Insert image "Insert image")Quote "Quote")Insert Media Embed "Insert Media Embed")Insert Code Snippet "Insert Code Snippet")Spoiler "Spoiler")InsertTable "Table")More "More") Press ALT 0 for help Related topics Distributing points among options (dynamic ranking)icon Build your typeform #### HELP before buyingicon Build your typeform #### Need a calculator for a sales goal formulaicon Build your typeform #### How to share your videoask Share your videoask #### Is it possible for typeform to do a cost calculation?icon Build your typeform Typeform integrations Typeform design Typeform logic jumps Typeform embed Still need help? 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188332	https://pubmed.ncbi.nlm.nih.gov/36297244/	Long-Term Outcomes in Children with Congenital Toxoplasmosis-A Systematic Review - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Long-Term Outcomes in Children with Congenital Toxoplasmosis-A Systematic Review Justus G Garweg12,François Kieffer3,Laurent Mandelbrot45,François Peyron6,Martine Wallon67 Affiliations Expand Affiliations 1 Swiss Eye Institute, Rotkreuz, and Uveitis Clinic, Berner Augenklinik, Zieglerstrasse 29, 3007 Bern, Switzerland. 2 Department of Ophthalmology, Inselspital, University Hospital, 3010 Bern, Switzerland. 3 Assistance Publique-Hôpitaux de Paris, Hôpital Armand Trousseau, Service de Néonatologie, 75012 Paris, France. 4 Assistance Publique-Hôpitaux de Paris, Hôpital Louis-Mourier Service de Gynécologie-Obstétrique, 178 rue des Renouillers, 92700 Colombes, France. 5 Inserm IAME-U1137, 75000 Paris, France. 6 Hospices Civils de Lyon, Hôpital de la Croix Rousse, Department of Parasitology and Medical Mycology, 69004 Lyon, France. 7 Walking Team, Centre for Research in Neuroscience in Lyon, 69500 Bron, France. PMID: 36297244 PMCID: PMC9610672 DOI: 10.3390/pathogens11101187 Item in Clipboard Review Long-Term Outcomes in Children with Congenital Toxoplasmosis-A Systematic Review Justus G Garweg et al. Pathogens.2022. Show details Display options Display options Format Pathogens Actions Search in PubMed Search in NLM Catalog Add to Search . 2022 Oct 15;11(10):1187. doi: 10.3390/pathogens11101187. Authors Justus G Garweg12,François Kieffer3,Laurent Mandelbrot45,François Peyron6,Martine Wallon67 Affiliations 1 Swiss Eye Institute, Rotkreuz, and Uveitis Clinic, Berner Augenklinik, Zieglerstrasse 29, 3007 Bern, Switzerland. 2 Department of Ophthalmology, Inselspital, University Hospital, 3010 Bern, Switzerland. 3 Assistance Publique-Hôpitaux de Paris, Hôpital Armand Trousseau, Service de Néonatologie, 75012 Paris, France. 4 Assistance Publique-Hôpitaux de Paris, Hôpital Louis-Mourier Service de Gynécologie-Obstétrique, 178 rue des Renouillers, 92700 Colombes, France. 5 Inserm IAME-U1137, 75000 Paris, France. 6 Hospices Civils de Lyon, Hôpital de la Croix Rousse, Department of Parasitology and Medical Mycology, 69004 Lyon, France. 7 Walking Team, Centre for Research in Neuroscience in Lyon, 69500 Bron, France. PMID: 36297244 PMCID: PMC9610672 DOI: 10.3390/pathogens11101187 Item in Clipboard Full text links Cite Display options Display options Format Abstract Even in the absence of manifestations at birth, children with congenital toxoplasmosis (CT) may develop serious long-term sequelae later in life. This systematic review aims to present the current state of knowledge to base an informed decision on how to optimally manage these pregnancies and children. For this, a systematic literature search was performed on 28 July 2022 in PubMed, CENTRAL, ClinicalTrials.gov, Google Scholar and Scopus to identify all prospective and retrospective studies on congenital toxoplasmosis and its long-term outcomes that were evaluated by the authors. We included 31 research papers from several countries. Virulent parasite strains, low socioeconomic status and any delay of treatment seem to contribute to a worse outcome, whereas an early diagnosis of CT as a consequence of prenatal screening may be beneficial. The rate of ocular lesions in treated children increases over time to 30% in European and over 70% in South American children and can be considerably reduced by early treatment in the first year of life. After treatment, new neurological manifestations are not reported, while ocular recurrences are observed in more than 50% of patients, with a mild to moderate impact on quality of life in European cohorts when compared to a significantly reduced quality of life in the more severely affected South American children. Though CT is rare and less severe in Europe when compared with South America, antenatal screening is the only effective way to diagnose and treat affected individuals at the earliest possible time in order to reduce the burden of disease and achieve satisfying outcomes. Keywords: congenital toxoplasmosis; follow-up; long-term outcomes; retinochoroiditis; treatment outcome. PubMed Disclaimer Conflict of interest statement Garweg acts as an advisor for several pharmaceutical companies (AbbVie, Bayer, Novartis and Roche) and participated in several international independent and industry-sponsored clinical studies without any bearing on this work. The other authors have no conflicts of interest relevant to this article to disclose. Figures Figure 1 PRISMA flow diagram. From: Page,… Figure 1 PRISMA flow diagram. From: Page, M.J.; McKenzie, J.E.; Bossuyt, P.M.; Boutron, I. Hoffmann,… Figure 1 PRISMA flow diagram. From: Page, M.J.; McKenzie, J.E.; Bossuyt, P.M.; Boutron, I. Hoffmann, T.C., Mulrow, C.D. et al. The PRISMA 2020 statement: an updated guideline for reporting systematic reviews. BMJ 2021, 372, n71. doi: 10.1136/bmj.n71. See this image and copyright information in PMC Similar articles The ocular manifestations of congenital infection: a study of the early effect and long-term outcome of maternally transmitted rubella and toxoplasmosis.O'Neill JF.O'Neill JF.Trans Am Ophthalmol Soc. 1998;96:813-79.Trans Am Ophthalmol Soc. 1998.PMID: 10360309 Free PMC article. Ophthalmic outcomes of congenital toxoplasmosis followed until adolescence.Wallon M, Garweg JG, Abrahamowicz M, Cornu C, Vinault S, Quantin C, Bonithon-Kopp C, Picot S, Peyron F, Binquet C.Wallon M, et al.Pediatrics. 2014 Mar;133(3):e601-8. doi: 10.1542/peds.2013-2153. Epub 2014 Feb 17.Pediatrics. 2014.PMID: 24534412 Ocular manifestations in congenital toxoplasmosis.Kodjikian L, Wallon M, Fleury J, Denis P, Binquet C, Peyron F, Garweg JG.Kodjikian L, et al.Graefes Arch Clin Exp Ophthalmol. 2006 Jan;244(1):14-21. doi: 10.1007/s00417-005-1164-3. Epub 2005 May 20.Graefes Arch Clin Exp Ophthalmol. 2006.PMID: 15906073 The congenital toxoplasmosis burden in Brazil: Systematic review and meta-analysis.Strang AGGF, Ferrari RG, do Rosário DK, Nishi L, Evangelista FF, Santana PL, de Souza AH, Mantelo FM, Guilherme ALF.Strang AGGF, et al.Acta Trop. 2020 Nov;211:105608. doi: 10.1016/j.actatropica.2020.105608. Epub 2020 Jun 29.Acta Trop. 2020.PMID: 32615081 New clinical and experimental insights into Old World and neotropical ocular toxoplasmosis.Pfaff AW, de-la-Torre A, Rochet E, Brunet J, Sabou M, Sauer A, Bourcier T, Gomez-Marin JE, Candolfi E.Pfaff AW, et al.Int J Parasitol. 2014 Feb;44(2):99-107. doi: 10.1016/j.ijpara.2013.09.007. Epub 2013 Nov 4.Int J Parasitol. 2014.PMID: 24200675 Review. See all similar articles Cited by Clinical Spectrum, Radiological Findings, and Outcomes of Severe Toxoplasmosis in Immunocompetent Hosts: A Systematic Review.Layton J, Theiopoulou DC, Rutenberg D, Elshereye A, Zhang Y, Sinnott J, Kim K, Montoya JG, Contopoulos-Ioannidis DG.Layton J, et al.Pathogens. 2023 Mar 31;12(4):543. doi: 10.3390/pathogens12040543.Pathogens. 2023.PMID: 37111429 Free PMC article.Review. In defense of children's brain: reshuffling the laboratory toolbox for the diagnosis of congenital toxoplasmosis.Montoya JG.Montoya JG.J Clin Microbiol. 2024 Jun 12;62(6):e0169723. doi: 10.1128/jcm.01697-23. Epub 2024 May 23.J Clin Microbiol. 2024.PMID: 38780287 Free PMC article. Animal venoms: a novel source of anti-Toxoplasma gondii drug candidates.Yang D, Liu X, Li J, Xie J, Jiang L.Yang D, et al.Front Pharmacol. 2023 May 3;14:1178070. doi: 10.3389/fphar.2023.1178070. eCollection 2023.Front Pharmacol. 2023.PMID: 37205912 Free PMC article.Review. Antenatal Screening for Toxoplasmosis and Rubella in Saudi Arabia: Assessing the Need for Screening.Al-Mughales J, Al-Rabia MW.Al-Mughales J, et al.J Multidiscip Healthc. 2023 Dec 6;16:3897-3905. doi: 10.2147/JMDH.S438895. eCollection 2023.J Multidiscip Healthc. 2023.PMID: 38084119 Free PMC article. Ocular Toxoplasmosis: Advances in Toxoplasma gondii Biology, Clinical Manifestations, Diagnostics, and Therapy.Miyagaki M, Zong Y, Yang M, Zhang J, Zou Y, Ohno-Matsui K, Kamoi K.Miyagaki M, et al.Pathogens. 2024 Oct 14;13(10):898. doi: 10.3390/pathogens13100898.Pathogens. 2024.PMID: 39452769 Free PMC article.Review. See all "Cited by" articles References Lago E.G., Endres M.M., Scheeren M.F.D.C., Fiori H.H. Ocular Outcome of Brazilian Patients With Congenital Toxoplasmosis. Pediatr. Infect. Dis. J. 2021;40:e21–e27. doi: 10.1097/INF.0000000000002931. - DOI - PubMed Delair E., Monnet D., Grabar S., Dupouy-Camet J., Yera H., Brézin A.P. Respective Roles of Acquired and Congenital Infections in Presumed Ocular Toxoplasmosis. Am. J. Ophthalmol. 2008;146:851–855. doi: 10.1016/j.ajo.2008.06.027. - DOI - PubMed Wallon M., Peyron F., Cornu C., Vinault S., Abrahamowicz M., Kopp C.B., Binquet C. Congenital Toxoplasma Infection: Monthly Prenatal Screening Decreases Transmission Rate and Improves Clinical Outcome at Age 3 Years. Clin. Infect. Dis. 2013;56:1223–1231. doi: 10.1093/cid/cit032. - DOI - PubMed Kodjikian L., Wallon M., Fleury J., Denis P., Binquet C., Peyron F., Garweg J.G. Ocular manifestations in congenital toxoplasmosis. Graefe’s Arch. Clin. Exp. Ophthalmol. 2006;244:14–21. doi: 10.1007/s00417-005-1164-3. - DOI - PubMed Gilbert R., Gras L., Wallon M., Peyron F., Ades E.A., Dunn D.T. Effect of prenatal treatment on mother to child transmission of Toxoplasma gondii: Retrospective cohort study of 554 mother-child pairs in Lyon, France. Int. J. Epidemiol. 2001;30:1303–1308. doi: 10.1093/ije/30.6.1303. - DOI - PubMed Show all 110 references Publication types Review Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Bern Open Repository and Information System Europe PubMed Central MDPI PubMed Central Full text links[x] MDPIFree PMC article [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. 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188333	https://researchoutreach.org/articles/divide-conquer-strategy-vehicle-routing-problem/	A divide-and-conquer strategy for the vehicle routing problem Search Home Publication Articles Arts & Humanities Behavioural Sciences Biology Business & Economics Earth & Environment Education & Training Health & Medicine Engineering & Technology Physical Sciences Thought Leaders Community Content Outreach Leaders About Us Our Company Our Clients FAQ Testimonials Our Services Researcher App Podcasts & Video Abstracts Blog Contact Us Search Subscribe To Our Free Publication Thank you for expressing interest in joining our mailing list and community. Below you can select how you’d like us to interact with you and we’ll keep you updated with our latest content. You can change your preferences or unsubscribe by clicking the unsubscribe link in the footer of any email you receive from us, or by contacting us at audience@researchoutreach.org at any time and if you have any questions about how we handle your data, please review our privacy agreement.. indicates required Email Address I would like to receive updates about: By selecting any of the topic options below you are consenting to receive email communications from us about these topics. [x] Health & Medicine [x] Biology [x] Behavioral Science [x] Earth & Environment [x] Engineering & Technology [x] Physical Sciences [x] Education & Training [x] Arts & Humanities [x] Thought Leaders [x] All The Above I consent to receiving promoted content: We are able share your email address with third parties (such as Google, Facebook and Twitter) in order to send you promoted content which is tailored to your interests as outlined above. If you are happy for us to contact you in this way, please tick below. [x] I consent to receiving promoted content. Would you like to learn more about our services? - [x] I would like to learn more about Research Outreach's services. We use MailChimp as our marketing automation platform. By clicking below to submit this form, you acknowledge that the information you provide will be transferred to MailChimp for processing in accordance with their Privacy Policy and Terms. Engineering & Technology December 5, 2022 A divide-and-conquer strategy for the vehicle routing problem Solving the vehicle routing problem is vital for distribution and transportation businesses needing to ensure timely distribution and minimise costs. The multivehicle routing problem is a complex variation involving multiple vehicles and numerous destinations for goods. Jiaqi Li, a graduate student at The University of Hong Kong, Yun Wang at Zhejiang Sci-Tech University, China, and Professor Ke-Lin Du at Concordia University, Canada, have vanquished this problem. The researchers have developed a new, improved genetic algorithm as a mathematical solution – with a novel ‘divide-and-conquer’ strategy inspired by dynamic programming. For optimal vehicle routing, ensuring timely distribution and reducing the cost of logistics (accurately delivering goods from the supplier to the customer) is crucial to the viability of distribution and transportation businesses. It was initially investigated by G B Dantzig and J H Ramser in 1959 and is known as ‘The Truck Dispatching Problem’. Behind this classical vehicle routing problem (VRP) is the need to uncover the optimal routes for a fleet of vehicles delivering to multiple locations, at the least possible cost to the distributor. It is relatively easy to understand the VRP; solving it is another matter. There is a more complex variation of the VRP called the multivehicle routing problem (MVRP). Solutions for the MVRP must factor in a fleet of vehicles delivering to a myriad of customers while also considering traveling-time delays arising from traffic congestion. All this must be achieved at the lowest cost. Fortunately, three researchers – Jiaqi Li, a graduate student at The University of Hong Kong, and colleagues Yun Wang from Zhejiang Sci-Tech University, China, and Professor Ke-Lin Du from Concordia University, Montreal, Canada – have defeated the conundrum with an innovative mathematical solution inspired by dynamic programming. The vehicle routing problem The VRP is a generalisation of the travelling salesperson problem, a classic optimisation problem that has been examined since the 1930s. It is based on a scenario where a salesperson must visit several cities. Their journey must finish back where they started, and they can visit each location only once. They can visit the cities in any order, but they need to travel the least possible distance. The problem is modelled as a network, with the cities represented by nodes and connected by paths weighted in terms of the time or distance required to travel between two cities. The VRP is a combinatorial integer programming problem. It involves minimising the total route cost and establishing the maximum number of stops each vehicle can make while keeping operating costs to a minimum. It considers factors such as the number of vehicles, vehicle capacity, time constraints, location of customers, order priority, speed limits, and traffic patterns. The VRP is traditionally solved using metaheuristics – strategies to guide the search process to find optimal or near-optimal solutions – in the form of an evolutionary or genetic algorithm (GA). GAs mimic biological evolution to solve optimisation problems. Using a natural selection process, the genetic algorithm repeatedly modifies a group of solutions. The VRP is an NP (nondeterministic polynomial time) Problem and is said to be NP-hard, so it is ‘at least as hard as any NP Problem’. That means an algorithm used to solve the VRP can be translated and used to solve any NP Problem. The multivehicle routing problem The MVRP extends the VRP to include mixed-fleet vehicles with time windows (specific time slots when visits must be made). Fleets can consist of any combination of vehicles using traditional fuel, electric vehicles, and plug-in hybrids. The goal of the MVRP is to discover a set of routes that multiple vehicles can use to serve numerous customers, which keeps total costs to a minimum but still takes traffic delays into account. Researchers have developed a new mathematical model to solve the MVRP using an improved genetic algorithm with a novel ‘divide-and-conquer’ strategy. Now researchers have developed a new mathematical model to solve the MVRP using an improved genetic algorithm with a novel ‘divide-and-conquer’ strategy. Inspired by dynamic programming, Li, Wang, and Du propose an distribution path optimisation method with two stages: a distribution sequence search and a distribution path search. Vanquishing the MVRP The objective of the researchers’ proposed model is to find the shortest possible route requiring the least number of vehicles while keeping costs to a minimum. Multiple vehicles are being used to distribute products to multiple demand nodes, so the optimal number of vehicles required and the route for each vehicle will have to be determined. Ensuring timely distribution and reducing the cost of logistics is crucial to the viability of distribution and transportation businesses. This is modelled with two objective functions (mathematical expressions whose values must be either maximised or minimised according to constraints). The first objective function involves minimising the total cost. The second objective function minimises the total length of the route. These are subject to the following constraints: a vehicle’s maximum load capacity cannot be exceeded; only one vehicle can fulfil a delivery at any node; all distribution jobs are completed by the assigned number of vehicles; only one vehicle can arrive or depart from a demand node; and, finally, there are no loops or duplicated sections within the optimised path or distribution sequence. Divide-and-conquer strategy When creating their new divide-and-conquer strategy, the researchers were inspired by dynamic programming, where an optimisation problem is broken down into simpler sub-problems. When there are a lot of demand nodes and road intersections, the problem is NP-hard, and it isn’t easy to uncover an optimal route. Instead, Li and colleagues break the problem into smaller sub-problems that they solve recursively. Initially, they find an optimal sequence of demand nodes. Then, using their improved genetic algorithm, they search for an optimal route between any two of the nodes and build the complete route step by step. An improved genetic algorithm Each solution in a genetic algorithm is referred to as an ‘individual’ and coded like a chromosome. A group of individuals form a ‘population’. Here, an individual is a route between two nodes. Individuals are selected from the population using a combination of roulette wheel selection and an elitist strategy. A roulette wheel is constructed based on each individual’s relative ‘fitness’ to select the individuals for the next generation – or population. Elitist selection retains the individuals with the best fitness values for the next generation. Selected individuals form a new population, and the process is repeated for all paths between nodes for a vehicle’s complete route. This process is repeated to find optimal distribution paths for each of the remaining vehicles. Real-life application The researchers verify the rationality and feasibility of their model with a simulation using data from a manufacturer in Hangzhou, China. The manufacturer delivers products to 40 customers. Analysing the results revealed that the new, improved genetic algorithm – employing the divide-and-conquer approach – converged on the optimal solution quicker than the traditional, simple genetic algorithm. Moreover, the improved genetic algorithm outperformed the traditional genetic algorithm for route feasibility, route topology, and total cost. The researchers’ new methodology for solving the MVRP can help to reduce the overall cost of a fast and convenient distribution service. In addition to increasing logistics efficiency for manufacturers, distributors, and transport companies, Li and his colleagues’ model can be useful for other applications, such as optimising manufacturing flow–shop scheduling. Personal Response What inspired you to develop the divide-and-conquer strategy? The divide-and-conquer strategy is the spotlight of our work. When we investigated route planning on a map, we found that for two locations in a large region there may be tens of thousands of nodes for route planning, and it is impossible to find an optimal or suboptimal route. For example, if we plan a route from a location in Hangzhou to a location in Beijing (a distance of over 1,000 kilometres), there are tens or hundreds of thousands of road intersections between the two locations; no matter what metaheuristics are used, it is impossible to make the route planner practical. We therefore designed the divide-and-conquer strategy, inspired by dynamic programming, to speed up a metaheuristic to effectively solve the problem. What are the next steps before your model is ready for real-world use? The divide-and-conquer strategy is practical, not patented, and it can be used by anyone in their real-world applications. A divide-and-conquer strategy for the vehicle routing problem was last modified: November 8th, 2023 Keywords: ‘divide-and-conquer’ strategy canada China Concordia University distribution dynamic programming genetic algorithm Jiaqi Li multivehicle routing problem MVRP NP Problem Professor Ke-Lin Du The Truck Dispatching Problem The University of Hong Kong transportation transportation businesses vehicle routing vehicle routing problem VRP Yun Wang Zhejiang Sci-Tech University Download this Article References Li, J, Wang, Y, Du, K-L, (2022) Distribution path optimization by an improved genetic algorithm combined with a divide-and-conquer strategy. Technologies, 10(4), 81. www.doi.org/10.3390/technologies10040081 Dantzig, GB, Ramser, JH, (1959) The Truck Dispatching Problem. Management Science, 6(1), 80–91. www.doi.org/10.1287/mnsc.6.1.80 Behind the Research Jiaqi Li Jiaqi Li has a BSc degree in mechanical engineering and is a graduate student in the Faculty of Engineering at The University of Hong Kong. Yun Wang is a faculty member at Zhejiang University, China. Professor Ke-Lin Du is an associate professor at Concordia University, Canada. Contact Details Contact Details Address Flat D, 13th Floor, Block C, Chong Yip Centre, 423-425 Queen’s Road West, Hong Kong E: jucky@connect.hku.hk T: +852 6672 1022 Close Research Objectives Li, Wang, and Du have worked out how to efficiently distribute goods from manufacturers’ distribution nodes to retailers’ demand nodes at the lowest cost possible. Funding The National Natural Science Foundation of China (Grant number: 51475434). Cite this Article DOI:10.32907/RO-133-37011672744 Create Harvard Reference Harvard Reference Jiaqi Li, Yun Wang, Ke-Lin Du (2022) A divide-and-conquer strategy for the vehicle routing problem. Research Outreach, 133. 10.32907/RO-133-37011672744 (Accessed YYYY/MM/DD) Close Creative Commons Licence (CC BY-NC-ND 4.0) This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. What does this mean? Share: You can copy and redistribute the material in any medium or format More Articles New pipelining method for heavy oil: Stabilising the flow of high viscosity fluids during transportation Physical Sciences Read This Article Vibration mitigation: New Force-Network based granular damping technology Physical Sciences Read This Article This feature article was created with the approval of the research team featured. This is a collaborative production, supported by those featured to aid free of charge, global distribution. 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188334	https://www.fool.com/investing/how-to-invest/annuity-due-vs-ordinary-annuity/	▲ S&P 500 +190% \| ▲ Stock Advisor +1059% Join The Motley Fool Accessibility Log In Help Accessibility Menu S&P 500 6,631.96 +0.5% +$31.61 DJI 46,142.42 +0.3% +$124.10 NASDAQ 22,470.73 +0.9% +$209.40 Bitcoin 116,491.00 -0.6% -734.10 AAPL $237.98 -0.4% -$1.02 AMZN $231.24 -0.2% -$0.38 GOOG $252.45 +1.0% +$2.60 META $779.83 +0.5% +$4.12 MSFT $508.48 -0.3% -$1.54 NVDA $176.28 +3.5% +$5.98 TSLA $416.87 -2.1% -$8.99 Daily Stock Gainers Daily Stock Losers Most Active Stocks Daily Stock Gainers Daily Stock Losers Most Active Stocks Home > Investing > How To Invest > Annuity Due Vs Ordinary Annuity The Difference Between Ordinary Annuity and Annuity Due By Catherine Brock – Updated Jan 13, 2025 at 5:12PM Key Points Ordinary annuities pay at the end of a period. Annuities due pay in advance or at the beginning of a period. Because of the difference in payment timing, the present value of an annuity due will be higher than that of an ordinary annuity with otherwise equal terms. Investor Alert: Our 10 best stocks to buy right now › Key findings are powered by ChatGPT and based solely off the content from this article. Findings are reviewed by our editorial team. The author and editors take ultimate responsibility for the content. How does an ordinary annuity differ from an annuity due? Both are contractually obligated payment series, but they differ in the timing of the payment. An ordinary annuity pays at the end of a period, while an annuity due pays in advance. Image source: Getty Images. Ordinary annuity What is an ordinary annuity? An ordinary annuity pays at the end of a period -- so, the payment covers the period that has already passed. The period can be any designated time frame, such as a month, quarter, or year. Note that "the end" of the period could be the first of the month. The date on the calendar is irrelevant. What's relevant is whether the payment covers the prior month or the following month. The annuity contract will specify this information, but the timing of the first payment can also be an indicator. Consider a traditional mortgage as an example. The first mortgage payment is due on the first of the month after you've owned the home for 30 days. That payment is in arrears, which makes the mortgage an ordinary annuity. Car loans work the same way. Your payments begin 30 days after the loan funds. Annuity due What is an annuity due? An annuity due is paid at the beginning of the period. Rent and lease agreements are examples. Your rent for September is due at the beginning of September -- not the end. Subscription fees for services like Netflix (NFLX -1.75%) and Amazon (AMZN -0.16%) Prime are also annuities due because you pay them ahead of the service period. Key differences Ordinary annuity versus annuity due: Key differences Payment timing and its effect on the annuity's present value are two key differences between an ordinary annuity and an annuity due. Timing As noted, the main difference between an ordinary annuity and an annuity due is whether the payment is made in arrears or in advance. Present value The timing of payments, in turn, affects the annuity's present value. Present value is what a future stream of payments is worth today. Understanding present value can help you evaluate an income annuity relative to its cost. First, know that the present value of any annuity will be less than the sum of the payments. This is because cash promised in the future is less valuable than cash in your hand today. Why? Because you can invest and grow cash on hand -- which you cannot do with cash promised. Present value formulas account for this by using an interest rate to discount those future payments. The present value of the ordinary annuity formula considers the dollar amount of each payment, the discount rate, and the number of payments. The present value of the annuity due formula uses the same inputs but adjusts for the earlier payment timing. Mathematically, that adjustment involves multiplying the result by the discount rate plus 1. You can see this by comparing the two present value formulas below. Note that "pmt" equals the payment amount, "r" equals the discount rate, and "n" is the total number of payments. Present value of ordinary annuity =pmt [(1–[1/(1+r)^n])/r] Present value of annuity due =pmt [(1–[1/(1+r)^n])/r] x (1+r) The takeaway is that an annuity due will have a higher present value than an ordinary annuity if all other factors are the same. Related investing topics #### How to Invest Money: A Step-by-Step Guide Before you put down hard-earned cash, start with a basic understanding of how investing works -- and how to avoid rookie mistakes. #### Best Index Funds and How to Invest in Them Index funds track a particular index and can be a good way to invest. Get a fast introduction to index funds here. #### How to Invest in Mutual Funds Mutual funds give investors exposure to lots of different kinds of investments. #### How to Invest in ETFs for Beginners Exchange-traded funds let an investor buy many stocks and bonds at once in a single transaction. Which is better? Ordinary annuity or annuity due: Which is better? Ordinary annuities are better for the payer, while annuities due are better for the payee. In other words, if you are paying the annuity, you'd rather pay later. Paying in arrears allows you to keep your funds invested longer -- or gives you more time to earn them via your paycheck. If you are receiving annuity income, an annuity due is preferred because you get the money sooner. Assuming monthly payments, an annuity due puts the cash in your hands one month earlier than an ordinary annuity. FAQ Ordinary annuity versus annuity due: FAQ What is the difference between an ordinary annuity and an annuity due? The timing of the payments is what makes an ordinary annuity differ from an annuity due. Ordinary annuity payments are made at the end of a period, which can be monthly, quarterly, or annually. Annuity due payments, on the other hand, are made at the beginning of the period. You pay your credit card bill at the end of the billing cycle, so it's an ordinary annuity. However, you pay rent, subscription fees, and insurance premiums in advance, making them annuities due. Annuities sold by insurance companies to provide retirement income can be structured as ordinary annuities or annuities due. What are the four types of annuities? The four main types of annuities are immediate, deferred, fixed, and variable. Immediate versus deferred annuities differ in when the payments begin. Fixed and variable annuities differ in risk and return. An immediate annuity begins producing income right away. You'd purchase an immediate annuity with a lump sum of cash. A deferred annuity starts generating payments at some future date, with taxes also deferred until that time. You can buy a deferred annuity with a single cash payment or a series of payments over time. Fixed annuities provide income based on a fixed interest rate. They are lower risk because the interest rate and payment amount don't change. Variable annuities produce income based on the performance of sub-accounts, which are usually stock or bond investment funds chosen by the annuitant. These have more return potential and more risk than fixed annuities. What is considered an ordinary annuity? An ordinary annuity can be any financial obligation that requires periodic payments made at the end of a period. Mortgages and car loans are ordinary annuities because you pay those in arrears, usually starting 30 or more days after the loan funds. About the Author Catherine grew up in Southern California wearing a lot of black and trying to perfect the art of sarcasm. Prior to joining The Fool as a contract writer, Catherine was climbing the corporate ladder in marketing roles and dabbling in too many side hustles. When she's not writing, she can be found riding a horse in the country or shopping online for clothes. XMFCatherineJB @catherinejbrock John Mackey, former CEO of Whole Foods Market, an Amazon subsidiary, is a member of The Motley Fool’s board of directors. Catherine Brock has no position in any of the stocks mentioned. The Motley Fool has positions in and recommends Amazon and Netflix. The Motley Fool has a disclosure policy. 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188335	https://prepp.in/question/to-avoid-plate-tearing-at-the-edge-in-a-riveted-jo-664db80f48b4bcbda2cda8f3	To avoid plate tearing at the edge in a riveted joint, what should be the relation between the margin (m) and the rivet diameter (d)? Hello, Guest Login / Register All Categories+ Test SeriesQuizzesPrevious Year PapersLive TestsLive QuizzesCurrent AffairsVideosNewsContact Us Get App All Exams Test series for 1 year @ ₹349 onlyEnroll Now ×× Home Design of Machine Elements Riveted Joint to avoid plate tearing at the edge in a riveted jo Question Design of Machine Elements To avoid plate tearing at the edge in a riveted joint, what should be the relation between the margin (m) and the rivet diameter (d)? m + d = 1.5 mm m - d = 1.5 mm m = 1.5 d d = 1.5 m Solution The correct answer is m = 1.5 d Understanding Margin in Riveted Joints to Prevent Tearing In the design of riveted joints, it is crucial to ensure that the joint can withstand the applied load without failing. One common mode of failure is the tearing of the plate material at the edge, particularly between the rivet hole and the edge of the plate. This tearing occurs when the material's tensile strength in this region is exceeded. What is Margin in a Riveted Joint? The margin (often denoted by 'm') in a riveted joint is defined as the distance from the center of the rivet hole to the nearest edge of the plate. This distance is important because it determines the amount of material available to resist the tearing force at the edge of the plate. Preventing Plate Tearing at the Edge To prevent the plate from tearing at the edge, there must be sufficient material between the rivet hole and the edge. A commonly accepted rule or thumb, based on empirical data and design standards, is used to specify this minimum margin distance. This rule relates the margin 'm' to the nominal diameter of the rivet 'd'. The relationship typically specified to avoid edge tearing is that the margin should be at least 1.5 times the rivet diameter. This provides an adequate amount of material to resist the shear and tensile stresses that can cause tearing near the edge. The relationship is expressed mathematically as: m=1.5 d m = 1.5 d m=1.5 d Where: m m m is the margin (distance from rivet center to plate edge) d d d is the nominal diameter of the rivet Following this guideline helps ensure that the strength of the plate material between the rivet hole and the edge is sufficient to prevent premature failure due to tearing under load. Other potential failure modes in a riveted joint include shearing of the rivets, crushing of the plate or rivet, and tearing of the plate between rivets, but the margin specifically addresses tearing at the edge. Conclusion To effectively prevent plate tearing at the edge in a riveted joint, the margin (m) must be appropriately chosen in relation to the rivet diameter (d). The standard design practice for this is to maintain a margin equal to 1.5 times the rivet diameter. Download PDF Was this answer helpful? 0 0 Important Questions from Riveted Joint If the tearing efficiency of a riveted joint is 60%, then ratio of rivet hole diameter to the pitch of rivets is _____. Design of Machine Elements View Answer The ratio of the diameter of rivet hole to the pitch of the rivet is 0.25, then the tearing efficiency of the joint is Design of Machine Elements View Answer The strength of a riveted joint would be _. Design of Machine Elements View Answer The strength of the unriveted or solid plate per pitch length is equal to Design of Machine Elements View Answer Which one of the following is not a rivet's primary component? Design of Machine Elements View Answer Need Expert Advice? Ask A Question Start Your Preparation with Prepp Mobile App Download the app from Google Play & App Store Download the app from Google Play & App Store Prepp Community Download Mobile App Government Exams & Jobs AFCAT CDS NDA UP Police UGC NET IAS Exam MPSC BPSC Railways RRB NTPC IBPS PO SBI CLERK SBI PO IBPS Clerk SSC CGL SSC CHSL CTET NTSE UPTET KVPY Previous Year Papers IAS Question Paper BPSC Question Paper IBPS PO Question Paper NDA Question Paper CDS Question Paper UGC NET Question Paper SSC CGL Question Paper RBI Grade B Question Paper RRB Question Paper SBI PO Question Paper Notes Current Affairs IAS Notes NDA Notes SBI PO Notes CDS Notes SSC CGL Notes SSC MTS Notes SSC CHSL Notes RRB ALP Notes RRB NTPC Notes About Us Contact Us Terms & Conditions Privacy Policy Prepp © 2025 Super Charge Your Preparation With Prepp+ Mock Test For All Government Exams Benefits of Prepp+ Subscription loading.. x
188336	https://chem.libretexts.org/Courses/College_of_Marin/CHEM_114%3A_Introductory_Chemistry/07%3A_Chemical_Reactions/7.08%3A_Acid%E2%80%93Base_and_Gas_Evolution_Reactions	Skip to main content 7.8: Acid–Base and Gas Evolution Reactions Last updated : May 19, 2019 Save as PDF 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations 7.9: Oxidation–Reduction Reactions Page ID : 98034 ( \newcommand{\kernel}{\mathrm{null}\,}) Template:HideTOC Learning Objectives Identify when a reaction will evolve a gas. Neutralization Reactions Acids and bases react chemically with each other to form salts. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen-ion-containing compound and the base is a hydroxide-ion-containing compound, water is also a product. The general reaction is as follows: acid + base→water + salt The reaction of acid and base to make water and a salt is called neutralization. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows: NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(ℓ)(7.8.1) with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows: 2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(ℓ)(7.8.2) Example 7.8.1: Neutralizing Nitric Acid Nitric acid (HNO3(aq)) can be neutralized by calcium hydroxide (Ca(OH)2(aq)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Solution Solutions to Example 7.8.1 \| Steps \| Explanation \| Equation \| \| Write the unbalanced equation. \| This is a double displacement reaction, so the cations and anions swap to create new products. \| Ca(OH)2(aq) + HNO3(aq) → Ca(NO3)2(aq) + H2O(ℓ) \| \| Balance the equation. \| Because there are two OH− ions in the formula for Ca(OH)2, we need two moles of HNO3 to provide H+ ions \| Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ) \| \| Additional step: identify the salt. \| \| The salt formed is calcium nitrate. \| Exercise 7.8.1 Hydrocyanic acid (HCN(aq)) can be neutralized by potassium hydroxide (KOH(aq)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Answer [\ce{KOH (aq) + HCN(aq) → KCN (aq) + H2O(ℓ)} \nonumber ] Gas Evolving Reactions A gas evolution reaction is a chemical process that produces a gas, such as oxygen or carbon dioxide. In the following examples, an acid reacts with a carbonate, producing salt, carbon dioxide, and water, respectively. For example, nitric acid reacts with sodium carbonate to form sodium nitrate, carbon dioxide, and water (Table 7.8.1): 2HNO3(aq)+Na2CO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l) Sulfuric acid reacts with calcium carbonate to form calcium sulfate, carbon dioxide, and water: H2SO4(aq)+CaCO3(aq)→CaSO4(aq)+CO2(g)+H2O(l) Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide, and water: 2HCl(aq)+CaCO3(aq)→CaCl2(aq)+CO2(g)+H2O(l) Figure 7.8.1 demonstrates this type of reaction: In this reaction setup, lime water, a dilute calcium hydroxide (Ca(OH)2) solution, is poured into one of the test tubes and sealed with a stopper. A small amount of hydrochloric acid is carefully poured into the remaining test tube. A small amount of sodium carbonate is added to the acid, and the tube is sealed with a rubber stopper. The two tubes are connected. As a result of the acid-carbonate reaction, carbon dioxide is produced and the lime water turns milky. Table 7.8.1: Types of Compounds That Undergo Gas-Evolution Reactions \| Reactant Type \| Intermediate Product \| Gas Evolved \| Example \| \| sulfide \| none \| H2S \| 2HCl(aq)+K2S→H2S(g)+2KCl(aq) \| \| carbonates and bicarbonates \| H2CO3 \| CO2 \| 2HCl(aq)+K2CO2→H2O(l)+CO2(g)+2KCl(aq) \| \| sulfites and bisulfites \| H2SO3 \| SO2 \| 2HCl(aq)+K2SO2→H2O(l)+SO2(g)+2KCl(aq) \| \| ammonia \| NH4OH \| NH3 \| (\ce{NH4Cl(aq) + KOH \rightarrow H2O (l) + NH3(g) + 2KCl (aq)}) \| The gas-evolving experiment lime water is illustrated in the following video: Video 7.8.1: Carbon Dioxide (CO2) & Limewater (Chemical Reaction). As the reaction proceeds, the limewater on the turns from clear to milky; this is due to the CO2(g) reacting with the aqueous calcium hydroxide to form calcium carbonate, which is only slightly soluble in water. When this experiment is repeated with nitric or sulfuric acid instead of HCl, it yields the same results: the clear limewater turns milky, indicating the production of carbon dioxide. Another method to chemically generate gas is the oxidation of metals in acidic solutions. This reaction will yield a metal salt and hydrogen gas. 2HCl(aq)+Zn(s)→ZnCl2(aq)+H2(g) Here, hydrochloric acid oxidizes zinc to produce an aqueous metal salt and hydrogen gas bubbles. Recall that oxidation refers to a loss of electrons, and reduction refers to the gain of electrons. In the above redox reaction, neutral zinc is oxidized to Zn2+, and the acid, H+, is reduced to H2(g). The oxidation of metals by strong acids is another common example of a gas evolution reaction. Contributors & Affiliations Boundless (www.boundless.com) Wikipedia (CC-BY-SA-3.0) Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations 7.9: Oxidation–Reduction Reactions
188337	https://www.studocu.com/en-us/messages/question/7424938/ozone-o3-is-produced-in-automobile-exhaust-by-the-reaction-represented-by-the-equation-no2g	[Solved] Ozone O3 is produced in automobile exhaust by the reaction - Chemistry SL - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent Ozone, O3, is produced in automobile exhaust by the reaction represented by the equation, NO2(g) + O2(g) → NO(g) + O3(g). What mass of ozone is predicted to form from the reaction of 2.0 g NO2 in a car's exhaust and excess oxygen? Element Molar mass Nitrogen 14.01 g/mol Oxygen 16.00 g/mol a. 4.2 g O3 b. 1.1 g O3 c. 1.8 g O3 d. 2.1 g O3 Chemistry SL My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Flvs Full-Time 9-12 Chemistry SL Question Ozone O3 is produced in automobile exhaust by the reaction Flvs Full-Time 9-12 Chemistry SL Question Anonymous Student 1 year ago Ozone, O3, is produced in automobile exhaust by the reaction represented by the equation, NO2(g) + O2(g) → NO(g) + O3(g). What mass of ozone is predicted to form from the reaction of 2.0 g NO2 in a car's exhaust and excess oxygen? Element Molar mass Nitrogen 14.01 g/mol Oxygen 16.00 g/mol a. 4.2 g O3 b. 1.1 g O3 c. 1.8 g O3 d. 2.1 g O3 Like 0 Related documents 8.03 Solutions Lab Report: Molarity Calculations & Results Chemistry Assignments 100%(8) 5.07 Percent Yield Lab Report: Magnesium Oxide Experiment Analysis Chemistry Assignments 75%(16) 05.04 Stoichiometry I - Module 5 Study Notes on Mole Ratios Chemistry Class notes 100%(1) 8.03 Solutions Lab: Detailed Report and Data Analysis for Chem Students Chemistry Assignments None 8.03 Solutions Lab Assignment: Molarity Calculations & Post-Lab Analysis Chemistry Assignments 100%(3) 05 06 Lab Report: Determining Percent Yield of MgO from Reactions Chemistry Assignments None 8.03 Solutions Lab Assignment: Overview and Conclusion Insights Chemistry Assignments None Answer Created with AI 1 year ago Solution To solve this problem, we need to use the concept of stoichiometry, which is the calculation of relative quantities of reactants and products in chemical reactions. First, we need to calculate the molar mass of NO2 and O3. Molar Mass Calculation The molar mass of NO2 is calculated as follows: 1 Nitrogen atom: 14.01 g/mol 2 Oxygen atoms: 2 16.00 g/mol = 32.00 g/mol So, the molar mass of NO2 = 14.01 g/mol + 32.00 g/mol = 46.01 g/mol The molar mass of O3 is calculated as follows: 3 Oxygen atoms: 3 16.00 g/mol = 48.00 g/mol So, the molar mass of O3 = 48.00 g/mol Stoichiometry Calculation From the balanced chemical equation, we can see that 1 mole of NO2 produces 1 mole of O3. So, the moles of NO2 in 2.0 g is calculated as follows: js 2.0 g NO2 (1 mol NO2 / 46.01 g NO2) = 0.043 mol NO2 Since 1 mole of NO2 produces 1 mole of O3, the moles of O3 produced will also be 0.043 mol. Finally, we can calculate the mass of O3 produced as follows: js 0.043 mol O3 (48.00 g O3 / 1 mol O3) = 2.1 g O3 So, the answer is 2.1 g O3 (option d). Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Chemistry SL IB (International Baccalaureate Diploma) 804 Documents Go to course 8 Exploring Ideal Gas Laws: Gizmos Lab Activities Guide Chemistry SL 100%(14) 5 Assignment II.2 Food Chain Gizmo Answers Finished WORK Chemistry SL 92%(62) 3 Isotopes Worksheet Review - Practice for Chemistry Class Chemistry SL 100%(7) 12 IB Chemistry Stoichiometry Review Solutions and Explanations Chemistry SL 100%(7) Discover more from: Chemistry SL IB (International Baccalaureate Diploma) 804 Documents Go to course 8 Exploring Ideal Gas Laws: Gizmos Lab Activities Guide Chemistry SL 100%(14) 5 Assignment II.2 Food Chain Gizmo Answers Finished WORK Chemistry SL 92%(62) 3 Isotopes Worksheet Review - Practice for Chemistry Class Chemistry SL 100%(7) 12 IB Chemistry Stoichiometry Review Solutions and Explanations Chemistry SL 100%(7) 29 400134426 Access BARS: Sintesis de la Técnica y Beneficios Chemistry SL 100%(7) 16 IB SL Bonding Practice Questions & Answers (KEY)Chemistry SL 100%(6) Related Answered Questions 3 months ago medicinal chemistry questions Chemistry SL 4 months ago Briefly describe how one can determine moles of Mn²⁺ in a steel sample using atomic absorption or absorption (UV-Vis). Chemistry SL 4 months ago c) Perform the Q-test for the following data: 0.223 0.217 0.224 0.195 0.221 0.221 Chemistry SL 5 months ago Can you help me devise IB Chemistry EE research questions? You can choose any topic within or outside the Chemistry HL syllabus. The chosen topic should allow for a quantitative study of a chemical system. It must not overlap with the Chemistry IA (Scientific Investigation). Topics involving the use of IR and NMR spectroscopy, and electron microscopy, for example, will be rejected as these techniques cannot be carried out in a school laboratory. Topics which are better explored as a Biology or Physics EE will also be rejected. Chemistry SL 5 months ago Position Functionality 3363.47 O-H stretch (alcohol or carboxylic acid) 2971.61 C-H stretch (alkane) 2719.97 C-H stretch (aldehyde) 2658.54 C-H stretch (aldehyde) 2524.65 O-H stretch (carboxylic acid) 2408.54 Overtone or combination band 2298.05 C≡N stretch (nitrile) 2195.74 C≡C stretch (alkyne) 1902.43 C≡C stretch (alkyne) 1767.16 C=O stretch (ketone, ester, carboxylic acid) 1657.78 C=C stretch (alkene) 1467.03 C-H bend (alkane) 1379.13 C-H bend (alkane) 1319.76 C-O stretch (alcohol, ether, ester). The molecular formula for this spectrum is c8h8o. What compound is it? Chemistry SL 5 months ago What percentage of employees earn less than $300.00 per week but at least $260.00 per week? 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188338	https://brainly.com/question/4154005	[FREE] Solve by completing the square: a^2 + 10a + 21 = 0 - brainly.com 2 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +68,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +28,5k Ace exams faster, with practice that adapts to you Practice Worksheets +6,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Solve by completing the square: a 2+10 a+21=0 2 See answers Explain with Learning Companion NEW Asked by makayla63 • 06/21/2017 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 6717235 people 6M 0.0 0 Upload your school material for a more relevant answer To solve the equation a 2+10 a+21=0 by completing the square, you need to follow a series of steps. The solutions to the equation are a = -3 and a = -7. Explanation To solve the equation a 2+10 a+21=0 by completing the square, follow these steps: Move the constant term to the other side of the equation: Divide the coefficient of the linear term by 2, then square the result: Add the result from step 2 to both sides of the equation: Factor the perfect square trinomial on the left side: Take the square root of both sides of the equation: Set up the equation with the positive and negative square root: Solve for a: Therefore, the solutions to the equation a 2+10 a+21=0 by completing the square are a = -3 and a = -7. Learn more about Completing the square here: brainly.com/question/4822356 SPJ3 Answered by RachelMeghanMarkle •13.1K answers•6.7M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 6717235 people 6M 0.0 0 Physics for AP® Courses 2e - Kenneth Podolak, Henry Smith Physics 2e - Paul Peter Urone, Roger Hinrichs The AP Physics Collection - Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson, Nathan Czuba, Sudhi Oberoi, Liza Pujji, Irina Lyublinskaya, Douglas Ingram Upload your school material for a more relevant answer To solve the equation a 2+10 a+21=0 by completing the square, we found that the solutions are a=−3 and a=−7. By rearranging the equation, adding to both sides to form a perfect square, and solving, we arrived at these results. This method effectively helps in finding the roots of the quadratic equation. Explanation To solve the equation a 2+10 a+21=0 by completing the square, follow these steps: Move the constant term to the other side of the equation: a 2+10 a=−21 Divide the coefficient of the linear term by 2, then square the result: The coefficient of the linear term (10) divided by 2 is 5. Squaring this gives us 25. Add this result to both sides of the equation: a 2+10 a+25=−21+25 Which simplifies to: a 2+10 a+25=4 Factor the perfect square trinomial on the left side: (a+5)2=4 Take the square root of both sides of the equation: a+5=±2 Set up the equations with the positive and negative square root: This leads to two equations: a+5=2 a+5=−2 Solve for a: From a+5=2: a=2−5=−3 From a+5=−2: a=−2−5=−7 Thus, the solutions to the equation a 2+10 a+21=0 are a=−3 and a=−7. Examples & Evidence For example, if you have an equation like x 2+6 x+8=0, you can also complete the square to find its roots. The process would be similar: rearranging, calculating (2 b)2, and then solving. Completing the square is a standard technique in algebra to solve quadratic equations, and proven methods can be found in many textbooks and math tutorials. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 262451170 people 262M 3.0 5 a^2 + 10a + 21 = 0 a^2 + 10a = -21 a^2 + 10a + 25 = -21 + 25 (a + 5)^2 = 4 sqrt (a+5)^2 = sqrt 4 a + 5 = (+-)2 a = -5 (+-) 2 a = -5 + 2 = -3 a = -5 - 2 = -7 solution is : a = -3 and a = -7 <== Answered by texaschic101 •7.9K answers•262.5M people helped Thanks 5 3.0 (4 votes) Advertisement ### Free Mathematics solutions and answers Community Answer 10a²-3a=18 Solve for a using the quadratic formula or completing the square. CAN SOMEONE PLEASE HELP ME Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Which graph matches the inequality? $\begin{aligned} \frac{a}{-5} & \geq 10 \ a & \leq-50 \end{aligned}$ Joe told Mickey he got an hourly raise at work and his new rate will be 10.25 p er h o u r.M i c k ey w an t s t o kn o ww ha t J o e′s h o u r l yr a t e w a s b e f ore hi sr ai se.I f r$ represents the amount of the raise, which expression represents his hourly rate before his raise? A. r−10.25 B. 10.25−r C. 10.25+r D. r+10.25 Find the domain and range for each relation: {(3,9),(1,5),(2,9),(5,11),(3,12)} The product of two consecutive odd integers is 195. Kalinda is 68.7 inches tall and wants to know how many inches taller she is than Terrence. If t represents Terrence's height in inches, which expression can she use to find the answer? A. 68.7−t B. 68.7 t C. t 68.7 D. t−68.7 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
188339	https://pmc.ncbi.nlm.nih.gov/articles/PMC9685223/	Serial high-sensitivity cardiac troponin testing for the diagnosis of myocardial infarction: a scoping review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice BMJ Open . 2022 Nov 22;12(11):e066429. doi: 10.1136/bmjopen-2022-066429 Search in PMC Search in PubMed View in NLM Catalog Add to search Serial high-sensitivity cardiac troponin testing for the diagnosis of myocardial infarction: a scoping review Hirotaka Ohtake Hirotaka Ohtake 1 Department of Emergency and General Internal Medicine, Fujita Health University, Toyoake, Aichi, Japan Find articles by Hirotaka Ohtake 1, Teruhiko Terasawa Teruhiko Terasawa 1 Department of Emergency and General Internal Medicine, Fujita Health University, Toyoake, Aichi, Japan Find articles by Teruhiko Terasawa 1,✉, Zhivko Zhelev Zhivko Zhelev 2 University of Exeter Medical School, University of Exeter, Exeter, UK Find articles by Zhivko Zhelev 2, Mitsunaga Iwata Mitsunaga Iwata 1 Department of Emergency and General Internal Medicine, Fujita Health University, Toyoake, Aichi, Japan Find articles by Mitsunaga Iwata 1, Morwenna Rogers Morwenna Rogers 3 NIHR CLAHRC South West Peninsula, University of Exeter, Exeter, UK Find articles by Morwenna Rogers 3, Jaime L Peters Jaime L Peters 4 Peninsula Technology Assessment Group (PenTAG), University of Exeter, Exeter, UK Find articles by Jaime L Peters 4,4, Chris Hyde Chris Hyde 5 Exeter Test Group, University of Exeter, Exeter, UK Find articles by Chris Hyde 5 Author information Article notes Copyright and License information 1 Department of Emergency and General Internal Medicine, Fujita Health University, Toyoake, Aichi, Japan 2 University of Exeter Medical School, University of Exeter, Exeter, UK 3 NIHR CLAHRC South West Peninsula, University of Exeter, Exeter, UK 4 Peninsula Technology Assessment Group (PenTAG), University of Exeter, Exeter, UK 5 Exeter Test Group, University of Exeter, Exeter, UK ✉ Correspondence to Dr Teruhiko Terasawa; terasawa@fujita-hu.ac.jp ✉ Corresponding author. Series information Original research Received 2022 Jul 12; Accepted 2022 Oct 30; Collection date 2022. © Author(s) (or their employer(s)) 2022. Re-use permitted under CC BY. Published by BMJ. This is an open access article distributed in accordance with the Creative Commons Attribution 4.0 Unported (CC BY 4.0) license, which permits others to copy, redistribute, remix, transform and build upon this work for any purpose, provided the original work is properly cited, a link to the licence is given, and indication of whether changes were made. See: PMC Copyright notice PMCID: PMC9685223 PMID: 36414302 Abstract Objectives We aimed to assess the diversity and practices of existing studies on several assays and algorithms for serial measurements of high-sensitivity cardiac troponin (hs-cTn) for risk stratification and the diagnosis of myocardial infarction (MI) and 30-day outcomes in patients suspected of having non-ST-segment elevation MI (NSTEMI). Methods We searched multiple databases including MEDLINE, EMBASE, Science Citation Index, the Cochrane Database of Systematic Reviews and the CENTRAL databases for studies published between January 2006 and November 2021. Studies that assessed the diagnostic accuracy of serial hs-cTn testing in patients suspected of having NSTEMI in the emergency department (ED) were eligible. Data were analysed using the scoping review method. Results We included 86 publications, mainly from research centres in Europe, North America and Australasia. Two hs-cTn assays, manufactured by Abbott (43/86) and Roche (53/86), dominated the evaluations. The studies most commonly measured the concentrations of hs-cTn at two time points, at presentation and a few hours thereafter, to assess the two-strata or three-strata algorithm for diagnosing or ruling out MI. Although data from 83 studies (97%) were prospectively collected, 0%–90% of the eligible patients were excluded from the analysis due to missing blood samples or the lack of a final diagnosis in 53 studies (62%) that reported relevant data. Only 19 studies (22%) reported on head-to-head comparisons of alternative assays. Conclusion Evidence on the accuracy of serial hs-cTn testing was largely derived from selected research institutions and relied on two specific assays. The proportions of the eligible patients excluded from the study raise concerns about directly applying the study findings to clinical practice in frontline EDs. PROSPERO registration number CRD42018106379. Keywords: Coronary heart disease, Ischaemic heart disease, Myocardial infarction, Cardiology STRENGTHS AND LIMITATIONS OF THIS STUDY This is the first scoping review that explored the diversity of study methodologies adopted in studies on testing with serial high-sensitivity cardiac troponin measurements in patients with suspected non-ST elevation myocardial infarction in the emergency department. The review method included a comprehensive literature search, duplicate assessment of eligibility, double extraction of data and descriptive synthesis through tables and graphs. The assessed specifications included the regions and sources of studies, targeted participants and their study flow, troponin assays, sampling algorithms and direct comparisons thereof, and definitions of outcomes. The quantitative results about the accuracy and clinical outcomes were not assessed because this was beyond the scope of the review. Introduction Acute coronary syndrome (ACS) is considered a major cause of death worldwide.1 2 Acute myocardial infarction (AMI) is a form of ACS, which represents permanent cellular damage in the affected myocardium due to ischaemia. AMI is clinically subcategorised into ST-segment elevation myocardial infarction (STEMI) and non-STEMI (NSTEMI). Each type of AMI has a unique prognosis, and their managements differ substantially. Since STEMI is an acute life-threatening condition, prompt reperfusion therapy is essential. In contrast, because the prognosis of NSTEMI varies depending on its aetiology, accurate diagnosis and risk stratification based on medical history, ECG findings and cardiac biomarker concentrations are of paramount significance in patients suspected of ACS.3 4 For the clinical management of NSTEMI, cardiac troponin (cTn) has been used as the mainstay of clinical diagnosis since 2000.5–8 To avoid unnecessary hospital admissions and expedite the diagnostic process, high-sensitivity cardiac Tn (hs-cTn), a group of more sensitive cTn assays,9 has been introduced into clinical practice since 2010. Although several primary studies and meta-analyses on the single measurement of hs-cTn reported their high sensitivity and specificity,10–12 several challenges persist. First, blood concentrations of hs-cTn troponin take 2–3 hours to increase, and they may not be detectable within 3 hours from the onset of AMI.3 13 Second, despite its high sensitivity, elevated concentrations of hs-cTn are observed in several clinical conditions other than AMI, including acute myocardial injury (eg, acute heart failure and tachyarrhythmia) and chronic myocardial injury (eg, structural heart disease and chronic heart failure).3 8 13 To differentiate these conditions, serial measurements of hs-cTn, that is, assessing the absolute and/or relative changes of repeated measurements, were proposed to increase the specificity for diagnosing acute MI. Based on several studies on serial hs-cTn testing algorithms with high sensitivity and high negative predictive value, the current clinical guidelines on NSTE-ACS recommend serial measurements of hs-cTn at presentation and after 1–3 hours.3 4 However, the comparative effectiveness of management strategies based on serial hs-cTn measurements has not been fully elucidated, because several alternative assays are clinically available and existing reports are from studies with different designs and inconsistent testing algorithms. Thus, this study aimed to explore the diversity of the methodologies used in primary studies on serial hs-cTn measurements in patients suspected of having ACS in the emergency department (ED). We constructed an evidence map of existing studies on serial hs-cTn testing for diagnosing NSTEMI and predicting 30-day clinical outcomes. We critically appraised the currently available evidence and highlighted the issues that need to be addressed in future research. Methods This study is a focused analysis performed in conjunction with a registered systematic review project (PROSPERO registration number CRD42018106379). The protocol for the original systematic review is available at This report followed the Preferred Reporting Items for Systematic reviews and Meta-Analyses (PRISMA) Extension for Scoping Reviews.14 Data search We searched Ovid MEDLINE, EMBASE, Science Citation Index and Cochrane Database of Systematic Reviews for studies published between 1 January 2006 and 17 November 2021 with no restrictions of language or publication status. The search terms included “chest pain”, “acute coronary syndrome,” “myocardial infarction,” “cardiac troponin”, “emergency room” and their synonyms.10 The full search strategy is available in online supplemental appendix. We excluded editorials, letters, comments, conference abstracts, review articles and meta-analyses. Also, we excluded studies assessing clinical prediction rules (eg, Global Registry of Acute Coronary Events (GRACE) Risk Score15). Supplementary data bmjopen-2022-066429supp001.pdf (496.5KB, pdf) Study eligibility We included prospective and retrospective studies that evaluated patients aged ≥18 years who were suspected of having NSTEMI in an ED and had two or more serial troponin measurements using an hs-cTn assay. Eligible were studies that reported the diagnostic accuracy of AMI and/or 30-day clinical outcomes. We only included single-gate design studies, that is, studies that consisted of a single group of subjects based on a single eligibility criteria.16 Studies that included mixed populations of patients—with suspected STEMI and NSTEMI—were included only when data for the patients with suspected NSTEMI was separately extractable. Studies that exclusively assessed patients with suspected STEMI were excluded. Two investigators double-screened the titles and abstracts and examined the full-text articles to assess eligibility. We defined hs-cTn as assays that satisfied the requirements of the International Clinical Federation of Clinical Chemistry and Laboratory Medicine (ie, <10% coefficient of variation at the 99th percentile and ≥50% measurable concentrations above the limit of detection for both males and females).9 Discrepancies were resolved by consensus. Data extraction The following data were extracted: (1) publication and study characteristics: authors, journal name, publication year, enrolment years, number of eligible and included patients, study design, the name of the study cohort(s), geographical region(s), participant age and use of ECG to exclude patients; (2) test characteristics: assays, the timing of blood sampling, cut-off values, algorithms adopted (binary testing algorithms for ruling out MI vs three-strata testing algorithms for stratifying patients into three different risk groups, ie, high-risk, intermediate-risk and low-risk groups commonly referred to as ‘rule-in,’ ‘observational zone,’ and ‘rule-out’ for MI diagnosis) and (3) reference standard characteristics: specific diagnostic criteria of MI, such as those defined in clinical guidelines and/or versions of the universal definition of MI, and the assessors of the final diagnoses. Operationalisation Our target population was a group of patients suspected of having NSTEMI who presented at the ED. We recorded the numbers of patients suspected of having NSTEMI who presented at the ED, patients who completed one or more study-specific serial testing algorithm(s), and patients who were assessed for test accuracy and/or 30-day clinical outcomes. A testing algorithm was specified based on the number and timing of hs-cTn measurements. The results for the algorithm were typically reported as a single value measured at presentation, together with an absolute or a relative difference between specific measurement time points (typically at presentation and a few hours later), which was categorised as delta or percent change in the hs-cTn concentrations. We classified the studies that involved two hs-cTn measurements into three groups, namely, 0 and 1 hour, 0 and 2 hours, and 0 and 3 hours, based on the blood sampling timing (in hours) of the first and second samples. Other algorithms involving three or more blood samples were grouped into a separate category, labelled as ‘others.’ Assays were classified by specific manufacturers, that is, Abbott (Abbott Laboratories, Illinois, USA), Roche (Roche Diagnostics, Basel, Switzerland), Siemens (Siemens Healthcare, Erlangen, Germany) and Beckman (Beckman Coulter, California, USA). Assays by other manufacturers were categorised as ‘miscellaneous.’ To assess the evidence, we used the study with the largest sample to avoid double-counting when multiple studies reported (partially) overlapping patient populations. For the study locations, we assumed that each specific research institution involved in the study assessed patients residing within its geographical region only. Comparative studies were defined as studies that adopted a paired design to assess multiple assays on the same study participants and directly compared the diagnostic accuracy for AMI or 30-day clinical outcomes. This review did not standardise the definition of AMI or the 30-day clinical outcomes and adopted the study-reported outcome definitions as specifically reported. Analyses We considered each publication to be the unit of analysis and performed descriptive analyses by using percentages or medians and ranges. We combined data as a weighted average only if the pertinent data were available for specific subgroups. The assessed design specifications included the regions and sources of studies, characteristics of targeted participants and their study flow, specific troponin assays used, sampling algorithms with their operational characteristics, direct comparisons of two or more algorithms and/or assays, and definitions of the index MI and 30-day outcomes. The volume of clinical evidence was assessed with graphs and tables using Stata V.17.0 (Stata). The graphical presentation of the study locations was constructed using Google Maps (Google, Mountain View, California, USA) and Mapcustomizer (available at Patient and public involvement We did not involve patients or the public in the preparation of this scoping review. Results Inclusion of primary studies Figure 1 shows the PRISMA flow diagram for this scoping review. Our search yielded 6838 articles; 6230 of them were excluded after examining the titles and abstracts. After excluding 549 perused full-text articles, we finally included 86 publications, including 72 reports on test accuracy and 52 on 30-day clinical outcomes (online supplemental appendix table 1).17–102 The excluded articles are listed in online supplemental appendix table 2. Figure 1. Open in a new tab The Preferred Reporting Items for Systematic reviews and Meta-Analyses (PRISMA) flow diagram. Study characteristics Eighty-three studies (97%) were based on prospectively collected blood samples, and the median sample size was 1391 (range 160–6403) (online supplemental appendix table 1). The most commonly assessed assays were manufactured by Abbot (n=41) and Roche (n=53), followed by those manufactured by Siemens (n=10) and Beckman (n=6). Three studies assessed point-of-care assays that met the definition of hs-cTn.76 80 102 Serial hs-cTn testing was predominantly assessed in Europe and North America and the participating institutions were limited to specific research centres (online supplemental appendix table 1, online supplemental appendix table 3 and online supplemental appendix figure 1). In contrast, only a few research centres per country from Australasia and Asia participated and provided pertinent data. Of the 86 cohorts that reported accuracy and/or 30-day outcome data using a specific assay and a sampling algorithm, only 42 (49%) were considered unique, which included 78 606 non-overlapping patients (online supplemental appendix table 3). Of the 42 studies, 12 (29%) studies that assessed the assays manufactured by Abbott involved unique cohorts (seven, three and eight reports on the 0 and 1-hour, 0 and 2 hours, and 0 and 3 hours protocols, respectively). Similarly, data from 25 of 42 studies (60%) that assessed the assays manufactured by Roche involved unique cohorts (16, 5 and 4 reports on the 0 and 1-hour, 0 and 2 hours, and 0 and 3 hours protocols, respectively). Nineteen studies compared two or more assays using a paired design (table 1). The most commonly reported comparison was between the assays by Abbott and Roche (17 studies on test accuracy and six on 30-day outcomes). A few studies have performed head-to-head comparisons involving the same patients. Two studies compared hs-cTn with an earlier generation non-hs cTn assay.17 51 Table 1. Direct comparisons of alternative high-sensitivity cardiac troponin assays Hs-cTn assays Abbott Roche Siemens Beckman 0 and 1-hour protocol Abbott 6 (10731)0 0 Roche 8 (13779)0 0 Siemens 2 (1235)1 (418)0 Beckman 1 (278)1 (278)1 (278) 0 and 2 hours protocol Abbott 2 (4063)0 1 (1118) Roche 5 (7497)0 1 (1118) Siemens 1 (313)1 (313)0 Beckman 1 (1118)1 (1118)1 (217) 0 and 3 hour protocol Abbott 1 (2945)0 0 Roche 4 (6419)0 0 Siemens 1 (1809)1 (1809)0 Beckman 1 (1110)1 (1110)1 (1110) Miscellaneous Abbott 0 0 0 Roche 1 (1735)0 0 Siemens 0 1 (830)0 Beckman 0 1 (830)1 (830) Open in a new tab For each algorithm (ie, 0 and 1-hour, 0 and 2 hours, and 0 and 3 hours protocols and a combined, miscellaneous protocol group described in the left-most column), cells in the lower left and upper right parts (separated by a right-lower diagonal line comprising black closed cells) show the volume of comparative evidence on diagnostic accuracy of acute myocardial infarction and 30-day clinical outcomes, respectively. Each cell represents a specific comparison between assays described in the leftmost column and the top row. The number of comparative studies that compared a specific pair of assays is followed by the number of assessed patients in parentheses. Hs-cTn, high-sensitivity cardiac troponin. Typically, the studies reported only the number of enrolled patients (ie, patients suspected of having NSTEMI who were eligible for and agreed to participate in the study), and the complete data on all clinically relevant patients (ie, the number of all patients suspected of having NSTEMI who presented at the ED) were missing. Fifty-two studies (60%) reported quantitative data on patients who failed to complete a study-specific serial testing algorithm(s) or patients whose diagnosis or 30-day clinical outcomes could not be established (online supplemental appendix table 1). Various proportions (median 20%; range, 0%–90%) of enrolled patients were excluded from the analysis, typically due to missing blood samples or the lack of a final diagnosis. Patient characteristics The mean or median participant age ranged from 53 to 73 years, and most studies involved patients in their 40s–70s (online supplemental appendix table 1). Only one study56 specifically focused on 0 and 1-hour protocol for a subgroup of elderly patients derived from three cohorts, specifically, patients aged 70 years or older. Of 86 studies, 17 (20%) excluded patients with chronic kidney disease (CKD) or those requiring regular haemodialysis. In contrast, only four studies specifically focused on 0 and 1-hour, 3 hours or 6–12 hours protocol for a subgroup of patients with renal dysfunction, which were derived from four cohorts.61 62 64 65 Only a single study specifically focused on 0 and 3-hour protocol for a subgroup of female patients only derived from three cohorts. No further studies that focused on these sub populations have been found through the update search and manual search based on the reference lists.52 Testing algorithms Forty-seven studies (55%) assessed the three-strata algorithms. These studies stratified patients into ‘rule-out (low-risk),’ ‘observational zone (intermediate-risk)’ and ‘rule-in (high-risk)’ groups according to two sets of diagnostic criteria based on the concentrations of hs-cTn at baseline, 1–3 hours, and/or the difference between the hs-cTn concentrations of the two samples. Other studies conventionally categorised the patients into two strata (ie, ‘rule-out (low-risk)’ and ‘rule-in (high-risk)’ groups) based on a single set of criteria. The majority of data were based on assays manufactured by either Abbott or Roche, and the 0 and 1-hour algorithm was the most frequently reported (figure 2). Figure 2. Open in a new tab Cumulative number of study participants assessed by specific testing protocols and assays for serial high-sensitivity cardiac troponin measurements. A number of cumulative participants are based on simple summations of all participants assessed in the relevant studies. These numbers may be overestimated due to overlapping inclusions of patients from same cohorts and/or institutions. Outcomes Eighty-four studies (98%) adopted the universal definition of MI (versions 2007, 2012 and/or 2018)6–8 to establish a diagnosis of AMI. A few of these studies also followed the guidelines proposed by the American College of Cardiology (ACC) (1/86, 1%),103 the ACC and the American Heart Association (AHA) guidelines (1/86, 1%),104 ACC and the European Society of Cardiology (ESC),105 and the ACC/AHA104 106 and ESC3 107 guidelines (4/86, 5%) in addition to either version of the universal definition. Two studies17 21 relied on the ACC guidelines103 alone. The adjudicators of the clinical diagnosis of AMI were cardiologists in 72 studies (84%). The studies variably reported 30-day outcomes; of the 51 studies that reported one or more 30-day outcomes, 11 (22%) reported all-cause mortality and 41 (80%) reported cardiac death, whereas 36 (68%) reported major adverse cardiac events, a composite outcome including AMI, as well as cardiac death or death from all causes. Other reported clinical outcomes observed within 30 days included urgent revascularisation, percutaneous coronary intervention, coronary artery bypass graft (19/51, 37%) and ventricular arrhythmia (11/51, 22%). Twenty-seven studies (53%) also reported long-term clinical outcomes, which included events that developed within up to 2 years.34–37 45 57 62 65 75 94 95 Discussion To the best of our knowledge, this is the first scoping review to comprehensively assess the reports on testing with serial hs-cTn measurements in patients with suspected NSTEMI in the ED. We summarised how studies measured serial hs-cTn and assessed its diagnostic accuracy for AMI and the 30-day clinical outcomes. Our results showed that most existing data were based on the Abbott cTnI or Roche cTnT assays, and the timing of the blood measurements and the diagnostic algorithms varied. The number of studies assessing these two assays using the 0 and 1-hour, 0 and 2 hours, and 0 and 3 hours protocols has been continuously increasing since 2011 when guidelines29 recommended serial hs-cTn measurements; fewer studies have assessed other assays and/or alternative algorithms. Limited data on patients with CKD or older adults, as well as data stratified by sex, were reported,62 64 which were deemed still under evaluation. Most studies followed the universal definition to diagnose the index MI. However, in addition to North America or Europe, only a limited number of research teams involving several specialty institutions in specific countries have contributed to the current evidence. Most importantly, the studies excluded variable proportions of eligible patients from the analysis due to missing blood samples or concrete final diagnoses. Strengths We comprehensively explored the existing evidence, focusing on how the studies were designed, analysed and reported and, the implications for clinical practice of the identified limitations and concerns. Previously reported systematic reviews focused on a two-strata rule-out strategy using unique serial measurements of hs-cTnT assay only108 or the 0 and 1-hour three-strata strategy only109 regardless of the assays assessed, and they did not perform a comprehensive field synopsis covering all relevant information. The objective of our scoping review is to describe the diversity in the adopted study methodologies together with their potential limitations following the standard scoping review methods. Therefore, this review should be an additional view that follows the recently published critical appraisal of the current evidence base,110 111 both of which will help identify the current evidence gaps as well as help design future studies. Limitation This scoping review performed a focused analysis on the reported study methodologies and did not address the primary objectives of the originally planned systematic review. Therefore, several limitations need to be discussed. First, we did not assess the quantitative results on accuracy and other clinical outcomes because this was beyond the scope of the present scoping review. Second, we focused on studies that assessed only test accuracy and 30-day outcomes. Therefore, data on studies that evaluated long-term outcomes have not been addressed. Third, since we excluded conference abstracts, this review may have missed newer relevant publications. Fourth, the recently developed clinical prediction rules (eg, History, ECG, Age, Risk factors and Troponin Score,112 Emergency Department Assessment of Chest Pain Score113 and GRACE Risk Score15) have consistently incorporated hs-cTn concentrations as a component variable.40 We excluded these studies that used hs-cTn as part of risk prediction models. Finally, several studies have reported that higher hs-cTn concentrations were associated with increased mortality in patients with suspected ACS; however, our review did not address this association.34 36 37 45 57 62 65 Clinical implications Recently, a pooled study conducted by Neumann et al proposed a high-accurate risk-assessment tool for NSTEMI using a serial hs-cTn measurements based on an international consortium of prospectively registered data from 15 cohorts. Of these, nine cohorts were also assessed in this scoping review, at least some portions of which should have contributed to the dataset deriving the risk-assessment tool. Given the methodological concerns raised in this review, depending on how the study methodologies in the rest of the data have been addressed and/or improved, uncertainties still exist in its individualised application in real-life clinical practice. First, ACS and MI are common in elderly populations as well as patients with CKD,114 115 while the eligible studies in our review mainly focused on middle-aged populations, and approximately one-third of the studies excluded patients with impaired renal function. In addition, hs-cTnI concentrations are generally higher in men than in women,116 which bases the use of sex-specific cut-offs recommended by the fourth universal definition of MI in contrast to the uniform cut-off adopted in the risk-assessment tool.117 The recent critical appraisal report111 is also in line with this individualised approach. Moreover, quality specifications of the assays are also relevant. For example, critical factors, which are appropriate for collecting and measuring samples and applying the results to clinical practices.118 Therefore, the optimal sample timings and cut-off values need to be validated on an individualised basis and account for age, sex and renal function under the appropriate quality control.119 120 Second, our review found that the existing data were largely based on two assays by two manufacturers (ie, assays manufactured by Abbott and Roche); the evidence is sparse for the others. Furthermore, evidence on hs-cTnT and hs-cTnI has limited data concerning direct comparative studies of assays. Therefore, comparative studies are needed since systematic differences between hs-cTnT and hs-cTnI as well as among hs-cTnI methods have been reported.121 Third, most of the included studies were conducted in specialised centres in Europe, North America or Australasia. A recent meta-analysis of diagnostic accuracy that focused on only the 0 and 1-hour algorithm109 pointed out that sensitivity was not universally high across cohorts, as reported in the primary studies in these specialised centres; reproducibility of the excellent results appeared to be limited for the studies from Asia. Given this observation, validation in these regions is required. Fourth, the studies included in our review missed variable proportions of clinically relevant patients who presented at EDs with suspected NSTEMI. This appears to stem, at least in part, from convenience sampling. The failure to apply gold-standard tests to all participants may also have been responsible for the excluded cases without an established diagnosis of the cause of chest pain, which is inevitable in real-life clinical settings. These methodological weaknesses would have distorted, at least to some extent, the typical disease spectrum of clinically relevant populations. Our review failed to address how this patient loss affected the study results. Conclusions Data on diagnostic test accuracy and short-term outcomes by serial hs-cTn measurements were largely derived from particular research institutions in Europe, North America or Australasia and based mainly on two specific assays. The exclusion of variable proportions of eligible patients, which was inevitable even in well-conducted prospective studies, raised concerns regarding the studies’ generalisability and direct applications in real-world ED clinical practice. Supplementary Material Reviewer comments bmjopen-2022-066429.reviewer_comments.pdf (133.3KB, pdf) Author's manuscript bmjopen-2022-066429.draft_revisions.pdf (2.4MB, pdf) Acknowledgments The authors thank Drs Chihiro Kato and Jun Shinohara for assisting with data extraction. Footnotes Contributors: TT and ZZ lead the protocol development. HO, TT, ZZ, MI, MR, JLP and CH drafted and revised the protocol. MR performed the literature searches. HO, TT, ZZ, JLP determined the eligibility of primary studies and acquired the data. HO and TT analysed the data. HO, TT, ZZ, MI, MR, JLP and CH interpreted the findings. HO and TT drafted the first version of the report. HO, TT, ZZ, MI, MR, JLP and CH critically read the manuscript and provided feedback for revision. HO, TT, ZZ, MI, MR, JLP and CH read and approved the final manuscript. HO and TT are the guarantors of this scoping review. Funding: This study was supported in part by a research grant from the Ministry of Education, Culture, Sports, Science and Technology (MEXT) of Japan (18K08902) and by the National Institute for Health Research (NIHR) Collaboration for Leadership in Applied Health Research and Care South West Peninsula (NIHR CLAHRC South West Peninsula; grant no, N/A). Disclaimer: The funders had no involvement in the study design, data collection and analysis, decision to publish, or preparation of the manuscript. The views expressed in this publication are those of the author(s) and not necessarily those of MEXT in Japan, the National Institute for Health Research, or the Department of Health and Social Care. Map disclaimer: The inclusion of any map (including the depiction of any boundaries therein), or of any geographic or locational reference, does not imply the expression of any opinion whatsoever on the part of BMJ concerning the legal status of any country, territory, jurisdiction or area or of its authorities. Any such expression remains solely that of the relevant source and is not endorsed by BMJ. Maps are provided without any warranty of any kind, either express or implied. Competing interests: None declared. Patient and public involvement: Patients and/or the public were not involved in the design, or conduct, or reporting, or dissemination plans of this research. Provenance and peer review: Not commissioned; externally peer reviewed. Supplemental material: This content has been supplied by the author(s). It has not been vetted by BMJ Publishing Group Limited (BMJ) and may not have been peer-reviewed. Any opinions or recommendations discussed are solely those of the author(s) and are not endorsed by BMJ. BMJ disclaims all liability and responsibility arising from any reliance placed on the content. Where the content includes any translated material, BMJ does not warrant the accuracy and reliability of the translations (including but not limited to local regulations, clinical guidelines, terminology, drug names and drug dosages), and is not responsible for any error and/or omissions arising from translation and adaptation or otherwise. Data availability statement All data relevant to the study are included in the article or uploaded as online supplemental information. Ethics statements Patient consent for publication Not applicable. Ethics approval Since it was a secondary analysis of publicly available data, no ethical review was required. References 1.Organization WH . Cardiovascular diseases (CVDs): key facts, 2017. 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Supplementary Materials Supplementary data bmjopen-2022-066429supp001.pdf (496.5KB, pdf) Reviewer comments bmjopen-2022-066429.reviewer_comments.pdf (133.3KB, pdf) Author's manuscript bmjopen-2022-066429.draft_revisions.pdf (2.4MB, pdf) Data Availability Statement All data relevant to the study are included in the article or uploaded as online supplemental information. 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188340	https://www.craftonhills.edu/current-students/tutoring-center/mathematics-tutoring/calc_integrals_cheat_sheet.pdf	Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Integrals Definitions Definite Integral: Suppose ( ) f x is continuous on [ ] , a b . Divide [ ] , a b into n subintervals of width x ∆ and choose i x from each interval. Then ( ) ( ) 1 lim i b a n i f x dx f x x →∞ = ∞ = ∆ ∑ ∫ . Anti-Derivative : An anti-derivative of ( ) f x is a function, ( ) F x , such that ( ) ( ) F x f x ′ = . Indefinite Integral : ( ) ( ) f x dx F x c = + ∫ where ( ) F x is an anti-derivative of ( ) f x . Fundamental Theorem of Calculus Part I : If ( ) f x is continuous on [ ] , a b then ( ) ( ) x a g x f t dt = ∫ is also continuous on [ ] , a b and ( ) ( ) ( ) x a d g x f t dt f x dx ′ = = ∫ . Part II : ( ) f x is continuous on[ ] , a b , ( ) F x is an anti-derivative of ( ) f x (i.e. ( ) ( ) F x f x dx = ∫ ) then ( ) ( ) ( ) b a f x dx F b F a = − ∫ . Variants of Part I : ( ) ( ) ( ) ( ) u x a d f t dt u x f u x dx ′ =     ∫ ( ) ( ) ( ) ( ) b v x d f t dt v x f v x dx ′ = −     ∫ ( ) ( ) ( ) ( ) [ ] ( ) [ ] ( ) ( ) u x v x u x v x d f t dt u x f v x f dx ′ ′ = − ∫ Properties ( ) ( ) ( ) ( ) f x g x dx f x dx g x dx ± = ± ∫ ∫ ∫ ( ) ( ) ( ) ( ) b b b a a a f x g x dx f x dx g x dx ± = ± ∫ ∫ ∫ ( ) 0 a a f x dx = ∫ ( ) ( ) b a a b f x dx f x dx = − ∫ ∫ ( ) ( ) cf x dx c f x dx = ∫ ∫ , c is a constant ( ) ( ) b b a a cf x dx c f x dx = ∫ ∫ , c is a constant ( ) b a cdx c b a = − ∫ ( ) ( ) b b a a f x dx f x dx ≤ ∫ ∫ ( ) ( ) ( ) b c b a a c f x dx f x dx f x dx = + ∫ ∫ ∫ for any value of c. If ( ) ( ) f x g x ≥ on a x b ≤ ≤ then ( ) ( ) b b a a f x dx g x dx ≥ ∫ ∫ If ( ) 0 f x ≥ on a x b ≤ ≤ then ( ) 0 b a f x dx ≥ ∫ If ( ) m f x M ≤ ≤ on a x b ≤ ≤ then ( ) ( ) ( ) b a m b a f x dx M b a − ≤ ≤ − ∫ Common Integrals k dx k x c = + ∫ 1 1 1 , 1 n n n x dx x c n + + = + ≠− ∫ 1 1 ln x x dx dx x c − = = + ∫ ∫ 1 1 ln a a x b dx ax b c + = + + ∫ ( ) ln ln u du u u u c = − + ∫ u u du c = + ∫e e cos sin u du u c = + ∫ sin cos u du u c = − + ∫ 2 sec tan u du u c = + ∫ sec tan sec u u du u c = + ∫ csc cot csc u udu u c = − + ∫ 2 csc cot u du u c = − + ∫ tan ln sec u du u c = + ∫ sec ln sec tan u du u u c = + + ∫ ( ) 1 1 1 2 2 tan u a a a u du c − + = + ∫ ( ) 1 2 2 1 sin u a a u du c − − = + ∫ Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. u Substitution : The substitution ( ) u g x = will convert ( ) ( ) ( ) ( ) ( ) ( ) b g b a g a f g x g x dx f u du ′ = ∫ ∫ using ( ) du g x dx ′ = . For indefinite integrals drop the limits of integration. Ex. ( ) 2 3 2 1 5 cos x x dx ∫ 3 2 2 1 3 3 u x du x dx x dx du = ⇒ = ⇒ = 3 3 1 1 1 :: 2 2 8 x u x u = ⇒ = = = ⇒ = = ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 8 5 3 1 1 8 5 5 3 3 1 5 cos cos sin sin 8 sin 1 x x dx u du u = = = − ∫ ∫ Integration by Parts : u dv uv vdu = − ∫ ∫ and b b b a a a u dv uv vdu = − ∫ ∫ . Choose u and dv from integral and compute du by differentiating u and compute v using v dv = ∫ . Ex. x x dx − ∫e x x u x dv du dx v − − = = ⇒ = = − e e x x x x x x dx x dx x c − − − − − = − + = − − + ∫ ∫ e e e e e Ex. 5 3 ln xdx ∫ 1 ln x u x dv dx du dx v x = = ⇒ = = ( ) ( ) ( ) ( ) 5 5 5 5 3 3 3 3 ln ln ln 5ln 5 3ln 3 2 x dx x x dx x x x = − = − = − − ∫ ∫ Products and (some) Quotients of Trig Functions For sin cos n m x x dx ∫ we have the following : 1. n odd. Strip 1 sine out and convert rest to cosines using 2 2 sin 1 cos x x = − , then use the substitution cos u x = . 2. m odd. Strip 1 cosine out and convert rest to sines using 2 2 cos 1 sin x x = − , then use the substitution sin u x = . 3. n and m both odd. Use either 1. or 2. 4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated. For tan sec n m x xdx ∫ we have the following : 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using 2 2 tan sec 1 x x = −, then use the substitution sec u x = . 2. m even. Strip 2 secants out and convert rest to tangents using 2 2 sec 1 tan x x = + , then use the substitution tan u x = . 3. n odd and m even. Use either 1. or 2. 4. n even and m odd. Each integral will be dealt with differently. Trig Formulas : ( ) ( ) ( ) sin 2 2sin cos x x x = , ( ) ( ) ( ) 2 1 2 cos 1 cos 2 x x = + , ( ) ( ) ( ) 2 1 2 sin 1 cos 2 x x = − Ex. 3 5 tan sec x x dx ∫ ( ) ( ) ( ) 3 5 2 4 2 4 2 4 7 5 1 1 7 5 tan sec tan sec tan sec sec 1 sec tan sec 1 sec sec sec x xdx x x x xdx x x x xdx u u du u x x x c = = − = − = = − + ∫ ∫ ∫ ∫ Ex. 5 3 sin cos x x dx ∫ ( ) 2 2 1 1 2 2 2 2 5 4 3 3 3 2 2 3 2 2 2 4 3 3 sin (sin ) sin sin sin cos cos cos sin (1 cos ) cos (1 ) 1 2 cos sec 2ln cos cos x x x x x x x x x x x u u u u u dx dx dx dx u x du du x x x c − − − + = = = = = − = − = + − + ∫ ∫ ∫ ∫ ∫ ∫ Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. 2 2 2 sin a b a b x x θ − ⇒ = 2 2 cos 1 sin θ θ = − 2 2 2 sec a b b x a x θ − ⇒ = 2 2 tan sec 1 θ θ = − 2 2 2 tan a b a b x x θ + ⇒ = 2 2 sec 1 tan θ θ = + Ex. 2 2 16 4 9 x x dx − ∫ 2 2 3 3 sin cos x dx d θ θ θ = ⇒ = 2 2 2 4 4sin 4cos 2 cos 4 9x θ θ θ = − = = − Recall 2 x x = . Because we have an indefinite integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute θ ’s and remove absolute value bars based on that and, if 0 if 0 x x x x x ≥  = − <  In this case we have 2 2cos 4 9x θ = − . ( ) ( ) 2 3 sin 2cos 2 2 2 4 9 16 12 sin cos 12csc 12cot d d d c θ θ θ θ θ θ θ θ = = = − + ⌠ ⌡ ∫ ∫ Use Right Triangle Trig to go back to x’s. From substitution we have 3 2 sin x θ = so, From this we see that 2 4 9 3 cot x x θ − = . So, 2 2 2 16 4 4 9 4 9 x x x x dx c − − = − + ∫ Partial Fractions : If integrating ( ) ( ) P x Q x dx ∫ where the degree of ( ) P x is smaller than the degree of ( ) Q x . Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in ( ) Q x Term in P.F.D Factor in ( ) Q x Term in P.F.D ax b + A ax b + ( ) k ax b + ( ) ( ) 1 2 2 k k A A A ax b ax b ax b + + + + + + L 2 ax bx c + + 2 Ax B ax bx c + + + ( ) 2 k ax bx c + + ( ) 1 1 2 2 k k k A x B A x B ax bx c ax bx c + + + + + + + + L Ex. 2 ( )( ) 2 1 4 7 13 x x x x dx − + + ∫ ( ) ( ) 2 2 2 2 ( )( ) 2 1 3 2 2 2 3 16 4 1 1 4 4 3 16 4 1 4 4 7 13 4ln 1 ln 4 8tan x x x x x x x x x x x x dx dx dx x x − + − − + + − + + + = + = + + = − + + + ∫ ∫ ∫ Here is partial fraction form and recombined. 2 2 2 2 4) ( ) ( ) ( )( ) ( )( ) 2 1 1 1 4 4 1 4 ( 7 13 Bx C x x x x x x x A x Bx C A x x + + + − − − + + − + + + = + = Set numerators equal and collect like terms. ( ) ( ) 2 2 7 13 4 x x A B x C B x A C + = + + − + − Set coefficients equal to get a system and solve to get constants. 7 13 4 0 4 3 16 A B C B A C A B C + = − = − = = = = An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : ( ) ( ) ( ) 2 2 7 13 4 1 x x A x Bx C x + = + + + − . Chose nice values of x and plug in. For example if 1 x = we get 20 5A = which gives 4 A = . This won’t always work easily. Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Applications of Integrals Net Area : ( ) b a f x dx ∫ represents the net area between ( ) f x and the x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The general formulas for the two main cases for each are, ( ) upper function lower function b a y f x A dx         = ⇒ = − ∫ & ( ) right function left function d c x f y A dy         = ⇒ = − ∫ If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. ( ) ( ) b a A f x g x dx = − ∫ ( ) ( ) d c A f y g y dy = − ∫ ( ) ( ) ( ) ( ) c b a c A f x g x dx g x f x dx = − + − ∫ ∫ Volumes of Revolution : The two main formulas are ( ) V A x dx = ∫ and ( ) V A y dy = ∫ . Here is some general information about each method of computing and some examples. Rings Cylinders ( ) ( ) ( ) 2 2 outer radius inner radius A π = − ( ) ( ) radius width / height 2 A π = Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use ( ) f x , ( ) g x , ( ) A x and dx. Vert. Axis use ( ) f y , ( ) g y , ( ) A y and dy. Horz. Axis use ( ) f y , ( ) g y , ( ) A y and dy. Vert. Axis use ( ) f x , ( ) g x , ( ) A x and dx. Ex. Axis : 0 y a = > Ex. Axis : 0 y a = ≤ Ex. Axis : 0 y a = > Ex. Axis : 0 y a = ≤ outer radius : ( ) a f x − inner radius : ( ) a g x − outer radius: ( ) a g x + inner radius: ( ) a f x + radius : a y − width : ( ) ( ) f y g y − radius : a y + width : ( ) ( ) f y g y − These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the 0 y a = ≤ case with 0 a = . For vertical axis of rotation ( 0 x a = > and 0 x a = ≤ ) interchange x and y to get appropriate formulas. Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Work : If a force of ( ) F x moves an object in a x b ≤ ≤ , the work done is ( ) b a W F x dx = ∫ Average Function Value : The average value of ( ) f x on a x b ≤ ≤ is ( ) 1 b avg a b a f f x dx − = ∫ Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b a L ds = ∫ 2 b a SA y ds π = ∫ (rotate about x-axis) 2 b a SA xds π = ∫ (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. ( ) ( ) 2 1 if , dy dx ds dx y f x a x b = + = ≤ ≤ ( ) ( ) 2 1 if , dx dy ds dy x f y a y b = + = ≤ ≤ ( ) ( ) ( ) ( ) 2 2 if , , dy dx dt dt ds dt x f t y g t a t b = + = = ≤≤ ( ) ( ) 2 2 if , dr d ds r d r f a b θ θ θ θ = + = ≤ ≤ With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit 1. ( ) ( ) lim t a a t f x dx f x dx →∞ ∞ = ∫ ∫ 2. ( ) ( ) lim b b t t f x dx f x dx − →−∞ ∞ = ∫ ∫ 3. ( ) ( ) ( ) c c f x dx f x dx f x dx − − ∞ ∞ ∞ ∞ = + ∫ ∫ ∫ provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a: ( ) ( ) lim b b a t t a f x dx f x dx + → = ∫ ∫ 2. Discont. at b : ( ) ( ) lim b t a a t b f x dx f x dx − → = ∫ ∫ 3. Discontinuity at a c b < < : ( ) ( ) ( ) b c b a a c f x dx f x dx f x dx = + ∫ ∫ ∫ provided both are convergent. Comparison Test for Improper Integrals : If ( ) ( ) 0 f x g x ≥ ≥ on [ ) , a ∞ then, 1. If ( ) a f x dx ∞ ∫ conv. then ( ) a g x dx ∞ ∫ conv. 2. If ( ) a g x dx ∞ ∫ divg. then ( ) a f x dx ∞ ∫ divg. Useful fact : If 0 a > then 1 a p x dx ∞ ∫ converges if 1 p > and diverges for 1 p ≤. Approximating Definite Integrals For given integral ( ) b a f x dx ∫ and a n (must be even for Simpson’s Rule) define b a n x − ∆= and divide [ ] , a b into n subintervals [ ] 0 1 , x x , [ ] 1 2 , x x , … , [ ] 1, n n x x − with 0 x a = and n x b = then, Midpoint Rule : ( ) ( ) ( ) ( ) 1 2 b n a f x dx x f x f x f x   ≈∆ + + +   ∫ L , i x is midpoint [ ] 1, i i x x − Trapezoid Rule : ( ) ( ) ( ) ( ) ( ) ( ) 0 1 2 1 2 2 2 2 b n n a x f x dx f x f x f x f x f x − ∆ ≈ + + + + + +     ∫ L Simpson’s Rule : ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 2 2 1 4 2 2 4 3 b n n n a x f x dx f x f x f x f x f x f x − − ∆ ≈ + + + + + +     ∫ L
188341	https://www.sciencedirect.com/topics/engineering/aluminum-electrolysis	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Furan Resins 7.6.6 Aluminum Electrolysis Most aluminum reduction cells in commercial use employ prefabricated carbon blocks as the cell lining and as the cathodic working surface. These blocks are formed into a liquid-tight container surfaced by filling the joints between the blocks with a ramming paste. The efficiency of sealing of the ramming paste is an important factor in determining the life and energy efficiency of a reduction cell, which depends to a great degree on the extent and rate of electrolytic penetration into the cell bottom . An ecofriendly cold ramming paste for an aluminum electrolysis was synthesized from electro-calcined anthracite, with artificial graphite as an aggregate and a furan resin as the binder material . The synthesized paste is an ecofriendly material and it has also some superior properties such as low electrical resistivity, high compressive strength, appropriate sodium penetration, and thermal expansion. View chapterExplore book Read full chapter URL: Review article Aluminum as energy carrier: Feasibility analysis and current technologies overview 2.3 Aluminum smelting Such standard methods of metal production as carbon reduction or electrolysis from aqueous solution are unsuitable for aluminum. Carbon reacts with aluminum producing aluminum carbide; in the process of electrolysis within aqueous solutions the hydrogen is reduced on the cathode instead of aluminum because aluminum is more electronegative. For aluminum electrolysis the cryolite (Na3AlF6) was established to be the best electrolyte. Cryolite does not contain metals, which can be precipitated on the cathode instead of aluminum. It has a density less than that of aluminum at operating temperatures: at 950–1000 °C the density of cryolite is less than 2.1 g/cm3, while aluminum density at these temperatures is about 2.3 g/cm3 that positively influences on aluminum sedimentation within the electrolyzer. Theoretically, cryolite isn’t consumed during the electrolysis. However, practically, at least 20 kg of cryolite materials (Na3AlF6, AlF3, NaF) per a kg of aluminum are added into electrolysis bath . Cryolite is not so abundant and for industrial purposes it is produced from fluorspar (CaF2). Molten cryolite dissociates into ions in the following way: Na3AlF6 → 3Na+ + AlF63−, Alumina dissociation can be described by the following equation: Al2O3 → 2Al3+ + 3O2−. Al3+ and Na+ ions are moved to the cathode, but mainly aluminum ions are discharged because they are more electropositive: Al3+ + 3e− → Al, AlF63− and O2− are moved to the anode, where oxygen ions are discharged: 2O2− − 4e− → O2. Overall process of aluminum electrolysis is described by the following equation: (2)Al2O3 → 2Al + 1.5O2. Eq. (2) corresponds to electrolysis, in which the anode is non-consumable – inert (based on platinum, ferrite, nitride or other). However, industrial process, the Hall-Héroult process, is based on carbon anodes. Electrolyzers based on inert anode still represent the field of research and development [74–81]. Hall-Héroult process can be described by the following equation: (3)Al2O3 + nC → 2Al + (3 − n)CO2 + (2n − 3)CO. Industrial anodes are composed of calcined coke (70–80 wt.%) and pitch (14–35 wt.%) as a binder . Two different anode types are currently used: self-baking and prebaked. Anode consumption rate ranges from 0.4 to 0.5 kg of carbon per a kg of aluminum depending on anode type and electrolysis conditions . Resulting anodic gas is composed of 30–50% CO and 50–70% CO2. About 8 kg of CO2-equivalent (including emissions from power plants) per a kg of aluminum are released when aluminum is electrolytically produced [82–84]. The use of inert anodes is intended to make the aluminum production more environmentally friendly. However, smelting based on inert anode requires more electrical energy than carbon-based smelting: theoretically, the minimum energy requirement for process (2) is 51% higher than the process (3) requirement . The necessary charge is defined from Faraday's law: (4) where q – electrochemical equivalent, A – atomic mass, Z – valency, F – Faraday's constant. In case of aluminum Eq. (4) gives the following: i.e., theoretically, 1 Ah should produce 0.335 g of aluminum. But, practically, the mass of aluminum produced is less then theoretical value. Electrolyzers with self-baking anodes have 88% current efficiency, electrolyzers based on prebaked anodes reach 91–94% current efficiency . Electrical energy intensity of aluminum varies from 13.4 kWh/kg for state-of-the-art electrolyzing technologies up to more than 20 kWh/kg for older plants . View article Read full article URL: Chapter Aluminum Production 2.5.2 Electrometallurgy of Aluminum 2.5.2.1 Introduction Aluminum is a highly reactive metal that forms a strong chemical bond with oxygen. Aluminum cannot be produced by the electrolysis of an aluminum salt dissolved in water because of the high reactivity of aluminum with the protons of water and the subsequent formation of hydrogen gas. As in aqueous solution, protons (H+) are preferentially reduced before Al3 + ions, leading to hydrogen evolution. Thus, the reduction of Al3 + requires to be performed by electrolysis in a molten aluminum salt in the absence of water. The direct reduction to aluminum with carbon, as used to produce iron, is not chemically possible, since aluminum is a stronger reducing agent than carbon. There is a possible indirect process, the carbothermic reduction to aluminum using carbon and Al2O3 which forms intermediate aluminum carbide (Al4C3) that can further yield aluminum metal at a temperature of about 2000 °C. This process uses less energy and yields less CO2 than the Hall–Héroult process, the major industrial process for aluminum extraction. However, this process has not yet been perfected and is still under development by companies in the aluminum industry. Therefore, aluminum must be extracted from a solution of purified alumina in a molten salt by electrolysis. The electrolytic process was discovered and patented in 1886 separately and almost simultaneously by Charles Martin Hall and Paul L.T. Héroult. Aluminum oxide is dissolved in molten cryolite, and the molten mixture is electrolyzed using carbon electrodes by passing a direct electric current through it producing pure aluminum metal. Therefore, it is called the Hall–Héroult process for manufacturing aluminum. The key discovery in the development of the aluminum process was that cryolite or sodium hexafluoroaluminate (Na3AlF6) was a suitable solvent for the dissolution and electrolysis of alumina. 2.5.2.2 Electrolyte Composition and Liquidus Temperature 2.5.2.2.1 Dissociation of Cryolite All molten salts have a high ionic character and generally disordered crystal structures where the volume has undergone an expansion of approximately 20%. The ions are free to move, and they may form ion pairs with opposite charges and electrostatic attractive forces. Sometimes these attractive forces can form weakly bound complex ions when there are mixtures of different anions or cations. Cryolite is one example of this, and when solid, the cationic lattice is only sodium ions, while the aluminum ions are ionically bonded to six fluoride ions forming the hexafluoroaluminate anion, . The mechanism and the degree of dissociation of hexafluoroaluminate ions have been developed from cryoscopy, density, viscosity, and Raman spectroscopy . Most investigators agree that cryolite completely ionizes to form hexafluoroaluminate () anions, which further dissociates to form tetrafluoroaluminate () as well as sodium (Na+) and fluoride (F−) ions according to the equations: (2.5.4) The hexafluoroaluminate ion then dissociates partly as a consequence of the melting process. (2.5.5) where α is the degree of dissociation of the hexafluoroaluminate ions, and for the cryolite composition, it is about 0.3. The aluminum fluoride added in excess of cryolite reacts with F− ions in order to form ions according to the following reaction: (2.5.6) These ions are generally considered as “complex,” but it is more correct in the molten media to refer to them as ion pairs, ion associations, or clusters. Sometimes they are calculated via association constants. At any one instant, a small fraction will have undergone complete dissociation and there will be a small, but finite concentration of discrete aluminum cations and oxygen anions. Under the influence of a potential gradient, which exists during electrolysis of molten salts, the dissociation of the complexes is enhanced. The fact that dissociation occurs is supported by measurement of cathode polarization in aluminum electrolysis and generally it is reported as being very small. The amount of polarization can be determined from the sodium content. The oxygen-containing species has the lowest concentration, and therefore this anode is more likely to undergo concentration polarization and depletion of the electroactive species at the interface. 2.5.2.2.2 Liquidus Temperature of Cryolite Phase diagrams indicate the rate of change in the depression of the liquidus temperature relative to the composition and the degree of dissociation of ionic species of the molten salts. Pure cryolite melts congruently at 1012 °C as indicated in the NaF–AlF3 phase diagram developed by Solheim and Sterten shown in Figure 2.5.2. A low melting point compound chiolite (Na5Al3F14) exists at high AlF3 and melts incongruently at 734 °C. A third compound, NaAlF4, is formed in the gas phase from the evaporation of cryolite. It can also be produced by chemical reactions in the solid state. Aluminum fluoride, as well as other compounds, is added to the cryolite melt to decrease the liquidus temperature and change several electrolyte properties that enhance the electrolytic process. The molten mixture of cryolite and additives in the electrolyte is commonly referred to as “bath.” The change in the liquidus temperature of cryolite versus AlF3 is nonlinear as it decreases sharply at higher AlF3 concentrations as demonstrated in Figure 2.5.2. The surplus AlF3 in molten cryolite is expressed as the weight ratio of NaF/AlF3, or alternatively as the molar ratio of NaF/AlF3. An excess concentration of AlF3 typically in the range 8–14% is maintained in the cryolite melt to improve the production efficiency of the electrolysis process. Electrical resistance within the anode–cathode distance (ACD) of the electrolyte provides sufficient heat to keep the cryolite molten. The chemistry of the electrolyte determines the operating temperature of the aluminum electrolysis cells and most commercial cells operate near 960 °C. The melting point of cryolite is lowered by the effect of the concentration of each additive on the electrolyte, for example: alumina = − 5.6 °C/% Al2O3, aluminum fluoride = 1.0–5.0 °C/% AlF3, calcium fluoride = − 2.9 °C/% CaF2, magnesium fluoride = − 3.8 °C/% MgF2, and lithium fluoride = − 8.7 °C/% LiF. Sterten and Mæland determined that the addition of Al2O3 to a cryolite melt lowers the liquidus temperature of the binary mixture, shown in the Na3AlF6–Al2O3 phase diagram in Figure 2.5.3, to the eutectic temperature of 966 °C at 10 wt.% Al2O3. Skybakmoen et al. determined by weight loss measurements the solubility of alumina in cryolite that describes the solidus line located on the right side of the phase diagram in Figures 2.5.3 and 2.5.4 at alumina concentrations higher than 10% and developed the alumina solubility equation: The decrease in Al2O3 solubility at higher concentrations of AlF3 is evident in the NaF–AlF3–Al2O3 shown in Figure 2.5.4. It is not practical to operate aluminum cells with AlF3 concentrations, greater than 12–13% due to the decrease in solubility of alumina in cryolite at the lower electrolyte temperatures that eventually results in undissolved alumina deposits in cells (sludge). An equation to calculate the solubility of alumina in cryolite melts containing combinations of AlF3, CaF2, MgF2, and LiF for all practical concentrations was published by Skybakmoen et al. where t is given in °C and the concentrations of components are given in wt.% while the concentration of AlF3 is given in excess wt.% relative to that in cryolite. where A = 2.464 − 0.007 (%AlF3) − 1.13 × 10− 5 (%AlF3)3 − 0.0385·(Li3AlF6)0.74 − 0.032·(CaF2) − 0.04 (%MgF2) + 0.0046·[(%AlF3)·(LiAlF6)]0.5 and B = − 5.01 + 0.11·(%AlF3) − 4.0 × 10− 5·(%AlF3)3 − 0.732·(Li3AlF6)0.4 + 0.085·[(AlF3)·(LiAlF6)]0.5. The phase diagram for the ternary mixture Na3AlF6–AlF3–Al2O3 shown in Figure 2.5.5 was first developed by Foster , and later it was revised by Skybakmoen et al. . The dotted line represents the maximum solubility of Al2O3 in the system. There is a 1.5% difference in the temperature of the alumina–cryolite multivariant line starting at 10% alumina. The compound chiolite (Na5Al3F14) melts at 730 °C. Foster found that the lowest liquidus temperature of the mixture, or ternary eutectic, was located at 37.3 wt.% AlF3, 3.2 wt.% Al2O3, and 59.5% Na3AlF6 and 684 °C. Compositions containing 20 wt.% or less AlF3 precipitated α-Al2O3 in the primary phase field of alumina while those containing 25% or more precipitated η-Al2O3. The electrolyte typically contains from 9% to 11% AlF3, 4% to 6% CaF2, and 1.5% to 4% Al2O3. The normal operating temperature of the electrolyte in aluminum cells is in the range 945–965 °C and has a corresponding liquidus temperature of 945–950 °C. The difference between the operating temperature and the liquidus temperature of the electrolyte is referred to as the “superheat.” Aluminum cells typically require about 5–10 °C superheat in order to have sufficient energy to dissolve alumina powder in the melt. The pool of molten aluminum metal pool in the cathode is about 300 °C above its freezing point, 660 °C. Because the electrolyte is so close to its freezing point, a ledge of frozen electrolyte is formed attached to the sidewalls of the cells, being one of the coldest zones of cells. The formation of the side-ledge protects the cell lining from the corrosive electrolyte. If the superheat is higher than normal for an extended period, it causes the protective layer of frozen cryolite ledge on the cell’s sidewalls to melt. The ledge plays an important role in regulating the heat balance and bath chemistry control of cells: The thickness of the cryolite sidewall ledge changes according to the thermal behavior of the cells. Whenever the cell superheat increases (difference between the bath temperature and the electrolyte liquidus temperature), some sidewall ledge melts, which increases cell heat losses. Similarly, whenever the superheat decreases, some bath freezes increasing the sidewall ledge thickness which decreases cell heat losses. • : The frozen sidewall ledge material is cryolite, containing almost no excess AlF3%. Therefore, the melting and freezing of almost pure cryolite ledge material has an impact on variations in the chemical composition of the total AlF3 concentration in the electrolyte. The composition of the cryolite electrolyte provides about 0.150 g/cm2 difference in density between molten electrolyte and liquid aluminum, thus ensuring a physical separation between the two liquids which reduces the reoxidation of the aluminum by the CO2 gas formed on the anode surfaces. 2.5.2.3 Dissolution of Alumina in Cryolitic Bath The relatively high maximum solubility of alumina in cryolitic bath, 13 wt.% at 1000 °C, results from the favorable stereochemistry of forming different stable oxyfluoride aluminate complex anions, and , with octahedral or tetrahedral coordination of the large O2 − and F− anions about the small Al3 + cations [2,7,8]. These solute polyhedrals are stable because of the nearly identical ionic diameters for oxygen and fluoride anions. Cryolite dissolves the alumina by a chemical reaction and, while numerous species have been proposed, the resulting complex anion is generally considered to be an aluminum fluoroaluminate (). Sterten established that the resulting complex anion depends on the electrolyte chemistry as shown in Figure 2.5.6. As alumina dissolves in cryolite at low alumina concentrations and high AlF3 concentrations, by forming oxyfluoride aluminate ions, with a 2:1 ratio of aluminum to oxygen by the equation: (2.5.7) At higher concentrations, alumina dissolves in cryolite by forming oxyfluoride aluminate ions with a 1:1 ratio of aluminum to oxygen by the equation: (2.5.8) Cells are generally operated with 2–4 wt.% Al2O3 in the electrolyte. Saturation ranges between 6% at 960 °C and 13% at 1000 °C. Alumina solubility varies largely with the excess % AlF3 in the electrolyte and temperature. The cryolite electrolyte has the ability to dissolve alumina, but of course only up to a certain percentage at normal operating temperatures. Above this level, the excess alumina concentration added to bath will be deposited undissolved onto the bottom of cells as sludge, disturbing the current flow and generating metal pad motion, which results in a decrease in current efficiency. Note that when the alumina concentration is below 10% and the temperature of the electrolyte decreases below 960 °C cryolite will freeze out on the coldest part of the cell, mainly on the sidewalls. However, if the alumina concentration is greater than 10%, then α-alumina freezes out of solution and it is difficult to redissolve in the melt. 2.5.2.3.1 Alumina Dissolution Mechanism Alumina additions to the cell electrolyte are called “feeding” and are typically performed by adding SGA to the top surface of the molten cryolite-based bath in the electrolysis cells, whereupon it is expected to dissolve and disperse rapidly in the bath. In actual practice with a few exceptions, SGA alumina is not used directly for aluminum electrolysis. Rather it is first sent to the plant’s dry scrubbing system where it is used as an adsorbent to capture gaseous and particulate fluorides from the cell exhaust gas. The reacted alumina, containing fluorides and moisture, is then transported to the electrolysis cells where it is added to the molten bath to produce aluminum. The adsorbed moisture and fluorides react and are released when added to the electrolyte, resulting in localized agitation and stirring of the molten bath and thus they assist the dissolution of alumina in cryolite. One of the difficulties experienced in adding SGA alumina or reacted alumina to cryolite melts in aluminum cells is that the alumina particles can form clumps or rafts that float on the top surface of the melt and then gradually sink, depending on the fines content, thus slowing the alumina dissolution process. The general solubility steps for the dissolution of alumina crystalline particles in cryolite melts in aluminum cells include: • : Addition: About 1–2 kg of alumina particles is dumped onto the surface of the molten cryolite melt by point feeders with clumping or raft formation on the top surface of the bath. • : Wetting: Alumina particles are wetted by the melt. • : Heating: Alumina particles are heated from about 100 to 960 °C. • : Crust: Cryolite crust formation and subsequent melting occur around alumina particles. • : Dissolution: Alumina particles dissolve into the boundary melt. • : Distribution: Melt impregnated with dissolved alumina from the addition area is transferred to every area of electrolyte in the cell. Alumina dissolution rates in cryolitic electrolytes, as shown in Figure 2.5.7, have been the subject of intensive study for several decades. Alumina dissolution rate can be affected by the physical, chemical, morphological and microstructure properties of the ore, the dynamics of the feeding process, the chemistry of the electrolyte, and the superheat of electrolyte in the cells. Models have been developed by Haverkamp and Welch for the dissolution of alumina powders in cryolite based on the rate transfer and diffusion. The shape of the curves generated gave a reasonable fit to experimental data on dissolution. It is evident that alumina “B” is initially slower than alumina “A” in the test, but they both eventually reach the same maximum concentration. The initial difference in alumina dissolution can increase the formation of deposits of undissolved alumina when feeding alumina very rapidly to aluminum cells that are operating at high amperage. Dando et al. in a study subjected SGA alumina samples to thermal pretreatments prior to adding to molten electrolyte, in order to study the interplay between HF evolution, raft formation and dissolution, formulated several conclusions upon feeding alumina to the surface of molten electrolyte: • : Moisture is released by the alumina in several stages: a rapid short-lived initial release from gibbsite in the SGA, followed by a rapid, broader rise in evolved moisture from alpha mineralization of the undissolved SGA, again followed by a sustained baseline evolution of moisture due to dissolution of the alumina. • : SGA is rapidly mineralized to its alpha phase by the moisture in the alumina combined with vapor-phase atmolite (NaAlF4). • : Cohesive rafts are formed by interparticle platelets. The data presented in this report suggest that the initial dispersion of alumina during feeding is one of the most important factors for preventing raft formation. Any clumping of SGA immediately following feed shots would promote raft formation, owing to the rapid mineralization, wetting, and interparticle platelet growth shown above. Raft formation promotes slow dissolution of the SGA by promoting formation of alpha phase rafts that subsequently sink to the bottom of the cell, promoting so-called sludge or muck formation. Furthermore, reductions in alumina dissolution rates will directly impact dissolved alumina availability and replenishment in the molten bath within the ACD of cells. Any delays in the rate of dissolved alumina replenishment will increase both anode effect frequency and duration. Given that aluminum smelting cells are operated at low dissolved alumina concentrations, 2–4%, and relatively low superheats (less than 10 °C), the dissolution characteristics of the SGA can significantly impact the operating performance of the cells. 2.5.2.4 The Industrial Production of Aluminum 2.5.2.4.1 Faraday’s Law and Aluminum Production According to Faraday’s law, one Faraday (26.80 A h) of electricity will theoretically deposit one gram equivalent (8.99 g) of aluminum. The theoretical quantity of aluminum produced per kiloampere (kA) is calculated by the equation, (2.5.9) A loss in aluminum production results mainly from recombination of anodic and cathodic products in the electrolyte. Also, some of the current is being consumed by parallel reactions that do not give net aluminum formation. The cell current efficiency is determined by the dividing actual Al production per unit time by the theoretical Al production for the same time period. For example, the actual aluminum production per day for one cell operating at 350 kA and 95% current efficiency is calculated: (2.5.10) Because the aluminum production depends on the magnitude of the electrical current, there has been a steady increase in amperage and size of industrial aluminum cells from 40 kA cells (7 m length) in the 1940s to 200–250 kA cells (9 m length) in the 1980s and 350–400 kA cells (20 m length) in the 2000s. The aluminum production per unit time primarily depends on three factors: the potline amperage, the current efficiency of the electrolysis process, and the number of operating cells. In this manner, the annual production capacity for a potline of 300 prebake cells operating at 350 kA and 95% current efficiency is 293,214 mt/year, as indicated in Figure 2.5.8. 2.5.2.4.2 Current Efficiency—Aluminum Back Reaction The amount of aluminum predicted by Faraday’s law is never obtained in practice. In parallel with the aluminum production reaction, a quantity of metal made at the cathode dissolves into the electrolyte at the boundary layer at the bath–metal interface and is transported to the reaction zone where it is oxidized, recognized in the industry as the back reaction, by CO2 gas forming CO gas and Al2O3 as products, as shown in Figure 2.5.9. Industrial aluminum cells prior to the 1970s operated from 85% to 88% current efficiency, while modern cells now can operate 95–96% current efficiency. The corresponding concentration of CO produced by the back reaction in these cells is normally about 10%. (2.5.11) The electrolytic production of aluminum from Al2O3 and the back reaction of aluminum to form CO gas and Al2O3 are expressed in the following equations: (2.5.12) (2.5.13) Combining these two reactions gives the overall aluminum cell reaction: (2.5.14) Dissolved metal must diffuse away from the aluminum interface before it can be oxidized. The loss in current efficiency due to the back reaction is directly related to the dissolution of the aluminum product into the melt at the boundary, transport to the reaction zone, and ultimate oxidation of the metal with dissolved CO2 near the anode surface, as shown in Figure 2.5.9. In order to minimize the dissolution of aluminum in the electrolyte, cells are purposely operated with a high excess % AlF3 and low dissolved alumina concentration in the melt; low electrolyte temperature, a balanced current distribution of all anodes, and a sufficiently wide interelectrode distance. The industry current efficiency of aluminum electrolysis for new cell technology has improved from about 82% in 1900 to 96% in 2000, as shown in Figure 2.5.10, while the current efficiency in the older cell technologies has lagged behind by about 5%. The abrupt increase in current efficiency in 1960–1965 was due to the advent of modern cell technologies and improved cell operating practices. It is interesting to note that current efficiency has not improved in aluminum cells over the past 10 years or so. Perhaps, we have reached the maximum obtainable current efficiency with the current industrial Hall–Héroult prebake cell design and technology? The maximum current efficiency for Søderberg cells is only 92% due to its poorer quality anode carbon, inferior magnetic design, and large gas bubbles that increases stirring and mixing at the bath–metal interface. The aluminum production per day for one aluminum electrolysis cell operating at 350 kA and 95% current efficiency is 2678 kg/day, which represents a 5%, or 121 kg/day, loss in aluminum per day. (2.5.15) The aluminum production per year for a modern potline of 300 cells operating at 350 kA and 95% current efficiency is: (2.5.16) (2.5.17) The production of 1 ton of aluminum typically requires 420 kg of carbon, 1920 kg Al2O3, and 16 kg AlF3, and the specific energy consumption is 13.20 kWh/kg Al, as shown in Figure 2.5.11. The theoretical carbon consumption is only 333 kg C/t Al. The excess carbon consumption is mainly due to air oxidation of the hot anodes in the hooded cells. The reaction occurs preferentially in the pitch binder matrix and leads to the physical loss of coke particles to generate dust that floats on top of the electrolyte. Thus, anodes are carefully covered with a mixture of crushed bath and alumina to prevent excessive oxidation. The unused top part of anodes, “butts,” are removed from the cells when new anodes are set, crushed, and recycled in the carbon plant to make new anode carbon. Furthermore, the theoretical alumina consumption is only 1.89 kg/t Al. The excess alumina usage is due to the impurities in the metallurgical grade alumina including water and a small amount of alumina dust may be lost during transportation and transferring to/from silos. 2.5.2.4.3 Industrial Cell Design 2.5.2.4.3.1 Magnetohydrodynamics Electrolysis cells in aluminum plants are arranged in long rows, called potlines where the cells are connected in an electrical series circuit with the transformer and rectifier systems located at one end of the potline. Such arrangements allow carrying out most cell operations by using overhead multipurpose cranes. A typical aluminum smelter consists of one to three potlines. Modern 350-kA prebake cells are positioned in a side-by-side arrangement in potlines, as shown in Figure 2.5.12. In most designs, the cells are situated above ground with solid aluminum bus bars located at ground level below them with current flow from one cell to another. The smelting process requires large amounts of electricity. A reliable and uninterrupted electrical power supply is a critical issue for aluminum smelters. Alternating or AC current supplied from the grid must be transformed into direct or DC current, which requires the use of large rectifiers, transformers, and sophisticated monitoring systems located adjacent to the potline building. The most inherent risk in aluminum production is a loss of electrical power. A failure of electricity supply lasting more than 2–3 h can cause the electrolyte in the cells to cool to the point where its electrical resistance is too great when power is restored resulting in shutdown of all cells. It is expensive and time consuming; usually it will take several months, to restart a frozen potline because the solidified aluminum and electrolyte must be physically broken out of the cells. Thus, significant business losses will be incurred due to the interruption in the event of a potline freezing. The remaining lifetime of the cathodes will be shortened due to the extra thermal stresses inevitable with shutdown and restart of cells. Magnetohydrodynamics (MHD) forces are generated in the aluminum metal caused by the interaction between the magnetic fields produced by the passage of high-amperage DC electrical current through nearby conductor bars and the electrical current flow in the aluminum metal in electrolysis cells. The large electric currents, 300–500 kA, in modern prebake cells pass through the electrically conducting liquids (electrolyte and metal) and generate powerful magnetic fields in the aluminum pool. These strong MHD forces increase metal velocity, metal wave height and frequency, and distortion of the aluminum–bath interface and thus require higher voltage and operate at lower current efficiency. These cells typically maintain a 25-cm deep pool of liquid aluminum in cells to minimize the impact of the MHD forces. A major achievement in the aluminum industry has been the development of advanced one-quarter size (3D) computer mathematical models of the interaction of MHD fields, forces, and resulting metal pool behavior in aluminum cells. This is exemplified by the model derived by Severo et al. that is used for cell design and retrofit studies, as shown in Figure 2.5.13. These programs are used to develop new magnetic compensation in cells by rearranging the bus bar conductor systems to reduce the magnitude of the magnetic fields inside the cells and thus allow significant increases in the potline amperage without losing aluminum productivity or requiring extra voltage. 2.5.2.4.3.2 ACD and Cell Voltage Electrolysis occurs in the electrolysis cell ACD or between the bottom horizontal surface of the anodes and the upper aluminum metal surface. The upper part of the aluminum cell shown in Figure 2.5.14 is referred to as the anode superstructure. The anodes, carbon blocks, are attached to long rods that are suspended and clamped to the anode beam located along the length of both sides of the anode superstructure. Electrical current enters the cell through the large aluminum conductor bus connected to the anode beam. The anode beam moves all the anodes up and down simultaneously, thus changing the ACD, typically 3.8–5.0 cm, from the bottom surface of the anodes to the top surface of the aluminum pool. The ACD is not measured, but the average value can be calculated. Changing the ACD changes the cell voltage and energy input into the cell; most of the cell heat is generated in the electrolyte in this zone. Steel current collector bars attached to the bottom cathode lining carry the electrical current from the cell to the next cell in the series. 2.5.2.4.3.3 Alumina Feeding The electrolytic process consumes alumina at a nearly constant rate. Alumina is fed to the aluminum cells at specific intervals at a rate equal to 1.92 times the aluminum production rate by point feeders located inside the anode superstructure. The computer process control system activates point feeders (crust breaker and alumina dosing devices) according to sophisticated algorithms to keep the alumina content in the electrolyte near the target value. The alumina concentration dissolved in the electrolyte cannot directly be measured; thus, the amount of alumina added to cells is regulated indirectly using an underfeed/overfeed computer control algorithm based on changes in the cell voltage due to the anode overvoltage at low and high alumina concentrations in the electrolyte. To add alumina to the bath, fast action pneumatic air cylinders activate crust breakers to open holes in the crust layer on top of the molten bath at two or more positions along the center line of the cell. Next, the point feeder adds 1.5–2 kg of alumina from a volumetric dispenser directly to 18–20 cm of molten bath where it dissolves. There are usually from two to six point feeders per cell depending on the aluminum metal production rate, which is determined by the potline amperage, as shown in Figure 2.5.15. The advantage of point feeders is that small quantities of alumina are added to the electrolyte at each break-and-feed. This method minimizes the risk of sludge formation in the center area of the cell, and there are minimal emissions of dust and fluorides during the alumina feeding operation. 2.5.2.4.3.4 Changing Anodes Prebake anodes are manufactured by compressing a mixture of petroleum coke aggregates and coal tar pitch binder into blocks typically 40–60 cm wide by 120–150 cm long and 50–60 cm high. Petroleum coke is used because of its high purity. These blocks are baked in anode-baking furnaces above 1100–1200 °C to convert the binder pitch into dense carbon. The carbon blocks are attached to an iron anode assembly by pouring molten cast iron into two to six holes in the anode block for fluted steel stub connections. An aluminum, or in some instances copper, rod is attached to the top of the anode assembly. The top section of the anode rods are clamped to the anode beam and thus carry the electrical current from the anode beam to the molten cryolite melt in the cells. Many smelters now manufacture or cut “slots” in the bottom surface of anodes to divert gas bubbles, thus reducing the electrical resistance in the electrolyte and reducing the total cell voltage of operating cells. From 18 to 40 prebaked carbon anodes are required per cell depending on the potline amperage to maintain the anode carbon current density in the range 0.7–1.2 A/cm2. The carbon anodes are consumed at a rate of about 1.9 cm/day, depending on current, by the electrolysis due to the reaction to produce CO2. One or two anodes are replaced each 24–48 h in the cell with new anodes after 21–28 days depending on the anode dimensions and amperage, i.e., the anodic current density. A crushed bath–alumina mixture is used to cover the top and sides of the new carbon anodes to avoid excessive oxidation and also to form a crust on top of the electrolyte to reduce heat losses and emission of fluorides from the electrolyte. Søderberg cells are another type of aluminum electrolysis technology with only one large anode per cell that is continuously produced by a self-baking process by the cell process heat. However, Søderberg cells are being phased-out in the aluminum industry now due to their lower production efficiency, inherent environmental problems, and the emission of polycyclic aromatic hydrocarbons (PAHs) from the large self-baking anodes. 2.5.2.4.3.5 Metal Tapping Positive aluminum-containing ions deposit continuously at the negatively charged cathode, the top surface of the aluminum pool, where they convert to liquid metal and thereby slowly raising the metal volume and level. A portion of the aluminum metal pool is removed every 24–48 h by siphoning the metal into a large crucible. The aluminum metal is weighed and sent to the cast house for further processing. The height of the metal level in the cells is maintained at a target value as the metal height impacts the heat loss out of the cell sidewalls because molten aluminum metal has a higher heat loss out the sides than molten cryolite. The metal depth also dampens the impact of the magnetic forces on the metal pool. 2.5.2.4.3.6 Cathode Lining The lower cathode components consist of a rectangular reinforced steel box lined on the inside with carbon, refractory bricks, and insulating materials. The cell is thermally insulated internally on the bottom and lower sides to reduce heat losses and maintain the optimum cell heat balance. As aluminum plants continue to increase potline amperage to produce more metal, it becomes necessary to increase the heat transfer out of the sidewall of cathodes in order to maintain the frozen cryolite ledge to protect the sidewall lining material. Cell cathodes are now being constructed using higher thermal conductivity silicon carbide sidewall blocks, steel fins attached along the outside sidewalls, and air pipes that blow compressed air along the sidewalls. The cell cathode linings have lifetimes generally from 1700 to 3000 days. Higher cathode life requires excellent construction and good quality materials. 2.5.2.5 Electrode Reactions for Aluminum Electrodes 2.5.2.5.1 Anode Reactions 2.5.2.5.1.1 Electrochemical Production of Carbon Dioxide In the electrochemical reaction to produce aluminum, the carbon anode is continuously consumed by the subsequent reaction with oxygen. There are two possible anode gas products, CO2 and CO, formed in the reactions: (2.5.18) (2.5.19) The reversible voltage potential at 1000 °C to produce carbon monoxide at the carbon electrode surface in reaction (2.5.18) is − 1.065 V compared with − 1.181 V to produce carbon dioxide in reaction (2.5.19). Thus, the production of CO in reaction (2.5.18) is thermodynamically favored. However, in practice, the reaction kinetics and polarization result in the generation of CO2 gas at the anode electrode surface. At a low anodic current density, oxygen forms a stable C–O surface compound, and at a low gas evolution it can detach slowly and produce CO gas, as shown in Step “a” of Figure 2.5.16. But at the normal anodic current density used in commercial aluminum cells, 0.70–1.2 A/cm2 and high alumina concentration, additional oxygen atoms react with the CO surface compound before it can detach forming an unstable CO2 stable surface compound that detaches rapidly producing CO2 gas, as shown in Step “b” of Figure 2.5.16. Desorption of CO from the carbon surface must be kinetically hindered. The formation of carbon dioxide at high current density decreases the effective surface coverage favorable for the formation of carbon monoxide, and therefore, at the high current densities the amount of carbon monoxide formed decreases. Welch and Richards measured the anodic overpotentials for the discharge of oxygen-containing anions on a carbon anode in molten cryolite–alumina mixtures, and the results are shown in Figure 2.5.17. It was determined that the rate-determining step is the two-electron transfer reaction in which oxygen-containing anions are discharged. The steepness of the polarization curves demonstrates how quickly it changes in modern prebake cells that operate at low alumina concentrations and higher current density due to increases in potline amperage. The polarization inflection for a change in electrode process at lower current intensity below 0.05 A/cm2 for all alumina concentrations and carbon types indicates the formation/desorption of CO gas at the electrode surface. A hysteresis before and after electrolysis indicates that there is a slowly desorbing intermediate on the surface which is consistent with the CO formation mechanism. Welch and Richards found that the anodic overvoltage follows the classical Tafel equation and concluded that the overvoltage was caused by slow transport of oxygen-carrying ions through the double layer. Welch indicated that anode carbon used for electrodes in aluminum electrolysis cells is disordered as the oxygen bonded to carbon atoms is in a strained five-membered ring, a poorly bonded carbon which is expected to have a higher rate of oxygen desorption from the anode surface. In practice, this has been proven out as more CO is evolved at low current density for poorly baked anodes which have a higher reactivity to carbon dioxide. 2.5.2.5.1.2 Parallel Reactions at the Anode 2.5.2.5.1.2.1 Electrochemical Production of Carbon Monoxide Hume et al. have shown that carbon monoxide is electrochemically produced and coevolved at areas on the carbon anode electrode at localized areas of low current density as illustrated by the diagram shown in Figure 2.5.18. Because of the simultaneous rapid acceleration in both the consumption rate and the rate of carbon monoxide formation, it is unlikely to be solely due to the Boudouard reaction. During this process, the current can be carried by two reactions, and this is supported by the anode polarization gradient. Thus, direct electrochemical formation of carbon monoxide becomes an important mechanism at low current densities. These conditions exist on the sides of anodes, which contribute up to one-third of the electroactive area during electrolysis, and depending on the depth of immersion of the electrodes, they can carry more than 10% of the total anode current. At much lower current density, Zoric et al. have illustrated that the current intensity decreases rapidly from 0.70 to 0.2 A/cm2 on the bottom corners of anodes and decreases from 0.2 to 0.08 A/cm2 on the sides of anodes. (2.5.20) The electrochemical formation of carbon monoxide by reaction involving the carbon electrode is one cause of excess carbon consumption. As seen from Equation (2.5.20), the electrochemical formation of carbon monoxide doubles the rate of carbon consumption. 2.5.2.5.1.2.2 Production of Carbon Monoxide by the Boudouard Reaction The endothermic reaction of carbon dioxide with carbon and the subsequent transformation into carbon monoxide is favored at higher temperatures and also doubles the carbon consumption rate according to the equation: (2.5.21) The Boudouard, or the carboxy reaction, is temperature dependent with the thermodynamically favored product being all CO at temperatures greater than 900 °C. However, the Boudouard reaction is not a major function in prebake aluminum cells as it appears that the polarized carbon anodes immersed in cryolite in aluminum electrolysis cells are protected from attack by CO2. It is possible that CO2 generated at the anode interface by electrochemical can attack the carbon anode, especially on the submerged sides of anodes that have a lower current density, as shown in Figure 2.5.18. The subsurface Boudouard reaction does occur to a greater extent within the anode pores in Søderberg cell anodes. These anodes have a significant higher permeability that allows CO2 to penetrate deeper into the pores of the anode. 2.5.2.5.1.2.3 Electrochemical Production of Perfluorocarbons 2.5.2.5.1.2.3.1 Anode Effects An anode effect is phenomena observed in many processes involving the electrolysis of molten salts and is not always well understood. It is a condition produced by polarization of the anode in the electrolysis of fused salts and characterized by a sudden increase in voltage and a corresponding decrease in amperage. The onset of the anode effect in Hall–Héroult aluminum cells is primarily due to the depletion of the oxygen-containing ionic species at the surface of carbon anodes causing an increase in anode polarization. Prior to the anode effect, the alumina concentration in typical modern cells decreases about 30% of its normal value. Thus, oxygen-containing ions will be arriving at the electrode surface at approximately two-thirds of the normal rate. Once a resistive layer is formed on the anode surface, it causes an increase in voltage at a constant current or a decrease in current at a constant voltage. Once an anode effect occurs in operating cells, its anodic current density has exceeded its critical current density (ic). The cell critical current density is primarily a function of the concentration of dissolved alumina, anode dimensions, and amperage. However, it is also influenced by anode immersion in electrolyte, electrolyte flow, gas bubbles, temperature, and anode spacing. The critical current density calculated by Thonstad and Richards for industrial cells is calculated by the following equation . (2.5.22) The decrease in alumina concentration also causes an initial deterioration of the wetting and increased gas bubble coverage, causing the current density at the active parts of the anode to increase leading to the risk of local alumina depletion, which results in an eventual anode effect. Fluorocompounds start being codischarged at the carbon anode surface forming carbon–fluoride intermediate compounds analogous to polytetrafluoroethylene, –{F2C–CF2}–, on the anode surface. These have a strong dewetting effect causing a rapid rise in voltage and increase the start of fluorides being oxidized and forming intermediate compounds at the anode surface with the discharge of perfluorocarbon gases, carbon tetrafluoromethane (CF4), and carbon hexafluoroethane (C2F6). Multiple anode cells are complicated in that each anode in the cell has a different anode current density due to difference in their individual ACD, causing anodes to have an unequal current distribution. Anodes also have different alumina concentrations in the anode–cathode spacing due to differences in their location with respect to the point source of alumina feeding and also differences in bath flow patterns in cells. The anode current distribution data shown in Figure 2.5.19 demonstrate that the anode effect may often start on only one of the prebake anodes prior to the onset of the anode effect, as indicated by the decrease in current on two anodes about 1–2 min prior to the actual anode effect. Once blockage of the electrical current by a film of CF4 gas occurs under one or more of the anodes, it significantly increases the current density on the remaining anodes resulting in an escalating anode effect sequence on all the prebake anodes. The anode effect is described as a blockage effect which inhibits the normal current flow between the anode and the electrolyte. A high percentage of current flow shifts from the bottom surface of anodes to the side surface, causing large fluctuations in current flow between anodes which induces instability in the aluminum metal pool. It takes from 5 to 10 min for the current distribution to become almost equal again on all of the anodes. Aluminum electrolysis cells operate at constant current; thus, the formation of a highly electrical resistive carbon–fluoride intermediate film on the bottom surface of carbon anodes causes the cell voltage to increase very rapidly from about 4.2 to > 30 V, causing the electrolyte temperature to increase to > 1000 °C, which is observed practically. This generates a lot of heat especially as localized arcing often occurs because of the film. Once the anode effect starts, it proceeds rapidly as indicated by the fast increase in cell voltage. (2.5.23) 2.5.2.5.1.2.3.2 Cell Gas Composition During Anode Effects Electrochemical reactions are still occurring during anode effects as this is the only way current can flow through the electrolyte. Tabereaux et al. measured the change in anode gas composition during anode effects, as shown in Figure 2.5.20. The gaseous mixture consists primarily of carbon monoxide, 60–70%, and carbon dioxide, 20–30%; the measured CF4 content from both prebake and Søderberg cells was from 16% to 20%, and the C2F6 generation was small, 0.0–0.05%. The increase in the carbon monoxide product is due to the higher voltages and changes in the electrode surface reactions. 2.5.2.5.1.2.3.3 Anode Effect Termination The cell remains under the influence of the anode effect until alumina is added, and the cell current is interrupted by a strong electrical short of the cell current by lowering the anodes until they make electrical contact with the aluminum metal waves or, alternatively, by inserting wooden poles under some of the anodes. These actions cause the current density on the anodes to decrease below the critical current density, which allows adherent gas bubbles formed at the anode surface to collapse or become detached. 2.5.2.5.1.2.3.4 Formation of COF2 The standard potential required for the decomposition of cryolite and the formation of CF4 is − 2.54 V and the potential for formation of C2F6 is − 2.78 V during anode effects, which are higher than the − 1.191 V potential necessary for the formation of CO2. The changeover from solely oxide ion discharge (forming CO or CO2) to codischarge of the fluoride intermediate, COF2, in reaction (2.5.24) occurs at a much lower anode potential of − 1.86 V. (2.5.24) The electrode potential for COF2 reaction was detected in the late 1970s. However, its existence was not confirmed until Dorreen and coworkers [19,20] detected the intermediate compound using mass spectroscopy. It is shown in Figure 2.5.21 that the COF2 formation occurs during electrolysis about 5 min prior to the full onset of the anode effect. After the rapid rise in cell voltage, the COF2 immediate compound decomposes at 960 °C to form CF4 during anode effects. The formation of COF2 and subsequent CF4 formation continues at a fixed rate during the anode effect and this agrees with the measurements of CF4 emissions by Tabereaux et al. from aluminum cells at smelters. Welch indicates that the sharp rise in voltage is due to the coevolution formation of the fluoro compound, COF2 intermediate, that passivates the anode surface (ΔG° = − 45.8 kJ at 960 °C). (2.5.25) Mechanistically, the formation of the COF2 product is likely to be similar to the formation of carbon dioxide, simply because of the high polarization, which is usually brought about by lowered rates of oxide arrival at the electrode surface through concentration polarization and fluoride ions codischarge. This can occur with either the surface edge or the bridge carbon atoms. 2.5.2.5.1.2.3.5 PFC Emission Rates A nearly universal definition of an anode effect has been established under the guidance of the International Aluminium Institute (IAI) to ensure that aluminum plants are using the same methodology when reporting PFC emissions. Anode effect and anode effect duration: An anode effect is typically considered to begin when the cell voltage exceeds a defined voltage threshold (6.0–8.0 V), and the anode effect is considered to end when the cell voltage drops below a second voltage threshold (6.0 V) and remains below this voltage level for a defined time (15 min). However, if the anode effect reoccurs within 15 min, as indicated by the cell voltage exceeding the threshold value, then it is considered as a repeat anode effect and is not counted as a new anode effect. The anode effect duration is the sum of the individual minutes when the cell is on anode effect. Anode effect duration refers to the total minutes per cell and day during which the voltage in the cells is above the threshold. The rate of PFC emission, kg CF4/min, has been established for different cell technologies based on actual PFC emission measurements from commercial cells in aluminum potlines for the duration over which cells have their anode effect as reported by Marks and Bayliss . (2.5.26) where AEF is the average anode effect frequency, AED is the average anode effect duration in minutes, and S is a technology-specific factor that is equivalent to the slope of the graph for (AEM·AED) versus kg CF4/t Al per AEM, which is the emission rate in kg CF4/AE minute. Specific default emission rate values have been determined based on plant measurements by the IAI for five different categories of cell technology. These technology-specific factors are used primarily as default values for aluminum smelters that have not measured their emissions rates. Aluminum smelters are encouraged to make their own PFC measurements in order to determine the specific slope value for their individual aluminum smelter. There are considerable variations (± 30%) in these factors. The cause for the wide variations in measured emission rates between plants that have the same cell technologies is not well understood. It is known that the low slope emission rate for the Søderberg cells is due to electric shorting between the aluminum metal waves, as indicated by large variations in cell voltage during anode effects, and the anode carbon during the anode effect and the PFC emissions stop during the shorting. In addition, it is known that PFC emissions actually begin on one or more individual anodes prior to the cell voltage reaching the AE threshold value. The method used to terminate the anode effects also to some extent influences the PFC emission rates as differences in emissions have been measured in plants using slow anode up/down pumping versus plants using fast aggressive anode down moves to terminate the anode effect quickly. 2.5.2.5.2 Cathode Reactions Measurements indicate that the smaller and more ionic mobile Na+ ions carry most of the current through the electrolyte. However, the deposition of aluminum is favored over sodium as the reversible decomposition is favorable. Thus, aluminum is the primary product produced at the top surface of the cathodic aluminum pool in the electrolysis cells. The most probable cathodic process involves a charge transfer at the cathode surface in which aluminum-containing anions (hexafluoroaluminate) are discharged to produce aluminum metal as well as F− ions to neutralize the charge of the current carrying Na+ ions . The most probable overall cathodic reactions are: (2.5.27) (2.5.28) Sodium is codeposited at the cathode along with aluminum, (2.5.29) Since the reaction involves an interfacial transfer of the Al3 + ions, the rate of aluminum deposition must be matched by the rate of transfer of Al3 +ions from the solution. Polarization effects at the cathode contribute much less to overvoltage than at the anode. The ion complexes and have higher ionic mobility than their anodic counterparts, which lowers the concentration polarization effect. In addition, there are no gas bubbles at the cathode which influence both resistance and concentration polarization. 2.5.2.5.2.1 Cathode Parallel Reactions 2.5.2.5.2.1.1 Production of Sodium at the Cathode The primary cathodic reaction in cryolite-based melts is the reduction of Al3 + containing species. The next most favored cathodic reaction is the deposition of sodium which forms an alloy with aluminum. Sodium is produced at the bath–metal interface due to the chemical reaction occurring when reaching thermodynamic equilibrium and sodium partitioning into the two phases based upon equilibrium constants for the system. (2.5.30) During electrolysis, the content of sodium increases with increasing cathodic current density. This effect is related to the cathodic concentration overvoltage which is mass transfer controlled. It is caused by the fact that the sodium ion is the carrier of current, while aluminum is being deposited. This leads to the formation of a concentration gradient in the boundary layer at the cathode, with enrichment of NaF and depletion of AlF3 at the interface. The addition of lithium fluoride has been shown to result in a decrease in the cathodic overvoltage. The cathode overvoltage reflects the concentration gradient with respect to the concentration of NaF and AlF3 at the bath–metal interface. Hence, there should be a relationship between the overvoltage and the sodium content in the aluminum metal in operating cells. The magnitude of the overvoltage is dependent on the cell design and on the convection pattern in the cell. Modern aluminum cells with good magnetic compensation and low flow rates at the bath–metal interface have a high sodium content, > 120 ppm in the aluminum cathode pool. Strong relationships exist between the sodium content in the aluminum pool in cell and the cell current efficiency. 2.5.2.5.2.1.2 Production of Other Metals at the Cathode Metal impurities, e.g., iron, silicon, manganese, copper, and vanadium, in the alumina and anode carbon dissolve in cryolite to form ions that are codeposited at the aluminum cathode surface. These metals are more noble than aluminum, i.e., they have reversible decomposition potentials that are more favorable for reduction than aluminum. Iron and silicon are specific concerns as they are the highest impurities present in alumina and coke and will eventually be contaminates in the aluminum pool at the cathode. (2.5.31) (2.5.32) Impurities in the electrolyte with multiple valence states, such as phosphorus, can be reduced at the cathode and then reoxidized at the anode, thus consuming electrical current without producing any aluminum. 2.5.2.5.2.1.3 Production of Alkali Metal at the Cathode The alkali and alkaline earth metals (lithium, magnesium, and calcium) dissolve in cryolite to form ions (LiF, MgF2, and CaF2) but are not reduced at the cathode because they are less noble than aluminum, and thus, their concentration increases in the electrolyte. The concentration of alkali and alkaline earth metal fluorides is normally well below the saturation concentration in cryolite. However, it has been found that some reduction of these metals does take place as an equilibrium concentration of lithium, magnesium, and calcium in the aluminum metal pool in cells relative to the concentration of these metals present in the cryolite electrolyte, indicating that the process may be mass transfer controlled, or a chemical equilibrium at the bath–metal interface. Given enough time to reach thermodynamic equilibrium, any metal fluoride will partition into the two phases based upon equilibrium constants for the system, for example, lithium: (2.5.33) Peterson and Tabereaux determined that the concentration of alkali and alkaline earth metal in aluminum relative to % alkali metal fluoride dissolved in the electrolyte and excess AlF3 concentration, when expressed as the CR ratio (% NaF/% AlF3) can generally be expressed as a second-order polynomial expression written as: (2.5.34) (2.5.35) (2.5.36) 2.5.2.6 Thermodynamics for Aluminum Electrolysis 2.5.2.6.1 Standard-State Gibbs Free Energy The driving force for the heterogeneous electrochemical reaction for the production of aluminum is the standard Gibbs energy gradient at the reaction interfaces, and it is calculated by the change in the standard Gibbs energy at equilibrium, ΔG°, for the reaction to form the products, Al and CO2, from the reactants, Al2O3 and C, at 1250 TK and where x is the fraction current efficiency according to the equation: (2.5.37) Substituting the individual numerical values of the standard Gibbs energy for the formation of compounds from the elements into the above equations gives the following equation for the standard Gibbs energy as a function of current efficiency . (2.5.38) Accordingly, at 100% current efficiency, the standard Gibbs energy for the formation of aluminum from alumina is 342.3 kJ/mol, and for every 10% increase in the reoxidation of aluminum in industrial cells, there is a 4 kJ/mol decrease in the standard Gibbs energy. 2.5.2.6.2 Reversible Decomposition Potential, Nernst Voltage The magnitude of the cell potential is a measure of the driving force behind a reaction. The larger the value of the cell potential, the further the reaction is from equilibrium and the sign of the cell potential indicates the direction in which the reaction must shift to reach equilibrium. Consider the electrochemical reaction between alumina and carbon to produce aluminum: (2.5.39) The fact that E° is negative indicates that the system does not favor the products of the reaction in the direction as written. The reaction will require about 1.20 V using a carbon electrode in order to proceed to form aluminum as a product. The Nernst equation describes the relationship between the cell potential at any moment in time and the standard-state cell potential. The standard cell potential of the reaction is directly proportional to the change in the standard Gibbs energy of the electrochemical reaction in the Nernst equation. (2.5.40) Substituting n = 12, F = 96,485 J/V equivalent and the numerical values for ΔG° as a function of TK for an alumina-saturated electrolyte, according to Haupin , gives the following equation for the reversible potential for aluminum production. (2.5.41) However, during normal cell operations, the bath is not saturated with alumina. Consequently, the Nernst equation is used to correct the standard potential for aluminum electrolysis for the actual activities for the reactants to give the reversible equilibrium potential ERev for aluminum electrolysis. At 960 °C, the equilibrium potential for cells at alumina saturation is 1.191 V for saturated electrolyte, ~ 8% Al2O3; however, the equilibrium potential is 1.222 V at ~ 2.6% Al2O3 alumina. (2.5.42) (2.5.43) where αA is the activity concentration of aluminum in the melt, is the activity of Al2O3 in the melt, and is the partial pressure of CO2. Substituting the appropriate values gives the dependent cell potential at reversible, nonstandard conditions: (2.5.44) 2.5.2.6.3 Change in Enthalpy The change in the standard enthalpy per mole, or per kg of aluminum for transference of the reactants at 298 TK to the reaction products at the electrolyte temperature TK for reaction (2.5.10) is 550 kJ/mol Al at 1000 °C. But the actual energy requirement for aluminum production depends on the cell current efficiency . Substituting the various energy expressions for the specified conditions, 970 °C and 8 wt.% alumina, gives the theoretical minimum energy requirements as a function of current efficiency, xAl. (2.5.45) For a typical 5% loss in current efficiency in aluminum cells, the overall enthalpy is 6.65 kWh/kg (or 478.5 kJ/mol Al); in addition, for each 5% change in current efficiency, the change in the standard enthalpy is 0.08 kWh/kg. The electrolyte in industrial aluminum cells is not saturated with alumina, thus its chemical activity is less than unity. This necessitates a correction term in the above calculations: , where R is the universal gas constant, T = 1250 TK, and = 0.3. The energy correction term amounts to about 0.2 kWh/kg Al (or 12 kJ/mol Al). A small fraction of CO2 reacts with carbon to make CO by the Boudouard reaction, but in general, the reaction is small and thus ignored in these calculations. 2.5.2.7 Energy Efficiency of Aluminum Cells The aluminum industry average energy consumption is presently about 14.0 kWh/kg Al compared with the theoretical minimum energy consumption of 6.42 kWh/kg Al for cells operating at 95% current efficiency. The actual energy efficiency for industrial aluminum cells is only about 50% due to the large heat losses to the surroundings. The specific energy consumption of aluminum electrolysis cells can be calculated from the total cell voltage and % current efficiency. (2.5.46) The aluminum industry has made major progress in reducing the energy consumption as modern industrial aluminum cells operate close 4.2 V and 95% current efficiency and 13.2 kWh/kg Al. The improvement is due largely to changes in cell designs that reduce the cell voltage components by increasing the size of the conductors and using materials that have a higher electrical conductivity. A few aluminum plants operate at voltage as low as 3.85 V, which reduces the cell energy consumption to 12.2 kWh/kg Al by reducing the metal instability by improving the magnetic compensation in cells and then decreasing the anode–cathode spacing lower than normal. The thermoelectric design of an aluminum cell is the aspect of the cell which has the most influence on the cell power consumption and it is also a key element affecting the life of the cell lining. The thermal balance of the cell is often the limiting factor which prevents the smelter to increase production by increasing the line amperage. Currently, the heat losses and corresponding ledge freeze profiles of one-quarter of the entire cell can be modeled using 3D thermoelectrical models. The relationship between the voltages corresponding to the enthalpy and cell heat losses was first introduced by Haupin in the voltage–energy diagram in Figure 2.5.22 for aluminum production with carbon . The total cell voltage, 4.20 V, is shown on the left scale split up into individual components and it is compared with the total corresponding energy consumption (14.25 kW), comprising the process energy (6.4 kW) and the cell heat losses (7.1 kW), as shown on the right scale. Additional energy (0.25 kW) is required to heat the anode cover material and anodes to operating temperature in cells. Operating an aluminum cell with noncarbon anodes results in a 2.8 kWh/kg Al higher energy consumption compared with carbon anodes. Tests to date indicate that inert anodes will not result in a lower, or higher, current efficiency than cells using carbon anodes. 2.5.2.7.1 Cell Voltage Components Because the ΔG° for the aluminum cell production reactions is positive, the E° values are negative. Likewise, by thermodynamic convention, cell voltage and all components of it are negative. However, engineers and cell operators in the primary aluminum industry consider the cell voltage to be positive; hence in this work, the sign of all cell voltages will be made positive. But because some of the voltage components are considered to be positive or negative depending on the scientific principle used, the absolute value is indicated by brackets. The overall cell voltage Vcell is the sum of individual cell ohmic voltages VIR and the sum of the cell and electrode nonohmic polarization voltages \|Epol\|, (2.5.47) The ohmic voltage drops in aluminum cells result from the potential difference when passing electrical current through the cell conductors (anode electrode, electrolyte, gas bubbles in the electrolyte, cathode electrode, and the connecting bus between cells) that have a measurable resistance. (2.5.48) The nonohmic polarization voltage components \|Epol\| of an aluminum electrolysis cell include the reversible equilibrium potential \|Erev\| and the polarization concentration and reaction overvoltages η at the electrode interfaces. Typically, voltage values for all the individual voltage components in an industrial aluminum cell are illustrated in Table 2.5.3. Table 2.5.3. Typical Voltage Components in an Aluminum Electrolysis Cell \| \| \| \| --- \| Empty Cell \| Cell Component \| V \| \| Van \| Anode voltage drop \| 0.30 \| \| Erev \| Reversible equilibrium voltage \| 1.20 \| \| ηaa \| Anode reaction overvoltage \| 0.50 \| \| ηca \| Anode concentration overvoltage \| 0.02 \| \| Vbub \| Gas bubble voltage drop \| 0.05 \| \| Vel \| Electrolyte voltage drop \| 1.50 \| \| ηca \| Cathode overvoltage \| 0.08 \| \| Vca \| Cathode voltage drop \| 0.40 \| \| Vext \| External busbar voltage drop \| 0.15 \| \| Vcell \| Total cell voltage \| 4.20 \| The individual electrochemical potentials and the voltage drops within the ACD of aluminum cells are shown in the expanded drawing shown in Figure 2.5.23. The voltage drops due to the electrical resistance of the electrolyte and gas bubble occur in the electrolyte and are a function of the electrical current in the cell. The electrochemical potentials occur at the boundary gradients at the electrode interfaces and are a function of chemical concentrations and temperature, and current intensity has an influence on the electrode overpotentials. The largest ohmic component in the cell is the voltage drop through the electrolyte, which is linearly dependent on the ACD. Under normal operating conditions, the ACD is between 4 and 5 cm. The ohmic voltage drop in the electrolyte can be calculated by the following equation: where ia is the amperage, К is the electrical conductivity (S/cm), ACD is the anode–cathode distance (cm), and db is the average bubble size (cm). The resistance of the bath voltage drop can be reduced by decreasing the interelectrode distance or by improving the electrical conductivity of the melt by changing the individual additives. The combined effect of temperature and AlF3 can be large. An equation to calculate bath conductivity (К, per ohm/cm) with concentrations in wt.% and T in Kelvin is given by the following equation . (2.5.49) 2.5.2.7.2 Electrode Polarization 2.5.2.7.2.1 Polarization Voltage Theoretically at the Nernst electrode potential, the rates of the reactions in the forward and reverse directions, expressed in terms of rate equations, are counterbalanced and result in no formation of product. Therefore, before the reaction can actually occur, the potential gradient at each of the electrodes must exceed the equilibrium Nernst potential. This extra voltage, which can be different at each electrode, is referred to as the electrode polarization or overvoltage. To get the reaction to go faster by carrying a high current density, the magnitude of the electrode potential gradient has to be increased by providing the required amount of polarization or overvoltage. The amount needed depends on the specific electrode reaction and the existing conditions. There is a very thin stagnant layer of electrolyte (boundary layer) at the surface of each electrode. Convection does not function in this layer. Ions must diffuse through it. This creates concentration gradients. Gas bubbling makes the layer much thinner at the anode than at the cathode. The total anodic overpotential is composed of two components: reaction overpotential, ηa, attributable to the energy necessary to drive the reaction controlling step at the given rate or prevailing current density (i), and the concentration polarization, ηc, due to any inhibition of transfer of oxyanions (diffusion, convection, geometrical or temperature constraints), leading to a concentration gradient across the boundary layer. The anode reaction overvoltage can be large, 0.40–0.56 V, depending upon the nature of the carbon, orientation of the anode, and current density, while concentration polarization is in the order of millivolts except near anode effect. 2.5.2.7.2.2 Back EMF Nonohmic Voltage When current is passed through the electrolysis cell, the overall voltage is a combination of the ohmic voltage drops due to the resistance of the ionic and electronically conducting materials plus the sum of the operating anode and cathode potentials. Immediately on interruption of the current during normal operation of the cell, the ohmic component of the voltage drops disappears but there is a residual voltage given by the sum of the electrode potentials at the moment this current is interrupted—this is commonly referred to as the back EMF. At the instance of current interruption, each electrode has reaction products absorbed on the surface as well as reaction intermediates that may be performed before release of the final product. The electrolytic cell is therefore able to acts like a battery and that is why it is referred to as the back EMF. It is the sum of the reversible decomposition potential and the anode and cathode overpotentials that exist for the conditions that are based on the preceding conditions. This is a nonohmic cell voltage, VBemf, even though its establishment may be current dependent given by the following equation: (2.5.50) where \|Erev\| is the reversible equilibrium potential and \|η\| is the sum of the cell polarization overvoltages. The overpotentials are generally current density dependent as well as dependent on temperature, alumina concentration, and structure of the anode carbon. Where some of the overpotential is due to concentration gradients between the bulk of electrolyte and that existing on the electrode surface, this will slowly self decay. However, the “battery” can hold a voltage or potential greater than the Nernst potential due to the presence of intermediate products from the overall electrolysis reaction. Typically, this voltage is between the Nernst potential of 1.2 and 1.6 V. Because the cells are interconnected in series, there can be a very high back voltage of the potline. The theoretical relationships for establishing the electrode polarizations that exist are usually expressed in terms of current density of the cells. These theoretical relationships are complex functions of the current, even though the value is dependent on process conditions rather than current. Hence, it is a nonohmic voltage. Historically, cells used to be operated at anode and cathode densities in the range 0.7–0.8 A/cm2, and in this range, the sum of the polarizations typically were approximately 0.45. Hence, if voltage measurements within that current band were linearly extrapolated to zero current, the intercept voltage would be approximately 1.65 V. It assumes a higher values at higher current densities used in modern cell operation. In order to smooth out cell voltages when the line current fluctuates (and fluctuations occur through rectification and process changes), the simplification has been made to normalize the voltage by calculating a pseudoresistance based on the assumption that the back EMF or combined combination of Nernst potential and electrode polarization is 1.65 V. In addition, there is a safety issue regarding the nonohmic voltage potential. Each cell continues to have an electrode potential even after the power is switched off. These cells behave as a charged battery, slowly decaying with time, because the anodes still have partially discharged chemical intermediates in a thin region on its surface. The total voltage potential can be quite large as it is the sum of the individual decaying voltages on each cell in the potline, for example: 300 cells × 1.5 V (average) = 450 V. Mohammed and coworkers reported that a slower decrease in potline amperage reduces the cell potential in cells when the potline power outages are planned, and also extra precautions are necessary when working on cells during power outages in potlines. 2.5.2.7.3 Cell Pseudoresistance and Cell Voltage Control In aluminum smelters, the computer process systems automatically control the cell voltage in aluminum electrolysis cells by moving the anodes up or down and thus change the ACD, depending on the target value and control bands. The control system also makes automatic adjustments in the cell voltage based on the variation in the actual voltage values or noise level. By Ohm’s law, the resistance of a circuit is equal to the voltage divided by current. However, the voltage in an aluminum cell changes whenever the potline amperage changes, for example, due to anode effects or other events. Thus, the computer process control system calculates a pseudoresistance that is used to actually control cells that corrects for errors in cell voltage control of cells during periods of current fluctuations by subtracting the total nonohmic voltage from the actual voltage values, thus making the change in cell pseudoresistance directly proportional to any change in potline amperage. The target set point for the pseudoresistance of an aluminum cell operating at 200,000 A and 4.20 V is calculated in the following manner: (2.5.51) (2.5.52) 2.5.2.7.4 Gas Bubbles Initially, nearly spherical CO2 bubbles nucleate and grow on the bottom horizontal surface of the carbon anodes. Buoyancy forces acting on the bubbles cause the small bubble to coalesce with nearby bubbles, and as they eventually start moving toward the edge of the anode, these bubbles engulf smaller bubbles adhering to the anode surface, sweeping the surface clean. These macrobubbles assume the shape of aerofoil in a cross section with the thicker front and a reduced thickness of the trailing end. Through instability caused by bubbles coalescing this generates a flow and a forward movement of the gas as it rolls along the horizontal surface prior to release from the edges of anodes (Figure 2.5.24). The presence of gas bubbles under the anode contributes to the overall bath resistance in the cross section of electrolyte in that zone because the gas is nonconducting and therefore the cross-sectional area of the electrolyte has been reduced. Accordingly, models have been developed for calculating the resistance effect based on the average layer thickness and fractional surface coverage of the surface. An expression for the additional voltage drop caused by gas bubbles that is applied in numerous cell models has been developed by Hyde and Welch which expresses the voltage drop in terms of the anode current density, electrolyte conductivity, average bubble layer thickness and fraction, is given by the following equation (2.5.53) where ia is the anodic current density (A/cm2), κ is the conductivity (1/Ωcm), db is the bubble thickness (cm), and ϕ is the fraction of anode covered by bubbles. Furthermore, Haupin verified that the analysis and equations by Hyde and Welch can also be used to calculate the average bubble layer thickness under the anodes by the following equation: (2.5.54) The average data reported for gas coverage under the anodes are in the range 40–80%, the thickness of the gas bubble layer is 0.5–0.7 cm and the bubble size is 100–120 cm. These relationships have been developed for electrodes that are predominantly horizontally orientated but the rate of release of the bubbles is also dependent on the path they have to travel and also the angle of the surface from the horizontal. With the advent of slotted anodes, these parameters have therefore changed and the bubble layer resistance has been reduced. 2.5.2.7.5 Aluminum Cell Overvoltages The kinetics for the production of reaction products requires a hierarchy electrode potential beyond that of the Nernst potential for maintaining the imposed current density (which is the measure of the reaction rate). This is often referred to as overvoltages or more correctly overpotentials. These potentials resulting from the electrode kinetics can be due to some or all of the following: • : Concentration gradients between the bulk of the electrolyte and those existing at the interfacial surface for the participating species. • : Surface intermediate reactions at the electrode where rearrangements take place to form the final product. These can include variations in surface coverage. This requires additional voltage at both electrode (overvoltages) surfaces to cause the reaction to proceed at the desired rate to produce products (i.e., current density). These overvoltages result from electrode kinetics due to concentration gradients and surface reactions at the electrodes and are affected by the electrolyte chemistry, current density, and temperature. The components of the cell overvoltage \|η\| are ηac = anodic concentration overvoltage, ηar = anodic reaction overvoltage, ηcc = cathode concentration overvoltage, and ηcr = cathode reaction overvoltage. The cathode reaction overvoltage is negligible in aluminum cells and thus it is omitted in further considerations. (2.5.55) The higher the current, the higher the steady state surface coverage, and this will depend on the rate constant, which in turn is dependent on the carbon structure and stability. The build-up of the surface immediately will introduce polarization by that concentration change, so the electrode potential has to increase to maintain the prevailing rate as high surface coverage is approached. 2.5.2.7.6 Environmental Issues Within the potroom, a combination of work practices such as anode covering and the automated alumina feeding system can generate alumina dust, because of the small particle size of the alumina and make-up electrolyte materials used. The emission of gaseous and particulate fluorides is of prime concern in aluminum smelters. In order to minimize the environmental impact of large capacity aluminum smelters to meet more stringent environmental requirements, the aluminum industry has made significant efforts to reduce fluoride emission on a kilogram (kg) per metric ton (t) of aluminum production basis. Aluminum cells are completely enclosed with removable covers to collect all the cell fluoride emissions of particulates and gas fumes. Aluminum smelters have highly efficient alumina dry scrubbing equipment, which removes up to 99% of all emissions from the pots. Dry scrubbers use the raw material alumina as the sorbent for removal of gaseous and particulate fluorides from the cell exhaust gases. The fluorides are chemisorbed on the surface of the alumina particles, which are then called secondary or reacted alumina. This material is stored in large silos and is later used as feed material to the cells. This means recycling of the captured fluorides and reduces the overall fluoride consumption significantly. As a result, current average levels of fluoride emission to atmosphere are as low as 0.5–0.6 kg of fluoride/t of aluminum for modern large capacity prebake smelters compared with 1 kg/t for older prebake smelters. It is even possible to reduce smelter fluoride emissions to the level of 0.3–0.4 kg/t with the development of new, innovative dual duct suction technology that increases the cell hooding/duct suction velocity by a factor greater than three, 5000–15,000 Nm3/h when covers are removed during anode changing. Also, the emission from hot spent anode butts accounts for 35% of the total emission in potrooms. The majority of HF emissions occur during the first 20 min of cooling. The solution was to put the hot anodes inside cooling boxes that retain the fluorides. Another major environmental impact of refining and smelting is the emissions of carbon tetrafluoride (CF4) and carbon hexafluoride (C2F6) which are strong greenhouse gases. Fluorides such as CF4 and C2F6 are of concern because of their high global warming potential due to their long lifetime in the atmosphere. The worldwide aluminum industry voluntarily reduced its PFC emission by 75% from 1990 to 2009 despite a 90% increase in the primary production of aluminum. Since 1990, the industry has reduced PFC emission from 96 million tons, carbon dioxide equivalent, to 22 million tons while increasing primary aluminum production from 19.5 million tons to 37 million tons. A 23% reduction in the greenhouse emission intensity was achieved by aluminum companies, from a total of 10.40 metric tons of CO2 equivalents/ton of aluminum in 2005 to 7.95 metric tons of CO2 equivalents/ton of aluminum in 2011. PAHs result from the manufacture of anodes containing coal tar pitch as the binder and is removed from the exhaust gases of the anode plant and anode-baking furnaces by scrubbing with alumina. Furthermore, aluminum plants have discontinued the use of hot ramming paste for cathode construction that contained coal tar pitch binder due to the development of low-temperature ramming paste that does not contain coal tar pitch. PAH is emitted into the potroom atmosphere from the anodes of Søderberg cells during the electrolytic process; the average level of measured emissions to air of total PAH is 0.06 g/t of aluminum. Modernized Søderberg cells use dry anode paste technology with lower pitch content, cooler anodes, and some aluminum smelters have doors on top of the anodes that reduce the PAH emissions. However, Søderberg cells are being discontinued in the aluminum industry due to their lower production efficiency and inherent environmental problems, particularly the emission of PAHs from the large self-baking anodes. Sulfur is a typical impurity in the anode coke with concentrations ranging from 1 to 4 wt.%. This is ultimately emitted with the exhaust gases as the carbon anodes are consumed at the electrolytic process. While it is initially released as COS, as it mixes with the other hot gases, it is ultimately oxidized to sulfur dioxide as it enters the gas collection ducting system. Depending on the concentration of sulfur in the anodes, some aluminum smelters remove SO2 by scrubbing the exhaust gas with either sea water or chemicals. Spent pot lining (SPL) is a waste generated by the aluminum industry during the manufacture of aluminum metal in electrolytic cells. Aluminum cells have a bottom steel cathode shell that is lined with carbon materials and brick insulation. The cathode lining materials eventually become saturated with cryolitic bath, which causes expansion, cracking, and chemical deterioration, requiring that the spent potlining be replaced after 5–10 years. Over the useful life of the linings, the carbon becomes impregnated with aluminum and fluorides, averaging 34% of the spent carbon lining, and cyanide compounds, 400 ppm. Contaminant levels in the refractories portion of linings that have failed are generally low. Aluminum smelter produces 40–60 kg of mixed solid wastes per ton of product with spent cathodes linings being the major fraction. These linings consist of 50% refractory material and 50% carbon. The SPL material is considered a hazardous waste in many countries because it contains significant quantities of absorbed fluorides along with traces of cyanide. The initial disposal of SPL was primarily in landfills. However in 1993, the United States Environmental Protection Agency banned its disposal in landfills. Sophisticated plants were then established specifically for the treatment and disposal of SPL in the United States and Canada. Globally, many aluminum smelters today send their SPL to be consumed as fuel in both cement and ferrosilicon furnaces. The calorific value of SPL depends on the carbonaceous content, which varies with the percentage of refractory materials. The burnability of cement raw mix in presence of SPL showed reduction in free lime due to mineralizing effect of fluoride present in SPL. The carbonaceous content of SPL is well burnt at higher temperature and provides additional heat to the system. 2.5.2.7.7 The Development of Inert Anodes Extensive research has been conducted in the aluminum industry to develop a material that can replace the consumable carbon anode used in the electrolysis cells so that the electrolysis product is pure oxygen. These efforts are driven by the increasing emphasis in industry on reducing carbon dioxide emissions. The use of inert anodes in the electrowinning of aluminum would act as a catalyst to oxidize oxygen ions in the melt to oxygen gas and produce aluminum by the following equation: (2.5.56) The focus of the research has been the attempt to find conducting or semi-conducting materials that do not dissolve in the electrolyte under the operating conditions and are stable in the presence of pure oxygen atmosphere. Hitherto, no material has been found that can give long-term stability in conventional cells. The most extensively researched material for the electrode surface has been nickel ferrites and different methods of manufacturing these electrodes have been used. The two extremes being using an oxidized surface of a metallic alloy and the other extreme is to form a cermet material by blending and reacting ceramic oxides. Besides the challenge of developing such electrode materials, a disadvantage of inert anodes is the reversible decomposition potential of alumina to produce aluminum is about 1 V higher with inert anodes compared with the carbon anode process (2.10 versus 1.20 V). This means that the cell voltage would be about 1.0 V higher for inert anodes if the overvoltages were the same. However, it may be possible to offset the higher voltage by changes in cell design that reduces the ACD. 2.5.2.7.8 Aluminum Casting Molten aluminum siphoned from the electrolysis cells is transferred in a crucible to the casting shop where it is poured into large gas or fuel-fired holding furnaces, where the metal is cleaned by skimming, and fluxing gas and/or salt fluxing to remove bath and other impurities. The purified molten aluminum is then cast into a variety of forms such as T-ingots and pig ingots to be sold as primary aluminum for remelting. Small 25-kg ingots produced by a chain conveyor casting machine is a common item that are usually sold to smaller companies to remelt and manufacture a wide variety of aluminum products. Many producers manufacture semi-finished cast products such as rolling ingot (slab) and strip, which have been alloyed to meet different fabricating requirements. Large 15–25-ton aluminum slabs that are 1–2 m wide and up to 8 m long are made by a direct-chill vertical casting process (DC) using a water-cooled mold for producing slabs for aluminum-rolling mills to fabricate aluminum can stock. Aluminum slabs are also produced by advanced electromagnetic casting (EMC) technology that eliminates the surface ripples produced by direct-chill casting. The molten metal is levitated by generating an electromagnetic field inside the mold and solidified without contact to a solid surface yielding an ingot free of surface imperfections that do not require scalping to remove rough edges, thus saving time and money in the manufacturing process. Aluminum billets (10–50 cm diameter round and 6–8 m long) are cast to produce extrusion products. These billets are heated in furnaces to homogenizing temperatures about 450–520 °C to eliminate the internal stress imposed on the aluminum bar, minimizing the deflection, improving the plasticity, and reducing the extruding resistance. Preheating or homogenizing reduces the chemical segregation of cast structures and improves their workability. Hot strip continuous casting is becoming common for manufacturing of products such as wire, foil, and pharmaceutical packaging. A continuous caster is a combination casting machine and rolling mill. Molten alloyed aluminum is continuously fed into the machine and emerges in solid form as a rod for a rod mill or as a roll stock for a sheet or foil mill. Aluminum metal produced by the Hall–Héroult electrolysis is 99% pure, but contains impurities of iron, silicon, etc. Aluminum can be further refined by using the three-layer electrolysis Hoopes cell. The three layers are separated from each other due to differences in density. The lower liquid layer consists of an alloy of impure aluminum with about 30% copper to increase the density to 3.4–3.7 g/cm2 that serves as anode. The middle liquid layer is the electrolyte consisting of cryolite and barium fluoride with a density of 2.7–2.8 g/cm2. The uppermost layer is the separated high-purity aluminum with a density of 2.3 g/cm2, which serves as cathode. During electrolysis, Al3 + ions from the middle layer migrate to the upper layer where they are reduced to aluminum. Pure aluminum tapped from the cell gives 99.99% pure aluminum. Higher purities of up to 99.9999%, or “six-nines” aluminum, can be obtained using additional zone-refining operations that traps impurities in a molten zone that moves gradually from one end to the other of specially prepared ingots. View chapterExplore book Read full chapter URL: Book2014, Treatise on Process Metallurgy: Industrial ProcessesAlton T. Tabereaux, Ray D. Peterson Review article Technologies for CO emission reduction and low-carbon development in primary aluminum industry in China: A review 2024, Renewable and Sustainable Energy ReviewsAngxing Shen, Jihong Zhang 6.1 Large-scale intelligent electrolytic cells The chemical reaction of aluminum electrolysis takes place in an electrolytic cell, which consists of a cell body, anode, and cathode. When the current passes through the electrolytic cell, the oxidation reaction occurs at the interface between the anode and solution, whereas the reduction reaction occurs at the interface between the cathode and solution. To produce molten aluminum (with CO2 emission), China's PAI has entered the fast lane of development in the 21st century, and production cell types of 160 kA, 240 kA, 300 kA, and 400 kA are being upgraded constantly . Nowadays, 600 kA high current intensity process is widely used in new primary aluminum projects in China, and their design and production management are maturing. In general, although an ultra-large electrolytic cell with a capacity of more than 500 kA is favored by enterprises, its superiority is not well reflected on the premise of increasing technical difficulty. The design of a large electrolytic cell increases the material consumption of busbar, which may offset the investment savings in large-scale applications . It is difficult to make a significant progress in the technical indices of large electrolytic cells. From the operational perspective, even the 500 kA cell can hardly reach the advanced indices of current efficiency higher than 95 % and DC power consumption lower than 12,600 kWh/t Al . Two pilot projects can show the potential for technological application: (1) : SY600: Aluminum Corporation of China completed the technical research and development (R&D) of a 600 kA ultra-large capacity aluminum electrolytic cell. The average DC power consumption per ton of aluminum of SY600 was 12,136 kWh, 884 kWh lower than the national average in 2012 . The average current efficiency was 92.77 %, and the average anode effect coefficient was 0.08 times/cell·day . If a 600 kA electrolytic cell is industrialized in 10 series (each series has a capacity of 500,000 t), it can save 4.42 billion kWh of electricity every year. Given the electricity price of 0.45 CNY/kWh, the power cost can be reduced by 1.989 billion CNY with considerable economic benefits. Based on the thermal power usage ratio of 80 % and carbon emission intensity of 887 g/kWh, each series of SY600 can reduce indirect carbon emissions by 313,600 t/yr, and the per-ton emission intensity of aluminum production can decline by approximately 627 kg (6.8 %). (2) : NEUI600kA: The Weiqiao Group constructed the first 600 kA electrolytic aluminum production line in China . The current efficiency of this cell reached 94.6 %, and the DC power consumption per ton of aluminum produced was 12,443 kWh (436 kWh lower than the national average in 2015). The anode effect coefficient was 0.01 times/cell·day. If the technology is industrialized in 10 series (each series has a capacity of 400,000 t), it can annually save 1.744 billion kWh of electricity. Based on the electricity price of 0.45 CNY/kWh, the electricity cost can be reduced by 784.8 million CNY with considerable economic benefits. Each series of NEUI600kA can reduce indirect carbon dioxide emissions by 120,000 t/yr, and the per-ton emission intensity of aluminum production can be reduced by approximately 300 kg (3.4 %). Chinese enterprises have established new electrolytic cell series with their current intensity increased to 660 kA. The electrolysis technology in China has played a leading role in the world and paved the way for the widespread application of the large-scale and intelligent electrolytic cell technologies. In the future, the development of the electrolytic cells in China should focus on solving the issue of safe production. First, ultra-high current intensity, high series voltage, high magnetic field, high temperature, and high labor intensity in PAI should be improved. Accidents, such as rectifier unit explosions, short-circuit explosions, depolarization explosions, and electrolytic cell leakage explosions should be strictly controlled [107,108]. Most electrolytic aluminum projects have been operational for more than five years, thus posing safety risks, such as material and equipment aging in the production series, which require interventions via test- and detection-related businesses and technical innovation. Despite the large scale of technological R&D, talent introduction, and business investments in the early stage of the 600 kA project, compared with 400 kA electrolytic cell, the 600 kA technology approximately reduces the land occupation by 26.5 %, investment per ton of aluminum by 13.2 %, and cost by 8 % as well as increases the unit area capacity of electrolytic cells by 10 % and the labor productivity by 25 %. If the application capacity of 600 kA technology reaches 10 Mt/yr, it can save 7.065 billion CNY in investment, 1.6 km2 land, 5500 workers, and 11.2 billion CNY in cost [104,109], with remarkable economic benefits. View article Read full article URL: Journal2024, Renewable and Sustainable Energy ReviewsAngxing Shen, Jihong Zhang Review article A survey of industrial applications of Demand Response 2016, Electric Power Systems ResearchMaryam H. Shoreh, ... João P.S. Catalão 3.1 Metals: aluminium and steel Electrolytic aluminium is a significant industry for DR implementation or energy saving. Fig. 3 illustrates the amount of electricity consumption of processes from mining to electrolysis in a typical aluminium production industry . As shown, electrolysis is considered as the dominant electricity consumption process of producing aluminium. Aluminium smelting is based on an electrolytic process for converting alumina to aluminium that is considered as the primary material for a vast majority of industries from car making to packaging can industries. Smelting consumes approximately 46% of the total energy consumed in U.S. manufacturing of aluminium. In a common aluminium smelter, 30–40% of the total cost of producing primary aluminium is devoted to electricity costs [24,25]. Compared with cement or steel manufacturing, it has a fairly simpler production process. One of its energy intensive processes is aluminium smelting pot; indeed, a good potential for DR is primarily found on the pot line. The smelting pot is electronically controlled and could precisely track automatic generation control signals . Aluminium smelting takes place in pots, where a DC electric current is passed through a cryolite bath in order to separate the aluminium from the oxygen and remains the molten metallic aluminium at the pot bottom. Aluminium oxide and electric current should be added continuously and aluminium metal is periodically extracted. Typically, smelting pots are operated at high direct current, as much as several hundred thousand amps, and very low voltage, often below 10 V. Potlines usually consist of hundreds of pots that are operated frequently to form a connected series load. Total power consumption for a potline can reach to as much as hundreds of MW’s and a smelter may consist of multiple potlines . Aluminium smelting pots operate at very high temperatures to maintain the aluminium that is produced in a molten metal form. The huge energy intensity of 15 MWh for every ton of aluminium is required for starting the smelting process up to the right temperature. During the operation, thermal balance, which is essential for the proper pot operation and can modify the power consumption, is achieved by regulating the input voltage of each pot [4,24]. Indeed, the power consumption of a pot line can be adjusted by changing the output voltage of the rectifier that supplies a DC current to the pot line. By doing so, the power consumption rate can be adjusted very quickly (within seconds) and accurately. In cases when the load reduction takes longer times, a recovery period should be defined in order to reach to previous normal or above normal specifications. Each pot has a large amount of thermal mass with a multi-hour thermal time constant that allows for instantaneous electric power variations that do not greatly impact the pot thermal balance. However, average power over a few hours would be critical. Another way to achieve power consumption flexibility and generate a larger amount of power change within a short time is to shut down an entire potline by switching a breaker. The smelter’s flexibility enabled in this way makes aluminium smelting an ideal DR resource . In that case, the duration of an interruption is more critical and can be sustained only for short periods, based on the particular plant restrictions. The interruption on a single potline may last from minutes to about two hours. Facilities with multiple potlines can rotate this interruption from line to line, enabling a longer total interruption. Base on the above explanations, aluminium electrolysis is categorized as a suitable process for load shedding rather than load shifting in DR programs, since the utilization levels for this process is seen to be 95–98% annually. Based on experiments the power demands of electrolysis process are allowed to be reduced by up to 25% for 4 h before an undesirable interruption occur. This amount of reduction (25%) is usually sold to the power market as positive tertiary capacity. The price for calling positive reserve energy in some cases is proposed as much as 1000 €/MWh, due to the high value of lost load which is defined based on the price of aluminium and the technical risks of destabilizing the electrolytic process . In Ref. , a stochastic optimization model is considered for aluminium smelters’ participation in both energy and spinning reserve day-ahead markets. The aim of this mixed-integer linear programming problem is to obtain the day-ahead bidding strategy for the smelters. The price scenarios as a representative for future prices, besides to smelting plants parameters are fed into the problem and the energy bidding curves, the optimum spinning reserve provision and the amount of potlines’ power consumption are obtained as a result of the running program. This model is validated by case studies, and the results show that by the proposed optimal scheduling the aluminium smelting plant would advantage from both electricity markets participation and aluminium production. Some of the results obtained from the bidding curves and the spinning reserve provision curves are as follows. In some hours of the scheduling day the plant is more conservative in terms of selling energy as the smelter asks for a very high price for selling the low amount of energy. Therefore, the smelter would focus on producing aluminium rather than selling energy during these hours. On the other hand, the bidding curves are more aggressive in some hours of the day in which the smelter bids significant amounts of energy into the market. Due to the higher energy prices during these hours, the smelter wills to be aggressive in selling energy. The spinning reserve provision results show that the smelter has the tendency to provide more reserve when the reserve prices are higher and provides little reserve at peak hours. It is due to the fact that the smelter would sell the higher amount of energy in these hours, as the energy prices are comparably higher than the spinning reserve prices. Thus, the potlines’ load during these peak hours is very low, allowing lower spinning reserve. The available spinning reserve capacity is upper bounded by the difference between the potline’s loading level and its minimum loading level. In other words, if the smelter lowers the potlines’ loading levels in order to sell energy, then lower amount of spinning reserve capacity would be remained. Contra wise to aluminium industry, steel manufacturing processes are known as one of the most complicated industries to schedule. It is considered as a large-scale, multilevel, and multiproduct industry which consists of parallel tools, complicated processes and energy limitations . Fig. 4 shows a simple diagram of steel supply chains. In this figure, the most energy intensive process which is the melt shop is bolded with a red line. Melting scrap steel is a large energy consuming process for producing steel. The energy intensity of the process is measured at nearly 0.525 MWh/t of steel produced . In this process, heat is generated by an electric arc furnace or by means of induction, which let the scrap metal start melting in the furnace. Although the process is able to halt instantly, the melting process has to resume again if the disruption exceeds half an hour, since the scrap metal begins to cool down after that. Extra costs would emerge if the process is disrupted for longer than half an hour. It is reported that melting scrap steel in this process lasts approximately 45 min and near 15 min to fill the furnace again for another round. The factory can halt the process entirely in order to sell the contracted power on the spot market or on the market for regulating energy. As mentioned in this report, up to the year 2011, approximately 50% of steel mills in Germany have pre-qualified their furnaces in the tertiary reserve market as positive capacity. Capacity prices tend to be reasonable and lower compared to prices of reserve energy, and this is based on the high value of lost loads in this process . Ref. , presents the scheduling problem of a steel plant by considering energy constraints. The suggested optimization model follows pre-contracted load profiles of a steel factory with the real energy and power consumption. The consumer will be penalized for under and overconsumption. It is shown that electricity cost minimization leads up to 12% savings in high capacity operation and up to 52% in low capacity one. In another study an optimization framework is presented to consider different load characteristics of a steel mill industry. These characteristics include interdependency between industrial units, multiple days, size and number of cycles, sequential operation, and simultaneous management of energy and material flow. The optimization is also carried out under different tariffs and in each case the results lead to higher profits. As an example, the enhancement is 45% under inclining block rates and 10% under TOU pricing. This improvement is based on not only optimization of power usage, but also optimization of material usage and the quantity of final product in this study. Therefore, different optimum energy usage levels are attained based on changes in pricing scenarios. Increases in scheduling horizon will also lead to increase in total profit as it becomes more time flexible. On the basis of the results in this case, the improvement in TOU pricing is over 50%, under CPP is around 40%, under day-ahead pricing is over 20%, and under peak pricing is over 10%. In this study it is concluded that for industrial units a late CPP warning is often very costly and that’s because the industry consumer does not tend to change its operation in a short notice. In another consideration in this study, the duration of batch cycles is considered as variables in the optimization problem that will lead to additional load flexibility and increase in profit. The numerical results show that the total energy consumption under fixed and variable batch cycles is 6.67 and 5.85 GWh, respectively. As a result, there is an increase in total profit as well. View article Read full article URL: Journal2016, Electric Power Systems ResearchMaryam H. Shoreh, ... João P.S. Catalão Chapter Nucleation and Growth of Metals 2008, Electrochemistry for Materials ScienceWaldfried Plieth 7.6.1 Aluminum deposition from a molten salt The classical process for winning aluminum is one developed by Hall and by Herault in 1886 and 1888.29 Aluminum oxide (Al2O3) of high purity is mixed with cryolite (Na3AlF6) in the ratio of 1:10. The mixture melts at approximately 950 °C. The preparation of the pure aluminum oxide, which is usually from bauxite, is the first step in winning aluminum. This process was developed by K.J. Bayer in 1892. The bauxite is treated with concentrated sodium hydroxide and reacted with the aluminum minerals to get sodium aluminate. After aluminum hydroxide precipitates from the aluminate solution, it is then calcinated at 1200–1300°C. The modern electrolysis cell is part of a long development. Schematic view of a cell is shown in Figure 7.18. The electrolysis is made at 4–5 V with 100,000 to 180,000 A. To produce 1000 kg of aluminum one needs 13.5 MW h electrical energy. In the melt part of the cryolite dissociates30 (7.46) The aluminum oxide and the ions in the cryolite melt form oxy-fluoride complexes, e.g., From kinetic experiments, the following mechanism of aluminum deposition at the liquid aluminum electrode was derived:31 (7.47a) (7.47b) (7.47c) In a technical deposition process one also has to look for the counter-electrode, which in aluminum electrolysis is a graphite electrode. The processes at the carbon electrode are described by the equations (7.48a) (7.48b) (7.48c) The electrolysis needs a large amount of energy. Increasing energy prices may open the way to alternative processes, e.g., to win aluminum from chloride melts. View chapterExplore book Read full chapter URL: Book2008, Electrochemistry for Materials ScienceWaldfried Plieth Chapter Aluminum Production 2014, Treatise on Process Metallurgy: Industrial ProcessesAlton T. Tabereaux, Ray D. Peterson 2.5.2.4.2 Current Efficiency—Aluminum Back Reaction The amount of aluminum predicted by Faraday’s law is never obtained in practice. In parallel with the aluminum production reaction, a quantity of metal made at the cathode dissolves into the electrolyte at the boundary layer at the bath–metal interface and is transported to the reaction zone where it is oxidized, recognized in the industry as the back reaction, by CO2 gas forming CO gas and Al2O3 as products, as shown in Figure 2.5.9. Industrial aluminum cells prior to the 1970s operated from 85% to 88% current efficiency, while modern cells now can operate 95–96% current efficiency. The corresponding concentration of CO produced by the back reaction in these cells is normally about 10%. (2.5.11) The electrolytic production of aluminum from Al2O3 and the back reaction of aluminum to form CO gas and Al2O3 are expressed in the following equations: (2.5.12) (2.5.13) Combining these two reactions gives the overall aluminum cell reaction: (2.5.14) Dissolved metal must diffuse away from the aluminum interface before it can be oxidized. The loss in current efficiency due to the back reaction is directly related to the dissolution of the aluminum product into the melt at the boundary, transport to the reaction zone, and ultimate oxidation of the metal with dissolved CO2 near the anode surface, as shown in Figure 2.5.9. In order to minimize the dissolution of aluminum in the electrolyte, cells are purposely operated with a high excess % AlF3 and low dissolved alumina concentration in the melt; low electrolyte temperature, a balanced current distribution of all anodes, and a sufficiently wide interelectrode distance. The industry current efficiency of aluminum electrolysis for new cell technology has improved from about 82% in 1900 to 96% in 2000, as shown in Figure 2.5.10, while the current efficiency in the older cell technologies has lagged behind by about 5%. The abrupt increase in current efficiency in 1960–1965 was due to the advent of modern cell technologies and improved cell operating practices. It is interesting to note that current efficiency has not improved in aluminum cells over the past 10 years or so. Perhaps, we have reached the maximum obtainable current efficiency with the current industrial Hall–Héroult prebake cell design and technology? The maximum current efficiency for Søderberg cells is only 92% due to its poorer quality anode carbon, inferior magnetic design, and large gas bubbles that increases stirring and mixing at the bath–metal interface. The aluminum production per day for one aluminum electrolysis cell operating at 350 kA and 95% current efficiency is 2678 kg/day, which represents a 5%, or 121 kg/day, loss in aluminum per day. (2.5.15) The aluminum production per year for a modern potline of 300 cells operating at 350 kA and 95% current efficiency is: (2.5.16) (2.5.17) The production of 1 ton of aluminum typically requires 420 kg of carbon, 1920 kg Al2O3, and 16 kg AlF3, and the specific energy consumption is 13.20 kWh/kg Al, as shown in Figure 2.5.11. The theoretical carbon consumption is only 333 kg C/t Al. The excess carbon consumption is mainly due to air oxidation of the hot anodes in the hooded cells. The reaction occurs preferentially in the pitch binder matrix and leads to the physical loss of coke particles to generate dust that floats on top of the electrolyte. Thus, anodes are carefully covered with a mixture of crushed bath and alumina to prevent excessive oxidation. The unused top part of anodes, “butts,” are removed from the cells when new anodes are set, crushed, and recycled in the carbon plant to make new anode carbon. Furthermore, the theoretical alumina consumption is only 1.89 kg/t Al. The excess alumina usage is due to the impurities in the metallurgical grade alumina including water and a small amount of alumina dust may be lost during transportation and transferring to/from silos. View chapterExplore book Read full chapter URL: Book2014, Treatise on Process Metallurgy: Industrial ProcessesAlton T. Tabereaux, Ray D. Peterson Chapter 13 International Symposium on Process Systems Engineering (PSE 2018) 2018, Computer Aided Chemical EngineeringQingxin Guo, ... Xin Gao 1 Introduction Aluminium is the most widely used nonferrous metal, the production process of aluminium includes bauxite mining, alumina production, electrolysis aluminium, castings, rollings, the production of consumer products and recycling. In the electrolysis aluminium production line, more than 99% pure molten aluminium is formed at the cathode deposited at the bottom of the electrolytic cell, and is tapped from the cell into a crucible by a vacuum siphoning system. The molten metal is transported and poured into a holding furnace and caster to be cast into ingots, extrusion or rolling ingots. In a typical aluminium smelter, there may be hundreds of electrolysis cells. Each crucible taps the metal from up to three cells that contain different purity of the molten aluminium. The capacity of the holding furnace is more than three times the crucible, so there can be several different crucibles feeding one holding furnace. The molten aluminium which is mixed and stabilized synchronously in one holding furnace is called a charge. Therefore, both crucible and holding furnace are batch production mode. Therefore, in the aluminium electrolysis process, the production schedule is to form the batch tapping of the cells into crucibles and arrange batch feeding of the crucibles into holding furnaces, considering the constraints of the electrolysis and cast. Ryan (1998) formulated the tapping of the cells problem as a set-partitioning model and solved it by the LP relaxation. Piehl (2000) used the revised simplex method and branch and bound with a constraint branching approach to solve the similar cell batching problem. Prasad et al. (2006) provided a MILP model for the scheduling of aluminium casts of different alloys with respect to the actual number of batches to be processed in multistage. For the formulation of the scheduling problem, we refer to Floudas and Lin (2004) and Harjunkoski et al. (2014). The remainder of this paper is organized as follows. In Section 2, a precedence-based model is presented for this problem. In Section 3, a novel unit-specific event based continuous-time mixed integer programming model are proposed to describe the problem. Section 4 reports the experimental results of the two models solved by CPLEX. Finally, Section 5 presents the conclusions. View chapterExplore book Read full chapter URL: Chapter Industrial sector energy efficiency 6.2.4 The Aluminum Subsector Aluminum is another important metal used widely across the economy. It is light, strong, and resistant to corrosion. As a result, aluminum is used extensively in transportation, packaging, building, construction, and electrical industries. Manufacturers produce aluminum primarily from the processing of the earth-mined ore, bauxite, which is converted alumina (aluminum oxide), and subsequently separated into aluminum and oxygen via electrolytic smelters in EAFs. According to the aluminum manufacturing giant Alcoa, it takes roughly 4 tonnes of bauxite to produce 1 tonne of aluminum. As with steel, primary aluminum production is highly energy intensive. The energy intensity of this process is greatly reduced by mixing new aluminum with plant scrap or recycled aluminum, which accounts for a growing portion of new production. (The International Aluminum Institute reports that in 2015, around 27 million tonnes of recycled aluminum were produced from old and traded new scrap, compared with 58 million tonnes of primary aluminum.) According to a 2007 LBNL report (Worrell et al.) on best practices in energy intensity in industrial subsectors, key energy-intensive elements of primary aluminum manufacturing include bauxite extraction, aluminum production, anode manufacturing, aluminum smelting, ingot casting, rolling, and extrusion. In contrast, the key elements of secondary manufacturing involve only melting and reshaping recycled aluminum, which is much less energy intensive. A simplified schematic overview of aluminum production from primary and secondary sources to finished products is shown in Fig. 6.6. From a best practices energy use/energy-intensity perspective, the primary aluminum manufacturing processes mentioned above require a substantial amount of raw fuel and electricity. Estimates of specific primary energy values, measured in terms of specific energy (GJ/t) or specific electricity (kWh/t) consumption, are tabulated in Table 6.4. These are based on estimates established by the LBNL, assuming the electricity is produced using conventional thermal processes with an operational efficiency of 33%. Table 6.4. Estimated Primary Specific Energy Consumption for Key Aluminum Production Processes \| \| \| \| \| --- --- \| \| Empty Cell \| Empty Cell \| Primary (GJ/t) \| Secondary (GJ/t) \| \| Alumina production \| Digesting (fuel) \| 12.1 \| \| \| \| Calcining kiln (fuel) \| 6.5 \| \| \| \| Electricity \| 4.3 \| \| \| Anode manufacturing \| Fuel \| 1 \| \| \| \| Electricity \| 0.64 \| \| \| Aluminum smelting \| Electricity \| 148.4 \| \| \| Ingot casting \| Electricity \| 1.06 \| \| \| Total \| \| 174 \| 7.6 \| Table created using data presented in Worrell et al. (2007). As presented in Table 6.4, electricity used in smelting is the most energy-intensive component in primary aluminum production. Because the LBNL analysis assumes electricity utilized from conventional power generation, it is possible to reduce energy consumption by improving the efficiency of the smelting process. According to the LBNL, the Hall–Heroult carbon anode aluminum electrolysis process is the basis for contemporary industrial aluminum smelting and the two chief electrolytic smelting technologies use cells with prebaked anodes and cells with baked in situ anodes, also known as Søderberg cells. Prebaked anodes are far less energy intensive and the subject of continued research and development. Based on thermodynamic limits, the theoretically achievable minimum energy requirement is about 6 kWh/kg of aluminum. State-of-the-art plants can consume as little as 13 kWh/kg of aluminum. Søderberg cell plants consume around 20 kWh/kg. In addition, aluminum production energy intensity can be reduced by improving electricity generation efficiency by using natural gas instead of coal, when possible. Also, in countries with abundant hydropower capacity, aluminum manufacturing facilities are often located nearby, which lowers the energy intensity significantly, due to fuel switching. Finally, using the United States as an example of energy reduction potential in the global aluminum manufacturing subsector, the massive reduction in energy intensity of the United States industry over the past 40 years has been the result of better, “inert” anode technology and increased utilization of scrap and recycled aluminum. Indeed, according to the US EIA, the energy needed to process and manufacture aluminum from recycled materials is 90% less than from bauxite ore. It will be important to follow how other nations improve energy consumption in the aluminum subsector, especially China, since according to the International Aluminum Institute in 2017, China produced 35.9 million of the 63 million tonnes of aluminum produced worldwide. View chapterExplore book Read full chapter URL: Chapter Aluminum Production 2014, Treatise on Process Metallurgy: Industrial ProcessesAlton T. Tabereaux, Ray D. Peterson 2.5.2.4.3 Industrial Cell Design 2.5.2.4.3.1 Magnetohydrodynamics Electrolysis cells in aluminum plants are arranged in long rows, called potlines where the cells are connected in an electrical series circuit with the transformer and rectifier systems located at one end of the potline. Such arrangements allow carrying out most cell operations by using overhead multipurpose cranes. A typical aluminum smelter consists of one to three potlines. Modern 350-kA prebake cells are positioned in a side-by-side arrangement in potlines, as shown in Figure 2.5.12. In most designs, the cells are situated above ground with solid aluminum bus bars located at ground level below them with current flow from one cell to another. The smelting process requires large amounts of electricity. A reliable and uninterrupted electrical power supply is a critical issue for aluminum smelters. Alternating or AC current supplied from the grid must be transformed into direct or DC current, which requires the use of large rectifiers, transformers, and sophisticated monitoring systems located adjacent to the potline building. The most inherent risk in aluminum production is a loss of electrical power. A failure of electricity supply lasting more than 2–3 h can cause the electrolyte in the cells to cool to the point where its electrical resistance is too great when power is restored resulting in shutdown of all cells. It is expensive and time consuming; usually it will take several months, to restart a frozen potline because the solidified aluminum and electrolyte must be physically broken out of the cells. Thus, significant business losses will be incurred due to the interruption in the event of a potline freezing. The remaining lifetime of the cathodes will be shortened due to the extra thermal stresses inevitable with shutdown and restart of cells. Magnetohydrodynamics (MHD) forces are generated in the aluminum metal caused by the interaction between the magnetic fields produced by the passage of high-amperage DC electrical current through nearby conductor bars and the electrical current flow in the aluminum metal in electrolysis cells. The large electric currents, 300–500 kA, in modern prebake cells pass through the electrically conducting liquids (electrolyte and metal) and generate powerful magnetic fields in the aluminum pool. These strong MHD forces increase metal velocity, metal wave height and frequency, and distortion of the aluminum–bath interface and thus require higher voltage and operate at lower current efficiency. These cells typically maintain a 25-cm deep pool of liquid aluminum in cells to minimize the impact of the MHD forces. A major achievement in the aluminum industry has been the development of advanced one-quarter size (3D) computer mathematical models of the interaction of MHD fields, forces, and resulting metal pool behavior in aluminum cells. This is exemplified by the model derived by Severo et al. that is used for cell design and retrofit studies, as shown in Figure 2.5.13. These programs are used to develop new magnetic compensation in cells by rearranging the bus bar conductor systems to reduce the magnitude of the magnetic fields inside the cells and thus allow significant increases in the potline amperage without losing aluminum productivity or requiring extra voltage. 2.5.2.4.3.2 ACD and Cell Voltage Electrolysis occurs in the electrolysis cell ACD or between the bottom horizontal surface of the anodes and the upper aluminum metal surface. The upper part of the aluminum cell shown in Figure 2.5.14 is referred to as the anode superstructure. The anodes, carbon blocks, are attached to long rods that are suspended and clamped to the anode beam located along the length of both sides of the anode superstructure. Electrical current enters the cell through the large aluminum conductor bus connected to the anode beam. The anode beam moves all the anodes up and down simultaneously, thus changing the ACD, typically 3.8–5.0 cm, from the bottom surface of the anodes to the top surface of the aluminum pool. The ACD is not measured, but the average value can be calculated. Changing the ACD changes the cell voltage and energy input into the cell; most of the cell heat is generated in the electrolyte in this zone. Steel current collector bars attached to the bottom cathode lining carry the electrical current from the cell to the next cell in the series. 2.5.2.4.3.3 Alumina Feeding The electrolytic process consumes alumina at a nearly constant rate. Alumina is fed to the aluminum cells at specific intervals at a rate equal to 1.92 times the aluminum production rate by point feeders located inside the anode superstructure. The computer process control system activates point feeders (crust breaker and alumina dosing devices) according to sophisticated algorithms to keep the alumina content in the electrolyte near the target value. The alumina concentration dissolved in the electrolyte cannot directly be measured; thus, the amount of alumina added to cells is regulated indirectly using an underfeed/overfeed computer control algorithm based on changes in the cell voltage due to the anode overvoltage at low and high alumina concentrations in the electrolyte. To add alumina to the bath, fast action pneumatic air cylinders activate crust breakers to open holes in the crust layer on top of the molten bath at two or more positions along the center line of the cell. Next, the point feeder adds 1.5–2 kg of alumina from a volumetric dispenser directly to 18–20 cm of molten bath where it dissolves. There are usually from two to six point feeders per cell depending on the aluminum metal production rate, which is determined by the potline amperage, as shown in Figure 2.5.15. The advantage of point feeders is that small quantities of alumina are added to the electrolyte at each break-and-feed. This method minimizes the risk of sludge formation in the center area of the cell, and there are minimal emissions of dust and fluorides during the alumina feeding operation. 2.5.2.4.3.4 Changing Anodes Prebake anodes are manufactured by compressing a mixture of petroleum coke aggregates and coal tar pitch binder into blocks typically 40–60 cm wide by 120–150 cm long and 50–60 cm high. Petroleum coke is used because of its high purity. These blocks are baked in anode-baking furnaces above 1100–1200 °C to convert the binder pitch into dense carbon. The carbon blocks are attached to an iron anode assembly by pouring molten cast iron into two to six holes in the anode block for fluted steel stub connections. An aluminum, or in some instances copper, rod is attached to the top of the anode assembly. The top section of the anode rods are clamped to the anode beam and thus carry the electrical current from the anode beam to the molten cryolite melt in the cells. Many smelters now manufacture or cut “slots” in the bottom surface of anodes to divert gas bubbles, thus reducing the electrical resistance in the electrolyte and reducing the total cell voltage of operating cells. From 18 to 40 prebaked carbon anodes are required per cell depending on the potline amperage to maintain the anode carbon current density in the range 0.7–1.2 A/cm2. The carbon anodes are consumed at a rate of about 1.9 cm/day, depending on current, by the electrolysis due to the reaction to produce CO2. One or two anodes are replaced each 24–48 h in the cell with new anodes after 21–28 days depending on the anode dimensions and amperage, i.e., the anodic current density. A crushed bath–alumina mixture is used to cover the top and sides of the new carbon anodes to avoid excessive oxidation and also to form a crust on top of the electrolyte to reduce heat losses and emission of fluorides from the electrolyte. Søderberg cells are another type of aluminum electrolysis technology with only one large anode per cell that is continuously produced by a self-baking process by the cell process heat. However, Søderberg cells are being phased-out in the aluminum industry now due to their lower production efficiency, inherent environmental problems, and the emission of polycyclic aromatic hydrocarbons (PAHs) from the large self-baking anodes. 2.5.2.4.3.5 Metal Tapping Positive aluminum-containing ions deposit continuously at the negatively charged cathode, the top surface of the aluminum pool, where they convert to liquid metal and thereby slowly raising the metal volume and level. A portion of the aluminum metal pool is removed every 24–48 h by siphoning the metal into a large crucible. The aluminum metal is weighed and sent to the cast house for further processing. The height of the metal level in the cells is maintained at a target value as the metal height impacts the heat loss out of the cell sidewalls because molten aluminum metal has a higher heat loss out the sides than molten cryolite. The metal depth also dampens the impact of the magnetic forces on the metal pool. 2.5.2.4.3.6 Cathode Lining The lower cathode components consist of a rectangular reinforced steel box lined on the inside with carbon, refractory bricks, and insulating materials. The cell is thermally insulated internally on the bottom and lower sides to reduce heat losses and maintain the optimum cell heat balance. As aluminum plants continue to increase potline amperage to produce more metal, it becomes necessary to increase the heat transfer out of the sidewall of cathodes in order to maintain the frozen cryolite ledge to protect the sidewall lining material. Cell cathodes are now being constructed using higher thermal conductivity silicon carbide sidewall blocks, steel fins attached along the outside sidewalls, and air pipes that blow compressed air along the sidewalls. The cell cathode linings have lifetimes generally from 1700 to 3000 days. Higher cathode life requires excellent construction and good quality materials. View chapterExplore book Read full chapter URL: Book2014, Treatise on Process Metallurgy: Industrial ProcessesAlton T. Tabereaux, Ray D. Peterson
188342	https://www.youtube.com/watch?v=XuLfcztpWJE	How to Convert Kilobytes, Megabytes and Gigabytes How-To with Mr Kumar 7030 subscribers 1860 likes Description 138503 views Posted: 6 Jan 2018 Learn how to convert between Kilobytes (KB), Megabytes (MB) and Gigabytes (GB). These are units of measurements used to measure storage in digital format. The video clearly shows the relationship between KB, MB and GB and explains how each can be converted to the other with minimal calculations. The factor used to convert between KB, MB and GB is 1024. 21 comments Transcript: Intro hello everyone in this video I'll show you how you can convert between kilobytes megabytes and gigabytes right so it's important to note that 1024 kilobytes equals one megabyte and 1024 megabytes equals 1 gigabyte right so the factor we use for conversion is 1024 right and I have made a diagram here to explain this better when we're converting ok so if we are converting from kilobyte to megabyte we need to divide by 1024 and if we are converting from megabyte to gigabyte we again need to divide the megabytes by 1024 on the other hand if we are working backwards if we had to convert gigabyte to megabytes we'll multiply the amount in gigabytes by 1024 and if we are converting megabytes to kilobytes again we'll multiply the amount we have in megabytes by 1024 now how do we decide whether we multiply or divide well it's very easy right all you need to remember is which unit is smaller and which is higher so kilobytes is small or smaller than megabytes and megabytes is smaller than gigabytes right they are smaller units of measurement so you start at kilobytes and then megabytes and then gigabytes so from kilobytes to megabytes if I'm going from a smaller unit to a larger unit I will always have to divide if I'm coming from a larger unit to a smaller unit for example if I'm converting from megabyte to kilobyte or gigabyte to megabyte I will always multiply right so from a larger unit of measurement I will always have to multiply if I am converting to a smaller unit of measurement right if I'm converting from a smaller unit of measurement to a larger unit of measure I will always have to divide right and the conversion factor that you use in order to convert between these three units of measurement is 1,024 now we'll do some examples is to understand better right so for example how many megabytes are there in 2048 kilobytes right so we convert in kilobytes to megabytes we need to divide by 1024 so the formula is megabytes is equal to kilobytes divided by 1024 and once we substitute the amount 2048 kilobytes divided by 1024 the answer we get is 2 megabytes right Converting Megabytes in this example how many gigabytes are there in 4096 megabytes right so we converting megabytes to gigabytes we need to divide by 1024 so the formula is gigabytes is equal to megabytes and divided by 1024 so we substitute the amount 4096 divided by 1024 will give us 4 gigabyte right so the answer is 4 gigabytes in this example how many Converting Gigabytes gigabytes are there in 20 thousand kilowatts right so we are converting kilobytes to gigabytes that means first we need to convert kilobytes to megabytes and then megabytes to gigabytes so we need to divide kilobytes twice by 1024 so the formula becomes kilobytes divided by 1024 x 1024 right because we are dividing kilobytes twice by 1024 so once you equate that and substitute the kilobytes amount you get 20,000 divided by 1,048,576 and once you equate that the answer is 0.01 9 gigabytes in this example convert five megabytes Converting megabytes to kilobytes to kilobyte threads are we going backwards now to convert megabytes back to kilobytes of the formula is kilobytes is equal to megabytes x 1024 so we'll substitute the amount we've got five megabytes x 1024 and the answer we get is five thousand 120 kilobytes so notice the answer or the amount five increases to 5120 because now we're converting a larger unit to a smaller unit right so we are multiplying in this example Converting gigabytes to megabytes convert three point five gigabytes to megabytes right again we are converting a larger unit to a smaller unit so the answer will obviously be higher than three point five so the formula is megabytes is equal to gigabytes x 1024 will substitute the amounts of 3.5 x 1024 gives us three thousand five hundred eighty four megabytes Converting gigabytes to kilobytes in this example convert 3.5 gigabytes to kilobytes right so now we are converting gigabytes to megabytes and then to kilobytes right the formula becomes kilobytes is equal to gigabytes x 1024 x 1024 and I've put the 1024 in brackets so we can calculate that first and then multiply it with the gigabyte amount and why would using to 1024 simply because we first need to convert it to megabytes and then convert it to kilobytes right so we multiplied by 1024 twice so once we substitute the amounts the formula becomes 3.5 x 1,048,576 and what's you equate that the answer we get is three million six hundred seventy thousand and 16 kilobytes right so that's how we convert gigabyte to kilobyte in conclusion in Conclusion order to convert kilobytes to megabytes we need to divide by 1024 and in order to convert megabytes to gigabytes we need to divide by 1024 if we have to convert gigabytes two megabytes we need to multiply by 1024 and if we have to convert megabytes to kilobytes again we have to multiply by 1024 if we had to convert kilobytes to gigabytes first we need to convert to megabytes and then gigabytes and the same if we are converting gigabytes to kilobytes right thank you very much for watching my video if you did like it please share and subscribe for more
188343	https://www.nyb.com/latest-news/2024/02/21/three-key-fan-laws-explained/	Products Quick Search SEARCH Quick Links Info about my fan. Talk to someone from nyb. Order replacement parts. Trying to find a product. Three Key Fan Laws Explained Ask A Question Request a Quote Home / Three Key Fan Laws Explained FAN LAWS EXPLAINED: WATCH OUR NEW VIDEO HERE! In this video, we review 3 major principles of fan performance called Fan Laws. These 3 Fan Laws explain how a fan’s airflow, pressure, and power consumption change in response to a change in fan operating speed within certain fixed systems. By a fixed system, we mean no variables in the system are changing that would skew the calculations discussed in this video, such as closing a damper or a blast gate. The formulas discussed in this video should be used as general approximations with an understanding that certain equipment, like dust collectors, do not follow fan laws. Lets start by reviewing some key terms: RPM, or, Revolutions Per Minute, quantifies the speed a fan’s wheel rotates. CFM, or, Cubic Feet per Minute, is a unit of measurement for the volume of airflow a fan moves as the wheel rotates. Volume is also commonly measured in Cubic Meters per Hour in metric units. BHP, or, Brake Horsepower, is a measure of power consumption. Power consumption is commonly measured in Horsepower or Kilowatts. SP stands for Static Pressure, and is commonly measured in inches of water gauge. Pressure is also commonly measured in Pascals when using the metric system. In the context of HVAC and industrial air movement, static pressure is the resistance to airflow. Because air has mass, it creates friction as a fan moves it through ducting or piping. That friction is resistance that a fan has to overcome to move air through a system. To overcome that resistance, a fan has to generate enough pressure to match the system’s static pressure. With these concepts in mind, we can define the first three fan laws. The first fan law describes how airflow changes in response to a change in speed, and the change is a 1 to 1 ratio. This can be conceptualized by thinking of the blades on a wheel as a set of equally sized shovels, each moving an equal amount of matter with each shovel full, or, rotation. By changing the speed, you change the frequency with which each shovel moves an equal amount of matter. In other words, it changes at a 1:1 ratio, which is what the first fan law shows us algebraically. The change in speed, or, new speed divided by old speed, is equal to the change in flow, or, new flow divided by old flow. As a numeric example, let’s say we have a fan that’s running at 1,000 RPM, moving 10,000 cubic feet per minute while generating 10 inches of water gauge in static pressure, at 100 brake horsepower. In our example, we’ll double the speed, from 1,000 rpm to 2,000 rpm. So, the equation is the new speed, 2,000 RPM, divided by the previous speed, 1,000 RPM, equals the new CFM, X, divided by the old CFM, 10,000, which solves for a new flow of 20,000 CFM. The second fan law describes how pressure changes in response to a change in speed. The equation is similar to the first fan law, but here the pressure changes as a function of the change in speed squared, meaning pressure is much more sensitive to a change in speed than flow. Mathematically, that’s: the quantity of the change in speed squared equals the change in pressure. We’ll use the same operating point as our first example, doubling the speed from 1,000 RPM to 2,000 RPM, to see how the 10-inch static pressure changes. The equation is the quantity of 2,000 RPM over 1,000 RPM, squared, equals the new pressure, X, divided by the old pressure, 10, which solves for a new static pressure of 40 inches of water gauge. The third fan law describes how brake horsepower changes in response to a change in speed. Another similar setup, but here the horsepower changes as a function of the change in speed cubed. This exponential impact on power consumption is useful in determining how much headroom a fan assembly has for speed increase, before exceeding the motor’s rated capability. In our example, the equation is, the quantity of 2,000 RPM over 1,000 RPM cubed, equals the new brake horsepower, X, divided by the old brake horsepower, 100, which solves for a new power consumption of 800 BHP. To summarize these 3 fan laws, flow changes proportionately to speed. Static pressure changes as a function of the change in speed squared. And brake horsepower changes as a function of the change in speed cubed. There are other fan laws, but these 3 fundamentals can help you quickly estimate the performance effects of changing speed, either during fan selection, or making changes in the field.
188344	https://youglish.com/pronounce/excreted/english	Excreted \| 340 pronunciations of Excreted in English Toggle navigation Login Sign up Daily Lessons Submit Get your widget Donate! for English▼ • Arabic • Chinese • Dutch • English • French • German • Greek • Hebrew • Italian • Japanese • Korean • Polish • Portuguese • Romanian • Russian • Spanish • Swedish • Thai • Turkish • Ukrainian • Vietnamese • Sign Languages Say it! All US UK AUS CAN IE SCO NZ [x] All - [x] United States - [x] United Kingdom - [x] Australia - [x] Canada - [x] Ireland - [x] Scotland - [x] New Zealand Close How to pronounce excreted in English (1 out of 340): Speed: arrow_drop_down Normal arrow_drop_up vertical_align_top emoji_objects settings ▼ ▲ ↻ ↻ U × Is it excreted in your kidneys? ••• [Feedback] [Share] [Save] [Record] [YouTube] [G. TranslateB. TranslateDeepLReverso▼] Google Translate Bing Translate DeepL Reverso Definition: Click on any word below to get its definition: is it excreted in your kidneys Discover more Hire voice coach online English language tutors Purchase English practice materials ESL textbooks Buy English phonetics guide Book English pronunciation workshop Translation services Buy English pronunciation course Hire English pronunciation coach Get English fluency program Nearby words: Having trouble pronouncing 'excreted' ? Learn how to pronounce one of the nearby words below: excited exciting except excellent exchange excuse exception excitement excellence excess exceptional excel exclusive exceptions exclusively excessive exchanges excluded excuses exceed exclusion exclaimed exclude exceedingly exceeded exceptionally exchanged exceeds excite excerpt Discover more Speech therapy tools Accent reduction coaching Virtual language tutors TOEFL IELTS preparation materials High-quality headphones Real-world speech examples English language dictionaries Daily pronunciation lessons Pronunciation learning app English vocabulary flashcards Phonetic: When you begin to speak English, it's essential to get used to the common sounds of the language, and the best way to do this is to check out the phonetics. Below is the UK transcription for 'excreted': Modern IPA: ɪksgrɪ́jtɪd Traditional IPA: ɪkˈskriːtɪd 3 syllables: "ik" + "SKREET" + "id" Test your pronunciation on words that have sound similarities with 'excreted': excrete excoriated excreta exceeded excited excluded excretion exported extruded executed accelerated exaggerated excavated expected iscariot Discover more Book English conversation classes Buy English phonetics guide Buy English pronunciation course Buy language learning app Online English classes English language tutors English pronunciation apps Hire voice coach online Find online English courses Foreign language courses Tips to improve your English pronunciation: Here are a few tips that should help you perfect your pronunciation of 'excreted': Sound it Out: Break down the word 'excreted' into its individual sounds "ik" + "skreet" + "id". Say these sounds out loud, exaggerating them at first. Practice until you can consistently produce them clearly. Self-Record & Review: Record yourself saying 'excreted' in sentences. Listen back to identify areas for improvement. YouTube Pronunciation Guides: Search YouTube for how to pronounce 'excreted' in English. Pick Your Accent: Mixing multiple accents can be confusing, so pick one accent (US or UK) and stick to it for smoother learning. Here are a few tips to level up your english pronunciation: Mimic the Experts: Immerse yourself in English by listening to audiobooks, podcasts, or movies with subtitles. Try shadowing—listen to a short sentence and repeat it immediately, mimicking the intonation and pronunciation. Become Your Own Pronunciation Coach: Record yourself speaking English and listen back. Identify areas for improvement, focusing on clarity, word stress, and intonation. Train Your Ear with Minimal Pairs: Practice minimal pairs (words that differ by only one sound, like ship vs. sheep) to improve your ability to distinguish between similar sounds. Explore Online Resources: Websites & apps offer targeted pronunciation exercises. Explore YouTube channels dedicated to pronunciation, like Rachel's English and English with James for additional pronunciation practice and learning. Discover more Book English pronunciation workshop Get professional language services English pronunciation apps Purchase English learning software Buy English pronunciation course English language tutors Hire voice coach online Language learning platforms Find English speaking partner Vocabulary building apps YouGlish for: Arabic Chinese Dutch English French German Greek Hebrew Italian Japanese Korean Polish Portuguese Romanian Russian Spanish Swedish Thai Turkish Ukrainian Vietnamese Sign Languages Choose your language: English Français Español Italiano Português Deutsch العربية HOME ABOUT CONTACT PRIVACY & TERMS SETTINGS API BROWSE CONTRIBUTE ×Close ■Definitions■Synonyms■Usages■Translations Translate to : Close More
188345	https://www.geeksforgeeks.org/maths/exterior-angle-theorem/	Exterior Angle Theorem Last Updated : 23 Jul, 2025 Suggest changes 4 Likes Exterior Angle Theorem is one of the foundational theorems in geometry, as it describes the relationship between exterior and interior angles in any triangle. An Exterior Angle is formed when any side of a polygon is extended to one side. In simple terms, an Exterior Angle is an angle that exists outside the boundaries of the polygon but shares a vertex with the polygon. This article explores the relationship between the exterior angle and the remote angle of a triangle, i.e., the Exterior Angle Theorem. We will cover various topics related to the Exterior Angle Theorem, including its statement, proof, and some applications as well. Also, we will learn the Exterior Angle Inequality Theorem, as it is somewhat related to the Exterior Angle Theorem. Table of Content What is Exterior Angle? What is Exterior Angle Theorem? Proof of Exterior Angle Theorem Exterior Angle Inequality Theorem Interior and Exterior Angles in a Triangle Applications of the Exterior Angle Theorem How to Find the Unknown Exterior Angles of Triangle? How to Find the Unknown Interior Angles of Triangle? Solved Examples on Exterior Angle Theorem Practice Problems on Exterior Angle Theorem What is Exterior Angle? The angle formed between a side of the polygon and by extending the adjacent side is called the exterior angle of the polygon. The exterior angle and its adjacent angle follow the linear property i.e., the sum of the exterior angle and its adjacent angle is 180 degrees. What is Exterior Angle in Triangle? The angle formed by extending one side of the triangle and its adjacent side is called as exterior angle of the triangle. In other words, the angle formed outside the triangle is known as Exterior Angle in Triangle. There are six exterior angles in a triangle. The exterior angle and the adjacent interior angle form a linear pair of angles i.e., the sum of exterior and corresponding interior angle is 180°. In the above figure, angle 1, 2 and 3 are interior angles and angle 4 is exterior angle of the triangle. What is Exterior Angle Theorem? The exterior angle theorem states that: When a side of the triangle is extended, then the exterior angle formed is equal to the sum of two opposite interior angles. In the above figure of triangle ABC and BC be the extended side where, ∠ACD is the exterior angle of the triangle and ∠ABC and ∠BAC are the two opposite interior angles. Then, by exterior angle theorem: ∠ACD = ∠ABC + ∠BAC Proof of Exterior Angle Theorem Consider a triangle PQR and QR be the extended side of the triangle. A line RS is drawn parallel to side PQ of the triangle as shown in the following figure. PQ \|\| RS and PR is the transversal and ∠RPQ and ∠PRS are pair of alternate interior angles. ∠RPQ = ∠PRS ⇒∠1 = ∠x PQ \|\| RS and QT is the transversal and ∠PQR and ∠SRT are corresponding angles. ∠PQR = ∠SRT ⇒ ∠2 = ∠y From above statements ∠1 + ∠2 = ∠x + ∠y From above figure ∠x + ∠y = ∠PRT From above two statements ∠1 + ∠2 = ∠PRT ⇒ ∠QPR + ∠PQR = ∠PRT [Exterior angle theorem] From the above table, the exterior angle ∠PRT is equal to the sum of the two opposite interior angles ∠QPR and ∠PQR in triangle PQR. Exterior Angle Inequality Theorem The exterior angle inequality theorem states that: The value of the exterior angle of a triangle is always greater than the value of either of the opposite angles of the triangle. In the above figure the ∠1, ∠2 and ∠3 are interior angles of triangle ABC and ∠4 is exterior angle of the triangle ABC. The interior angles ∠1 and ∠2 is the opposite interior angles w.r.t exterior angle ∠4. According to the Exterior Angle Inequality Theorem, the exterior angle ∠4 is greater than the either of the opposite interior angles ∠1 and ∠2. ∠4 > ∠1 and ∠4 > ∠2 Interior and Exterior Angles in a Triangle As the name suggests, interior and exterior angles refer to the angles in a triangle that are either inside or outside the geometric shape. Apart from this, there are some more differences between these two types of angles in a triangle, and these differences are listed in the following table: \| Type of Angle \| Definition \| Measurement \| --- \| Interior Angle \| An angle formed inside a triangle between two sides \| Sum of all interior angles in a triangle is always 180°. For an equilateral triangle, each angle in a triangle is 180° divided by 3 i.e., 60 ° or varying in other types of triangles. \| \| Exterior Angle \| An angle formed outside a triangle when one side is extended \| Sum of an exterior angle and the adjacent interior angle is always 180 °. The measurement of an exterior angle is equal to the sum of the measurements of the two non-adjacent interior angles. \| Applications of the Exterior Angle Theorem The applications of the exterior angle theorem are: To Find the Unknown Exterior Angles of Triangle To Find the Unknown Interior Angles of Triangle Let's discuss these cases in detail with the help of examples. How to Find the Unknown Exterior Angles of Triangle? To find the unknown exterior angle of the triangle we use the exterior angle theorem. If the value of two remote interior angles are given we can easily find the value of exterior angles. Example: Given a triangle with an exterior angle and remote interior angles. The value of the two remote interior angles are 30° and 65°. Find the value of the exterior angle. Solution: By the exterior angle theorem Exterior angle = sum of remote interior angles. ∠e = ∠i1 + ∠i2 ⇒ The value of exterior angle = 30° + 65° ⇒ The value of exterior angle = 95° How to Find the Unknown Interior Angles of Triangle? To find the unknown exterior angle of the triangle we use the exterior angle theorem. If the value of two remote interior angles are given we can easily find the value of exterior angles. Example: Given a triangle with an exterior angle and remote interior angles. The value of the one of the remote interior angles and exterior angle is are 80° and 165°. Find the value of the other remote interior angle. Solution: By the exterior angle theorem Exterior angle = Sum of Remote Interior Angles. ⇒ ∠e = ∠i1 + ∠i2 ⇒ 165° = 80° + ∠i2 ⇒ ∠i2 = 165° - 80° The value of other interior angle = 85° Read More, Triangles in Geometry Types of Triangles Area of Triangle Triangle Inequality Theorem Congruence of Triangles Types of Angles Solved Examples on Exterior Angle Theorem Example 1: Find the value of the exterior angle from the below figure: Solution: By the exterior angle theorem ∠ABC + ∠BAC = ∠ACD ⇒ e = 50° + 67° ⇒ e = 117° Example 2: Find the value of i using the information from the following figure: Solution: By the exterior angle theorem ∠ABC + ∠BAC = ∠ACD ⇒ 40° + i = 170° ⇒ i = 170° - 40° ⇒ i = 130° Example 3: Find the value of x Solution: By exterior angle theorem ∠ABC + ∠BAC = ∠ACD ⇒ 4x + 6x = 110° ⇒ 10x = 110° ⇒ x = 11° Example 4: Find the value of x Solution: Since the exterior angle and adjacent interior has linear property ∠ACB + ∠ACD = 180° ⇒ x + 130° = 180° ⇒ x = 180°- 130° ⇒ x = 50° Example 5: Find the values of interior angles. Solution: By the exterior angle theorem ∠ACD = ∠A + ∠B ⇒ 130 = y + 2 + y + 10 ⇒ 2y + 12 = 130 ⇒ 2y = 118 ⇒ y = 59° The interior angles are 61° and 69° Practice Problems on Exterior Angle Theorem Problem 1: In a triangle, one of the exterior angles measures 110 degrees and both remote angles are equal. What are the measures of the two remote interior angles? Problem 2: If the measure of one of the remote interior angles of a triangle is 45° and other is 70° then what is the measure of the corresponding exterior angle? Problem 3: In a triangle, the measure of all exterior angles is 120°. Find the measures of the interior angles. Problem 4: A triangle has exterior angles measuring 70°, 80°, and 110°. Find the measures of the three corresponding interior angles. Problem 5: Two angles of a triangle measure 40° and 65°. Find the measure of the third angle and corresponding Exterior Angle as well. Problem 6: If an exterior angle of a triangle is 80∘and one of the interior angles is 40∘ find the measure of the third interior angle. Problem 7: In a triangle, the exterior angle at vertex C is 120∘. If the interior angles at A and B are 45∘ and 15∘ respectively determine if the triangle is valid. Problem 8: A triangle has an exterior angle measuring 150∘. Determine the measures of the two interior angles of the triangle. Problem 9: If an exterior angle of a triangle is 105∘ and the measure of one of the interior angles is 35∘ find the measure of the remaining interior angle. Problem 10: In a triangle, the exterior angle at vertex A is 130∘. Find the measure of the other two interior angles of the triangle. Conclusion The Exterior Angle Theorem states that an exterior angle of the triangle is equal to the sum of the two remote interior angles. This theorem is fundamental in the understanding triangle properties and is used to the solve various geometric problems involving the triangles. By applying this theorem we can easily find unknown angles in the triangle verify the validity of triangles and understand the relationship between the exterior and interior angles. The Mastering this concept is crucial for the solving more complex geometry problems and proving the various geometric properties. 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188346	https://www.chegg.com/homework-help/questions-and-answers/van-t-hoff-equation-relates-equilibrium-constant-reaction-change-temperature-assuming-enth-q8455663	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: The van’t Hoff equation relates the equilibrium constant of a reaction to change in temperature assuming that the enthalpy of a reaction (?HR°) is constant as a function of temperature. The assumption of constant enthalpy is generally valid over a short temperature range. The van’t Hoff equation relates the equilibrium constant of a reaction to change in temperature assuming that the enthalpy of a reaction (?HR°) is constant as a function of temperature. The assumption of constant enthalpy is generally valid over a short temperature range. Derivation and discussion of the van’t Hoff equation is given in Krauskopf and Bird, Chapter 8, pg. 203- 210. Using the van't Hoff equation, calculate the standard enthalpy of reaction for the solubility of O2 in water using its equilibrium solubility at 10°C (from Problem #1) and the equilibrium value at 25°C (which you will need to look up). Thank you. This AI-generated tip is based on Chegg's full solution. Sign up to see more! Determine the natural logarithm of the ratio of the Henry's Law constants at the two given temperatures. Applying Henry's Law constant: ln(588 atm-L/mol… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
188347	https://mathresearch.utsa.edu/wiki/index.php?title=Proportionality_vs._Linearity	Proportionality vs. Linearity From Department of Mathematics at UTSA Jump to navigation Jump to search 1 Propotionality 1.1 Direct proportionality 1.1.1 Examples 1.2 Inverse proportionality 1.3 Hyperbolic coordinates 2 Linearity 2.1 In mathematics 2.1.1 Linear polynomials 2.1.2 Boolean functions 3 Licensing Propotionality The variable y is directly proportional to the variable x with proportionality constant ~0.6. The variable y is inversely proportional to the variable x with proportionality constant 1. In mathematics, two varying quantities are said to be in a relation of proportionality, multiplicatively connected to a constant; that is, when either their ratio or their product yields a constant. The value of this constant is called the coefficient of proportionality or proportionality constant. If the ratio () of two variables (x and y) is equal to a constant (k = ), then the variable in the numerator of the ratio (y) can be product of the other variable and the constant (y = k ⋅ x). In this case y is said to be directly proportional to x with proportionality constant k. Equivalently one may write ⋅ y; that is, x is directly proportional to y with proportionality constant . If the term proportional is connected to two variables without further qualification, generally direct proportionality can be assumed. If the product of two variables (x ⋅ y) is equal to a constant (k = x ⋅ y), then the two are said to be inversely proportional to each other with the proportionality constant k. Equivalently, both variables are directly proportional to the reciprocal of the respective other with proportionality constant k ( = k ⋅ and y = k ⋅ ). If several pairs of variables share the same direct proportionality constant, the equation expressing the equality of these ratios is called a proportion, e.g., = ⋯ = k (for details see Ratio). Direct proportionality Given two variables x and y, y is directly proportional to x if there is a non-zero constant k such that The relation is often denoted using the symbols "∝" (not to be confused with the Greek letter alpha) or "~": : or For the proportionality constant can be expressed as the ratio It is also called the constant of variation or constant of proportionality. A direct proportionality can also be viewed as a linear equation in two variables with a y-intercept of 0 and a slope of k. This corresponds to linear growth. Examples If an object travels at a constant speed, then the distance traveled is directly proportional to the time spent traveling, with the speed being the constant of proportionality. The circumference of a circle is directly proportional to its diameter, with the constant of proportionality equal to π. On a map of a sufficiently small geographical area, drawn to scale distances, the distance between any two points on the map is directly proportional to the beeline distance between the two locations represented by those points; the constant of proportionality is the scale of the map. The force, acting on a small object with small mass by a nearby large extended mass due to gravity, is directly proportional to the object's mass; the constant of proportionality between the force and the mass is known as gravitational acceleration. The net force acting on an object is proportional to the acceleration of that object with respect to an inertial frame of reference. The constant of proportionality in this, Newton's second law, is the classical mass of the object. Inverse proportionality Inverse proportionality with a function of y = 1/x The concept of inverse proportionality can be contrasted with direct proportionality. Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same. As an example, the time taken for a journey is inversely proportional to the speed of travel. Formally, two variables are inversely proportional (also called varying inversely, in inverse variation, in inverse proportion, in reciprocal proportion) if each of the variables is directly proportional to the multiplicative inverse (reciprocal) of the other, or equivalently if their product is a constant. It follows that the variable y is inversely proportional to the variable x if there exists a non-zero constant k such that or equivalently, Hence the constant "k" is the product of x and y. The graph of two variables varying inversely on the Cartesian coordinate plane is a rectangular hyperbola. The product of the x and y values of each point on the curve equals the constant of proportionality (k). Since neither x nor y can equal zero (because k is non-zero), the graph never crosses either axis. Hyperbolic coordinates The concepts of direct and inverse proportion lead to the location of points in the Cartesian plane by hyperbolic coordinates; the two coordinates correspond to the constant of direct proportionality that specifies a point as being on a particular ray and the constant of inverse proportionality that specifies a point as being on a particular hyperbola. Linearity Linearity is the property of a mathematical relationship (function) that can be graphically represented as a straight line. Linearity is closely related to proportionality. Examples in physics include the linear relationship of voltage and current in an electrical conductor (Ohm's law), and the relationship of mass and weight. By contrast, more complicated relationships are nonlinear. Generalized for functions in more than one dimension, linearity means the property of a function of being compatible with addition and scaling, also known as the superposition principle. The word linear comes from Latin linearis, "pertaining to or resembling a line". In mathematics In mathematics, a linear map or linear function f(x) is a function that satisfies the two properties: Additivity: f(x + y) = f(x) + f(y). Homogeneity of degree 1: f(αx) = α f(x) for all α. These properties are known as the superposition principle. In this definition, x is not necessarily a real number, but can in general be an element of any vector space. A more special definition of linear function, not coinciding with the definition of linear map, is used in elementary mathematics (see below). Additivity alone implies homogeneity for rational α, since implies for any natural number n by mathematical induction, and then implies . The density of the rational numbers in the reals implies that any additive continuous function is homogeneous for any real number α, and is therefore linear. The concept of linearity can be extended to linear operators. Important examples of linear operators include the derivative considered as a differential operator, and other operators constructed from it, such as del and the Laplacian. When a differential equation can be expressed in linear form, it can generally be solved by breaking the equation up into smaller pieces, solving each of those pieces, and summing the solutions. Linear algebra is the branch of mathematics concerned with the study of vectors, vector spaces (also called 'linear spaces'), linear transformations (also called 'linear maps'), and systems of linear equations. For a description of linear and nonlinear equations, see linear equation. Linear polynomials In a different usage to the above definition, a polynomial of degree 1 is said to be linear, because the graph of a function of that form is a straight line. Over the reals, a linear equation is one of the forms: where m is often called the slope or gradient; b the y-intercept, which gives the point of intersection between the graph of the function and the y-axis. Note that this usage of the term linear is not the same as in the section above, because linear polynomials over the real numbers do not in general satisfy either additivity or homogeneity. In fact, they do so if and only if b = 0. Hence, if b ≠ 0, the function is often called an affine function (see in greater generality affine transformation). Boolean functions Hasse diagram of a linear Boolean function In Boolean algebra, a linear function is a function for which there exist such that : , where Note that if , the above function is considered affine in linear algebra (i.e. not linear). A Boolean function is linear if one of the following holds for the function's truth table: In every row in which the truth value of the function is T, there are an odd number of Ts assigned to the arguments, and in every row in which the function is F there is an even number of Ts assigned to arguments. Specifically, f(F, F, ..., F) = F, and these functions correspond to linear maps over the Boolean vector space. In every row in which the value of the function is T, there is an even number of Ts assigned to the arguments of the function; and in every row in which the truth value of the function is F, there are an odd number of Ts assigned to arguments. In this case, f(F, F, ..., F) = T. Another way to express this is that each variable always makes a difference in the truth value of the operation or it never makes a difference. Negation, Logical biconditional, exclusive or, tautology, and contradiction are linear functions. Licensing Content obtained and/or adapted from: Proportionality (mathematics), Wikipedia under a CC BY-SA license Linearity, Wikipedia under a CC BY-SA license Retrieved from " Navigation menu Personal tools Log in Namespaces Page Discussion Variants Views Read View source View history Navigation Main page Recent changes Random page Help about MediaWiki Tools What links here Related changes Special pages Printable version Permanent link Page information Cite this page
188348	https://www.nature.com/articles/s41598-020-68456-7	Prevalence of sacroiliitis and acute and structural changes on MRI in patients with psoriatic arthritis \| Scientific Reports Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles News & Comment Collections Subjects Follow us on Facebook Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal About Scientific Reports Contact Journal policies Guide to referees Calls for Papers Editor's Choice Journal highlights Open Access Fees and Funding Publish with us Publish with us For authors Language editing services Open access funding Submit manuscript Sign up for alerts RSS feed nature scientific reports articles article Prevalence of sacroiliitis and acute and structural changes on MRI in patients with psoriatic arthritis Download PDF Download PDF Article Open access Published: 14 July 2020 Prevalence of sacroiliitis and acute and structural changes on MRI in patients with psoriatic arthritis Marcio Vale BragaORCID: orcid.org/0000-0003-4827-67461, Samily Cordeiro de OliveiraORCID: orcid.org/0000-0002-0023-71502, Antonio Helder Costa VasconcelosORCID: orcid.org/0000-0002-9223-60821, Jailson Rodrigues Lopes3, Carlos Leite de Macedo FilhoORCID: orcid.org/0000-0002-2497-10773, Lysiane Maria Adeodato RamosORCID: orcid.org/0000-0001-9047-46371& … Carlos Ewerton Maia RodriguesORCID: orcid.org/0000-0003-1367-67821,2,3 Show authors Scientific Reportsvolume 10, Article number:11580 (2020) Cite this article 24k Accesses 29 Citations 7 Altmetric Metrics details Abstract Sacroiliac joint involvement is one of the earliest manifestations of psoriatic arthritis (PsA). Magnetic resonance imaging (MRI) is a useful tool in the early diagnosis of axial disease due to its sensitivity for detecting acute and chronic changes associated with sacroiliitis. In this study, we evaluated the prevalence of sacroiliitis, acute and structural image changes on MRI in PsA patients and identified predictive clinical, laboratory and disease activity factors. Cross-sectional study on PsA patients submitted to MRI of the sacroiliac joints. The scans were evaluated by two blinded radiologists and the level of agreement was calculated (kappa). Clinical, disease activity and quality-of-life indices (DAS28, BASDAI, PASI, MASES, HAQ, CRP, ESR) were estimated. The sample consisted of 45 PsA patients with a mean age of 50.1 ± 11.5 years. The prevalence of sacroiliitis was 37.8% (n = 17), 47% of which was unilateral. The kappa coefficient was 0.64. Only 5 (29.4%) of the 17 patients with sacroiliitis on MRI had back pain. The most prevalent acute and chronic changes on MRI were, respectively, subchondral bone edema (26.7%) and enthesitis (20%), periarticular erosions (26.7%) and fat metaplasia (13.3%). CRP levels were higher among sacroiliitis patients (p = 0.028), and time of psoriasis was positively associated with chronic lesions (p = 0.006). Sacroiliitis on MRI was highly prevalent in our sample of PsA patients. Raised CRP levels were significantly associated with sacroiliitis, and longer time of psoriasis was predictive of chronic sacroiliitis lesions. Most sacroiliitis patients displayed no clinical symptoms. Similar content being viewed by others Comparison of sacroiliac CT findings in patients with and without ankylosing spondylitis aged over 50 years Article Open access 20 October 2023 Predominant ligament-centric soft-tissue involvement differentiates axial psoriatic arthritis from ankylosing spondylitis Article 02 November 2023 Machine learning identifies right index finger tenderness as key signal of DAS28-CRP based psoriatic arthritis activity Article Open access 19 December 2023 Introduction Psoriatic arthritis (PsA) is an immune-mediated chronic inflammatory joint pathology, one of several spondyloarthropathies. Potentially destructive and progressive, it affects 20–30% of patients with cutaneous psoriasis. The clinical spectrum of PsA includes peripheral arthritis, axial disease, enthesitis or dactylitis, and skin and/or nail involvement1,2,3. According to clinical or radiographic evidence, the prevalence of axial disease is 25–70% in patients with peripheral PsA4,5 and 7–17% in patients without peripheral involvement6. Magnetic resonance imaging (MRI) is an excellent diagnostic tool for axial disease in spondyloarthropathies due to its sensitivity for detecting inflammatory lesions. It is currently used with the Assessment of Spondyloarthritis International Society (ASAS) classification criteria for axial spondyloarthritis, identifying the acute and chronic changes of sacroiliitis, and plays an important role in the follow-up of patients treated with immunobiological drugs. According to some authors, radiological involvement of the sacroiliac joint is one of the earliest signs of PsA, preceding clinical manifestations in approximately 30% of patients7. Another study revealed associations between radiological sacroiliitis and clinical PsA patterns and concluded that 25% of the patients had radiological sacroiliitis (asymmetric 73%, clinical history of back pain 100%). Sacroiliitis was significantly associated with peripheral joint erosion (p = 0.043), high psoriasis activity and severity scores (PASI) (p = 0.041) and early onset of PsA (p ≤ 0.001)2. However, only little has been published on MRI-detected changes of sacroiliitis in PsA patients or on factors predictive of sacroiliitis in this population8. Thus, in this study we evaluated the prevalence of sacroiliitis, acute and structural image changes on MRI in patients with PsA and identified predictive clinical, laboratory and disease activity factors. Materials and methods Study approval In this observational, cross-sectional and quantitative study, semi-structured questionnaires were administered to patients attending the rheumatology service of the General Hospital of Fortaleza (HGF), Brazil, between August and October 2018. Submitted through an online national research database (Plataforma Brasil), the study protocol was approved by the research ethics committee of the HGF and filed under entry #2837133. All patients gave their written informed consent, and all study activities were conducted in accordance with the approved protocols and guidelines. Patients The inclusion criteria were: diagnosis of PsA according to the CASPAR classification9, availability of MRI scans of the sacroiliac joints, age over 18 years, and compliance with follow-up at the outpatient rheumatology service. Data collection Information was collected on sociodemographic and clinical variables, and the MRI scans were analyzed by two independent radiologists blinded to all clinical and laboratory data. The MRI scans were performed during the study period, regardless of the presence of low back pain symptoms. The sociodemographic variables included sex, age (years), level of schooling, city of origin, address, and household income. The clinical variables included the use of NSAIDs, disease-modifying anti-rheumatic drugs and immunobiological products. The patients were also scored with the disease activity score of 28 joints (DAS28)10, PASI10, the Maastricht ankylosing spondylitis enthesis score (MASES)10 and the Bath ankylosing spondylitis disease activity index (BASDAI)10. In addition, the health assessment questionnaire (HAQ)10 was administered. DAS28 was based on CRP. Three ranges of DAS28 were considered: low (2.6–3.1), moderate (> 3.2–5.0) and high (> 5.1). PASI (range 0–72) was calculated using GRAPPA for IOS and Android, whereas MASES (range 0–13) was determined by summing up points. BASDAI was dichotomized into inactive (< 4) and active (≥ 4), while HAQ was calculated for 8 domains, each with a range of 0–3, with higher scores indicating greater deficiency. Blood drawn from the patients was submitted to determination of erythrocyte sedimentation rate (ESR) and C-reactive protein (CRP). HLA-B27 was performed whenever available. All patients were submitted to MRI (Philips Achieva 1.5T) of the sacroiliac joints, with or without endovenous administration of paramagnetic contrast agents (gadolinium), in the oblique coronal plane (parallel to the long axis of the sacrum) and axial plane in the following sequence: T1-weighted in turbo spin-echo (TSE) (TR/TE, 600/minimum), STIR (TR/TE, 4.000/60), T2-weighted with fat suppression (TR/TE, 4000/50), and enhanced T1-weighted in TSE (TR/TE, 600/minimum) with fat suppression. The latter images were obtained after endovenous administration of 0.1 mmol gadobenate dimeglumine per kg of body weight. The equipment was configured to 20 cm field of view, 4 mm slice thickness and 2 excitations. Imaging lasted 7 min (STIR), 4 min (fat-suppressed T2 TSE), 6 min (T1 TSE) or 6 min (fat-suppressed enhanced T1 TSE). Sacroiliitis was classified as symmetric or asymmetric. The sacroiliac joints were evaluated for signs of involvement of the joint space, surrounding bone, bone marrow adjacent to the joint in each image sequence, enhancement, subchondral bone marrow edema, synovitis, enthesitis, capsulitis, cartilage abnormalities, periarticular erosions, fat infiltration of the subchondral bone marrow, and ankylosis. Chronic disease was suspected based on low-signal changes in T1 and T2-weighted images, periarticular erosions, subchondral bone sclerosis, joint space narrowing, bone bridges, and ankylosis. Active inflammatory disease was diagnosed based on high-signal erosions in STIR and fat-suppressed T2-weighted images, subchondral edema, synovitis, enthesitis, capsulitis and enhancement inside or adjacent to the joint. In our evaluation of the MRI images we employed the definition of sacroiliitis in axial spondyloarthritis given by the ASAS/OMERACT MRI group11, according to which active inflammatory lesions in sacroiliac joints are described as bone marrow edema (STIR images) or osteitis (post-gadolinium T1 images)11. Statistical analysis All analyses were performed with the software IBM SPSS Statistics v. 23.0. The data and results were managed with Excel 2018 spreadsheets and graphs. Qualitative variables were expressed as absolute and relative frequencies, while quantitative variables were expressed as mean values ± standard deviation or median values (1st–3rd interquartile range). The normality of the quantitative data was verified with the Shapiro–Wilk test, while the binomial test was used to evaluate differences in the prevalence of sacroiliitis between our study and the literature. In the bivariate analysis, quantitative variables were compared using the Mann–Whitney test, while qualitative variables were compared with the chi-squared test or Fisher’s exact test. We also calculated the percentage variation between the groups (duration of arthritis and psoriasis with acute and chronic changes in sacroiliitis). The inter-examiner agreement was determined by calculating kappa coefficients (weak 0–0.2, acceptable 0.21–0.40, moderate 0.41–0.60, strong 0.61–0.80, almost perfect 0.81–0.99, and perfect 1.00)12. The strength of the observed associations was expressed as prevalence ratios and their respective confidence intervals (95% CI). The level of statistical significance was set at 5% (p< 0.05). Results Clinical-epidemiological aspects of PsA patients The sample consisted of 45 PsA patients aged 50.1 ± 11.5 years on the average who met the inclusion criteria. The two sexes were nearly equally represented (female 51.1% vs. male 48.9%), the vast majority were classified as non-white (95.6%), the predominant monthly income bracket (53.3%) was 0–3 minimum wages (at the time of writing, the Brazilian minimum wage corresponded to USD ~ 270), and the average time of schooling was 9.5 ± 3.8 years. (Table 1). No significant association was found between the prevalence of sacroiliitis on MRI and any age range or sociodemographic variable (Table 2). Table 1 Clinical and epidemiological characteristics of the sampled patients. Full size table Table 2 Clinical and laboratory findings of patients with and without sacroiliitis on MRI. Full size table The prevalence of the different forms of joint involvement was 68.9% (n = 31) for symmetric polyarticular arthritis, 22% (n = 10) for asymmetric oligoarthritis, and 22.2% (n = 10) for spondylitis. Among other clinical manifestations, the most prevalent was nail involvement (n = 31; 68.9%), followed by enthesitis (51.1%; n = 23), dactylitis (40%; n = 18) and uveitis (4.4%; n = 2) (Table 1). No significant association was found between the prevalence of sacroiliitis on MRI and any of the clinical spectra (Table 2). Clinical, disease activity and quality-of-life indices in PsA patients The following mean PsA activity scores were observed: DAS28-CRP 3.40 ± 1.6, DAS28-ESR 2.91 ± 2.0, and BASDAI 3.22 ± 2.1. Most patients had low HAQ (1.06 ± 0.7), PASI (2.00 ± 2.9) and MASES (1.90 ± 3.2) scores (Table 1). PsA patients with and without sacroiliitis on MRI did not differ significantly with regard to clinical, disease activity or quality-of-life indices (Table 2). Treatment With regard to the use of medication at the time of the evaluation, most patients used no drugs, while 26.7% (n = 12) used NSAIDs, 35.6% (n = 16) received immunobiologicals, and 8.9% (n = 4) used corticoids. Historically, 73.3% (n = 33) had never used corticoids. Evaluation of sacroiliitis on MRI In our sample of 45 PsA patients, the prevalence of sacroiliitis on MRI was 37.8% (n = 17), 47.0% (n = 8) of which was unilateral. Only one fifth (22%; n = 10) presented symptoms of inflammatory lumbar pain. Low back pain was observed in 29.4% (n = 5) of the patients with sacroiliitis. The most prevalent acute structural change observed on MRI was subchondral bone edema (26.7%; n = 12), followed by enthesitis (20%; n = 9), capsulitis (17.8%; n = 8) and synovitis (8.8%; n = 4) (Fig.1, Table 3). The most prevalent chronic structural change was periarticular erosion (26.7%; n = 12), followed by fat metaplasia (13.3%; n = 6), bone sclerosis (11.1%; n = 5) and bone bridge/ankylosis (2.2%; n = 1) (Fig.2, Table 3). The agreement between the two radiologists regarding the diagnosis of sacroiliitis was strong (kappa = 0.640). Figure 1 Findings compatible with acute sacroiliitis on MRI of the sacroiliac joints. Source: the authors (2019). Coronal STIR sequence: high signal intensity consistent with synovitis (white arrows) on the left and capsulitis (blue arrows) on the right (a). Coronal STIR sequence: high signal intensity bilaterally consistent with bone marrow edema (white arrows) (b). Coronal T1 post-contrast with fat suppression: high signal intensity bilaterally consistent with bone marrow edema (white arrows) and suggestive of capsulitis (blue arrows) on the right and enthesitis on the left (green arrows) (c). Coronal STIR sequence: high signal intensities on the right compatible with bone marrow edema (white arrows) and enthesitis (blue arrows) (d). R: right side, L: left side. Full size image Table 3 Structural changes on MRI of the sacroiliac joints of PsA patients. Full size table Figure 2 Findings compatible with chronic sacroiliitis on MRI of the sacroiliac joints. Source: the authors (2019). Coronal T1 sequence: bilateral reduction of the intra-articular space in the sacroiliac joints consistent with bone bridges (white arrows), and bilateral high signal intensity suggestive of fat metaplasia (blue arrows) (a). Coronal T1 sequence: low signal intensity bilaterally consistent with bone erosion (white arrows) and bone sclerosis (blue arrows) (b). Coronal T1 sequence: on the left high signal intensity consistent with fat metaplasia (white arrows) and low signal intensity suggestive of bone erosion (green arrows) and bone sclerosis (blue arrows); similar findings on the right, with lower signal intensity (c). Coronal STIR sequence: low signal intensity bilaterally consistent with bone erosion (white arrows) (d). R: right side , L: left side. Full size image Time of disease (psoriasis and arthritis) in patients with and without sacroiliitis on MRI The time of disease in arthritis patients was 4–12 years (median 6). The corresponding value for psoriasis patients was 6–17 years (median 9). A significant association was observed between time of psoriasis and acute and chronic changes of sacroiliitis (Table 4). The median time of arthritis differed between the groups (acute vs chronic) by 41.7%, while the median time of psoriasis differed by 260% between acute and chronic changes in sacroiliitis. Table 4 Association between time of disease (psoriasis and arthritis) and acute and chronic changes of sacroiliitis according to the Mann–Whitney test. Full size table Inflammatory markers and HLA-B27 in PsA patients All patients were tested for CRP and ESR. Sacroiliitis and non-sacroiliitis patients differed significantly with regard to CRP, but not with respect to ESR (Table 2). Discussion To our knowledge, this is the first Brazilian study to evaluate the prevalence of sacroiliitis on MRI in PsA patients and correlate positivity for sacroiliitis with acute and chronic changes and clinical, laboratory and epidemiological variables. The joint analysis of these variables made it possible to identify predictors of sacroiliitis in our sample. Early diagnosis of axial involvement on MRI, prior to the emergence of structural damage (which is diagnosed with an average delay of 8–11 years13,14,15), can help prevent loss of quality of life from limited physical function, psychological distress and inability to perform daily activities due to pain and discomfort associated with joint and skin symptoms16,17. Loss of quality of life can also be prevented with emerging biological therapy18. Sociodemographically, our sample was similar to that of another Brazilian cohort of 175 PsA patients19 with regard to age and male/female ratio, however the proportion of the non-white racial type was much higher in our study (95.6% vs. 28%). A similar discrepancy was observed in relation to a recent study from the US involving 1,567 patients of which only 9% were non-white20. This may be explained by differences in sample size or by regional differences (e.g., Southeastern vs. Midwestern Brazil). The broad demographic spectrum of the Brazilian population is the most likely explanation for the particularly non-white ethnic composition of our sample. The prevalence of sacroiliitis on MRI in our sample of PsA patients (37.8%) matches studies reporting prevalences of axial disease in PsA between 25 and 50%2, but is somewhat higher than the prevalence of sacroiliitis in PsA patients reported in another Brazilian study (24.4%)19. The higher prevalence observed in the current study may be explained by our recruitment from a referral center for patients with more advanced and long-standing disease, or by regional, genetic and environmental factors. Interestingly, inflammatory lumbar pain was only observed in 29.4% of the patients with sacroiliitis and the most prevalent findings on MRI were subchondral bone edema and enthesitis (acute changes) and periarticular erosions and fat metaplasia (chronic changes), proving that changes on MRI may occur in the absence of clinical symptoms, and reinforcing the role of MRI in the prevention of distant structural damage and in the evaluation of therapy for axial spondyloarthritis. Despite the regional and ethnic peculiarities of our sample, our results are supported by the literature showing that patients with sacroiliitis do in fact display early structural damage in the sacroiliac joints on MRI before the emergence of clinical symptoms21,22,23. In fact, another study documented clinical features of sacroiliitis in 14 (33%) of 42 patients with normal MRI scans and in 10 (38%) of 26 patients with abnormal scans (normal vs abnormal scans, p = 0.7). Moreover, the presence of sacroiliitis on MRI was associated with restricted spinal movements (p = 0.004) and the duration of PsA (p = 0.04), but no association between sacroiliitis on MRI and BASDAI or BASFI or HLA-B27 was observed, leading the authors to conclude that clinical assessment of sacroiliitis and HLA-B27 are poor predictors of sacroiliitis diagnosed by MRI in PsA patients24. Finally, a study using MRI reported inflammatory and chronic changes to defined areas in the sacroiliac joints in patients with early-stage versus late-stage PsA and concluded that whereas the iliac and the sacral side of the sacroiliac joints are almost equally affected, the dorsocaudal synovial part of the joint is involved significantly more often than the ventral part, especially in early disease. Sacroiliac enthesitis is not a special feature of early sacroiliac inflammation25. In our sample, the time of psoriasis (12.5 ± 9.6 years) was significantly associated with the presence of chronic sacroiliitis damage (p = 0.006), matching the results of another Brazilian study reporting a mean psoriasis duration of 15.4 ± 11.7 years26. In a controlled study involving patients with cutaneous psoriasis but no arthritis or spondylitis, the evidence of subclinical axial disease with sacroiliac involvement on MRI was limited. The authors admitted their findings were insufficient to warrant routine MRI screening of patients with long-standing psoriasis27. A relevant finding of this study was the non-significance of sacroiliac involvement (53.0% displayed bilateral sacroiliitis), in disagreement with the literature. Thus, two studies evaluating the type and frequency of sacroiliitis in psoriasis patients found, respectively, that 26% of 133 psoriasis patients presented sacroiliac involvement, more than half of which was bilateral27, and that 25% of 70 psoriasis patients had sacroiliitis, less than half of which was bilateral2. Moreover, the presence of sacroiliitis was not significantly associated with oligoarthritis, mutilating arthritis, dactylitis, nail involvement or enthesitis. However, an earlier study found a higher prevalence of uveitis in patients with than without sacroiliitis (8.5% vs. 1.4%, p = 0.003), contrasting with our findings (50% vs. 50%, p = 1.000)2. In this study, the increased CRP levels resulting from disease activity were significantly associated with the presence of sacroiliitis (p = 0.028). An earlier study2 also found significant correlations between CRP and sacroiliitis in PsA patients (p = 0.006), but no other disease activity index was significantly associated with sacroiliitis (DAS28-CRP p = 0.358, DAS28-ESR p = 0.251, BASDAI p = 0.389). The same was true for the quality-of-life index (HAQ p = 0.420) and clinical scores (PASI p = 0.495, MASES p = 0.341). Likewise, current PASI values were non-significant (p = 0.495), matching the literature (p = 0.23)2. Approximately 15% of patients with peripheral PsA may be expected to develop axial disease in the course of 10 years. The possible risk factors include severe peripheral arthritis, early-onset arthritis, increased ESR, presence of HLA-B27, severe ungual dystrophy and inflammatory bowel disease28,29. Sacroiliitis has been shown to be significantly associated with severe peripheral arthritis (presence of erosions) and early-onset arthritis2. On the other hand, the risk factors for sacroiliac involvement also include severe psoriasis associated with HLA-B0801 positivity and reduced HLA-B27 frequency, suggesting HLA-B27 is a less important predictor of sacroiliitis. In our study, sacroiliitis was not associated with ESR (p = 0.494) or with clinical predictors such as joint involvement pattern (p> 0.05) and enthesitis (p = 0.848). A similar lack of correlation between sacroiliitis and enthesitis was reported in another recent study30. Our study has some limitations: (i) the relatively small sample of patients recruited from a single center, (ii) the unavailability of HLA-B27 testing in some cases, (iii) the cross-sectional study design, making it more difficult to draw conclusions regarding the ability of the selected clinical and laboratory variables to predict sacroiliitis in PsA patients, and (iv) the absence of analysis of the possible impact of racial type on PsA diagnosis, therapy and outcome. Sacroiliitis on MRI was highly prevalent in our sample of PsA patients, but most sacroiliitis patients displayed no clinical symptoms. Raised CRP levels were significantly associated with sacroiliitis, and the time of psoriasis was predictive of chronic sacroiliitis lesions. Prospective studies on larger samples, including HLA-B27 testing, are necessary to identify predictive clinical, laboratory and image changes on MRI in PsA patients. References Lee, M. P. et al. Patterns of systemic treatment for psoriatic arthritis in the US: 2004–2015. Arthritis Care Res70, 791–796 (2018). ArticleGoogle Scholar Haroon, M. et al. Clinical and genetic associations of radiographic sacroiliitis and its different patterns in psoriatic arthritis. Clin Exp Rheumatol35, 270–276 (2017). PubMedGoogle Scholar Huynh, D. Q. & Kavanaugh, A. Psoriatic arthritis: Current therapy and future approaches. Rheumatology (Oxford)54, 20–28 (2014). ArticleGoogle Scholar Queiro, R. et al. Patients with psoriatic arthritis may show differences in their clinical and genetic profiles depending on their age at psoriasis onset. Clin. Exp. Rheumatol.30, 476–480 (2012). PubMedGoogle Scholar Chandran, V. et al. Axial psoriatic arthritis: Update on a longterm prospective study. J. Rheumatol.36, 2744–2750 (2019). ArticleGoogle Scholar Niccoli, L. et al. Frequency of iridocyclitis in patients with early psoriatic arthritis: A prospective, follow up study. Int. J. Rheum. Dis.15, 414–418 (2012). ArticleGoogle Scholar Baraliakos, X., Coates, L. C. & Braun, J. The involvement of the spine in psoriatic arthritis. Clin. Exp. Rheumatol.33, 31–35 (2015). Google Scholar McQueen, F., Lassere, M. & Østergaard, M. Magnetic resonance imaging in psoriatic arthritis: A review of the literature. Arthritis Res. Ther.8, 207 (2006). ArticleGoogle Scholar Taylor, W. et al. Classification criteria for psoriatic arthritis: Development of new criteria from a large international study. Arthritis Rheum.54, 2665–2673 (2006). ArticleGoogle Scholar Mease, P. J. Measures of psoriatic arthritis. Arthritis Care Res.63, 64–85 (2011). ArticleGoogle Scholar Rudwaleit, M. et al. The development of Assessment of SpondyloArthritis international Society classification criteria for axial spondyloarthritis (part II): Validation and final selection. Ann Rheum Dis.68, 777–783 (2009). ArticleCASGoogle Scholar McHugh, M. L. Interrater reliability: The kappa statistic. Biochem. Med. (Zagreb)22, 26–282 (2012). Google Scholar McHugh, M. L., Lambert, R. G. W., Østergaard, M. & Jaremko, J. L. Magnetic resonance imaging in rheumatology. Magn. Reson. Imaging Clin. N. Am.26, 599–613 (2018). ArticleGoogle Scholar Sengupta, R. et al. Short-term repeat magnetic resonance imaging scans in suspected early axial spondyloarthritis are clinically relevant only in HLA-B27-positive male subjects. J. Rheumatol.45, 202–205 (2018). ArticleGoogle Scholar Hayashi, M. et al. Superiority of magnetic resonance imaging over conventional radiography in the early diagnosis of psoriatic arthritis. J. Dermatol.44, e232–e233 (2017). ArticleGoogle Scholar Koo, J. Population-based epidemiologic study of psoriasis with emphasis on quality of life assessment. Dermatol. Clin.14, 485–496 (1996). ArticleCASGoogle Scholar Walsh, J. A. et al. Work productivity loss and fatigue inpsoriatic arthritis. J Rheumatol41, 1670–1674 (2004). ArticleGoogle Scholar Siba, P. et al. Management of psoriatic arthritis: Early diagnosis, monitoring of disease severity and cutting edge therapies. J. Autoimmun.76, 21–37 (2017). ArticleGoogle Scholar Ranza, R. et al. Prevalence of psoriatic arthritis in a large cohort of Brazilian patients with psoriasis. J. Rheumatol.42, 829–834 (2015). ArticleGoogle Scholar Mease, P. J. et al. Clinical characteristics, disease activity, and patient-reported outcomes in psoriatic arthritis patients with dactylitis or enthesitis: Results from the corrona psoriatic arthritis/spondyloarthritis registry. Arthritis Care Res.69, 1692–1699 (2017). ArticleCASGoogle Scholar Eshed, I., Hermann, K. G. A., Zejden, A. & Sudoł-Szopińska, I. Imaging to differentiate the various forms of seronegative arthritis. Semin. Musculoskelet. Radiol.22, 189–196 (2018). ArticleGoogle Scholar Sahin, N. et al. Is there a role for DWI in the diagnosis of sacroiliitis based on ASAS criteria?. Int. J. Clin. Exp. Med.8, 7544–7552 (2015). PubMedPubMed CentralGoogle Scholar Bredella, M. A. et al. MRI of the sacroiliac joints in patients with moderate to severe ankylosing spondylitis. Am. J. Roentgenol.187, 1420–1426 (2006). ArticleGoogle Scholar Williamsonn, L. et al. Clinical assessment of sacroiliitis and HLA-B27 are poor predictors of sacroiliiitis diagnosed by magnetic resonance imaging in psoriatic arthritis. Rheumatology (Oxford)43, 85–88 (2004). ArticleGoogle Scholar Muche, B. et al. Anatomic structures involved in early- and late-stage sacroiliitis in spondylarthritis: A detailed analysis by contrast-enhanced magnetic resonance imaging. Arthritis Rheum.48, 1374–1384 (2003). ArticleCASGoogle Scholar Bratu, V. A. et al. Do patients with skin psoriasis show subclinical axial inflammation on MRI of the sacroiliac joints and entire spine?. Arthritis Care Res. (Hoboken)71, 1109–1118 (2019). ArticleGoogle Scholar Kaçar, C. et al. Sacroiliac joint involvement in psoriasis. Rheumatol. Int.30, 1263–1266 (2010). ArticleGoogle Scholar Gladman, D. D. Clinical features and diagnostic considerations in psoriatic arthritis. Rheum. Dis. Clin. North Am.41, 569–579 (2015). ArticleGoogle Scholar Chandran, V., Tolusso, D. C., Cook, R. J. & Gladman, D. D. Risk factors for axial inflammatory arthritis in patients with psoriatic arthritis. J. Rheumatol.37, 809–815 (2010). ArticleGoogle Scholar Moshrif, A. et al. (2017) Subclinical enthesopathy in patients with psoriasis and its association with other disease parameters: A power Doppler ultrasonographic study. Eur. J. Rheumatol.4, 24–28 (2017). ArticleGoogle Scholar Download references Acknowledgements This research received no specific grant from any funding agency in the public, commercial or not-for-profit sectors. Author information Authors and Affiliations Postgraduate Program, University of Fortaleza (Unifor), Fortaleza, Brazil Marcio Vale Braga,Antonio Helder Costa Vasconcelos,Lysiane Maria Adeodato Ramos&Carlos Ewerton Maia Rodrigues Federal University of Ceará, Fortaleza, Ceará, Brazil Samily Cordeiro de Oliveira&Carlos Ewerton Maia Rodrigues General Hospital of Fortaleza, Fortaleza, Brazil Jailson Rodrigues Lopes,Carlos Leite de Macedo Filho&Carlos Ewerton Maia Rodrigues Authors 1. Marcio Vale BragaView author publications Search author on:PubMedGoogle Scholar 2. Samily Cordeiro de OliveiraView author publications Search author on:PubMedGoogle Scholar 3. Antonio Helder Costa VasconcelosView author publications Search author on:PubMedGoogle Scholar 4. Jailson Rodrigues LopesView author publications Search author on:PubMedGoogle Scholar 5. Carlos Leite de Macedo FilhoView author publications Search author on:PubMedGoogle Scholar 6. Lysiane Maria Adeodato RamosView author publications Search author on:PubMedGoogle Scholar 7. Carlos Ewerton Maia RodriguesView author publications Search author on:PubMedGoogle Scholar Contributions All authors were involved in drafting the article or revising it critically for important intellectual content, and all authors approved the final version to be submitted for publication. C.E.M.R. had full access to all study data and takes responsibility for the integrity of the data. Study conception and design: M.V.B., L.M.A.R., C.E.M.R. Acquisition of data: M.V.B., J.R.L., C.L.M.F., C.E.M.R. Analysis and interpretation of data: M.V.B., S.C.O., A.H.C.V., C.E.M.R. Corresponding author Correspondence to Carlos Ewerton Maia Rodrigues. Ethics declarations Competing interests The authors declare no competing interests. Additional information Publisher's note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this license, visit Reprints and permissions About this article Cite this article Braga, M.V., de Oliveira, S.C., Vasconcelos, A.H.C. et al. Prevalence of sacroiliitis and acute and structural changes on MRI in patients with psoriatic arthritis. Sci Rep10, 11580 (2020). Download citation Received: 16 December 2019 Accepted: 25 June 2020 Published: 14 July 2020 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Subjects Medical research Psoriatic arthritis This article is cited by Sacroiliac and spine imaging in spondyloarthritis: Does phenotype or sex matter? Gabriel Caetano Pereira Natalia Pereira Machado Valderílio Feijó Azevedo Advances in Rheumatology (2024) How Are We Addressing Axial Psoriatic Arthritis in Clinical Practice? Xabier Michelena Clementina López-Medina Xavier Juanola Rheumatology and Therapy (2024) MRI in axial spondyloarthritis: understanding an ‘ASAS-positive MRI’ and the ASAS classification criteria Torsten Diekhoff Robert Lambert Kay Geert Hermann Skeletal Radiology (2022) Musculoskeletal Imaging for Dermatologists: Techniques in the Diagnosis and Management of Psoriatic Arthritis Alice B. Gottlieb Catherine Bakewell Joseph F. Merola Dermatology and Therapy (2021) Differentiating Psoriatic Arthritis from Osteoarthritis and Rheumatoid Arthritis: A Narrative Review and Guide for Advanced Practice Providers William Saalfeld Amanda M. Mixon Eileen J. Lydon Rheumatology and Therapy (2021) Download PDF Associated content Collection Psoriatic disease Sections Figures References Abstract Introduction Materials and methods Results Discussion References Acknowledgements Author information Ethics declarations Additional information Rights and permissions About this article This article is cited by Advertisement Figure 1 View in articleFull size image Figure 2 View in articleFull size image Lee, M. P. et al. Patterns of systemic treatment for psoriatic arthritis in the US: 2004–2015. Arthritis Care Res70, 791–796 (2018). ArticleGoogle Scholar Haroon, M. et al. Clinical and genetic associations of radiographic sacroiliitis and its different patterns in psoriatic arthritis. Clin Exp Rheumatol35, 270–276 (2017). PubMedGoogle Scholar Huynh, D. Q. & Kavanaugh, A. Psoriatic arthritis: Current therapy and future approaches. Rheumatology (Oxford)54, 20–28 (2014). ArticleGoogle Scholar Queiro, R. et al. Patients with psoriatic arthritis may show differences in their clinical and genetic profiles depending on their age at psoriasis onset. Clin. Exp. Rheumatol.30, 476–480 (2012). PubMedGoogle Scholar Chandran, V. et al. Axial psoriatic arthritis: Update on a longterm prospective study. J. Rheumatol.36, 2744–2750 (2019). ArticleGoogle Scholar Niccoli, L. et al. Frequency of iridocyclitis in patients with early psoriatic arthritis: A prospective, follow up study. Int. J. Rheum. Dis.15, 414–418 (2012). ArticleGoogle Scholar Baraliakos, X., Coates, L. C. & Braun, J. The involvement of the spine in psoriatic arthritis. Clin. Exp. Rheumatol.33, 31–35 (2015). Google Scholar McQueen, F., Lassere, M. & Østergaard, M. Magnetic resonance imaging in psoriatic arthritis: A review of the literature. Arthritis Res. Ther.8, 207 (2006). ArticleGoogle Scholar Taylor, W. et al. Classification criteria for psoriatic arthritis: Development of new criteria from a large international study. Arthritis Rheum.54, 2665–2673 (2006). ArticleGoogle Scholar Mease, P. J. Measures of psoriatic arthritis. Arthritis Care Res.63, 64–85 (2011). ArticleGoogle Scholar Rudwaleit, M. et al. The development of Assessment of SpondyloArthritis international Society classification criteria for axial spondyloarthritis (part II): Validation and final selection. Ann Rheum Dis.68, 777–783 (2009). ArticleCASGoogle Scholar McHugh, M. L. Interrater reliability: The kappa statistic. Biochem. Med. (Zagreb)22, 26–282 (2012). Google Scholar McHugh, M. L., Lambert, R. G. W., Østergaard, M. & Jaremko, J. L. Magnetic resonance imaging in rheumatology. Magn. Reson. Imaging Clin. N. Am.26, 599–613 (2018). ArticleGoogle Scholar Sengupta, R. et al. Short-term repeat magnetic resonance imaging scans in suspected early axial spondyloarthritis are clinically relevant only in HLA-B27-positive male subjects. J. Rheumatol.45, 202–205 (2018). ArticleGoogle Scholar Hayashi, M. et al. Superiority of magnetic resonance imaging over conventional radiography in the early diagnosis of psoriatic arthritis. J. Dermatol.44, e232–e233 (2017). ArticleGoogle Scholar Koo, J. Population-based epidemiologic study of psoriasis with emphasis on quality of life assessment. Dermatol. Clin.14, 485–496 (1996). ArticleCASGoogle Scholar Walsh, J. A. et al. Work productivity loss and fatigue inpsoriatic arthritis. J Rheumatol41, 1670–1674 (2004). ArticleGoogle Scholar Siba, P. et al. Management of psoriatic arthritis: Early diagnosis, monitoring of disease severity and cutting edge therapies. J. Autoimmun.76, 21–37 (2017). ArticleGoogle Scholar Ranza, R. et al. Prevalence of psoriatic arthritis in a large cohort of Brazilian patients with psoriasis. J. Rheumatol.42, 829–834 (2015). ArticleGoogle Scholar Mease, P. J. et al. Clinical characteristics, disease activity, and patient-reported outcomes in psoriatic arthritis patients with dactylitis or enthesitis: Results from the corrona psoriatic arthritis/spondyloarthritis registry. Arthritis Care Res.69, 1692–1699 (2017). ArticleCASGoogle Scholar Eshed, I., Hermann, K. G. A., Zejden, A. & Sudoł-Szopińska, I. Imaging to differentiate the various forms of seronegative arthritis. Semin. Musculoskelet. Radiol.22, 189–196 (2018). ArticleGoogle Scholar Sahin, N. et al. Is there a role for DWI in the diagnosis of sacroiliitis based on ASAS criteria?. Int. J. Clin. Exp. Med.8, 7544–7552 (2015). PubMedPubMed CentralGoogle Scholar Bredella, M. A. et al. MRI of the sacroiliac joints in patients with moderate to severe ankylosing spondylitis. Am. J. Roentgenol.187, 1420–1426 (2006). ArticleGoogle Scholar Williamsonn, L. et al. Clinical assessment of sacroiliitis and HLA-B27 are poor predictors of sacroiliiitis diagnosed by magnetic resonance imaging in psoriatic arthritis. Rheumatology (Oxford)43, 85–88 (2004). ArticleGoogle Scholar Muche, B. et al. Anatomic structures involved in early- and late-stage sacroiliitis in spondylarthritis: A detailed analysis by contrast-enhanced magnetic resonance imaging. Arthritis Rheum.48, 1374–1384 (2003). ArticleCASGoogle Scholar Bratu, V. A. et al. Do patients with skin psoriasis show subclinical axial inflammation on MRI of the sacroiliac joints and entire spine?. Arthritis Care Res. (Hoboken)71, 1109–1118 (2019). ArticleGoogle Scholar Kaçar, C. et al. Sacroiliac joint involvement in psoriasis. Rheumatol. Int.30, 1263–1266 (2010). ArticleGoogle Scholar Gladman, D. D. Clinical features and diagnostic considerations in psoriatic arthritis. Rheum. Dis. Clin. North Am.41, 569–579 (2015). ArticleGoogle Scholar Chandran, V., Tolusso, D. C., Cook, R. J. & Gladman, D. D. Risk factors for axial inflammatory arthritis in patients with psoriatic arthritis. J. Rheumatol.37, 809–815 (2010). ArticleGoogle Scholar Moshrif, A. et al. (2017) Subclinical enthesopathy in patients with psoriasis and its association with other disease parameters: A power Doppler ultrasonographic study. Eur. J. Rheumatol.4, 24–28 (2017). 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188349	https://satheeneet.iitk.ac.in/article/physics/physics-puzzles-and-brain-teasers/	Notifications Puzzles And Brain Teasers Physics Puzzles and Brain Teasers Physics of Motion Word Search The topic you’ve mentioned, “Physics of Motion Word Search,” seems to be a combination of two different concepts: “Physics of Motion” and “Word Search.” Let’s break them down separately. Physics of Motion: This is a fundamental concept in physics, often referred to as “kinematics.” It deals with the study of objects in motion, their speed, velocity, acceleration, and the forces acting upon them. The laws governing motion were first comprehensively presented by Sir Isaac Newton and are known as Newton’s Laws of Motion. They include: Newton’s First Law (Law of Inertia): An object at rest tends to stay at rest, and an object in motion tends to stay in motion, unless acted upon by an external force. Newton’s Second Law: The force acting on an object is equal to the mass of the object multiplied by its acceleration (F=ma). Newton’s Third Law: For every action, there is an equal and opposite reaction. Word Search: This is a type of puzzle game where a grid of letters is presented, and the player’s task is to find specific words within this grid. The words can be arranged horizontally, vertically, or diagonally, and can also be backwards or forwards. When you combine these two concepts, “Physics of Motion Word Search” would likely be a word search puzzle that includes terms related to the physics of motion. This could be an educational tool used to help students familiarize themselves with key terms and concepts in this area of physics. The words to be found might include terms like “velocity,” “acceleration,” “force,” “inertia,” “mass,” “gravity,” “friction,” and “momentum,” among others. Electricity Crossword The term “Electricity Crossword” likely refers to a crossword puzzle that is themed around the topic of electricity. This could be a fun and educational tool used to teach students about various concepts and terms related to electricity. Here’s a deeper explanation of some potential terms that might appear in such a crossword: Current: This is the flow of electric charge. It’s measured in amperes (A). There are two types of electric current: direct current (DC) and alternating current (AC). Voltage: Also known as electric potential difference, this is the force that pushes electric charge around a circuit. It’s measured in volts (V). Resistance: This is a measure of the difficulty to pass an electric current through a conductor. It’s measured in ohms (Ω). Conductor: A material that allows electric charge to move through it easily. Metals like copper and silver are good conductors. Insulator: A material that doesn’t allow electric charge to move through it easily. Rubber and glass are examples of insulators. Circuit: A closed path that electric current follows. Ohm’s Law: A fundamental concept in electricity, it states that the current passing through a conductor between two points is directly proportional to the voltage across the two points. Capacitor: A device used in an electric circuit that stores energy in an electric field. Inductor: A component in an electric circuit that stores energy in a magnetic field. Transformer: A device that increases or decreases the voltage of alternating current. Semiconductor: A material whose electrical conductivity is between that of a conductor and an insulator. Silicon is a common semiconductor material. Diode: A semiconductor device that allows current to flow in one direction only. Transistor: A semiconductor device used to amplify or switch electronic signals and electrical power. These are just a few examples of the terms related to electricity that might appear in a crossword puzzle. The goal of such a puzzle would be to help students learn and remember these terms and concepts. Physics Brain Teasers Physics brain teasers are puzzles or problems that challenge your understanding of physics concepts and principles. They are designed to test your ability to apply physics knowledge in a creative and critical way. They often involve real-world scenarios or hypothetical situations where you need to use physics to solve a problem or explain a phenomenon. Physics brain teasers can cover a wide range of topics, from classical mechanics to quantum physics. They might ask you to calculate the speed of a falling object, explain why the sky is blue, predict the behavior of particles in a quantum state, or determine the trajectory of a projectile. Here are a few examples of physics brain teasers: The Bullet and the Feather: If you were to drop a bullet and a feather at the same time from the same height in a vacuum, which would hit the ground first? This teaser tests your understanding of gravity and air resistance. The Boat and the Lake: If you have a boat floating in a lake and it has a heavy anchor on board, what happens to the water level of the lake if you throw the anchor overboard? This teaser tests your understanding of buoyancy and displacement. The Hot Air Balloon: A hot air balloon is in a closed room. If the balloon rises, does the room’s temperature increase, decrease, or stay the same? This teaser tests your understanding of thermodynamics and gas laws. The Quantum Cat: According to quantum mechanics, a cat in a box could be both alive and dead at the same time until someone opens the box to check. How is this possible? This teaser tests your understanding of quantum superposition and the observer effect. Solving physics brain teasers requires a good understanding of physics principles, logical thinking, and sometimes a bit of mathematical skill. They are a fun and challenging way to deepen your understanding of physics and improve your problem-solving abilities. A considerable amount of water is gushing out through a large pipe which narrows at the outlet. At which point does the water flow the fastest? The speed of water flow in a pipe is governed by the principle of continuity, which is a consequence of the conservation of mass. This principle states that the mass flow rate (the mass of fluid passing through a cross-section per unit time) must remain constant throughout a pipe if there is no loss or gain of fluid. In simpler terms, what goes in must come out. The mass flow rate is given by the product of the cross-sectional area of the pipe (A), the fluid density (ρ), and the fluid velocity (v). This can be expressed as ρAv = constant. In the case of water flowing through a pipe, the density of water remains constant. Therefore, the product of the area and velocity must remain constant. This means that if the cross-sectional area of the pipe decreases, the velocity of the water must increase to keep the product constant. So, in a pipe that narrows at the outlet, the water flow is fastest at the narrowest point, i.e., at the outlet. This is because the cross-sectional area at the outlet is smaller than at any other point in the pipe, and so the velocity of the water must be greater to ensure that the mass flow rate remains constant. This principle is also the basis for the operation of a Venturi meter, a device used to measure the flow rate of a fluid in a pipe. The meter works by creating a pressure difference as the fluid speeds up in a narrow section of the pipe, and this pressure difference can be used to calculate the flow rate. An adult man and his six-year-old daughter are swinging at the park. They are on separate, identical swings. The man has four times the mass of the child. Who swings faster? The speed at which a person swings on a swing set is not determined by their mass. This is due to the principle of the pendulum, which states that the period of a pendulum (the time it takes for one complete swing) is determined by the length of the pendulum, not its mass. This principle is derived from the physics of simple harmonic motion. A swing set is essentially a pendulum. When you sit on a swing and move back and forth, you are acting as the ‘bob’ or weight on the end of the pendulum. The ropes or chains that hold the swing up are the ‘arm’ of the pendulum. The period of a pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. As you can see, mass does not factor into this equation. Therefore, assuming that the man and his daughter are swinging on identical swings (meaning the length of the ‘pendulum’ is the same), and they both start from the same angle (meaning they both pull their swings back to the same height before letting go), they will swing at the same speed, regardless of their differing masses. However, the force exerted by the man on the swing will be greater due to his larger mass, and he will stretch the chains or ropes more than his daughter does. This could make his swing slightly longer, which would make his period slightly longer (meaning he swings slower). But this effect would likely be very small, especially on a well-built swing set. In conclusion, the man and his daughter, despite their difference in mass, will swing at approximately the same speed if they are on identical swings and start from the same angle. Why do astronauts feel light in space? Astronauts feel light in space due to the phenomenon of microgravity. When astronauts are in space, they are in a state of constant free fall towards the Earth, but they never reach it because of their horizontal velocity. This is similar to what happens when you throw an object horizontally - it falls towards the ground but also moves forward. If you throw it hard enough, it will keep falling towards the ground but never reach it because the Earth is curved and the object keeps missing it. This is essentially what an orbit is. In this state of constant free fall, astronauts feel as if they are weightless, or “light,” because there is no solid surface to stop their fall and create the reaction force we interpret as weight. This is not because there is no gravity in space. In fact, the force of gravity in low Earth orbit is almost as strong as it is on the Earth’s surface. The feeling of weightlessness is due to them continuously falling towards the Earth but never reaching it. This sensation is often referred to as zero gravity, but a more accurate term is microgravity, because the force of gravity is not actually zero in space. It’s just that the effects of gravity are not felt in the same way as they are on Earth. This is why astronauts can float around inside their spacecraft, and why they have to exercise regularly to prevent muscle and bone loss - their bodies are not experiencing the regular stresses and strains that come with moving against the force of gravity. How do ships float? Ships float based on the principle of buoyancy discovered by the Greek mathematician Archimedes. This principle, also known as Archimedes’ Principle, states that an object submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object. In the case of ships, they are designed to displace a large volume of water. Even though they are made of materials like steel, which is much denser than water, the shape of a ship causes it to displace a large amount of water before it becomes fully submerged. This displacement of water creates an upward buoyant force. When a ship is placed in water, it will sink into the water until the weight of the water it displaces is equal to the weight of the ship. If the ship weighs less than the maximum volume of water it can displace, it will float. If it weighs more, it will sink. The hull, or body of the ship, is designed to be hollow and contains air-filled spaces. This design increases the overall volume of the ship without significantly increasing its weight, allowing it to displace more water and thus float. The stability of the ship is also important. The center of gravity of the ship must be as low as possible. This is achieved by placing heavy components such as the engine and fuel at the bottom of the ship. This ensures that the ship remains upright and does not capsize. In summary, ships float because they are designed to displace a large volume of water, which creates enough upward buoyant force to counteract the weight of the ship. The design and distribution of weight within the ship also ensure its stability on the water. Learn how to make a Pattern Puzzle! Creating a pattern puzzle is a fun and educational activity that can help develop critical thinking and problem-solving skills. It involves creating a sequence or arrangement of objects, numbers, shapes, or colors that follow a certain rule or pattern. The goal is to figure out the pattern and predict what comes next or fill in the missing elements. Here’s a step-by-step guide on how to create a pattern puzzle: Choose the Type of Pattern: The first step in creating a pattern puzzle is to decide what type of pattern you want to use. This could be a simple repeating pattern (ABAB), an increasing pattern (ABCABC), a decreasing pattern (CBACBA), a numerical sequence (2, 4, 6, 8), or a more complex pattern involving shapes, colors, or other elements. Create the Pattern: Once you’ve chosen the type of pattern, the next step is to create the pattern. This involves arranging the elements in the chosen sequence or pattern. For example, if you’ve chosen a simple repeating pattern, you might arrange colored blocks in a sequence of red, blue, red, blue, and so on. Break the Pattern: After creating the pattern, the next step is to break the pattern at a certain point. This could involve removing one or more elements from the pattern, or leaving a space where an element should be. The goal is to create a puzzle that the solver must figure out by identifying the pattern and predicting what comes next. Provide Clues: Depending on the complexity of the pattern, you might need to provide some clues to help the solver figure out the pattern. This could involve providing a hint about the type of pattern (e.g., “This is a repeating pattern”), or giving a clue about the missing element (e.g., “The missing element is a color that is not already in the pattern”). Test the Puzzle: Finally, before presenting the puzzle to others, it’s a good idea to test it yourself or have someone else test it to make sure it’s solvable and the clues are helpful. If the puzzle is too easy or too hard, you might need to adjust the pattern or the clues. Creating a pattern puzzle can be a fun and challenging activity for both kids and adults. It’s a great way to develop critical thinking and problem-solving skills, and it can also be a fun way to learn about different types of patterns and sequences. 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188350	https://math.stackexchange.com/questions/1953187/on-function-names	notation - On function names - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more On function names Ask Question Asked 8 years, 11 months ago Modified8 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This may seem like a stupid question, but with regards to functions is f(x)f(x) a function name? We write f(x)f(x) for a function, where the actual algebraic formula on the right is the "machine". If we wrote something like sin x sin⁡x or x+3 x+3 without using y=…y=… or f(x)=…f(x)=…, could we say that something like sin x sin⁡x or x+3 x+3 is a function? functions notation Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 4, 2016 at 8:35 Parcly Taxel 106k 21 21 gold badges 123 123 silver badges 209 209 bronze badges asked Oct 4, 2016 at 8:26 MathLearnerMathLearner 143 7 7 bronze badges 1 4 f f is the name the function. f(x)f(x) is the result of f f evaluated at x x.BHT –BHT 2016-10-04 08:30:26 +00:00 Commented Oct 4, 2016 at 8:30 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Yes. Formally speaking, a function is a relation between an input value and an output value, such that one input gives no more than one output. There are different ways to provide function definitions, one of them being by the use of symbolic expressions, such as sin(x)sin⁡(x) or x+3 x+3. In these, sin sin denotes a "predefined" function, while x+3 x+3 uses an operator (++) which is actually a predefined function of two arguments. You can indeed assign a name to functions, for instance by means of the notation f(x):=x+3 f(x):=x+3 or f(x):=sin(x).f(x):=sin⁡(x). Then you can refer to the function f f, and its evaluation at x x, namely f(x)f(x). But the RHS expressions themselves are functions (of the variable(s) they mention), even if they don't get a specific name. x x denotes both the variable x x and the so-called "identity" function, which trivially maps x x to x x. A constant such as 3 3 also denotes a "constant" function, which maps all values of the variable (specified from context) to 3 3. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 4, 2016 at 8:44 answered Oct 4, 2016 at 8:38 user65203 user65203 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. As Ethan said, f f is the name of the function and f(x)f(x) is the number that f f gives you as output when you give x x to f f as input. But, people are frequently sloppy about this. In particular, you will often hear people refer to "the function sin(x)sin⁡(x)" or "the function x+3 x+3", for example, because it would be cumbersome to make a more correct statement such as "the function f:R→R f:R→R defined by f(x)=x+3 f(x)=x+3 for all x∈R x∈R". Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 4, 2016 at 8:37 littleOlittleO 54.2k 8 8 gold badges 107 107 silver badges 187 187 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The name of the function is f f. But there's another convention that comes into play, namely that when speaking about a function, instead of giving it a name we can just state an expression that tells how to compute the value of the function we're speaking. Thus, we can speak about "the function x 2+3 x x 2+3 x" for example, without giving it an explicit name. Now, since "f(x)f(x)" is actually an expression that tells us to compute the value of the function named f f, saying "f(x)f(x)" is also, by the above convention, a way to speak about that function. "f(x)f(x)" is not the name of the function, but is nevertheless a way to refer to it. The take-home lesson from this is that in practice whether people say f f or f(x)f(x) (when it's clear that they're speaking about a function, rather than the value of that function at some already known x x), does not make any difference to what they mean. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 4, 2016 at 8:43 hmakholm left over Monicahmakholm left over Monica 292k 25 25 gold badges 441 441 silver badges 706 706 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functions notation See similar questions with these tags. 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SponsoredTodocama - Set of 2 Large Bath Towels 100% Cotton 550 GSM Extra Soft Highly Absorbent and Quick Dry 70 x 140 cm Light Grey -------------------------------------------------------------------------------------------------------------------------- 4.5 out of 5 stars2,575 Price, product page€20.48€20.48 Save more with Subscribe & Save Save 5% on any 4 or more €33.98 delivery Fri 10 Oct Or €37.80 delivery Fri 3 Oct More results +2 AZZORO Gaveno Cavalia Set of 10 Toronto Cotton Towels with 4 Faces, 4 Hands, 2 Bath Towels, White, 28 x 23 x 20 cm, 10 Units ---------------------------------------------------------------------------------------------------------------------------- 4.2 out of 5 stars251 Price, product page€44.67€44.67(€4.47€4.47/count) Save more with Subscribe & Save €32.81 delivery Sat 11 Oct Or €36.52 delivery Fri 3 Oct Add to basket Top Towel - 4 Bath Towels - White Hotel Towels - Large Bath Towels - 100% Cotton - White - 50 x 100 cm ------------------------------------------------------------------------------------------------------ 3.9 out of 5 stars29 Price, product page€17.33€17.33(€8.67€8.67/count)RRP: €22.21 RRP: €22.21€22.21 Save more with Subscribe & Save €31.34 delivery Wed 15 Oct Or €34.90 delivery Fri 3 Oct Add to basket +11 Pamposh Satin Stripe Duvet Cover Set Double Duvet Cover Set Hotel Premium Duvet Cover Bedding Set Super Soft Hypoallergenic Microfiber Bedding ---------------------------------------------------------------------------------------------------------------------------------------------- Options: 4 sizes 4 sizes 4.2 out of 5 stars365 Safer chemicals Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX STANDARD 100 Learn more about OEKO-TEX STANDARD 100 OEKO-TEX STANDARD 100 OEKO-TEX® STANDARD 100 certified products require every component of a textiles production including all thread, buttons, and trims to be tested against a list of more than 1,000 regulated and unregulated chemicals which may be harmful to human health. The assessment process is globally standardised, independently conducted and updated at least once per year based on new scientific information or regulatory requirements. Certification number 2019OK1750 Discover more products with sustainability features.Learn more See options No featured offers available €19.97(1 new offer) +8 Gabel 00 00614 040 060 43 Hand Towels, White, 40 x 60 cm -------------------------------------------------------- Options: 2 sizes 2 sizes 4.4 out of 5 stars1,959 Price, product page€10.57€10.57(€3.52€3.52/count)Median: €11.18 Median: €11.18€11.18 Save more with Subscribe & Save Save 5% on any 4 or more €17.70 delivery Sat 11 Oct Or €21.01 delivery Fri 3 Oct Safer chemicals Made in Italy Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX STANDARD 100 Learn more about OEKO-TEX STANDARD 100 OEKO-TEX STANDARD 100 OEKO-TEX® STANDARD 100 certified products require every component of a textiles production including all thread, buttons, and trims to be tested against a list of more than 1,000 regulated and unregulated chemicals which may be harmful to human health. The assessment process is globally standardised, independently conducted and updated at least once per year based on new scientific information or regulatory requirements. Certification number 085779.O Discover more products with sustainability features.Learn more Made in Italy Shop products that have been wholly produced or have undergone their last substantial transformation in Italy. Discover more about “Made in Italy”, a label synonymous throughout the world with refined materials, attention to detail, and creativity.Learn more Add to basket +22 PETTI Artigiani Italiani - 3-Piece Soft and Breathable Single Bed Set with Top Sheet, Fitted Bottom Sheet and 1 Pillowcase Teal ------------------------------------------------------------------------------------------------------------------------------- Options: 5 sizes 5 sizes 3.8 out of 5 stars755 Price, product page€24.51€24.51 Save 5% on any 4 or more €25.77 delivery Thu 16 Oct Or €36.19 delivery Fri 3 Oct Made in Italy Made in Italy Shop products that have been wholly produced or have undergone their last substantial transformation in Italy. Discover more about “Made in Italy”, a label synonymous throughout the world with refined materials, attention to detail, and creativity.Learn more Add to basket +14 Komfortec Set of 4 Hand Towels 50 x 100 cm and 4 Bath Towels 70 x 140 cm, Terry Towelling and Large Towel Set, 100% Cotton, Soft, Quick Drying, Anthracite ---------------------------------------------------------------------------------------------------------------------------------------------------------- 4.1 out of 5 stars1,352 Price, product page€43.36€43.36(€10.84€10.84/count) Save more with Subscribe & Save €48.06 delivery Sat 11 Oct Or €59.21 delivery Fri 3 Oct Safer chemicals Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX STANDARD 100 Learn more about OEKO-TEX STANDARD 100 OEKO-TEX STANDARD 100 OEKO-TEX® STANDARD 100 certified products require every component of a textiles production including all thread, buttons, and trims to be tested against a list of more than 1,000 regulated and unregulated chemicals which may be harmful to human health. The assessment process is globally standardised, independently conducted and updated at least once per year based on new scientific information or regulatory requirements. Certification number 2019OK1134 Discover more products with sustainability features.Learn more Add to basket Herding Cot Bed Linen Teddy Bear Pillowcase 35 x 40 cm with Hotel Cover, Duvet Cover 80 x 80 cm with Hotel Cover, 100% Cotton, Renforcé --------------------------------------------------------------------------------------------------------------------------------------- 4.6 out of 5 stars213 Price, product page€15.53€15.53 €21.09 delivery Fri 10 Oct Or €28.07 delivery Fri 3 Oct Safer chemicals Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX STANDARD 100 Learn more about OEKO-TEX STANDARD 100 OEKO-TEX STANDARD 100 OEKO-TEX® STANDARD 100 certified products require every component of a textiles production including all thread, buttons, and trims to be tested against a list of more than 1,000 regulated and unregulated chemicals which may be harmful to human health. The assessment process is globally standardised, independently conducted and updated at least once per year based on new scientific information or regulatory requirements. Certification number A96-0065 Discover more products with sustainability features.Learn more Add to basket Egyptian Cotton "800'' TC Hotel" Plain Dyed Bedding Set (Single Duvet Cover) ---------------------------------------------------------------------------- 4.2 out of 5 stars26 Price, product page€46.76€46.76 Only 2 left in stock. Add to basket 100% Egyptian Cotton 400 Thread Count Duvet Cover Set Hotel Bedding White Double -------------------------------------------------------------------------------- Options: 4 sizes 4 sizes 4.2 out of 5 stars104 Price, product page€53.35€53.35 Only 9 left in stock. Add to basket OSVINO Single Duvet Cover 135 x 200 cm and 1 Pillowcases 50 x 75 cm White and Blue Microfibre Duvet Cover Set Hotel Stripe Soft Comfortable ------------------------------------------------------------------------------------------------------------------------------------------- Options: 3 sizes 3 sizes 4.0 out of 5 stars2,720 See options No featured offers available €38.41(1 new offer) +5 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Todocama - 2 x Large Bath Towels Set, 100% Cotton, 550 GSM Extra Soft, Highly Absorbent and Quick Drying - Beige 70 x 140 cm ---------------------------------------------------------------------------------------------------------------------------- Options: 3 sizes 3 sizes 4.5 out of 5 stars2,575 Price, product page€20.48€20.48 Save more with Subscribe & Save €33.98 delivery Fri 10 Oct Or €37.80 delivery Fri 3 Oct Add to basket +8 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement PETTI Artigiani Italiani - Double Bed Sheet Set with Double Satin Frill, 4 Pieces, 180 x 200 cm, Percale Cotton Sheets - Fitted Sheet, Top Sheet and 2 Pillowcases, Cream ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Options: 3 sizes 3 sizes 3.5 out of 5 stars33 Price, product page€65.49€65.49 RRP: €73.69 RRP: €73.69€73.69 €36.52 delivery Fri 3 Oct Only 9 left in stock (more on the way). Made in Italy Made in Italy Shop products that have been wholly produced or have undergone their last substantial transformation in Italy. Discover more about “Made in Italy”, a label synonymous throughout the world with refined materials, attention to detail, and creativity.Learn more Add to basket +3 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Top Towel - Set of 2 Towels and 2 Bath or Shower Towels - Hand Towel Set - 100% Cotton - 400gsm ----------------------------------------------------------------------------------------------- 3.3 out of 5 stars6 Price, product page€24.48€24.48(€12,240.00€12,240.00/kg) Save more with Subscribe & Save €32.81 delivery Sat 11 Oct Or €36.52 delivery Fri 3 Oct Add to basket Disposable Bed Sheets Travel King Size Hotel Disposable Bed Sheets with Duvet Cover and Pillow Case Disposable Bedding for Business Travel Hotel Essentials (4PCS) (1 Pack) --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 4.0 out of 5 stars135 Currently unavailable. See options Funky Gadgets Beige King Size Duvet Cover Set Satin Microfiber Luxury Hotel Bedding Set with Pillow Cases Breathable Ultra Soft Reversible ------------------------------------------------------------------------------------------------------------------------------------------ Options: 3 sizes 3 sizes 4.5 out of 5 stars37 See options No featured offers available €24.57(2 used & new offers) Sapphire Collection Egyptian Cotton 800'' TC Hotel Bedding Set White (Super King) --------------------------------------------------------------------------------- Options: 3 sizes 3 sizes 4.5 out of 5 stars869 Price, product page€72.86€72.86 Add to basket Herding Little Dreamer Bed Linen Set, Made in Green by OEKO TEX, Pillowcase Approx. 40 x 60 cm, Duvet Cover Approx. 135 x 100 cm, Button Closure, 100% Cotton/Flannel, White/Beige ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 4.7 out of 5 stars10 Price, product page€19.63€19.63 5% off coupon applied Save 5% with voucher €18.77 delivery Sat 11 Oct Or €21.16 delivery Fri 3 Oct Safer chemicals +2 more Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX MADE IN GREEN Learn more about OEKO-TEX MADE IN GREEN OEKO-TEX MADE IN GREEN Made In Green by Oeko-Tex certified products have been manufactured using processes with a reduced environmental impact and under socially responsible working conditions. By using a unique product ID, each product can be easily traced by entering the certification number on the Label Check on the Made in Green website, where more information about the production facilities is provided. Certification number M21LKECH3 ### Manufacturing practices Manufactured using processes that reduce the risk of negative environmental impact. As certified by OEKO-TEX MADE IN GREEN Learn more about OEKO-TEX MADE IN GREEN OEKO-TEX MADE IN GREEN Made In Green by Oeko-Tex certified products have been manufactured using processes with a reduced environmental impact and under socially responsible working conditions. By using a unique product ID, each product can be easily traced by entering the certification number on the Label Check on the Made in Green website, where more information about the production facilities is provided. Certification number M21LKECH3 ### Worker well-being Manufactured on farms or in facilities that protect the rights and/or health of workers. As certified by OEKO-TEX MADE IN GREEN Learn more about OEKO-TEX MADE IN GREEN OEKO-TEX MADE IN GREEN Made In Green by Oeko-Tex certified products have been manufactured using processes with a reduced environmental impact and under socially responsible working conditions. By using a unique product ID, each product can be easily traced by entering the certification number on the Label Check on the Made in Green website, where more information about the production facilities is provided. Certification number M21LKECH3 Discover more products with sustainability features.Learn more Add to basket +4 IHOMAGIC Laundry Basket with Wheels 40L – Slim Folding Laundry Basket, Rectangular Standing Laundry Basket, Thin Laundry Buckets for Hotel, Family Storage (Beige) ------------------------------------------------------------------------------------------------------------------------------------------------------------------ 4.4 out of 5 stars2,612 See options No featured offers available €33.38(1 new offer) +1 Rehan & Co Double Fitted Sheet Grey 100% Egyptian Cotton 300 Thread Count Breathable Hotel Bedding 30CM Deep ------------------------------------------------------------------------------------------------------------ Options: 4 sizes 4 sizes 4.2 out of 5 stars25 See options No featured offers available €22.78(1 new offer) N/X Biancheria da letto per hotel, casa degli ospiti, ingresso e colazione, kit di stampa in cotone da 4 pezzi, letto da 1,2 m, caffetteria da 3 pezzi ------------------------------------------------------------------------------------------------------------------------------------------------------ See options No featured offers available €13.49(1 new offer) Herding babybest Flower Bunny Bed Linen Set Pillowcase 40 x 60 cm Duvet Cover Approx. 100 x 135 cm with Button and Hotel Flap 100% Cotton Flannelette ----------------------------------------------------------------------------------------------------------------------------------------------------- 4.2 out of 5 stars3 Price, product page€19.63€19.63 €20.74 delivery Sat 11 Oct Or €23.39 delivery Fri 3 Oct Add to basket OSVINO Floral Bed Runner Bed Linen Scarf Anti--Fade Microfibre Bed Runner Bed Decoration for Home Hotel ------------------------------------------------------------------------------------------------------- Floral Options: 4 sizes 4 sizes 4.6 out of 5 stars87 See options No featured offers available €35.99(1 new offer) 4 Pcs Premium Disposable Bedding Set Includes One Flat Sheet, Duvet Cover, Two Pillow Shams Portable, Ready to Use, Breathable for Hotel, Camping, Travel, Hospital ------------------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€20.86€20.86 Add to basket Generico Carrello per la Biancheria Resistente con Telaio in Acciaio Inossidabile - Cesto per la Biancheria Oxford Impermeabile su Ruote, Carrello Portatile e Cesto per la Biancheria per casa, Hotel ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ See options No featured offers available €313.99(1 new offer) +7 other colours/patterns Hygienic Bedding Set Duvet Cover and Pillow Cases Breathable Durable Hotel and Guest House Disposable Bedding Set ----------------------------------------------------------------------------------------------------------------- Currently unavailable. See options +12 IHOMAGIC Set of 2 Large Travel Laundry Bag with Drawstring, Foldable Dirty Laundry Bag, Portable Dirty Laundry Bag with Smile for Hotel, Bedroom, Dark Blue ----------------------------------------------------------------------------------------------------------------------------------------------------------- 4.5 out of 5 stars3,924 Price, product page€22.63€22.63(€11.32€11.32/count) Add to basket +10 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Caleffi Cotton Bidet and Face Towel Set, Wash-resistant Towels, Includes 1 Face Towel 55 x 100 cm and 1 Guest Bidet Towel 40 x 60 cm, 2 Piece Towels, Anthracite ---------------------------------------------------------------------------------------------------------------------------------------------------------------- 4.6 out of 5 stars46 100+ bought in past month Price, product page€10.98€10.98 RRP: €11.47 RRP: €11.47€11.47 Save more with Subscribe & Save Save 5% on any 4 or more €19.68 delivery Sat 11 Oct Or €23.53 delivery Fri 3 Oct Add to basket +124 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement PETTI Artigiani Italiani - Cotton Sheets with Pillowcases in Digital Print, Double Bed Sheet, Swans, 100% Made in Italy ----------------------------------------------------------------------------------------------------------------------- Options: 5 sizes 5 sizes 3.8 out of 5 stars359 Price, product page€28.61€28.61 RRP: €32.70 RRP: €32.70€32.70 €32.81 delivery Wed 15 Oct Or €36.52 delivery Fri 3 Oct Only 7 left in stock (more on the way). Made in Italy Made in Italy Shop products that have been wholly produced or have undergone their last substantial transformation in Italy. Discover more about “Made in Italy”, a label synonymous throughout the world with refined materials, attention to detail, and creativity.Learn more Add to basket +5 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Todocama 4-Piece Sheet Set, Fitted Sheet - Top - Two Pillowcases 50 x 80 cm (Bed 105-105 x 190/200 cm, White) ------------------------------------------------------------------------------------------------------------- Options: 7 sizes 7 sizes 4.2 out of 5 stars3,965 Price, product page€18.02€18.02 Median: €19.26 Median: €19.26€19.26 €24.99 delivery Sat 11 Oct Or €33.48 delivery Fri 3 Oct Add to basket +5 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Todocama - Thin Quilted Bedspread with Diamond Pattern for Spring, Summer, Autumn and Winter, 100% Extra Soft Microfibre, Multi-purpose Bedspread, Bed 105 - 200 x 260 cm, White -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Geometric Options: 5 sizes 5 sizes 4.3 out of 5 stars680 Price, product page€23.76€23.76 Median: €25.40 Median: €25.40€25.40 €32.81 delivery Wed 15 Oct Or €36.52 delivery Fri 3 Oct Add to basket Herding Babybest Hot Air Balloon Bed Linen Pillowcase Approx. 40 x 60 cm Duvet Cover Approx. 100 x 135 cm with Hotel Zipper and Closure, 100% Cotton/Renforcé ------------------------------------------------------------------------------------------------------------------------------------------------------------- Options: 2 sizes 2 sizes 4.5 out of 5 stars2 Price, product page€19.63€19.63 €20.29 delivery Wed 15 Oct Or €27.41 delivery Fri 3 Oct Only 3 left in stock (more on the way). Add to basket Set di biancheria da letto non tessuto per hotel e viaggi include lenzuolo copripiumino e federa 3/4Pieces Set biancheria da letto in tessuto non tessuto --------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€7.89€7.89 Save 5% at checkout FREE delivery 27 Oct - 5 Nov Add to basket Mirweon Set di biancheria da letto per hotel, 3/4 pezzi, include lenzuolo, copripiumino e federa per viaggi, viaggi d'affari, biancheria da letto morbida e confortevole ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ Price, product page€8.16€8.16 Save 5% at checkout FREE delivery 9 - 31 Oct Add to basket Qsvbeeqj Set di biancheria da letto da viaggio in tessuto non tessuto per hotel e ostelli include lenzuolo copripiumino federa morbida e confortevole biancheria da letto ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€7.82€7.82 €7.50 delivery 22 - 30 Oct Add to basket Set di biancheria da letto in tessuto non tessuto per hotel e viaggi, include lenzuolo copripiumino e federa 3/4 pezzi letto usa e getta ---------------------------------------------------------------------------------------------------------------------------------------- Price, product page€9.06€9.06 Save 8% at checkout FREE delivery 9 - 30 Oct Add to basket +8 hermet Winner Microfibre Double Bed Sheet Set, White, Top Sheet, Fitted Sheet, Pair of Pillowcases -------------------------------------------------------------------------------------------------- Options: 3 sizes 3 sizes 3.9 out of 5 stars1,192 50+ bought in past month Safer chemicals Sustainability features This product has sustainability features recognised by trusted certifications. ### Safer chemicals Made with chemicals safer for human health and the environment. As certified by OEKO-TEX STANDARD 100 Learn more about OEKO-TEX STANDARD 100 OEKO-TEX STANDARD 100 OEKO-TEX® STANDARD 100 certified products require every component of a textiles production including all thread, buttons, and trims to be tested against a list of more than 1,000 regulated and unregulated chemicals which may be harmful to human health. The assessment process is globally standardised, independently conducted and updated at least once per year based on new scientific information or regulatory requirements. Certification number 2018OK0249 Discover more products with sustainability features.Learn more See options No featured offers available €24.99(2 new offers) Kalttoy Set di biancheria da letto in tessuto non tessuto per hotel e viaggi, include copripiumino 3/4 pezzi ------------------------------------------------------------------------------------------------------------ Price, product page€8.29€8.29 FREE delivery 15 Oct - 5 Nov Add to basket Herding Babybest Safari Flannel Bedding Set, Pillowcase 40 x 60 cm with Hotel Fold, Duvet Cover 100 x 135 cm with Button Closure, Cotton/Flannel ------------------------------------------------------------------------------------------------------------------------------------------------ 3.0 out of 5 stars1 Price, product page€19.63€19.63 €20.74 delivery Wed 15 Oct Or €23.39 delivery Fri 3 Oct Only 2 left in stock. Add to basket Prasacco 20 Pcs Disposable Pillow Covers Non-Woven Pillow Covers for Travel Hotel Bedroom Home Bedding (81 x 50cm) ------------------------------------------------------------------------------------------------------------------ 4.0 out of 5 stars13 Price, product page€18.93€18.93(€0.95€0.95/count) Add to basket VICASKY 1 x Disposable Adult Sleeping Bag, Portable Bedding for Hotels and Travel, Protects Dirt, Offers -------------------------------------------------------------------------------------------------------- Price, product page€10.99€10.99 €5.61 delivery 8 - 29 Oct Add to basket Herding Biancheria da letto per bambini babybest, Cuddle Crew, federa 40x60 cm circa, copripiumino 100x135 cm circa, con fascia e cerniera hotel, 100% cotone/Renforcé ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- 5.0 out of 5 stars3 Price, product page€19.63€19.63 €19.18 delivery Sat 11 Oct Or €27.15 delivery Fri 3 Oct Add to basket Sweeaau Set di biancheria da letto usa e getta portatile in tessuto non tessuto bianco, copriletto, trapunta, doppia federa, per viaggi, affari, hotel, viaggi, viaggi, federe usa e getta ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ Price, product page€10.30€10.30 Save 5% at checkout FREE delivery 27 Oct - 5 Nov Add to basket Lussuoso runner da letto ricamato, con bandiera, per hotel, casa, matrimonio, biancheria da letto, asciugamano, federa (stile 1,50 x 180 cm) -------------------------------------------------------------------------------------------------------------------------------------------- Currently unavailable. See options CHAKAE Set di biancheria da letto per hotel, 3/4 pezzi, include lenzuolo, copripiumino e federa per viaggi e viaggi d'affari, soluzione temporanea per biancheria da letto -------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€11.99€11.99 Save 5% at checkout €5.99 delivery 22 - 31 Oct Add to basket Xuancai Set di biancheria da letto usa e getta portatile in tessuto non tessuto bianco, copriletto, trapunta, doppia federa, per viaggi, affari, hotel, viaggi, viaggi, federa usa e getta da viaggio ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€11.85€11.85 Save 5% on any 2 or more €1 delivery 9 - 31 Oct Add to basket Obotsnoi Set di biancheria da letto usa e getta portatile in tessuto non tessuto bianco, copriletto, trapunta, doppia federa, per viaggi, affari, hotel, viaggi, viaggi, federe usa e getta mediche --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€11.27€11.27 Save 10% on any 2 or more FREE delivery 22 - 30 Oct Add to basket Erwin Müller Hotel Linen Pillow Case Tobago with Envelope Closure, Cotton, Sekt, 65 cm x 100 cm ----------------------------------------------------------------------------------------------- 5.0 out of 5 stars1 Currently unavailable. See options Marmoday Asciugamani Viso Capelli Biancheria Da Bagno Asciugatura Rapida Morbido Altamente Assorbente Asciugamani Set Salone Casa Hotel Spa 4 Pack -------------------------------------------------------------------------------------------------------------------------------------------------- Currently unavailable. See options +1 other colour/pattern Set di biancheria da letto semplice soft comodo chimico in fibra sottile artigianato foglio letto set per la casa di hotel d'affari (Bianco) -------------------------------------------------------------------------------------------------------------------------------------------- Price, product page€33.81€33.81 Save 5% at checkout €10.97 delivery 9 - 31 Oct Add to basket GBZXANG Runner da letto con motivo a righe, lussuoso, multicolore, per hotel, casa, biancheria da letto, asciugamano a coda (G, 45 x 45 cm, federa 2 pezzi) ----------------------------------------------------------------------------------------------------------------------------------------------------------- Currently unavailable. See options SUMMITDRAGON Set di biancheria da letto da viaggio, con copripiumino e federa per un facile utilizzo, set da 4 pezzi usa e getta per hotel ------------------------------------------------------------------------------------------------------------------------------------------ Price, product page€15.33€15.33 Save 5% at checkout €1.99 delivery 9 - 31 Oct Add to basket KASFDBMO Non-woven Bedding Set Hotel Travel Bedding Set Includes Duvet Cover Sheet and Pillowcase 4 Piece Set ------------------------------------------------------------------------------------------------------------- Price, product page€15.06€15.06 Save 5% at checkout €0.50 delivery 27 Oct - 4 Nov Add to basket +9 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Caleffi 3X Large Bath Towel Set, Size 55x105cm, 100% Cotton - Soft and Absorbent, Face and Daily Use, High Quality Shower Towel, Large Towel -------------------------------------------------------------------------------------------------------------------------------------------- 4.2 out of 5 stars57 Price, product page€15.65€15.65(€5.22€5.22/count) Save 5% on any 4 or more €20.38 delivery Sat 11 Oct Or €29.86 delivery Fri 3 Oct Add to basket +26 other colours/patterns Sponsored Sponsored You’re seeing this ad based on the product’s relevance to your search query. Learn more about this advertisement Single Bed Sheet Set Natural Colour Cotton, Made in Italy – Complete Bed Set, Fitted Sheet, Top Sheet 90 x 200 cm and Pillowcase with Inner Flap, Petroleum/Bottle ------------------------------------------------------------------------------------------------------------------------------------------------------------------ Options: 3 sizes 3 sizes 4.2 out of 5 stars1,587 Price, product page€25.98€25.98 Median: €26.96 Median: €26.96€26.96 €22.47 delivery Made in Italy Made in Italy Shop products that have been wholly produced or have undergone their last substantial transformation in Italy. Discover more about “Made in Italy”, a label synonymous throughout the world with refined materials, attention to detail, and creativity.Learn more Add to basket This advertisement showcases a versatile cleaning brush with specialized attachments designed for hard-to-reach areas. The tool effectively cleans sink drains, toilet hinges, tile grout, and kitchen fixtures, removing built-up dirt and grime with precision. Its slim design allows access to narrow spaces that traditional cleaning tools cannot reach. Debug info copied. SponsoredSxhyf Cleaning Brushes - Household Cleaning Brush, Multifunctional Gap Cleaning Brush, Cleaning Tools for Home, Kitchen, Bathroom, Window (Black) ------------------------------------------------------------------------------------------------------------------------------------------------- 4.4 out of 5 stars15,998 1K+ bought in past month Previous 1 2 3 7 Next Brands related to your search Sponsored Soft, durable towels, 100% cotton Soft, durable towels, 100% cotton ZOLLNER shower towels - Hotel quality ZOLLNER shower towels - Hotel quality Transform your bathroom with our rug sets Transform your bathroom with our rug sets Do you need help? 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188352	https://flexbooks.ck12.org/cbook/ck-12-algebra-ii-with-trigonometry-concepts/section/8.11/primary/lesson/solving-logarithmic-equations-alg-ii/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 8.11 Solving Logarithmic Equations Written by:Lori Jordan \| Kate Dirga Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 "I'm thinking of a number," you tell your best friend. "The number I'm thinking of satisfies the equation log10x2−logx=3." What number are you thinking of? Solving Logarithm Equations A logarithmic equation has the variable within the log. To solve a logarithmic equation, you will need to use the inverse property, blogbx=x, to cancel out the log. Let's solve the following logarithmic equations. log2(x+5)=9 There are two different ways to solve this equation. The first is to use the definition of a logarithm. log2(x+5)=929=x+5512=x+5507=x The second way to solve this equation is to put everything into the exponent of a 2, and then use the inverse property. 2log2(x+5)=29x+5=512x=507 Make sure to check your answers for logarithmic equations. There can be times when you get an extraneous solution.To check, plug in 507 for x : log2(507+5)=9→log2512=9 3ln(−x)−5=10 First, add 5 to both sides and then divide by 3 to isolate the natural log. 3ln(−x)−5=103ln(−x)=15ln(−x)=5 Recall that the inverse of the natural log is the natural number. Therefore, everything needs to be put into the exponent of e in order to get rid of the log. eln(−x)=e5−x=e5x=−e5≈−148.41 Checking the answer, we have 3ln(−(−e5))−5=10→3lne5−5=10→3⋅5−5=10 log5x+log(x−1)=2 Condense the left-hand side using the Product Property. log5x+log(x−1)=2log[5x(x−1)]=2log(5x2−5x)=2 Now, put everything in the exponent of 10 and solve for x. 10log(5x2−5x)=1025x2−5x=1005x2−5x−100=0x2−x−20=0(x−5)(x+4)=0x=5,−4 Now, check both answers. log5(5)+log(5−1)=2log5(−4)+log((−4)−1)=2log25+log4=2 log(−20)+log(−5)=2log100=2 -4 is an extraneous solution. In the step log(−20)+log(−5)=2, we cannot take the log of a negative number, therefore -4 is not a solution. 5 is the only solution. Examples Example 1 Earlier, you were asked to determine what number you are thinking of if the number satisfies the equation log10x2−logx=3. We can rewrite log10x2−logx=3 as log10x2x=3 and solve for x. log10x2x=3log10x=310log10x=10310x=1000x=100 Therefore, the number you are thinking of is 100. Example 2 Solve: 9+2log3x=23. Isolate the log and put everything in the exponent of 3. 9+2log3x=232log3x=14log3x=7x=37=2187 Example 3 Solve: ln(x−1)−ln(x+1)=8. Condense the left-hand side using the Quotient Rule and put everything in the exponent of e. ln(x−1)−ln(x+1)=8ln(x−1x+1)=8x−1x+1=e8x−1=(x+1)e8x−1=xe8+e8x−xe8=1+e8x(1−e8)=1+e8x=1+e81−e8≈−1.0007 Checking our answer, we get ln(−1.0007−1)−ln(−1.0007+1)=8, which does not work because you cannot take the log of a negative number. Therefore, there is no solution for this equation. Example 4 Solve: 12log5(2x+5)=2. Multiply both sides by 2 and put everything in the exponent of a 5. 12log5(2x+5)=2log5(2x+5)=45log5(2x+5)=542x+5=6252x=620x=310 Review Use properties of logarithms and a calculator to solve the following equations for x. Round answers to three decimal places and check for extraneous solutions. log2x=15 log12x=2.5 log9(x−5)=2 log7(2x+3)=3 8ln(3−x)=5 4log33x−log3x=5 log(x+5)+logx=log14 2lnx−lnx=0 3log3(x−5)=3 23log3x=2 5logx2−3log1x=log8 2lnxe+2−lnx=10 2log6x+1=log6(5x+4) 2log12x+2=log12(x+10) 3log23x−log2327=log238 Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)29/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. 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188353	https://www.cjkt.com/chapter/play_video?video_id=926	智慧校园平台 [教师去登录] \| \| \| \| 未读通知全部标为已读 \| APP端下载 \| 首页初中数学平面几何初步—线平面几何初步—角相交线与平行线三角形初步全等三角形等腰三角形等边三角形直角三角形和勾股定理勾股定理的应用解直角三角形四边形与平行四边形特殊平行四边形相似三角形圆的基本性质圆的相关图形及性质圆中的角与圆有关的位置关系圆与圆的位置关系实数上—有理数的相关概念有理数的加减有理数的乘除有理数的乘方与混合运算实数下—平方根与立方根代数式及整式的加减幂运算及整式的乘除因式分解分式分式方程应用二次根式一元一次方程一元一次方程应用二元一次方程（组）不等式与不等式组一元一次不等式应用题平面直角坐标系一次函数一元二次方程反比例函数二次函数数据的收集与整理数据的分析概率与统计中考数学思想方法上中考数学思想方法下图形的变换面积问题与面积法压轴题—动点产生的定值与最值经典语文光影魔术作文系列上光影魔术作文系列下中考暴力作文写作传世名作初体验人物之百态人生看历史风云变识文体品佳作品味诸子百家铭记美丽河山虚词上虚词下性格稳定的实词比较变态的实词变色龙实词之名词变色龙实词之动词一变色龙实词之动词二变色龙实词之动词三文言文翻译古代诗歌鉴赏趣味英语零基础超级音标场景词汇—核心名词一场景词汇—核心名词二场景词汇—核心动词场景词汇—核心形容词神奇动物在哪里？谁偷窥了我的日常？零点奇迹之名词时态—基础到进阶零点奇迹之冠词零点奇迹之代词零点奇迹之数词零点奇迹之形容词零点奇迹之动词零点奇迹之易混淆动词辨析零点奇迹之副词零点奇迹之宾语从句初中易错词汇情感交际类词汇学业竞争类词汇职业自然旅游类词汇人物修饰类词汇词汇进阶第一季词汇进阶第二季初中物理长度和时间的测量运动的描述运动的快慢温度与温标温度计的使用与计算熔化与凝固汽化现象上—蒸发汽化现象下—沸腾液化、升华与凝华现象物态变化的拓展质量和天平密度及其特性密度公式的应用混合物质的密度空心物体的密度液体密度的测量固体密度的测量压力的概念与计算压强和压强比较法压强公式的拓展切割后的压强问题液体压强公式及应用直柱形与非直柱形容器中的液体压强物浸液问题上物浸液问题下连通器及应用 U形管中的混合液压问题帕斯卡定律及应用大气压强的存在与托里拆利实验大气压强的应用气体压强的计算流体压强及应用浮力的产生与判断阿基米德原理与应用物体浮沉条件与应用浮力的计算与综合应用上浮力的计算与综合应用下电荷光学与声音力与运动压力与浮力简单机械功与能电学探秘电磁攻略初中化学物质的变化与性质走进化学实验室构成物质的微粒化学元素化学式和方程式气体性质和制取实验探究酸碱盐之酸酸碱盐之碱酸碱盐之盐酸碱盐之金属生物入门高中数学集合上集合下函数的概念和表示复合函数解析式求法分段、组合函数及映射函数的单调性单调性的应用函数的奇偶性函数的周期性函数的综合应用基本初等函数指数及指数函数对数运算及应用对数函数及应用对数方程的解法反函数与幂函数函数的应用三角函数任意角与弧度制任意角的三角函数同角三角函数的基本关系诱导公式三角函数的图象三角复合函数的图象与应用正切函数的图象与性质三角函数的综合应用平面向量平面向量的概念平面向量的线性运算共线向量及其应用小学数学一年级数字找规律图形找规律巧数图形上二年级排队问题等量代换火柴棒问题运筹学初步思维趣题巧数图形下三年级加减法计算乘除法计算乘除法的巧算一笔画问题平均数问题周期问题植树问题数列问题和倍问题差倍问题和差问题盈亏问题巧算24点上四年级数论一基本鸡兔同笼问题基本还原问题年龄问题相遇问题追及问题多次相遇问题火车过桥问题流水行船问题五年级小数计算列方程解应用题面积巧求法等积变形与推广格点多边形的面积逻辑问题不定方程和方程组方程组的解法与技巧容斥原理计数问题加法原理与应用乘法原理与应用加乘原理与应用抽屉原理上抽屉原理下六年级分数计算分数计算裂项法带分数与循环小数的计算分数的比较分数的变化时钟问题排列组合公式及应用排列组合的技巧概率牛吃草问题巧算24点比和比例百分数问题浓度问题工程问题面积中的比例共角定理与相似模型直线型面积的高级技巧行程问题六年级圆的基本公式与应用曲线型面积的求法圆的滚动问题圆柱的认识套餐套餐初中物理伴学套餐初中数学伴学套餐小学思维数学伴学套餐高中全科套餐小学全科套餐小学科学启蒙套餐高中数学必修4 文言文经典篇目精讲中考冲刺综合套餐初中超级状元·银牌套餐初中超级状元·金牌套餐金榜题名英语零点奇迹语法套餐小学趣味思维数学光影作文系列套餐高中数学必修1 初中数学满分套餐初中超级状元·铜牌套餐初中理科状元初中语数英夺魁精英初中物理初中化学学习轨迹心灵成长心灵成长安全教育存银票，赢平板电脑 7500万奖学金大放送请先登录再领取去登陆哎呀，你获得了银票快去玩玩立即购买加入购物车￥98 原价：98 选课中心高中数学单调性的应用视频列表视频列表试听视频 1、函数单调性的应用—求最值或值域上 12:02 2、函数单调性的应用—求最值或值域下 08:34 3、抛物线的底线上—二次函数在实数集上的值域讨论 07:13 4、抛物线的底线下—二次函数恒为正或负条件的综合应用 07:58 5、抛物线最值问题一—二次函数在闭区间上的最值 09:38 6、抛物线最值问题二—定值问题和含参问题 14:00 7、抛物线最值问题三—最大值和最小值并存的问题 06:41 8、值域的限定上—恒成立问题 05:21 9、值域的限定中—存在解问题 06:04 10、值域的限定下—参变互换 09:26 11、组合函数的应用之对勾函数上 11:37 12、组合函数的应用之对勾函数下 12:59 13、判断函数值或自变量的大小关系上—自变量和单调性求解函数值 11:06 14、判断函数值或自变量的大小关系中—单调性固定的问题 05:58 15、判断函数值或自变量的大小关系下—单调性不固定的问题 09:36 16、前两类二次分式函数上—简单的求值域问题 09:50 17、前两类二次分式函数下—换元法解决含常数项的值域问题 08:21 18、第三类二次分式函数一—分离常数法求值域 08:11 19、第三类二次分式函数二—判别式法求值域 08:09 20、第三类二次分式函数三—判别式法的注意事项 08:24 21、第三类二次分式函数四—判别式法的综合应用 06:41 分享给朋友：难度：进阶 \| 1016人点赞 128825人已学习 \| 视频有问题？课后练习 0/14 综合试题 0/4 课程详情视频列表购买说明 3天无理由退款一年有效期 3天无理由退款：退款将以超级币形式退至您的超级课堂学习账户，便于您重新选购其他课程。恶意退款将被冻结账号。一年有效期：自购买之日起，有效期内可反复观看视频，并可至我的题库温习所有练习，有效期内若更新视频可以免费享有。课程简介在这个章节，我们将继续函数单调性的学习，研究单调性在题目中的应用方法。因为单调性反应了函数值随自变量变化的规律，所以对求解最值及值域是有帮助的。比如对于某一单调区间，端点值就是区间的最值。此外，超级课堂在这个章节还会着重研究二次函数这一具有固定单调性的函数，以及复杂的对勾函数、分式函数的的性质及应用。视频列表函数单调性的应用—求最值或值域上 12:02 AI出题下载题目做题0/12 1、单调性的给出方式主要有五种：直接给出；由图象给出；由定义法给出；由定义法的变形给出；给出解析式，自己判断单调性2、了解最值的定义，还有它的几何意义，即函数图象最高点或最低点的纵坐标3、单调性和最值的关系。对于任意函数的一个单调区间[a，b]，这个区间的最值一定由端点取得函数单调性的应用—求最值或值域下 08:34 AI出题下载题目做题0/4 1、在函数的单调闭区间上，端点函数值就是最值。如果是开区间，则可能没有最值2、如果在区间上没有单调性，可分为先增后减和先减后增两种情况，单调性发生变化的那个顶点是最大或最小值。另外一个最值由端点比较得出抛物线的底线上—二次函数在实数集上的值域讨论 07:13 AI出题下载题目做题0/17 1、主要内容是二次函数在实数集上的值域问题，可以用顶点坐标公式或配方法来求2、由此引出二次式恒为正或恒为负的问题，只要把二次式看成二次函数，通过开口方向，和判别式，就能得出恒为正或恒为负的条件抛物线的底线下—二次函数恒为正或负条件的综合应用 07:58 AI出题下载题目做题0/14 1、反映在不等式的解集方面，如果恒为正，则$ax^{2}+bx+c>2、 0$的解集就是$R$，而$ax^{2}+bx+c≤0$的解集就是空集3、如果恒为负，则$ax^{2}+bx+c<4、 0$的解集就是$R$，而$ax^{2}+bx+c≥0$的解集就是空集5、注意讨论某些题目中的二次项系数可能为$0$的情况抛物线最值问题一—二次函数在闭区间上的最值 09:38 AI出题下载题目做题0/14 1、主要讨论了二次函数$f(x)=ax^{2}+bx+c(a\neq 0)$在闭区间$[m,n]$上的最值的分类方法2、当$a＞0$时，求最小值分区间左中右，求最大值分中点左右3、当$a＜0$时，恰好相反，求最大值分区间左中右，求最小值分中点左右抛物线最值问题二—定值问题和含参问题 14:00 AI出题下载题目做题0/12 1、当函数解析式与区间都完全确定时，值域也完全确定，不需要分类讨论，直接利用图象就能看出来2、如果解析式或区间中含有参数，对称轴与区间的相对位置关系通常是不确定的，需要按具体的位置关系分类讨论，不要忘了借助图象帮你分析具体的分类方法抛物线最值问题三—最大值和最小值并存的问题 06:41 AI出题下载题目做题0/15 1、通过一道典型例题解决同时包含闭区间[m，n]上的最大值与最小值的题目值域的限定上—恒成立问题 05:21 AI出题下载题目做题0/15 1、在某闭区间上，函数值大于或大于等于$k$恒成立，和最小值大于或大于等于$k$，可以相互推导2、同理，函数值小于或小于等于$k$恒成立，和最大值小于或小于等于$k$，可以相互推导值域的限定中—存在解问题 06:04 AI出题下载题目做题0/13 1、某闭区间上，函数值大于或大于等于$k$有解，和最大值大于或大于等于$k$，可以相互推导2、同理，函数值小于或小于等于$k$有解，和最小值小于或小于等于$k$，可以相互推导值域的限定下—参变互换 09:26 AI出题下载题目做题0/14 1、参变互换的技巧就是用于解决已知参数所在区间，求自变量$x$的取值范围的问题2、函数的值域在某个区间内恒成立的条件，要分类求最大值与最小值，有时由条件可以省去一种甚至几种分类，然后再通过列不等式组求解组合函数的应用之对勾函数上 11:37 AI出题下载题目做题0/27 1、两类对勾函数，其中第一类，$ab$异号的情况，你只要记住各自的单调性2、第二类对勾函数，即$ab$同号的情况。你需要记住三点：(1)图像形状，(2)顶点横坐标，(3)两条渐近线，一条是$y$轴，一条是$y=ax$ 组合函数的应用之对勾函数下 12:59 AI出题下载题目做题0/23 1、学习和对勾函数相关的三类题型：求单调区间的题，求值域的题，由值域求参数的题。注意利用恒等变形和换元法转化成对勾函数的技巧判断函数值或自变量的大小关系上—自变量和单调性求解函数值 11:06 AI出题下载题目做题0/9 1、“自变量大小关系”“函数值大小关系”“单调性”这三者之间可以知二求三2、熟悉了第一类题型，由“自变量大小关系”和“单调性”判断“函数值大小关系” 判断函数值或自变量的大小关系中—单调性固定的问题 05:58 AI出题下载题目做题0/25 1、讲解了由“函数值大小关系”和“单调性”判断“自变量大小关系”的单调性固定的题型2、要注意的就是千万不要忽略定义域的限定，通常需要三个不等式判断函数值或自变量的大小关系下—单调性不固定的问题 09:36 AI出题下载题目做题0/18 1、讲解了由“函数值大小关系”和“单调性”判断“自变量大小关系”的单调性不固定的题型2、要注意的就是要分单调区间来讨论，还要充分利用函数图像来帮助分析单调性前两类二次分式函数上—简单的求值域问题 09:50 AI出题下载题目做题0/14 1、对于第一类二次分式函数，分子为非零常数，分母为二次式。运用复合函数求值域的方法即可：由内向外，逐层求值域2、对于第二类二次分式函数的第一种形式：分子或分母只含一次项，变形技巧为拆分解析式和分子分母同除以$x$，本质就是通过代数变形，变成含有对勾函数的复合函数前两类二次分式函数下—换元法解决含常数项的值域问题 08:21 AI出题下载题目做题0/16 1、介绍了第二类二次分式函数中一次式含有常数项的值域求法2、主要技巧为先用换元法，再拆分解析式和分子分母同除以$x$，本质还是通过代数变形，变成含有对勾函数的复合函数第三类二次分式函数一—分离常数法求值域 08:11 AI出题下载题目做题0/11 1、介绍了第三类二次分式函数求值域的第一种技巧，分离常数法2、分离常数后，再分子分母同除以$x$，按复合函数去求整体值域第三类二次分式函数二—判别式法求值域 08:09 AI出题下载题目做题0/11 1、介绍了第三类二次分式函数值域的第二种求法，判别式法2、关键就两步操作：一转化并整理方程；二讨论并综合范围3、要注意两点：一是讨论系数；二是能约分时不可用第三类二次分式函数三—判别式法的注意事项 08:24 AI出题下载题目做题0/9 1、形如$f(x)=\frac{(ax^{2}+bx+c)}{(dx^{2}+ex+f)}$$(ad\neq 0)$这样的二次分式函数求值域，要注意，当二次分数函数可以约分时，是不能用判别式法的2、这节课所举例子的定义域，都是自然定义域，如果人为限定了x的范围，就要牵涉到一元二次方程根的分布问题，除了判别式，还要考虑其他因素第三类二次分式函数四—判别式法的综合应用 06:41 AI出题下载题目做题0/10 1、用一道经典例题展示了判别式法的综合应用2、判别式法求二次分式函数的值域是学霸才能精通的技巧，在它的帮助下，你才能秒杀所有的二次分式函数单调性的应用综合练习下载题目做题0/4 我要评论发表评论表情热门评论第1页 < > 共1页已购买： 128825人最新购买 1 测试 2 超级学员1464258 3 超级学员 4 超级学员1611235 5 超级学员1725606 6 超级学员2790848 7 超级学员2 8 杜金泽13 9 超级学员4285551 10 超级学员4372199 猜你需要免费试学视频X18 习题X259 共 18集，已更新第 18集集合上 198353人在学￥ 78 ￥ 78 免费试学视频X16 习题X259 共 16集，已更新第 16集集合下 129017人在学￥ 78 ￥ 78 免费试学视频X14 习题X171 共 14集，已更新第 14集￥ 0 ￥ 0 免费试学视频X23 习题X308 共 23集，已更新第 23集￥ 72 ￥ 72 0 将优惠码000000FQA复制并发送给好友使用邀请码“000000FQA”首次购买课程可直减 100超级币。兴趣产生时，教育自然开始，点击查看详情
188354	https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/2.10/primary/lesson/algebraic-solutions-to-one-variable-inequalities-alg-i-hnrs/	Algebraic Solutions to One Variable Inequalities \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Start Practice 2.10 Algebraic Solutions to One Variable Inequalities FlexBooks 2.0> CK-12 Algebra 1 Concepts - Honors> Algebraic Solutions to One Variable Inequalities Written by:Brenda Meery \|Kaitlyn Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Jul 02, 2025 The Morgan Silver Dollar is a very valuable American dollar minted between 1978 and 1921. When placed on a scale with twenty 1-gram masses, the scale tips toward the Morgan Dollar. Draw a picture to represent this scenario and then write the inequality. Algebraic Solutions to One Variable Inequalities Linear equations are of the form a x+b=0, where a≠0. With linear equations there is always an equals sign. Linear inequalities are mathematical statements relating expressions by using one or more inequality symbols <,>,≤, or ≥. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to. Recall that one method for solving an equation is to use the balance method. To solve for a+2=5, you would draw the following balance: If you were to use the balance method to solve the linear inequality version, it would look more like this: a+2>5 Think about it this way. It is like having someone heavier on the (a+2) side of the balance and someone light on the (5) side of the balance. The (a+2) person has a weight greater than (>) the (5) person and therefore the balance moves down to the ground. The rules for solving inequalities are basically the same as you used for solving linear equations. If you have parentheses, remove these by using the distributive property. Then you must isolate the variable by moving constants to one side and variables to the other side of the inequality sign. You also have to remember that whatever you do to one side of the inequality, you must do to the other. The same was true when you were working with linear equations. One additional rule is to reverse the sign of the inequality if you are multiplying or dividing both sides by a negative number. It is important for you to remember what the symbols mean. Always remember that the mouth of the sign opens toward the larger number. So 8>5, the mouth of the > sign opens toward the 8 so 8 is larger than 5. You know that’s true. 6 b−5<300, the mouth opens toward the 300, so 300 is larger than 6 b−5. Let's practice by solving the following inequalities: 15<4+3 x Remember that whatever you do to one side of the inequality sign, you do to the other. 15−4<4−4+3 x Subtract 4 from both sides to isolate the variable 11<3 x Simplify 11 3<3 x 3 Divide by 3 to solve for the variable x>11 3 Simplify Do a quick check to see if this is true. 11 3 is approximately 3.67. Try substituting 0, 3, and 4 into the equation. 15<4+3 x 15<4+3 x 15<4+3 x 15<4+3(0)15<4+3(3)15<4+3(4)15<4 F a l s e 15<13 F a l s e 15<16 T r u e The only value out of 0, 3, and 4 where x>11 3 is 4. If you look at the statements above, it was the only inequality that gave a true statement. 2 y+3>7 Use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other. 2 y+3−3>7−3 Subtract 3 from both sides to isolate the variable 2 y>4 Simplify 2 y 2>4 2 Divide by 2 to solve for the variable y>2 Simplify Do a quick check to see if this is true. Try substituting 0, 4, and 8 into the equation. 2 y+3>7 2 y+3>7 2 y+3>7 2(0)+3>7 2(4)+3>7 2(8)+3>7 3>7 F a l s e 11>7 T r u e 19>7 T r u e The values out of 0, 4, and 8 where y>2 are 4 and 8. If you look at the statements above, these values when substituted into the inequality gave true statements. −2 c−5<8 Again, to solve this inequality, use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other. −2 c−5+5<8+5 Add 5 to both sides to isolate the variable−2 c<13 Simplify−2 c−2<13−2 Divide by -2 to solve for the variable Note that when you divide by a negative number, the inequality sign reverses. c>−13 2 Simplify Do a quick check to see if this is true. −13 2 is equal to –6.5. Try substituting –8, 0, and 2 into the equation. −2 c−5<8−2 c−5<8−2 c−5<8−2(−8)−5<8−2(0)−5<8−2(2)−5<8 11<8 F a l s e−5<8 T r u e−9<8 T r u e The values out of –8, 0, and 2 where c>−13 2 are 0 and 2. If you look at the statements above, these values when substituted into the inequality gave true statements. Examples Example 1 Earlier, you were told that w hen a Morgan Silver Dollar is placed on a scale with twenty 1-gram masses, the scale tips toward the Morgan Dollar. Draw a picture to represent this scenario and then write the inequality. Let m be the weight of a Morgan dollar. Since the weight of the Morgan dollar is greater than 20 g, the mouth of the inequality sign would open towards the variable, m. Therefore the inequality equation would be: m>20 Example 2 Solve the inequality: 4 t+3>11 t>2. Here are the steps: 4 t+3>11 4 t+3−3>11−3 Subtract 3 from both sides to isolate the variable 4 t>8 Simplify 4 t 4>8 4 Divide by 4 to solve for the variable t>2 Example 3 Solve the inequality: 2 z+7≤5 z+28 z≥−7. Here are the steps: 2 z+7≤5 z+28 2 z−2 z+7≤5 z−2 z+28 Subtract 2 z from both sides to get variables on same side of the inequality sign.7≤3 z+28 Simplify 7−28≤3 z+28−28 Subtract 28 from both sides to isolate the variable−21≤3 z Simplify 3 z 3≥−21 3 Divide by 3 to solve for the variable z≥−7 Example 4 Solve the inequality: 9(j−2)≥6(j+3)−9 j≥9. Here are the steps: 9(j−2)≥6(j+3)−9 9 j−18≥6 j+18−9 Remove parentheses 9 j−18≥6 j+9 Combine like terms on each side of inequality sign 9 j−6 j−18≥6 j−6 j+9 Subtract 6 j from both sides to get variables on same side of the inequality sign.3 j−18≥9 Simplify 3 j−18+18≥9+18 Add 18 to both sides to isolate the variable 3 j≥27 Simplify 3 j 3≥27 3 Divide by 3 to solve for the variable j≥9 Review Solve for the variable in the following inequalities. a+8>4 4 c−1>7 5−3 k<6 3−4 t≤−11 6≥11−2 b e 5−3>−1 1 5(r−3)<−1 1 3(f+2)<4 p+3 4≥−2 1 2(5−w)≤−3 3(2 x−5)<2(x−1)+3 2(y+8)+5(y−1)>6 2(d−3)<−3(d+3) 3(g+3)≥2(g+1)−2 2(3 s−4)+1≤3(4 s+1) Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. 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188355	https://fiveable.me/ap-stats/unit-6/setting-up-test-for-difference-two-population-proportions/study-guide/sJt2F9NwsQ2gihYEHGjP	printables 📊AP Statistics Unit 6 Review 6.10 Setting Up a Test for the Difference of Two Population Proportions 📊AP Statistics Unit 6 Review 6.10 Setting Up a Test for the Difference of Two Population Proportions Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 📊AP Statistics Unit & Topic Study Guides Unit 6 Overview: Inference for Categorical Data: Proportions 6.1 Introducing Statistics: Why Be Normal? 6.2 Constructing a Confidence Interval for a Population Proportion 6.3 Justifying a Claim Based on a Confidence Interval for a Population Proportion 6.4 Setting Up a Test for a Population Proportion 6.5 Interpreting p-Values 6.6 Concluding a Test for a Population Proportion 6.7 Potential Errors When Performing Tests 6.8 Confidence Intervals for the Difference of Two Proportions 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions 6.10 Setting Up a Test for the Difference of Two Population Proportions 6.11 Carrying Out a Test for the Difference of Two Population Proportions Another way to check a statistical claim is to perform a significance test for the difference in two population proportions. As with any significance test, we have to write hypotheses, check our conditions and then calculate and conclude. Still lost? Let's do a refresher! A statistical significance test is used to determine whether the difference between two population proportions is statistically significant, or whether it could have occurred by chance. To perform a significance test for the difference in two population proportions, you need to first write your null and alternative hypotheses. The null hypothesis states that there is no difference between the two population proportions, while the alternative hypothesis states that there is a difference. Next, you need to check that the conditions for the test are met. These include having a large enough sample size and having a random and independent sample. Once you have checked the conditions, you can calculate the test statistic and determine the p-value. The p-value is the probability of obtaining a test statistic as extreme as the one observed, given that the null hypothesis is true. If the p-value is less than the significance level (usually 0.05), you can reject the null hypothesis and conclude that the difference between the two population proportions is statistically significant. If the p-value is greater than the significance level, you cannot reject the null hypothesis and must conclude that the difference is not statistically significant. Hypotheses and Parameters The first thing we need to do when setting up a significance test for the difference in two population proportions is to write out our hypotheses. Our null hypotheses will always have our two population proportions being equal, while our alternate has them either greater than, less than or not equal to each other. It is also important in this stage of setting up the test to identify what p1 and p2 represent. We have to define our parameters so the reader knows what we are truly comparing. more resources to help you study practice questionscheatsheetscore calculator Conditions We also must check our conditions for inference. The same three conditions apply as did for confidence intervals with one little small change in the normal check. (1) Random Probably the most important condition is that we need to be sure that both of our samples come from random samples. If we don't take a random sample from our population, then our findings suffer from sampling bias and we are stuck and we can't generalize our findings to our population. (2) Independence To check that our sample is independent, we need to make sure that both of our populations are at least 10 times that of our samples. Also, if we are dealing with a randomized experiment, the random assignment of treatments classifies our samples as independently selected. (3) Normal When dealing with proportions, we always check our normal condition by using the Large Counts Condition, which states that our expected successes and failures is at least 10. With a 2 proportion z test, we have to combine our proportions to create a combined p-hat. This is what we use to find our expected failures and successes. Then we have to verify that each of our expected failures and successes are at least 10. This is because we are using a pooled sample. In this test, you combine the two samples into a single "pooled" sample and calculate a single proportion for the combined sample. The test statistic is then calculated based on the difference between the two proportions and the pooled sample proportion. Example Let's return to our MJ vs. Lebron problem from earlier... again. Recall that MJ made 836/1623 shots and Lebron made 622/1493 shots. Instead of testing this claim with a confidence interval, let's test it using a 2 Prop Z Test to verify our results. Hypotheses and Parameters Another great idea when writing our hypotheses is to use meaningful subscripts such as MJ and L that clarify which proportion matches which population. Conditions Random: Even though the problem never stated that they were random (and we discussed the problems with this in Unit 6.9) we are going to assume it is random. Independent: It is reasonable to believe (and obviously true) that MJ took at least 16, 230 shots in his career and Lebron took at least 14,930 shots in his career, so the samples are independent. Normal: This is the one that will be a bit different. First, we have to calculate our pooled p-hat. Using the formula above, we get 0.468 Next, we have to check our large counts condition using this pooled p-hat. 1623 (0.468) > 10 ✔️ 1623 (0.532) > 10 ✔️ 1493 (0.468) > 10 ✔️ 1493 (0.532) > 10 ✔️ Now that we have checked conditions, we are ready to calculate and test our claim. 🎥 Watch: AP Stats - Inference: Hypothesis tests for Proportions Frequently Asked Questions How do I set up null and alternative hypotheses for comparing two proportions? For a two-sample test of proportions you always start with the null that the populations are the same (difference = 0): H0: p1 = p2 or H0: p1 − p2 = 0. The alternative depends on your question: - Two-sided: Ha: p1 ≠ p2 (tests for any difference). - One-sided: Ha: p1 > p2 or Ha: p1 < p2 (tests a directional claim). On the AP exam you’ll use a two-sample z-test for proportions (when conditions are met). Under H0 you form the pooled proportion p̂c = (x1 + x2)/(n1 + n2) and use the pooled SE: sqrt[p̂c(1−p̂c)(1/n1 + 1/n2)] to compute the z statistic. Before testing check independence (random/independent samples, 10% rule if without replacement) and the success–failure condition using the pooled p̂c: n1p̂c, n1(1−p̂c), n2p̂c, n2(1−p̂c) ≥ 5 (or 10 per your class). For a short walkthrough and examples see the Topic 6.10 study guide ( For extra practice, try problems on the Unit 6 page ( or the general practice bank ( What's the difference between a one-sample z-test and a two-sample z-test for proportions? One-sample z-test for a proportion tests one population proportion against a hypothesized value: H0: p = p0. The test statistic is (p̂ − p0) / sqrt[p0(1−p0)/n] and you check independence (random sample, 10% condition) and success–failure using p0 (or p̂ for CI). Two-sample z-test compares two independent proportions: H0: p1 = p2 (or p1 − p2 = 0) and Ha can be one- or two-sided (p1 ≠ p2, p1 > p2, p1 < p2). Under H0 you use the pooled proportion p̂c = (x1 + x2)/(n1 + n2) to compute the pooled standard error: SE = sqrt[p̂c(1−p̂c)(1/n1 + 1/n2)]. Check independence (two independent random samples or experiment + 10% for each) and success–failure using the pooled counts: n1p̂c, n1(1−p̂c), n2p̂c, n2(1−p̂c) ≥ 5 (or 10). AP tip: the CED expects a two-sample z-test for comparing proportions (Topic 6.10). Review pooled/stat formula and conditions at the Fiveable study guide ( and practice problems ( When do I use a two-sample z-test for difference of proportions vs other tests? Use the two-sample z-test for the difference of two population proportions when you have two independent random samples (or a randomized experiment) and you want to test H0: p1 = p2 (or p1 − p2 = 0). Under H0 you use the pooled proportion p̂c = (x1 + x2)/(n1 + n2) to compute the pooled standard error and z statistic. Before using the test check the CED conditions: independence (random samples, 10% rule when sampling without replacement) and approximate normality via the success–failure condition using the pooled p̂c—i.e., n1 p̂c, n1(1−p̂c), n2 p̂c, n2(1−p̂c) all ≥ about 5 (or 10 for stricter practice). When not to use it: if samples aren’t independent, counts are small (use Fisher’s exact or exact methods), or you’re doing one proportion (use one-sample z-test) or comparing many categories (chi-square). For CI for p1 − p2 you use the two-sample z-interval formula (unpooled SE). For more AP-aligned detail and examples see the Topic 6.10 study guide ( and the Unit 6 overview ( For extra practice try the problems at ( I'm confused about the pooled proportion formula - how do I calculate p̂c? p̂c (the pooled proportion) is the combined estimate of the common proportion under H₀: p₁ = p₂. Compute it two equivalent ways: - p̂c = (n₁·p̂₁ + n₂·p̂₂) / (n₁ + n₂) - or, if you know counts of successes, p̂c = (X₁ + X₂) / (n₁ + n₂) You use p̂c when doing a two-sample z-test for p₁ − p₂ because under the null you assume the two populations have the same p, so you pool the data to estimate that common p. Then the pooled standard error is sqrt[ p̂c(1−p̂c) (1/n₁ + 1/n₂) ] and z = (p̂₁ − p̂₂) / (pooled SE). Don’t forget the AP conditions: samples independent, 10% rule if sampling without replacement, and the success–failure checks using p̂c (n₁p̂c, n₁(1−p̂c), n₂p̂c, n₂(1−p̂c) ≥ 5 or 10). See the Topic 6.10 study guide ( and Unit 6 overview ( for examples and practice ( What does it mean when the null hypothesis is p₁ - p₂ = 0? H₀: p₁ − p₂ = 0 means you’re claiming the two population proportions are equal—there’s no difference or effect between group 1 and group 2. In practice that means your test is set up so the center of the sampling distribution for p̂₁ − p̂₂ is 0 under the null. For AP Stats you’d use a two-sample z-test for proportions (CED VAR-6.I.1), compute the pooled proportion p̂c = (X₁+X₂)/(n₁+n₂) when assuming H₀, and use the pooled standard error √[p̂c(1−p̂c)(1/n₁ + 1/n₂)] to get the z statistic. Before you trust the z test, check independence (random, independent samples and 10% rule) and the success–failure condition using n₁p̂c, n₁(1−p̂c), n₂p̂c, n₂(1−p̂c) ≥ 5 or 10 (CED VAR-6.J). For a quick refresher on setup and pooled SE, see the Topic 6.10 study guide ( If you want practice problems, Fiveable also has lots at ( How do I check the conditions for a two-sample proportion test step by step? Step-by-step check for a two-sample z-test for p1 − p2 (assume H0: p1 = p2): 1. State H0 and Ha. For example H0: p1 − p2 = 0 (p1 = p2). Choose one- or two-sided Ha (p1 > p2, p1 < p2, or p1 ≠ p2). 2. Independence: confirm data come from two independent random samples or a randomized experiment. If sampling without replacement, check each sample size ≤ 10% of its population (n1 ≤ 0.10N1 and n2 ≤ 0.10N2). 3. Pooled proportion (use for the test): p̂c = (x1 + x2) / (n1 + n2), where x1 = n1·p̂1 and x2 = n2·p̂2. 4. Success–failure (normality): compute n1·p̂c, n1·(1−p̂c), n2·p̂c, n2·(1−p̂c). Each should be ≥ the rule-of-thumb (commonly 5, sometimes 10) so the sampling distribution of p̂1 − p̂2 is approximately normal. 5. If all checks pass, use z = (p̂1 − p̂2 − 0) / sqrt[p̂c(1−p̂c)(1/n1 + 1/n2)] to get p-value. If not, use a simulation/bootstrap or exact method. For AP alignment and examples, see the Topic 6.10 study guide ( and Unit 6 overview ( Practice more problems at ( Why do we need to check that n₁p̂c ≥ 5 and all those other conditions? You check n₁p̂c, n₁(1−p̂c), n₂p̂c, n₂(1−p̂c) ≥ 5 (or ≥10) because the two-sample z-test for p₁−p₂ relies on the sampling distribution being approximately normal. Under H₀: p₁ = p₂ you pool the samples to get p̂c = (n₁p̂₁ + n₂p̂₂)/(n₁ + n₂); pooling gives the best estimate of the common proportion and sets the correct standard error for the z statistic. If any of those counts are too small, the normal approximation is poor and the z-test (and its p-value) can be misleading. You also must check independence: samples should be random/independent and, when sampling without replacement, n ≤ 10% of the population (for each sample). These checks are required by the CED (VAR-6.J) for valid inference on the AP exam. For a quick refresher and examples, see the Topic 6.10 study guide ( and try practice questions ( What's the formula for the two-sample z-test for difference of proportions? Use a two-sample z-test for proportions when you want to test H0: p1 = p2 (or p1 − p2 = 0). Under H0 you pool the samples: - pooled proportion: p̂c = (x1 + x2) / (n1 + n2), where x1 = n1 p̂1 and x2 = n2 p̂2. - pooled standard error: SE = sqrt[ p̂c(1 − p̂c) (1/n1 + 1/n2) ]. - test statistic: z = (p̂1 − p̂2) / SE (since H0 assumes difference = 0). Also check conditions from the CED: independent random samples (and 10% rule when sampling without replacement) and success–failure using the pooled proportion: n1 p̂c, n1(1−p̂c), n2 p̂c, n2(1−p̂c) should be ≥ 5 (or ≥10 if your teacher wants the stricter rule) so the sampling distribution is approximately normal. For more AP-aligned review see the Topic 6.10 study guide ( and the Unit 6 page ( For extra practice, try problems at ( When do I write Ha: p₁ > p₂ vs Ha: p₁ ≠ p₂ in my alternative hypothesis? Pick Ha based on the question you’re trying to answer—the alternative should match the claim or research question. - If you want to detect any difference (either group could be higher), use two-sided: Ha: p1 ≠ p2. This is common on the exam when they ask “different” or don’t indicate direction (CED VAR-6.H.3). - If you have a directional claim (you expect one proportion to be larger or smaller), use one-sided: Ha: p1 > p2 or Ha: p1 < p2. Use Ha: p1 > p2 when you’re testing whether population 1’s success rate is greater than population 2’s (CED VAR-6.H.3). - Don’t pick direction after seeing data—the direction must be decided before collecting/analyzing data. Also remember the null is H0: p1 = p2 (or p1 − p2 = 0) and, for a two-sample z-test, you’ll use the pooled proportion when H0 is assumed (CED VAR-6.H.1, VAR-6.I.1, VAR-6.J.1b). For a quick review on wording and examples, see the Topic 6.10 study guide ( For more practice, Fiveable has lots of practice questions ( I don't understand what "independent random samples" means for proportion tests. “Independent random samples” means the two groups you compare were chosen separately so one sample doesn’t affect the other. For a two-sample z-test for p1 − p2 the CED requires either (a) two independent random samples from each population or (b) a randomized experiment that assigns units to two treatments. Practically that means you didn’t sample the same people twice (paired data) and you didn’t pick one sample based on the other. Why it matters: the standard error formula and pooled proportion (p̂c = (x1+x2)/(n1+n2)) assume independence. If samples aren’t independent you must use a paired method or a different test. Also check the 10% condition when sampling without replacement (n1 ≤ 10% N1 and n2 ≤ 10% N2) and the success–failure checks (n1 p̂c, n1(1−p̂c), n2 p̂c, n2(1−p̂c) ≥ 5 or 10) for approximate normality, per the CED. Want more practice or a quick refresher? See the Topic 6.10 study guide ( and thousands of practice problems ( How do I know if my sample sizes are big enough for the normal approximation? Check two things. First, independence: your data must come from two independent random samples or a randomized experiment, and if sampling without replacement each sample size should be ≤10% of its population (the 10% condition). Second, the sampling distribution must be approximately normal (success–failure condition). For a two-sample test you use the pooled proportion under H0: p̂c = (n1 p̂1 + n2 p̂2)/(n1 + n2). Then check that all four counts n1 p̂c, n1(1−p̂c), n2 p̂c, and n2(1−p̂c) are ≥ a cutoff (AP uses “typically 5 or 10”—many teachers use ≥5; some prefer ≥10 for extra caution). If they’re met, the two-sample z-test (pooled z) is OK; if not, don’t rely on the normal approx. For a quick refresher see the Topic 6.10 study guide ( and other Unit 6 resources ( For extra practice try the AP question bank ( What's the 10% rule and why do I need to check n₁ ≤ 10% N₁? The 10% rule is a quick check for independence when you sample without replacement: each sample size should be no more than 10% of its population (n₁ ≤ 0.10·N₁ and n₂ ≤ 0.10·N₂). Why? When you remove many units from a finite population, later draws aren’t independent of earlier ones. If each sample is ≤10% of the population, the dependence is negligible and you can treat observations as (approximately) independent—an AP requirement for inference (CED VAR-6.J.1.a.ii). On the AP exam you must verify independence before using the two-sample z-test for proportions. If you can’t meet the 10% rule, you either use sampling with replacement, adjust variance with a finite-population correction, or acknowledge the independence issue and avoid standard z-methods. For more on checks and the pooled-proportion z-test, see the Topic 6.10 study guide ( and grab extra practice at ( Can someone explain what pooled proportion means and why we use it? Pooled proportion is the single best estimate of the common population proportion you assume under the null H0: p1 = p2. You compute it as p̂c = (x1 + x2) / (n1 + n2) (equivalently (n1 p̂1 + n2 p̂2)/(n1 + n2)). We use p̂c when doing a two-sample z-test for p1 − p2 because the null says both populations share one p, so the standard error should be based on that common value. The pooled standard error is sqrt[p̂c(1−p̂c)(1/n1 + 1/n2)] and the z statistic is (p̂1 − p̂2)/SEpooled. Why this matters for AP: the CED requires you to assume p1 = p2 under H0 and to check the success–failure condition using the pooled count (n1 p̂c, n1(1−p̂c), n2 p̂c, n2(1−p̂c) ≥ 5 or 10). If those hold and samples are independent (and ≤10% rule if without replacement), the pooled z-test is appropriate. For a refresher, see the Topic 6.10 study guide ( For more review and lots of practice problems, check the Unit 6 page ( and the practice bank ( I keep getting confused about when to use pooled vs unpooled - which one is for proportion tests? Use pooled when you’re doing a hypothesis test that assumes the two population proportions are equal under H₀ (H₀: p₁ = p₂ or p₁ − p₂ = 0). In that case you combine successes to get the pooled proportion p̂c = (x₁ + x₂)/(n₁ + n₂), use p̂c in the success–failure check (n₁p̂c, n₁(1−p̂c), n₂p̂c, n₂(1−p̂c) ≥ 5 or 10) and in the pooled SE: sqrt[p̂c(1−p̂c)(1/n₁ + 1/n₂)]. That’s the AP-approved approach for two-sample z-tests (CED VAR-6.I/J). Use the unpooled (separate p̂₁ and p̂₂) when you’re building a confidence interval for p₁ − p₂ or when the test does NOT assume p₁ = p₂—CIs always use the separate SE: sqrt[ p̂₁(1−p̂₁)/n₁ + p̂₂(1−p̂₂)/n₂ ]. Also always check independence (random/independent samples and 10% condition). For a quick review see the Topic 6.10 study guide ( and practice questions ( How do I write the hypotheses if I want to test whether one proportion is greater than another? Pick an order for your groups (call them p1 = proportion for group 1, p2 = proportion for group 2). For a one-sided test that group 1 has a larger proportion than group 2 you write: H0: p1 = p2 (equivalently p1 − p2 = 0) Ha: p1 > p2 (equivalently p1 − p2 > 0) This is a one-sided (right-tailed) two-sample z-test for proportions. Under H0 you’ll use the pooled proportion p̂c = (x1 + x2)/(n1 + n2) to compute the standard error and z statistic. Before testing, check independence (random/independent samples and 10% condition) and the success–failure condition using the pooled counts (n1 p̂c, n1(1−p̂c), n2 p̂c, n2(1−p̂c) usually ≥ 10). For AP-aligned review and examples see the Topic 6.10 study guide ( and extra practice (
188356	https://mathoverflow.net/questions/69244/encoding-n-natural-numbers-into-one-and-back	Skip to main content Asked Modified 1 year, 8 months ago Viewed 6k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I want to encode n natural numbers into one natural number. Also, I should be able to decode it back. I tried Gödel's encoding scheme, but it takes a lot of space (doesn't fit into a double) and requires a lot of computation. Is there a better scheme? Thanks in advance. it.information-theory Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Dec 1, 2016 at 7:24 Rodrigo de Azevedo 2,63333 gold badges1818 silver badges3737 bronze badges asked Jul 1, 2011 at 10:31 BagurBagur 7511 silver badge22 bronze badges 4 1 By what standard are you going to measure 'better'? A random guess on my uneducated behalf hints to me that given any bounded set of n numbers (say by 2K, n arbitrary), the complexity of encoding these as a single number will probably escape the bound. What if you are trying to encode 2K−n−1,2k−n,…,2K−1? David Roberts – David Roberts ♦ 2011-07-01 10:46:50 +00:00 Commented Jul 1, 2011 at 10:46 9 A trivial counting argument tells you that it is impossible to encode n m-bit integers into less than nm bits. Anyway, this does not look like research-level math question, see the faq. Emil Jeřábek – Emil Jeřábek 2011-07-01 10:48:27 +00:00 Commented Jul 1, 2011 at 10:48 I do not understand why this question was downvoted. I find an interesting question whose answer can be useful for programming. Roland Bacher – Roland Bacher 2011-07-01 16:23:29 +00:00 Commented Jul 1, 2011 at 16:23 7 Without more restrictions, this doesn't seem to be a research level mathematics problem. In addition, this question could be rewritten to be much clearer and to be informative. I can immediately think of several encoding schemes, but I don't know what Gödel's encoding scheme was for Nn, or exactly why that was unsatisfactory. If you want such an encoding scheme for programming, you may want additional properties such as that the usual operations (e.g., addition, comparison) on the results have some meaning on the decoded number, at least for some restricted set of inputs. Douglas Zare – Douglas Zare 2011-07-01 17:16:39 +00:00 Commented Jul 1, 2011 at 17:16 Add a comment \| 8 Answers 8 Reset to default This answer is useful 12 Save this answer. Show activity on this post. One can give an explicit bijection fn:Nn→N. In the case n=4 it is f4(p,q,r,s)=⎛⎝⎜p+q+r+s+34⎞⎠⎟+⎛⎝⎜p+q+r+23⎞⎠⎟+⎛⎝⎜p+q+12⎞⎠⎟+p The general case follows the obvious pattern. We can order Nn as follows: if a1+⋯+an<b1+⋯+bn we declare that a<b, but if we have a tie according to this test then we instead compare a1+⋯+an−1 with b1+⋯+bn−1, and so on. Then fn(a)=\|{b∈Nn:b<a}\|. To find f−14(k) you first find the largest u such that ⎛⎝⎜u+34⎞⎠⎟≤k, then the largest v such that ⎛⎝⎜u+34⎞⎠⎟+⎛⎝⎜v+23⎞⎠⎟≤k, and so on. This is not quite as good as a formula but at least it is a fairly straightforward algorithm. One can check that f4(p,q,r,s)<264 provided that p+q+r+s<145053, so 64 bit integers are enough for a reasonable range of values. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Jul 1, 2011 at 11:58 answered Jul 1, 2011 at 11:35 Neil StricklandNeil Strickland 57.6k77 gold badges142142 silver badges263263 bronze badges 3 2 That's cute! But in practice, if you have a permanent restriction to 64 bits, and n=4, you probably may as well use my scheme (below) with B=216. It's worth mentioning that, with that value of B, you can rapidly encode and decode using binary shift operations. James Cranch – James Cranch 2011-07-01 12:38:43 +00:00 Commented Jul 1, 2011 at 12:38 3 We might as well add where this is coming from geometrically. For n=2, it's the inverse of the usual enumeration of N2 where one skims along diagonals parallel to x+y=0 in succession, and the starting point on each diagonal is on the x-axis. Similarly, for n=3, it's inverse to an enumeration which proceeds by filling out each successive triangle parallel to x+y+z=0. Todd Trimble – Todd Trimble 2011-07-01 16:27:44 +00:00 Commented Jul 1, 2011 at 16:27 (Well, I guess Neil pretty much said that in different words.) Todd Trimble – Todd Trimble 2011-07-01 16:29:42 +00:00 Commented Jul 1, 2011 at 16:29 Add a comment \| This answer is useful 9 Save this answer. Show activity on this post. I'd just like to point out that more than one of the answers above can be viewed as follows. The problem becomes easy if, instead of working with natural numbers, you work with finite strings over some fixed alphabet. To code a finite sequence of strings, just concatenate them, inserting some punctuation to show where one string ends and the next begins (or assuming some a priori information on the input that can determine the break points without punctuation). To solve the problem for natural numbers rather than strings, identify numbers with strings in one of the well-known ways, for example the standard expansion in some base b. The counting argument mentioned in some previous answers shows that you can't do significantly better than this. (In some cases you can do a little better by being clever about the punctuation.) In particular, Snark's interpretation of the question, asking for even a single integer to be compressed by coding, is implausible. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jul 1, 2011 at 14:32 Andreas BlassAndreas Blass 74.7k88 gold badges196196 silver badges297297 bronze badges Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. There is a series of good answers on StackOverflow and Quora, where I learned these go by the name pairing functions (related to dovetailing in computer science), can be generalized to any infinite sets, and iterated to encode arbitrary-length tuples as @Hugo illustrates. In particular Szudzik's function is relatively efficient in computation and representation: ElegantPair(x,y)={y2+xx2+x+yx≠max(x,y)x=max(x,y) and its inverse: ElegantUnpair(z)={(z−⌊z√⌋2,⌊z√⌋)(⌊z√⌋,z−⌊z√⌋2−⌊z√⌋)z−⌊z√⌋2<⌊z√⌋z−⌊z√⌋2≥⌊z√⌋ Which looks like this: Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited May 23, 2017 at 12:37 CommunityBot 122 silver badges33 bronze badges answered Dec 1, 2016 at 5:41 Doctor JDoctor J 14133 bronze badges Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. Imagine the sequence of numbers that covers the two dimensional grid as follows: This gives you a very simple and compact way of encoding two numbers into one. Suppose that a is the row, b is the column and N is the cell value at coordinates (a,b). You can easily verify that the encoding formula is just: N=(a+b)2+3a+b2 And the decoding, given N, is given by: s=⌊2N−−−√⌋ a=2N−s2−s2 b=s−a Where s is just an auxiliary variable, for conveniency, and ⌊⌋ is the floor operator, that keeps only the integer part of a real number. Now, you can extend this idea recursively. If you want to encode 3 numbers, encode the first with the second, and then encode the result with the third, to obtain a single integer. The same generalizes to k numbers. Just keep on going. To decode you apply a similar reasoning. If you know that you expect k numbers, perform the decoding k-1 times, successively. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Jan 12, 2024 at 11:28 Daniele Tampieri 6,80088 gold badges3434 silver badges4949 bronze badges answered Apr 5, 2014 at 12:49 HugoHugo 14122 bronze badges Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. To a finite set A⊂N assign the natural number ∑a∈A2a. This map is bijective and requires little computation. Of course this method is only good when the numbers to encode are different. If you want to encode a finite sequence of natural numbers into one, then I suggest the following. Write each number dyadically and and change the leading "1" into a "2". Then concatenate these strings and interpret the resulting sequence of digits 0,1,2 in base 3. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Jul 1, 2011 at 11:01 answered Jul 1, 2011 at 10:43 GH from MOGH from MO 111k88 gold badges312312 silver badges424424 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. If you want to encode n natural numbers a1,…,an, with 0≤ai<B, then encode it as a1+a2B+⋯+anBn−1. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jul 1, 2011 at 10:56 James CranchJames Cranch 3,0742424 silver badges3131 bronze badges 1 This is the type of optimized encodings I had in mind with better informations about the type of (finite) sequences to encode. I remember an atari computer where the pointers were 32 bits... but only 24 of them were used to jump, so the eight others were used to pass a little more data to the functions :-) Julien Puydt – Julien Puydt 2011-07-01 13:32:55 +00:00 Commented Jul 1, 2011 at 13:32 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I'm not sure GH's answer is what was asked : as I understand it, the question is about an injection Nn→N for n≥2. The easy way is to note p1,…,pn the first n prime numbers and consider : (d1,…,dn)↦pd11…pdnn. Decoding is pretty easy in this case because you have to factor an integer knowing the prime numbers which appear. EDIT: as I (and EJ 17 seconds before me) noted, this doesn't answer the question. So let me try another answer : generally no, it isn't possible to do better ; there's a definite limit of information you can encode in a given number of bits. But in more specific contexts, it is possible to do much better : for example for a pair of integers of the form (m,m+1), you can get away with the same size to encode the pair as encoding m! So if you give us more specifics on the tuples you want to encode, then you'll get more interesting answers. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Jul 1, 2011 at 10:56 answered Jul 1, 2011 at 10:49 Julien PuydtJulien Puydt 2,07211 gold badge2222 silver badges2828 bronze badges 5 This is Gödel’s encoding, which the OP explicitly said they do not want. Emil Jeřábek – Emil Jeřábek 2011-07-01 10:51:12 +00:00 Commented Jul 1, 2011 at 10:51 Sigh... reading the question again, I'm not answering correctly either... Julien Puydt – Julien Puydt 2011-07-01 10:51:30 +00:00 Commented Jul 1, 2011 at 10:51 What's wrong with my answer? I gave a simple way of encoding an arbitrary sequence of natural numbers into a single natural number. The encoding takes little space and decoding is easy. GH from MO – GH from MO 2011-07-01 11:03:04 +00:00 Commented Jul 1, 2011 at 11:03 @GH Well, the question was about a semi-efficient encoding, and your method makes encoding a single integer take as many bits as the integer itself already. Of course, you'll notice that my first answer was even worse -- in fact I typed as an answer what I thought was the worse answer (both inefficient and excluded by the question)... thinking about something and typing another sometimes doesn't give good results... Julien Puydt – Julien Puydt 2011-07-01 13:30:52 +00:00 Commented Jul 1, 2011 at 13:30 @Snark: Fair enough. My second method (added after your comment, I think) is more efficient. GH from MO – GH from MO 2011-07-01 15:08:30 +00:00 Commented Jul 1, 2011 at 15:08 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. What about Morton numbers ? They are obtained by interleaving the bits of one or more source numbers. If your source numbers are sufficiently small you can decode it back. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jul 1, 2011 at 16:49 Ghassen HamrouniGhassen Hamrouni 12355 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions it.information-theory See similar questions with these tags. By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
188357	https://www.pocketmath.net/math-softwares/radicals/bbc-bitesize-maths-gcse-nth.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 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\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| Home \| \| Solving Absolute Value Inequalities \| \| Quadratic Equations \| \| Real Numbers and Notation \| \| The Distance Formula \| \| Properties and Facts of Addition \| \| Multiplying Complex Numbers \| \| Factoring Trinomials by Grouping \| \| Representing Simple Arithmetic Symbolically \| \| Distributive Rule \| \| Solving Equations by 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I am stuck on this assignment that I have to submit in the next couple of days and I can’t seem to find a way to solve it. You see, my professor has given us this homework on bbc bitesize maths gcse nth term non linear sequences, function composition and least common denominator and I just can’t make head or tail out of it. I am thinking of going to some private tutor to help me solve it. If someone can give me some suggestions, I will very appreciative. \| \| \| Back to top \| \| \| \| \| \| \| oc_rana Registered: 08.03.2007 From: egypt,alexandria \| \| \| \| --- \| \| Posted: Friday 29th of Dec 09:56 \| \| \| \| I know how annoying it can be if you are not getting anywhere with bbc bitesize maths gcse nth term non linear sequences. It’s a bit hard to help you out without more information of your problems . But if you rather not get a tutor, then why not just use some piece of software and see what you think. There are so many programs out there, but one you should think about would be Algebrator. It is pretty useful plus it is not expensive . \| \| \| Back to top \| \| \| \| \| \| \| Voumdaim of Obpnis Registered: 11.06.2004 From: SF Bay Area, CA, USA \| \| \| \| --- \| \| Posted: Saturday 30th of Dec 08:56 \| \| \| \| I didn’t encounter that Algebrator software yet but I heard from my friends that it really does assist in solving algebra problems. Since then, I noticed that my classmates don’t really have a hard time answering some of the problems in class. It might really have been effective in improving their solving skills in algebra. I can’t wait to use it someday because I think it can be very effective and help me have a good grade in math . \| \| \| Back to top \| \| \| \| \| \| \| Vnode Registered: 27.09.2001 From: Germany \| \| \| \| --- \| \| Posted: Sunday 31st of Dec 20:26 \| \| \| \| adding numerators, function definition and function range were a nightmare for me until I found Algebrator, which is really the best math program that I have ever come across. I have used it through several algebra classes – College Algebra, Remedial Algebra and Remedial Algebra. Simply typing in the algebra problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my math homework would be ready. I truly recommend the program. \| \| \| Back to top \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| All Right Reserved. Copyright 2005-2025 \| \| \| \|
188358	https://www.practiceproblems.org/problem/Discrete_Math/Combinatorics/Counting_Arrangements_of_People_in_a_Line	Counting Arrangements of People in a Line - Discrete Math Combinatorics \| PracticeProblems.org Skip to Content PracticeProblems.org PracticeProblems.org Toggle Menu Search Problems...⌘K Counting Arrangements of People in a Line Home \| Discrete Math \| Combinatorics \| Counting Arrangements of People in a Line CombinatoricsEasyVideo How many ways can 10 people, consisting of 6 men and 4 women, stand in line without any restrictions? This problem falls under the domain of combinatorics, which deals with the counting, arrangement, and combination of objects. In this case, the problem is asking us to find the number of ways we can arrange a total of 10 people into a line. Since there are no restrictions—such as specific people needing to be next to each other—the solution involves a straightforward application of factorials, a fundamental operation in combinatorics. The number of ways to arrange a set of items, where the order of arrangement counts, is given by the factorial of the number of items. For this problem, since we want to arrange 10 individuals, the solution involves calculating 10 factorial, often denoted as 10!. This is a basic yet crucial concept in combinatorial mathematics that students will encounter frequently. It paves the way for understanding more complex permutations and combinations with restrictions, which are also central topics in this field. Understanding how to compute simple permutations helps students build intuition for more advanced problems, such as those involving constraints like adjacency, separation, or the inclusion of identical items. It is essential for developing strategies to tackle a wide array of problems students might face in discrete mathematics and related courses. Posted by Gregory 2 months ago Related Problems Evaluate Binomial Coefficient Evaluate the binomial coefficient inom{7}{5}. CombinatoricsEasyVideo Evaluate Binomial Coefficient with Equal n and r Evaluate the binomial coefficient when n n n and r r r are the same, for example, (9 9)\binom{9}{9}(9 9). CombinatoricsEasyVideo Arranging Men and Women in Line Without Adjacent Women How many ways can 6 men stand in line such that no two women are next to each other if there are 4 women in total? CombinatoricsMediumVideo Counting Binary Strings with Constraints How many binary strings contain exactly 5 zeros and 14 ones where each zero must be followed immediately by two ones? CombinatoricsMediumVideo Home • Privacy • Terms of Service © 2025 PracticeProblems.org
188359	https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick?srsltid=AfmBOopMQ4YqriyJbcpSjEBbklYFP6JxDfrhX6B14jqjwHzaedhIyVCS	Art of Problem Solving Simon's Favorite Factoring Trick - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Simon's Favorite Factoring Trick Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Simon's Favorite Factoring Trick Simon's Favorite Factoring Trick (SFFT) is often used in Diophantine equations where factoring is needed, applicable for Diophantine equations of the form , for integer constants , , and , where there is a constant on one side of the equation and on the other side and a product of variables with each of those variables in linear terms. Simon's Favorite Factoring Trick is named after AoPS user ComplexZeta, or Dr. Simon Rubinstein-Salzedo. Contents [hide] 1 Statement 2 Applications 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also 4.1 External Links Statement Let's put it in general terms. We have an equation , where , , and are integer constants. Simon's Favorite Factoring Trick states that this equation can be factored into the equation For example, is the same as: Here is another way to look at it. Consider the equation .Let's start to factor the first group out: . How do we group the last term so we can factor by grouping? Notice that we can add to both sides. This yields . Now, we can factor as . This is important because this keeps showing up in number theory problems. Let's look at this problem below: Determine all possible ordered pairs of positive integers that are solutions to the equation . (2021 CEMC Galois #4b) Let's remove the denominators: . Then . Take out the : (notice how I artificially grouped up the terms by adding ). Now, (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem. Look at all factor pairs of 20: . The first factor is for , the second is for . Solving for each of the equations, we have the solutions as . Applications This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually and are variables and are known constants. Sometimes, you have to notice that the variables are not in the form and . Additionally, you almost always have to subtract or add the and terms to one side so you can isolate the constant and make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. When coefficient of is not , you can sometimes achieve an equation that can be factored by dividing the coefficient off of the equation. Problems Introductory Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? (Source) Intermediate AoPS Online If has a remainder of when divided by , and has a remainder of when divided by , find the value of the remainder when is divided by . are integers such that . Find . (Source) A rectangular floor measures by feet, where and are positive integers with . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair ? (Source) Olympiad The integer is positive. There are exactly ordered pairs of positive integers satisfying: Prove that is a perfect square. (Source) See Also Number Theory Diophantine equation Algebra Factoring External Links Video on AoPS Retrieved from " Categories: Number theory Theorems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188360	https://www.mashupmath.com/graphing-a-cubic-function	How to Graph a Function in 3 Easy Steps — Mashup Math 0 items $0 How to Graph a Cubic Function Your complete guide to graphing a cubic function using 3 easy steps Graphing a cubic function is an algebra skill that requires you to draw a graph that represents a 3rd-degree function on the coordinate plane. Unlike graphing a parabola that represents a 2nd-degree function, a cubic function graph has its own unique shape and characteristics. This complete guide will explore and answer the following: What does a cubic function graph look like? How to graph a cubic function in 3 steps Graphing a cubic function examples Cubic functions are polynomials that are 3rd-degree functions and appear a lot when studying polynomials. After you’ve learned about graphing linear and quadratic functions, cubic functions are the next obvious step. Graphing a cubic function has similarities to linear and quadratic functions, however, cubic functions have unique characteristics that will be explored in this guide. By the end of this guide, you'll have the skills to graph cubic functions and gain a deeper understanding of their behavior. Let’s start by comparing the parent graphs of the linear function graph (1st-degree), quadratic function graph (2nd-degree), and cubic-function graph (3rd-degree) as shown in Figure 01 below. Characteristics of a Cubic Function Graph Cubic functions are functions of polynomials with the highest degree of 3. A general cubic function can be given as f(x) = ax^3 + bx^2 +cx + d, where a, b,c, and d are arbitrary numbers and a does not equal 0. The properties that cubic functions share with linear and quadratic functions are: The domain is the set of all real numbers The range is the set of all real numbers Since both the domain and range span the set of real numbers, there are no vertical or horizontal asymptotes The maximum number of roots (solutions to f(x) = 0) is the same as the degree of the function. Linear functions have one root, quadratics have a maximum of 2 roots, and 3rd-degree functions, cubic functions, have a maximum of 3 roots. Some features that distinguish cubic functions from linear and quadratics are: A cubic function graph has either one or three real roots (x-intercept/s) A cubic function graph may have two critical points, a local maximum, and a local minimum A cubic function graph has a single inflection point Figure 02 shows the end result of graphic a cubic function with equation f(x)=x^3-4x^2+5. Notice that the cubic function graph as three real roots (x-intercepts) and two critical points (a local maximum and a local minimum). How to Graph a Cubic Function When graphing a cubic function it is essential to determine the following key features and behaviors of the function. x and y -intercepts: The x-intercepts, also known as the roots of the function, can be determined by finding the solutions to f(x) = 0. The y-intercept is the value of the function when x = 0, i.e f(0). Critical points: The local maximum and minimum points, also known as stationary points, are points on the graph where the gradient of the function is 0. To obtain these points, you need to obtain the derivative f’(x) and solve for f’(x) = 0. Point of inflection: This is the point where the concavity of the curve changes direction. In a cubic function given by f(x) = , the inflection point is given by x=-b/3a Figure 03 below illustrates examples of critical points and an inflection point on a cubic function graph. End behaviors: The tendency of the function as x approaches the extremes. With odd-degree polynomials such as cubic functions, the end behaviors can be obtained through the coefficient of the x^3 term. Figure 04 below shows the end behavior of the graph of a cubic function with a leading coefficient that is positive or negative. To further visualize the effect of the sign of the leading coefficient on a cubic function graph, take a look at Figure 05 below. Notice the difference in the behavior of the graph when the leading coefficient is a positive value versus when it is a negative value. Now that you are familiar with the characteristics of the graph of a cubic function, including roots, critical points, the inflection point, and end behavior, let’s take a step-by-step approach to a few examples of graphing a cubic function using a simple 3-step process. Graphing a Cubic Function Example #1 Example #1: Graph f(x) = 2x^3 + 3x^2 - 8x +3 To graph a cubic function like the one given in this first example, you can use the following 3-step method: Step 1: Identify the intercepts Step 2: Determine the critical points Step 3: Draw the Curve and Extend Let’s go ahead and apply these three steps to the given function f(x) = 2x^3 + 3x^2 - 8x +3 as follows: Step 1: Identify the intercepts To find the y-intercept, figure out the value of the function when x=0 as follows: f(0)=2(0)^3 +3(0)^2 - 8(0) + 3 ➝ f(0)= 0 + 0 + 0 + 3 ➝ f(0)= 3 So, the y-intercept of the cubic function is at 3. To find the x-intercepts, figure out the value of the function when f(x)=0 as follows: 2x^3 + 3x^2 - 8x +3 = 0 It is easier to determine the roots of this cubic function by first factorizing the equation as shown below: (x-1)(x+3)(2x-1)=0 Now the solutions to the equation become more apparent and can be obtained by equating each factor to 0. x-1 = 0 ➝ x=1 x+3 = 0 ➝ x=-3 2x -1 = 0 ➝ x=0.5 So, the x-intercepts of the function are at 1, 3, and 0.5. As shown in Figure 06, after completing Step 1 by finding the y-intercept and the x-intercepts, you know that the cubic function graph will pass through the points (0,3), (-3,0), (0.5,0), and (1,0). Now let’s continue onto the next step. Step 2: Determine the Critical Points First, let’s work out the derivative of f(x) by applying the power rule as follows: f(x) = 2x^3 + 3x^2 - 8x +3 f’(x)= 6x^2 +6x -8 At the stationary points, the derivative (gradient) is 0. Therefore, to obtain the x-coordinates of the stationary points we can equate the derivative f’(x) to 0 by setting f’(x)=0 and solving for x as follows: 6x^2 +6x -8 = 0 You can solve for x by using the quadratic formula: x = [−b ± √(b2 − 4ac)]/2a By using the quadratic formula, you can conclude that the x-coordinates at the critical points are x≈0.758 and x≈-1.758 Next, let’s substitute the x-coordinates calculated into f(x) and obtain the y-coordinates of the stationary points as follows: f(0.758) = 2(0.758)^3 + 3(0.758)^2 - 8(0.758) +3 ➝ f(0.758) ≈ -0.469 f(-1.758) = 2(01.758)^3 + 3(-1.758)^2 - 8(-1.758) +3 ➝ f(0.758) ≈ 15.469 Therefore, the critical points are at approximately (0.758 , -0.469) and (-1.758 , 15.469) As shown in Figure 07, in addition to the y-intercept and x-intercepts of the function, you also have the critical points plotted. Notably, (0.758 , -0.469) is a local minimum and (-1.758 ,15.469) is a local maximum. Now you are ready to draw your graph and complete the problem. Step 3: Draw the Curve and Extend The final step is to draw a smooth curve that passes across all points. Before you extend the lines of the graph, it is a good practice to check the end behaviors. Since the leading coefficient (2) of the function is positive the following will be true: Next, extend the line as needed to cover the relevant domain and range of the function. You have just completed the cubic function graph for f(x) = 2x^3 + 3x^2 - 8x +3, as shown in Figure 09 above. Before we move onto to working through one more graph a cubic function example, let’s quickly recap how we completed Example #1 using the 3-step method as shown in Figure 10 below. Figure 10: Visual recap of the 3-step method used to complete the function graph for Example #1. Graphing a Cubic Function Example #2 Example #2: Graph f(x) = -x^3 + 8 The steps for graphing a cubic function like the one shown in Example #2 are the same three steps that we used in the previous example. Step 1: Identify the intercepts To find the y-intercept, calculate the value of the function when x=0 as follows: f(0)= -(0)^3 + 8 ➝ f(0)= 0 + 8 ➝ f(0)= 8 So, the y-intercept of the cubic function is at 8. To find the x-intercepts, figure out the value of the function when f(x)=0 as follows: -x^3 + 8 = 0 ➝ x^3 = 8 Hence, the x-intercepts are x = 2, x =-1+√3i , and x = -1-√3i But, since you are graphing a cubic function on the real plane, you can ignore the two complex roots. Thus, the graph will contain a single real x-intercept at x = 2. So, the x-intercept of the function is 2. As shown in Figure 11, after completing the first step by finding the y-intercept and the x-intercepts, you know that the cubic function graph will pass through the points (0,8) and (2,0). Step 2: Determine the Critical Points First, let’s work out the derivative of f(x) by applying the power rule as follows: f(x) = -x^3 + 8 f’(x)= -3x^2 At the stationary points, the derivative (gradient) is 0. Therefore, to obtain the x-coordinates of the stationary points we can equate the derivative f’(x) to 0 by setting f’(x)=0 and solving for x as follows: -3x^2 = 0 ➝ x = 0 Next, let’s substitute the x-coordinates calculated into f(x) and obtain the y-coordinates of the stationary points. f(x) = -x^3 + 8 f(0) = -(0)^3 + 8 ➝ f(0) = 8 Therefore, the stationary point is (0, 8). Next, let’s work out the inflection point of the curve. The x-coordinate of the inflection point is given by x=-b/3a ➝ x=-0/-3 = 0 The x-value at the inflection point: x=0 The y-coordinate of the inflection point can be obtained by calculating f(0), which you already know is equal to 8. The y-value at the inflection point: x=8 Therefore, the inflection point is at (0,8). In this example, the inflection point is identical to the stationary point obtained above. This means that the cubic function f(x) doesn’t have any local maximum or local minimum points but rather a single inflection point at (0,8) Step 3: Draw the Curve and Extend Now you are just about ready to construct your cubic function graph. You have already identified two distinct points that the cubic function will pass through (x-intercept and inflection point). To determine a few additional points that the graph will pass through and improve the accuracy of the curve that you will, you can find several other x and y coordinate pairs. Choose random values of x and calculate the corresponding value of f(x). For example, to calculate the y-value when x=-1: f(x) = -x^3 + 8 f(-1) = -(-1)^3 + 8 = 9 ➝ f(-1) = 9 Therefore, the graph will pass through the point (-1,9) You can repeat this process for additional x-values to figure out more points that the cubic function graph will pass through. The function table shown in Figure 13 includes the three points on the curve that we have already found—(0,8), (2,0), and (-1,9), plus two additional points: (-2,16) and (1,7). Finally, plot all of the points that you have found and then draw a smooth curve that passes through all of them. Since the leading coefficient (-1) of the function is negative, remember to follow the end behaviors accordingly. The graph in Figure 14 below shows the completed graph of the cubic function f(x)=-x^3 + 8. Graphing a Cubic Function: Conclusion The process to graph a cubic function can be summarized to following the following simple 3-step method: Identify the y-intercept and the x-intercept(s) Determining the critical points Draw the curve and extend By applying these simple steps, anyone can accurately graph a cubic function and gain insights into its behavior. Mastering the art of graphing cubic functions not only strengthens your understanding of polynomials, but it also provides valuable tools for analyzing real-world situations and making informed decisions. Keep Learning: How to Graph a Parabola in 3 Easy Steps (Quadratic Functions) Parent Functions and Parent Graphs Explained How to Graph ANY Function in 3 Easy Steps (Rational, Logarithmic, Exponential, etc.) Examples: Which Graph Represents a Function? The Vertical Line Test Explained What is the Formula for Slope? (and how to use it) More Math Resources You Will Love: Featured 10 Fun Math Projects for All Grade Levels Read More → How to Divide Fractions in 3 Easy Steps Read More → 10 Super Fun Math Riddles for Kids! (with Answers) Read More → How to Turn a Percent into a Fraction in 3 Steps Read More → How to Turn a Fraction into a Percent in 2 Steps Read More → How to Convert CM to Inches in 3 Easy Steps Read More → There's a Woman in a Boat Riddle—Answer and Explanation Read More →
188361	https://core.ac.uk/download/pdf/289974595.pdf	Trabajo Fin de Máster El objeto directo preposicional y los verbos de percepción visual: factores que favorecen la presencia/ausencia de a delante del objeto directo en español actual Autora María Elena Cebrián Auré Directora Dra. María Antonia Martín Zorraquino Facultad de Filosofía y Letras 2012/2013 brought to you by CORE View metadata, citation and similar papers at core.ac.uk provided by Repositorio Universidad de Zaragoza 3 ÍNDICE INTRODUCCIÓN 6 PRIMERA PARTE. REVISIÓN CRÍTICA DE LOS ESTUDIOS DIACRÓNICOS Y SINCRÓNICOS SOBRE LA PREPOSICIÓN A ANTE OBJETO DIRECTO EN ESPAÑOL 1. Los estudios sobre el origen y la evolución del uso de a delante del objeto directo en español 12 1.1. Origen y evolución del uso de a ante objeto directo en español: del latín al español actual 13 1.2. Los estudios sobre la presencia/ausencia de a ante objeto directo en castellano medieval 20 2. La presencia de a ante objeto directo en español actual 32 2.1. El objeto directo preposicional: definición y cuestiones problemáticas 32 2.2. Diferentes propuestas para explicar los casos problemáticos del objeto directo preposicional 35 2.2.1. La preposición a como marca diferenciadora de función en el objeto preposicional (frente a otras funciones primarias) 35 2.2.2. La presencia del objeto directo preposicional en dependencia de ciertas características del elemento nominal objeto directo 38 2.2.2.1. Nombres propios de persona o animal y pronombres personales: ámbito de obligatoriedad de la presencia de la preposición a 39 2.2.2.2. Nombres propios de nación o ciudad: algunas vacilaciones en la presencia/ausencia de la preposición a 40 2.2.2.3. Nombres comunes de persona: ámbito de vacilación en la 41 4 presencia/ausencia de la preposición a 2.2.2.4. Nombres comunes de animal y cosas personificadas: ámbito de vacilación en la presencia/ausencia de la preposición a 43 2.2.2.5. Grupos nominales indefinidos: ámbito de especial vacilación en la presencia/ausencia de la preposición a 44 2.2.3. El objeto directo preposicional en dependencia de ciertas características del núcleo verbal 46 2.2.4. El objeto directo preposicional en dependencia de ciertas características del sujeto del verbo que lo rige 56 SEGUNDA PARTE. ESTUDIO DEL USO DE LA PREPOSICIÓN A ANTE OBJETO DIRECTO CON VERBOS DE PERCEPCIÓN VISUAL EN ESPAÑOL ACTUAL 3. Sobre las características del corpus objeto de estudio 59 3.1. Datos cuantitativos y cualitativos de los ejemplos analizados 59 3.2. Caracterización de los verbos objeto de estudio 63 4. Los verbos de percepción visual y el objeto directo de persona 71 4.1. El objeto directo de persona no colectiva con preposición a 71 4.2. El objeto directo de persona no colectiva sin preposición a 73 4.3. El objeto directo de persona colectiva con preposición a 75 4.4. El objeto directo de persona colectiva sin preposición a 77 5. Los verbos de percepción visual y el objeto directo de lugar 81 5.1. El objeto directo de lugar con preposición a 81 5.2. El objeto directo de lugar sin preposición a 83 5 5.3. Los nombres de astros y constelaciones 84 6. Los verbos de percepción visual y el objeto directo referido a un animal o a varios animales 87 7. Los verbos de percepción visual y el objeto directo de cosa 90 7.1. El objeto directo de cosa sin preposición a 90 7.2. El objeto directo de cosa con preposición a 92 7.2.1. Algunos casos con el verbo contemplar 92 7.2.2. Algunos aspectos relevantes de los giros con el verbo mirar 94 CONCLUSIONES 99 BIBLIOGRAFÍA 105 APÉNDICE i 6 INTRODUCCIÓN El presente trabajo aborda el estudio del llamado objeto directo preposicional en español actual y los factores que favorecen su presencia con un grupo de verbos determinado. Se trata de un tema de la gramática española que hace un tiempo llamó mi atención a raíz de las dudas que me planteaban algunos usuarios del español en el desarrollo de mi profesión como traductora. El interés que despertó en mí este fenómeno lo convirtió en un tema muy atractivo para la realización de un trabajo de fin de máster como el que nos ocupa, pues me brindaría la posibilidad de profundizar en el problema y de entender mejor su funcionamiento, lo cual me permitiría dar respuesta a las preguntas que motiva. La revisión crítica de la bibliografía esencial sobre el tema ha dejado clara la dificultad y problematicidad que entraña. El objeto directo preposicional en español actual todavía no se encuentra totalmente resuelto o gramaticalizado. La complejidad de este fenómeno tan característico del español se puede expresar en los siguientes términos: por un lado, a pesar de que esta construcción no se puede explicar de manera sistemática, existen casos que responden clara y sistemáticamente a unas normas concretas; por otro lado, la cantidad y variedad de ejemplos que no encuentran respuesta en dichas normas ha dado lugar a que se hayan postulado diferentes explicaciones, sin que ninguna de ellas aclare de forma general todas las excepciones, de modo que nos encontramos con explicaciones múltiples para los múltiples casos de vacilación. Por ello, para realizar nuestro trabajo, hemos decidido acotar el análisis a un conjunto de ejemplos con verbos pertenecientes a una misma familia semántica. Así, hemos podido reunir un corpus en el que todos los verbos tienen una o más características en común, que nos sirva o sirvan como nexo para establecer conclusiones al final de nuestra investigación. Los ejemplos de nuestro corpus propio han sido extraídos de la base de datos del CREA de la Real Academia Española. Se trata de ejemplos datados entre los años 1990 y 1995 en España y abarcan todo tipo de textos y de temas. Estos ejemplos se recogen en el “Apéndice” que adjuntamos al final de este trabajo. 7 A la vista de las postulaciones encontradas en la bibliografía, nos hemos planteado una serie de preguntas que hemos intentado responder a través del análisis de nuestros ejemplos. Estas preguntas se dividen en tres grupos, que corresponden a las tres categorías principales en las que la bibliografía revisada divide los factores o condiciones que favorecen la presencia o la exclusión de la preposición a delante del objeto directo. La primera categoría corresponde a las características del objeto directo que favorecen la presencia de la preposición. En primer lugar, nos hemos planteado si el carácter [±humano] del elemento que funciona como objeto directo determina significativamente la presencia de a en nuestro corpus. En particular, nos hemos centrado en el carácter [±determinado], pues los gramáticos han observado vacilaciones en los casos que hacen referencia a personas no determinadas, y en el carácter [±colectivo] de los objetos directos [+humanos], pues comprobamos que algunos autores han postulado que la individuación puede influir o ha influido en la presencia de a delante del objeto directo a lo largo de la historia del fenómeno. En segundo lugar, hemos analizado si las propuestas que algunos autores han destacado con respecto a los nombres geográficos –ámbito que ha mostrado vacilaciones a lo largo de la historia del fenómeno‐ se cumplen en nuestro corpus. En tercer lugar, hemos estudiado si los objetos directos que hacen referencia a animales se ajustan también al carácter [± familiar] del objeto, postulación que plantean algunos autores. Y, por último, nos hemos planteado hasta qué punto los objetos directos [‐animados] aportan fijeza a la exclusión de la preposición con nuestro grupo de verbos. La segunda categoría que suscita problemas, dentro del tema que nos ocupa, se refiere a los verbos que rigen el objeto directo y a los factores que hacen que estos exijan o no la presencia de a. A este respecto nos hemos planteado si el aspecto léxico del verbo en cuestión tiene alguna influencia sobre la presencia de la preposición a y si este aspecto puede verse modificado por la presencia de un objeto directo preposicional. Por último, con respecto a la tercera categoría problemática –y discutida‐, que corresponde a las características del sujeto que favorecen la presencia de la preposición a delante del objeto directo, nos hemos 8 preguntado si la agentividad o no agentividad del sujeto tiene o no alguna influencia en la presencia de la preposición a con los ejemplos de nuestro corpus. Nuestro trabajo se ha dividido en dos partes. La primera de ellas está dedicada a la revisión crítica de la bibliografía sobre el tema. Esta primera parte de nuestro trabajo ha requerido un gran esfuerzo y ha resultado muy extensa, debido, sobre todo, a la complejidad que presenta nuestro objeto de estudio. Además, hemos considerado que un primer trabajo de investigación como el que nos ocupa debe servirnos para probar nuestra capacidad de analizar críticamente la bibliografía que trata del tema sometido a estudio, sobre todo, porque esto nos preparará para, en el futuro, poder emprender la ardua tarea de realizar una tesis doctoral. En la segunda parte de nuestro trabajo –más personal‐ hemos tratado de dar respuesta a las preguntas que nos hemos planteado como punto de partida de este trabajo. Gracias al análisis crítico de nuestra bibliografía, a nuestro corpus y a nuestras propias reflexiones, hemos podido obtener las conclusiones que explicamos a lo largo del análisis y que recogemos al final del mismo. Son unas conclusiones que consideramos modestas, pero satisfactorias, ya que reflejan que hemos podido probar la validez de algunas de las postulaciones realizadas por los estudiosos y plantear postulaciones propias para un grupo reducido y concreto de verbos. No nos hemos adscrito a una teoría lingüística o a una metodología lingüística definidas, pero nos ajustamos a los principios generales de la gramática descriptiva. Este trabajo revela una cierta adscripción a la gramática funcional, pero consideramos categorías lingüísticas apenas analizadas en este enfoque y presentes en las gramáticas descriptivas más recientes, como los capítulos respectivos de Esther Torrego y de Elena de Miguel de 1999, recogidos en la Gramática descriptiva de la lengua española (1999), o como la Nueva gramática de la lengua española (2009) de la Real Academia Española. Por tanto, no hemos adoptado una postura estricta y restringida a un enfoque, sino que hemos preferido realizar una revisión bibliográfica que nos dotara de las herramientas 9 necesarias para abordar nuestro propio análisis de los ejemplos y responder a las preguntas que nos hemos planteado en nuestro estudio. Antes de concluir, tenemos que aclarar que, a lo largo de este trabajo, hemos utilizado indistintamente las siguientes expresiones o fórmulas: “objeto directo preposicional”, “presencia/ausencia de a ante el objeto directo”, “a delante del objeto directo”, “a en el objeto directo”, etc. Esto se debe a que no hemos podido llegar a una conclusión definitiva sobre el estatuto de a en el objeto directo preposicional. El mismo término objeto directo preposicional refleja un cierto posicionamiento por parte de los lingüistas que se han ocupado del tema; es decir, esta denominación implica que a representa un elemento integrado en el constructo funcional designado, mientras que fórmulas como “la preposición a ante el objeto directo” o “la preposición a delante del objeto directo” implican que a se considera una preposición que marca a ciertos tipos de objeto directo. En el caso del objeto preposicional, a se comporta de modo diferente a como lo hace ella misma y otras preposiciones (con las que puede alternar) en otras funciones oracionales como, por ejemplo, en la de adyacente o complemento circunstancial (cf. voy /a, hacia, hasta, por (...) / los campos de mi vecino frente a vio a dos niños, vio dos niños o vio /hacia, sobre, por/ dos niños). En la bibliografía revisada son muchos los autores que utilizan la expresión “la preposición a ante el objeto directo” o “la preposición a delante del objeto directo”, etc.; sin embargo, Esther Torrego (1999) considera al signo a como una partícula y Alarcos (1994) lo denomina “marca funcional”, papel o estatuto que dicho autor otorga, en general, a las preposiciones. Está claro que la a del objeto directo preposicional no se comporta, según hemos señalado, como lo hace al introducir complementos circunstanciales, pues, en estos casos, puede alternar con otras preposiciones y refleja un contenido direccional específico. En el objeto directo preposicional, la preposición a ha sufrido un claro proceso de gramaticalización que la convierte en una mera marca funcional y que la integra en el propio objeto directo; sin embargo, parece poder alternar con ∅, es decir con su no presencia o ausencia. Esta posibilidad alternativa nos inclina a mantener, de momento, la entidad preposicional de a en 10 el objeto directo preposicional. No podemos profundizar más en esta cuestión. Por eso, a lo largo del trabajo utilizaremos indistintamente las fórmulas o expresiones citadas más arriba, puesto que, en la bibliografía consultada, todas ellas están presentes. Queda, pues, como una cuestión abierta que habremos de abordar, sin duda, en nuestra tesis doctoral y para la que, por cierto, contamos con trabajos relativamente recientes de investigadores que se hallan en la Universidad de Zaragoza y que son autores de estudios de gran envergadura. (Nos referimos especialmente a Horno Chéliz, 2002). Por último, queremos dedicar unas palabras de agradecimiento a las personas que han hecho posible la realización de este estudio. En primer lugar, a la directora de este trabajo, la doctora María Antonia Martín Zorraquino, por su paciencia, por sus correcciones, por confiar en mi capacidad para realizar este trabajo y por haber encontrado el tiempo para ayudarme a completarlo. También quiero dar las gracias a los profesores del Máster en Estudios Hispánicos: Lengua y Literatura y al coordinador del mismo, el doctor Vicente Lagüéns Gracia, que han realizado, a nuestro juicio, una gran selección de contenidos y han sabido ser exigentes con los alumnos, y a todos los compañeros, que han hecho que fuera una grata experiencia. Igualmente, deseo dar las gracias a los miembros del Tribunal, por dedicar su tiempo a leer este trabajo y a ofrecerme sus observaciones, que serán, sin duda, de gran interés y utilidad para mí. También quiero expresar mi gratitud, de todo corazón, a mi familia y amigos, por su apoyo y su ayuda; a mis padres, por haberme inculcado el valor del esfuerzo y por haberse preocupado siempre de lo más importante: la educación; y especialmente a Daniel, por haber invertido su tiempo para que yo pudiera realizar este trabajo y algún día pueda defender mi propia tesis doctoral, por su apoyo incondicional y por hacer que la vida sea mejor. Finalmente y sobre todo, gracias a ti, Iris, porque con tu nacimiento has hecho que la vida tenga otro significado y me has dado la motivación para que acabe este trabajo y me siga planteando desafíos como los que te planteará a ti este mundo, del que espero saques lo mejor. Gracias. PRIMERA PARTE. REVISIÓN CRÍTICA DE LOS ESTUDIOS DIACRÓNICOS Y SINCRÓNICOS SOBRE LA PREPOSICIÓN A ANTE OBJETO DIRECTO EN ESPAÑOL 12 1. LOS ESTUDIOS SOBRE EL ORIGEN Y LA EVOLUCIÓN DEL USO DE A DELANTE DEL OBJETO DIRECTO EN ESPAÑOL La presencia de a ante objeto directo en español constituye un fenómeno muy interesante que nuestra lengua no comparte con otras lenguas románicas, como el italiano o el francés, o con otras lenguas también indoeuropeas, pero del tronco anglo‐germánico, como el inglés o el alemán. No se trata de un fenómeno exclusivo del español, pues el rumano, entre otras lenguas, también cuenta con un objeto directo preposicional, en su caso introducido por la preposición p(r)e (cf. Pensado, 1995: 14‐15)1. Carmen Pensado (op. cit., 15, n. 5)2, por ejemplo, señala una serie de variedades lingüísticas que conocen fenómenos similares vinculados fundamentalmente con los objetos directos más animados, aunque con diferentes grados de intensidad, como el macedorrumano, algunos dialectos del sur de Italia, el italiano estándar coloquial, el corso, el portugués, el sardo, el gallego, el gascón y el catalán, entre otros3. También otros autores han encontrado fenómenos paralelos en lenguas más lejanas como el hebreo, el eslavo o el persa moderno (cf. Pensado, op. cit., 16). El origen del fenómeno todavía no está claro y tampoco lo está su funcionamiento en español, que presenta grandes variaciones que podemos tratar de interpretar, pero para las que no tenemos una respuesta clara y única. Por eso es abundante la bibliografía que se ocupa de las causas que determinan el empleo de dicha preposición ante el objeto directo en español y que trata de identificar el origen y los primeros ejemplos de su uso, así como la evolución del 1 Para las referencias citadas a lo largo de todo el trabajo, remitimos a la bibliografía incluida al final del mismo. 2 En El complemento directo preposicional (1995), Carmen Pensado ofrece una panorámica de las aproximaciones más recientes al complemento directo preposicional a través de una serie de obras que enfocan el tema desde diferentes perspectivas: “la descripción de su uso, tanto en español contemporáneo como en el de época medieval; su valor semántico actual y cómo llegó a constituirse; el problema de su origen y su motivación”. El primer trabajo que recoge el libro, que es obra de la misma autora, constituye un estado de la cuestión sobre el complemento directo preposicional en español que incluye además una bibliografía comentada sobre el mismo. 3 Otros autores que señalan las variedades lingüísticas en las que se encuentra este fenómeno o un fenómeno similar son Lapesa (1964: 76) y Martín Zorraquino (1976: 556, n.1): en el Centro y Mediodía de Italia, en Sicilia y Cerdeña, en retrorromano, triestino, algunos dialectos provenzales, en el francés de Friburgo y Bruselas y sobre todo en portugués y en catalán, aunque con menos frecuencia que en español. 13 mismo hasta el presente. En este capítulo presentamos una síntesis de los trabajos más relevantes que analizan el uso de a ante objeto directo en español desde una perspectiva histórica. El objeto directo preposicional no es un fenómeno de reciente aparición. Su presencia se ha constatado a lo largo de la historia del español con más o menos vacilaciones y encuentra su origen en el latín. Para acercarnos a la evolución diacrónica del objeto directo preposicional, nos hemos apoyado, en primer lugar, en los datos que aporta el Diccionario histórico de la lengua española de la Real Academia Española (1960‐1996)4 sobre la preposición a cuando funciona como introductora del objeto directo y en el trabajo de Rafael Lapesa (1964), que nos ofrece, entre otros contenidos, un resumen de cómo cada caso del latín evoluciona y da lugar a ciertas construcciones de nuestra lengua5. Además, también hemos revisado las contribuciones de Carlos Folgar (1988), José María García Martín (1992), María Antonia Martín Zorraquino (1976), Carmen Monedero Carrillo de Albornoz (1978 y 1983) y Carmen Pensado (1995). Y, por supuesto, hemos tenido muy en cuenta la Historia de la lengua española de Rafael Lapesa (1981)6. 1.1. Origen y evolución del uso de a ante objeto directo en español: del latín al español actual Lapesa (1981) nos ofrece una visión sintetizadora del origen y desarrollo del uso de a ante objeto directo en español7. Según el autor, (op.cit., § 22, 94), el uso de a 4 En adelante nos referiremos a dicha obra de la Real Academia Española con la abreviatura Dicc. Hco. 5 “La simplificación de la flexión nominal latina con desuso de la distinción casual constituye uno de los procesos más transcendentales para la formación de las lenguas románicas” (cf. Lapesa, 1964: 55). Lapesa, en su trabajo “Los casos latinos: restos sintácticos y sustitutos en español”, analiza este proceso, que comienza a gestarse ya en los orígenes mismos del latín y que se sirve de dos procedimientos fundamentales en español: los “restos sintácticos” y la sustitución de los casos flexivos por otras construcciones que suplieran la información funcional que aportaban aquellos. 6 Citamos, como habrá podido deducirse, por la última edición de esta obra, publicada por vez primera en 1942. 7 Existen otras obras y hay otros autores que nos ofrecen también una visión global de este fenómeno en español. Anterior es, por ejemplo, la aportación del Dicc. Hco. de la Real Academia Española, que nos ofrece una extensa, minuciosa y matizada presentación de la evolución del uso de a ante el objeto directo, bajo la voz a. También anterior es el trabajo de Lapesa “Los casos 14 delante del objeto directo se documenta ya en el latín vulgar y se cuenta entre las particularidades del latín hispánico. Lapesa (op.cit., § 22, 94) explica que las lenguas iberorrómanicas y los dialectos del sur de Italia mostraban varios rasgos morfológicos y sintácticos en común, como son el sistema y las formas de los tres demostrativos y el “empleo de la preposición a ante objeto directo que designe persona individuada («si vvisto a ffrátimo?» ‘¿has visto a mi hermano?’)”. Durante la etapa que Lapesa denomina del español arcaico8, el uso de a ante el objeto directo era ya general con los pronombres tónicos y los nombres propios de persona (“a ti adoro”, “salvest a Daniel”, Cantar de Mio Cid), pero con los nombres comunes de persona y los propios geográficos fluctuaba según existieran o no individualizadores, énfasis, mayor o menor carga afectiva, etc.: “a quatro matava” y “matarás el moro”; “gañó a Valençia” y “el que Valençia gañó” (Cantar de Mio Cid) (cf. Lapesa, op.cit., § 56, 187). Durante el Siglo de Oro, la inserción de a ante el acusativo de persona y cosa personificada se extiende. Juan de Valdés reprueba la omisión de a en “el varón prudente ama la justicia”, y afirma que “puede tener dos entendimientos: o que el varón prudente ame a la justicia, o que la justicia ame al varón prudente, porque sin la a parece que estén todos los nombres en el mesmo caso”9. No obstante, Lope de Vega usa: “no disgustemos mi abuela”, y Quevedo: “acusaron los escribas y fariseos la mujer adúltera”. Se observa una clara vacilación entre ambos usos, sujeta probablemente a la elección de cada autor (cf. Lapesa, op.cit., § 97, 341‐ 342). En síntesis, el uso de a ante objeto directo de persona se extiende y consolida a lo largo del tiempo. En el español medieval y clásico había alternancia en su uso latinos: restos sintácticos y sustitutos en español” de 1964. En él, explica cómo se ha llegado a diversas estructuras del español actual partiendo de los casos latinos y de las funciones sintácticas que expresaban. Una de estas estructuras es la del objeto directo preposicional en español actual. 8 Dadas las dificultades que entraña la periodización y a pesar de que haya autores que no acepten las etapas que propone Lapesa, en esta síntesis nos acogemos a ellas. 9 La cita procede del Diálogo de la lengua de Juan de Valdés, cuya primera edición puede ubicarse en 1535. La he tomado de Lapesa, 1981: § 97, 341‐ 342, y de Monedero Carrillo de Albornoz, 1978: 261, quienes remiten a la edición de la obra de Valdés en Clásicos Castellanos, Espasa Calpe, Madrid, 1964. 15 delante de nombre de persona y de lugar. En los siglos XVI y XVII predomina el uso de a, pero todavía es frecuente su omisión. Desde el siglo XVIII la ausencia de a es rara fuera de la compleja casuística aún hoy vigente (cf. Lapesa, 1964: 77‐ 78). El origen de este fenómeno en las lenguas románicas, y en particular en español, se remonta y se explica a partir de la desaparición y sustitución de los casos de la flexión latina. La declinación latina ofrecía seis casos diferentes en los que se englobaban tipos muy variados de relaciones. Tal y como nos explica Lapesa (1964: 57‐59), ya en los orígenes del latín se empezaron a introducir preposiciones para especificar qué tipo de relación representaba la desinencia de cada caso. También desde la literatura arcaica latina construcciones con preposición + acusativo y ablativo invadían el terreno del dativo, el ablativo y el genitivo. Las preposiciones aportaban una mayor precisión que hacía inútil a veces la distinción de los casos. A lo largo del tiempo, el orden de la oración se hizo menos libre y el uso de las preposiciones más intenso. Las desinencias casuales eran cada vez menos necesarias. En las áreas iberorrománicas, cada sustantivo quedó con una forma única para cada número sin valor casual: la del acusativo. Si quedó algún resto de otro caso en algún sustantivo, no estaba vinculado con la función casual de origen. En este proceso de sustitución de los casos latinos por construcciones preposicionales que expresasen las funciones que expresaban aquellos, Lapesa observa que hay varias estructuras de origen latino que favorecen la existencia de dos objetos directos diferentes formalmente en español: de cosa y de persona, el segundo introducido por la preposición a. En primer lugar, algunos verbos latinos podían construirse tanto con dativo como con acusativo sin preposición como, por ejemplo, adulāri, auscultāre, curāre, imitāri, etc. (cf. Lapesa, 1964: 75). En español actual, los verbos correspondientes se utilizan como transitivos con objeto directo sin preposición para cosas y con preposición a para personas: “adular la vanidad” frente a “adular al poderoso” o “escuchar sus palabras” frente a “escuchar al desdichado” 16 (ibídem). Otros verbos latinos que también admitían tanto acusativo como dativo, como adjutāre, assistĕre y servīre, también presentan en español actual una construcción transitiva en la que el complemento directo lleva a incluso ante complemento de cosa: “asistir a las necesidades de los pobres”, “servir a una causa justa”, etc. (cf. Lapesa, 1964: 76). Además, algunos verbos intransitivos en latín, como oboedīre, resistĕre y respondĕre, se emplean en español actual con acepciones transitivas, pero muestran variación en la presencia o ausencia de la a: “obedecer las órdenes” frente a “obedecer a las órdenes” o “resistir las amenazas” frente a “resistir a las amenazas” (ibídem). Con “responder” domina el ejemplo “responder a las preguntas”, aunque también se dan ejemplos sin preposición (ibídem). La presencia de a se observa en las diferentes etapas de la transición entre la construcción latina y la actual en ejemplos como “Que el sennor de la tierra deve aiudar a los iueces por prender los malfechores” (en Fuero Juzgo, c1260, apud Dicc. Hco., s.v. a2, § 84) ante objeto de persona, o como “Los árboles y el viento / al sueño ayudan con su movimiento” (en Garcilaso, Égloga II, apud Lapesa, 1964: 76) ante objeto de cosa. En segundo lugar, algunos verbos latinos como docēre, celāre, poscĕre o rogāre regían doble acusativo de persona y de cosa (cf. Lapesa, 1964: 82‐83). Esta construcción se ha sustituido en español actual por un régimen de objeto directo de cosa e indirecto de persona: “enseñar algo a alguien”, “ocultar algo a alguien”, “preguntar algo a alguien” o “rogar algo a alguien”. Durante la transición entre la construcción latina y la que se da en español actual, el proceso todavía no se había consumado en el siglo XII, pero estaba ya muy avanzado: en el Cantar de Mio Cid, cuando rogar aparece con los dos complementos, de persona y de cosa, en forma pronominal, el de persona aparece en dativo (“rogar gelo emos”) y, si el complemento de persona es un nombre propio, aparece siempre introducido por la preposición a (“ruego a San Pedro que me ayude”) (cf. Lapesa, op.cit., 83). Otros verbos latinos con doble acusativo eran aquellos en los que el verbo regía, además del objeto directo, un predicado de este. Era el caso de eligĕre, nomināre o iudicāre (cf. Lapesa, op.cit., 83). En su transición hasta el español actual, con frecuencia se encuentra la preposición a ante el objeto directo si es nombre de 17 persona: “e fizieron cabdiello della a Galba” (en Alfonso X, Primera Crónica General de España, apud Lapesa, op.cit., 83‐84); y el pronombre de acusativo si es un pronombre átono: “Que la ficiese Dios fablante e uidient” (en Berceo, Vida de Santo Domingo de Silos, apud Lapesa, op.cit., 84). En español actual, esta estructura suele presentar un régimen directo introducido por la preposición a, a veces incluso si es de cosa, y un predicativo: (“elegir a alguien presidente”, “llamar a algo de una determinada manera”). En tercer lugar, tenemos aquellos verbos latinos que regían una oración subordinada con un verbo en infinitivo cuyo sujeto no era el mismo que el de la oración principal. Eran verbos de mandato, consentimiento, prohibición, percepción y causativos en general: imperāre, permittĕre, concedĕre, facĕre, vidēre, etc. (cf. Lapesa, 1964: 85). El sujeto del infinitivo subordinado aparecía en acusativo o dativo según el régimen del verbo principal. Esta estructura se ha heredado en español actual y, en ella, el sujeto del infinitivo aparece con preposición si es de persona (“hacer a alguien llorar”, “dejar a alguien marchar”). Si el sujeto del infinitivo es de cosa, puede aparecer con la preposición o sin ella (“ver (a) los trenes pasar”). En el proceso de transición de esta estructura desde el latín hasta hoy, se han dado vacilaciones y transgresiones de las reglas imperantes (cf. Lapesa, op.cit., 85‐86), probablemente debidas a la complejidad de las combinaciones: si el sujeto del verbo subordinado era de cosa, aparecía normalmente como objeto directo sin preposición y, si era un pronombre átono, en acusativo (“Fazen los coraçones de los omnes rauiar de duelo”, en Alfonso X, Primera Crónica General de España, apud Lapesa, op.cit., 85), aunque también se encuentran casos con la preposición a (“Verán a las estrellas caer de su logar”, en Berceo, Signos del juicio final, apud Lapesa, op.cit., 85); si el objeto del verbo subordinado era de persona, dependía de si el infinitivo tenía objeto directo propio o no: si no lo tenía, el sujeto del infinitivo hacía la función de objeto directo del verbo principal y aparecía sin preposición (“Aquí veríedes quexarse ifantes de Carrión”, en Cantar de Mio Cid, apud Lapesa, op.cit., 85) y, si era pronombre, en acusativo (“Por nulla ionglería no lo faríen reyir”, en Berceo, Vida de Santo Domingo de Silos, apud Lapesa, op.cit., 85). Si el infinitivo tenía objeto directo, su sujeto era un objeto indirecto del verbo principal e iba introducido 18 por a o estaba representado por un pronombre de dativo (“Tu muert oí conssejar a infantes de Carrión”, en Cantar de Mio Cid, apud Lapesa, op.cit., 85), aunque también hay ejemplos de transgresión de estas fórmulas, por ejemplo, nos encontramos casos con infinitivo sin objeto directo propio, pero cuyo sujeto, en cuanto objeto directo del verbo subordinante, lleva la preposición a, como, por ejemplo, en “Vido venir a Diego y Fernando” (ibídem). Son muchos los autores que han estudiado el sentido que tuvo la aparición de este fenómeno. La mayoría de ellos considera que la aparición originaria de esta construcción está vinculada a los objetos de persona, es decir, que el uso de a se explica como consecuencia de haberse creado una categoría especial para el objeto de persona. Diez, Brauns y Hills10 lo consideran un procedimiento para evitar confusiones entre sujeto y objeto personal, pues la flexibilidad a la hora de ordenar las palabras dentro de la oración puede dar lugar a ambigüedades y dobles interpretaciones11. Para Meyer‐Lübke (Grammatik der Romanischen Sprachen, III, 1890‐1901: § 350, 371, apud Lapesa, op.cit., 78), se trata de una sustitución del dativo de interés latino para separar las categorías de seres animados e inanimados. El español separa gramaticalmente las categorías de seres animados (interesados en la acción del verbo) e inanimados (alcanzados por ella), de modo que podemos deducir que, para este autor, el objeto directo de persona tiene una mayor implicación o interés en la acción del verbo que el de cosa. Kalepky (“Präpositionale Passivobjekte im Spanischen, Portugiesischen und Rumänischen”, ZRPh, XXXVII, 1913: 358‐364, apud Lapesa, op.cit., 78‐79) ve en la preposición a del objeto directo preposicional un uso locativo de dirección con el sentido de ‘hacia’, es decir, encuentra en la a una cierta idea de dirección. Spitzer (“Rum. p(r)e, span. a vor persönlichem Akkusativobjekt”, ZRPh, XLVIII, 1928: 10 Los autores citados aparecen comentados por Lapesa 1964: 78 y por Martín Zorraquino, 1976: 555. Hemos tomado las referencias y los análisis de los textos aludidos de ambos estudiosos. Los textos originales son Diez, Grammatik der Romanischen Sprachen, 5ª edición de 1882; Brauns, Über den präpositionalen Akkusativ im Spanischen mit gelegentlich Berücksichtigung anderer Sprachen, publicada en Hamburg, 1909, y Hills, “The accusative a”, Hispania, III, 1920. 11 Un resumen similar al que encontramos en Lapesa, 1964, nos lo ofrece también Martín Zorraquino, 1976, en las páginas 555 y 556; artículo que también se ha revisado para la confección de este trabajo y de este apartado. 19 423‐432, apud Lapesa, op.cit., 79) interpreta que el objeto directo de persona tiene una esfera de autonomía que el de cosa no tiene. El uso de a viene a indicar que esa esfera de autonomía se ha roto por una agresión, pues los verbos más antiguos que aparecen con ad o a son verbos de acción violenta. Por último, Hatcher (“The Personal Accusative in Spanish. The Use of a as a Designation of the Personal Accusative in Spanish”, MLN, LVII, 1942: 421‐429, apud Lapesa, op.cit., 79) encuentra que el uso de a en el Cantar de Mío Cid tiene un sentido más de respeto hacia la persona representada por el objeto que de agresión. Frente a estas teorías, existen otras que sostienen que la construcción no surge para crear una categoría especial para los objetos de persona, sino que surge de la acción de otros factores y más tarde adquiere la función de oponer el objeto personal al de cosa. Meier (“Sobre as origens do acusativo preposicional nas línguas românicas”, Ensaios de Filologia Românica, Lisboa, 1948: 115‐164, apud Lapesa, op.cit., 79) parte de los pronombres personales átonos que comparten una misma forma para el dativo y el acusativo (me, te) en casi toda la Romania. Según su hipótesis, la indiferenciación entre las formas de acusativo y dativo se propagó a los pronombres tónicos, que llevaban la a (procedente del ad latino) para las formas del dativo y que pasó a acompañarlos también cuando hacían de objeto directo. Después, esta a se propagó a los nombres propios y comunes de persona en función de objeto directo por diferentes vías: paralelismo de construcción cuando un nombre común aparecía coordinado con uno propio o con un pronombre personal (“Reciben a Minaya e a las dueñas e a las otras compañas”, Cantar de Mio Cid), para indicar si el término de una comparación debía entenderse como objeto o como sujeto (“Dos cavalleros quel aguardan cum a señor”, Cantar de Mio Cid), vacilación entre el régimen transitivo e intransitivo de ciertas construcciones herederas del latín, etc. Por último, según Reichenkron (“Das präpositionale Akkusativ‐Objekt im ältesten Spanisch”, RF, LXIII, 1951: 432‐397, apud Lapesa, op.cit., 80‐81), el empleo de a ante objeto directo de persona obedeció en un principio a motivaciones rítmicas. La a venía a suplir en los nombres propios y en los pronombres tónicos la función rítmica del artículo 20 ante los apelativos12. Al usarse ante nombres propios y pronombres, adquirió un nuevo valor y pasó a utilizarse delante de los apelativos comunes con artículo o determinativo cuando designaban “individuos portadores de nombre propio” (cf. Lapesa, op.cit., 81). El signo a se propagó también a través de construcciones en las que el doble acusativo latino fue sustituido por un objeto directo y uno indirecto y de aposiciones, contagiada de un término a otro: “Verán a sus esposas, a don Elvira y a doña Sol”. A los casos de a procedentes de ad se sumaron otros procedentes de ab, en los cuales un ablativo agente originario acabó siendo interpretado como objeto directo: “Uos uedes a Munno Salido assí me desondrar” (cf. Lapesa, op.cit., 80‐81)13. La extensión del uso de a dio lugar a que dicha preposición perdiera su valor rítmico inicial y adoptara su valor actual. 1.2. Los estudios sobre la presencia/ausencia de a ante objeto directo en castellano medieval La preposición a –o su predecesora ad– ante objeto directo o acusativo aparece ya en documentos preliterarios latinos y su uso se va extendiendo con el transcurso del tiempo hasta su estadio actual. A pesar de que las hipótesis y los estudios recogidos en el apartado anterior de este trabajo resultan muy valiosos para tratar de explicar cómo apareció la construcción en nuestra lengua, parece necesario revisar estudios más minuciosos realizados a partir de datos cuantitativos que permitan determinar el grado de probabilidad del uso de a ante determinados objetos directos en etapas definidas de la historia del idioma para explicar cómo se dio dicha expansión (cf. Martín Zorraquino, 1976: 558). Cuantificar y analizar la presencia/ausencia de a en determinados textos de diferentes épocas puede aclarar las causas que favorecieron el uso de la misma. Los estudios de Martín Zorraquino (1976), Monedero Carrillo de Albornoz (1978 12 Esta función rítmica serviría para manifestar la independencia del objeto directo frente a su verbo, pues (el artículo) suponía una separación átona de la palabra acentuada que lo precedía: “Que mataras el moro e fizieras barnax” supone una independencia entre objeto y verbo que “Para arrancar moros e seer segudador” no tiene, pues verbo y objeto forman una representación unitaria. 13 Reichenkron aplica la misma explicación a la a que precede a nombres geográficos, sin necesidad de suponer que estos estuvieran personalizados. 21 y 1983), Folgar (1984) y García Martín (1992) analizan la presencia/ausencia de a ante objetos directos que hacen una referencia personal ‐pues ya hemos visto que la presencia de esta preposición está estrechamente vinculada a los objetos directos personales‐ y ante objetos directos de lugar ‐ya que este ha sido un ámbito de muchas fluctuaciones‐ fundamentalmente en el Cantar de Mio Cid, pero también en otras obras desde el siglo XII hasta el XV. Martín Zorraquino (1976) realiza un estudio cuantitativo de la presencia/ausencia de la preposición a delante de un objeto‐complemento directo en el Cantar de Mio Cid. Con sus datos se puede calcular la probabilidad de la presencia de la preposición a ante determinados objetos directos y tratar de precisar las condiciones que favorecían la presencia/ausencia de la misma. Según los datos obtenidos, el ámbito de mayor regularidad es el de los pronombres personales, pues los tónicos aparecen siempre introducidos por la preposición a, mientras que con los átonos nunca comparece (cf. Martín Zorraquino, op.cit., 559). Los datos que obtiene son significativos para detectar rasgos que favorecen la presencia de la preposición con nombres que hacen referencia a personas. Con nombres propios de persona, la preposición aparece 76 veces (como en “Mató a Búcar”)14, frente a los 3 casos que aparecen sin ella (como en “matar el moro Avengalvón”), de modo que la probabilidad de aparición es superior al 90% de los casos. La presencia de la preposición a también es frecuente con otros referentes de persona en singular (nombres comunes, pronombres posesivos, relativos, indefinidos, etc., como en “al uno dizen Ojarra e al otro Yéñego”), con los que aparece en 21 ocasiones frente a 14 casos de ausencia (como en “veremos vuestra mugier”), aunque la probabilidad de la aparición se limita al 60% de los casos. Por último, el uso es mucho menos frecuente con este tipo de referentes de persona en plural: la preposición a aparece 47 veces (“a aquestos dos mandó el Campeador”), frente a 83 casos de ausencia (“mató treinta e 14 Para la referencia a los versos del Cantar, remitimos al texto de Martín Zorraquino (1976) – aquí solo nos interesa mostrar los ejemplos que ella muestra‐. 22 quatro”); de modo que la probabilidad de su presencia se reduce a alrededor del 30% de los casos (cf. Martín Zorraquino, op.cit., 560‐565). Según Martín Zorraquino (op.cit., 565), “la preposición aparece, pues, ‐al menos en los materiales del Cantar‐ como marca individualizadora del objeto directo personal de la frase”. El uso de la preposición parece constante ante los pronombres personales tónicos y los nombres propios de persona y animal, que hacen referencia a individualidades concretas, y ante los nombres comunes de persona su presencia es más frecuente ante el singular que ante el plural (cf. Martín Zorraquino, op.cit., 565). Confirma así la siguiente afirmación de Lapesa: “No todo objeto de persona lleva a en español sino sólo el que designa un ente personal o grupo de personas vistas en su individualidad concreta o como suma de individualidades concretas” (cf. Lapesa, 1964: 77). Carmen Monedero Carrillo de Albornoz (1983) estudia la presencia de la preposición a ante nombres referentes de persona y plantea una teoría diferente sobre la presencia/ausencia de la preposición a ante objeto personal. En lugar de limitar su estudio al Cantar de Mio Cid, toma un corpus mucho más amplio con más de 4000 ejemplos que le proporcionan una perspectiva de conjunto de la presencia/ausencia de la preposición15. Monedero rechaza la teoría de Reichenkron16, pues “sigue a M. Pidal en su edición crítica de Mio Cid, donde la aparición del artículo ha sido regulada por el editor conforme a los criterios lingüísticos de la época” (cf. Monedero, op.cit., 241‐242). También rechaza la teoría de Meier17 según la cual, el nacimiento y la extensión del acusativo preposicional parte de los pronombres personales y sigue después con los propios y los apelativos de persona, pues parece que no se cumple de igual 15 Las obras que incluye en su corpus son: “Auto de los Reyes Magos, Disputa del Alma y el Cuerpo del XII, Razón de Amor, Santa María Egipciaca, Libre dels tres reys d’orient, Roncesvalles, Libro de Apolonio, Biblia Escurialense, Historia Troyana (sólo de la parte en prosa), 102 capítulos de la Primera Crónica General (Cid e Infantes de Lara), Elena y María, algo de la Crónica Particular de San Fernando y Crónica de Veinte Reyes, Conde Lucanor y Claros Varones de Castilla, aparte de los repertorios de Keniston, Cuervo y Diccionario Histórico de la RAE” (cf. Monedero, 1983: 242). Monedero tiene en cuenta para su estudio obras que no son uniformes en cuanto a estilo y género, ni tampoco en cuanto al origen geográfico. 16 Ver el capítulo 1.1., páginas 19 y 20, del presente trabajo. Esta teoría se resume en Lapesa, 1964: 80‐81 y en Martín Zorraquino, 1976: 556. 17 Ver el capítulo 1.1., página 19, del presente trabajo. Esta teoría se resume en Lapesa, 1964: 79 y en Martín Zorraquino, 1976: 556. 23 modo, por ejemplo, en el Cantar de Mio Cid, donde hemos visto que la preposición se da sin excepción delante de los pronombres personales tónicos, que en los documentos notariales y en los fueros, en los que la presencia de la preposición también abunda “sin que sea total y sin excepción la a ante pronombre tónico”: “yo dona Maria Pedrez offresco e do mi misma a Dios” (cf. Monedero, op.cit., 288‐291)18. Para Monedero, el uso de la preposición a podría ser más bien un recurso orientado al “destacamiento enfático de algo, lo mismo en el habla que en estos escritos, aunque no se llegara al abuso de los juristas (...). Este mismo énfasis sigue en el CMC [Cantar de Mio Cid], donde está ligado a la deixis juglaresca, a la expresividad (...) y a procedimientos estilísticos” (cf. Monedero, op.cit., 291). Frente a los rasgos léxico‐semánticos [±humano], [±animado], [±número] del estudio de Martín Zorraquino, Monedero alude a un aspecto más pragmático de la aparición de la preposición: el intento de destacar o dar énfasis a un objeto determinado. Carmen Monedero considera que factores como la igualdad de las formas de los pronombres átonos de primera y segunda persona (me y te) para el dativo y el acusativo, el paso de verbos intransitivos del latín a transitivos en romance (como rogar o pedir), el doble régimen de ciertos verbos (como ayudar o servir) y el fenómeno del leísmo serían favorecedores de la presencia de a, pero no determinantes. Estos fenómenos pudieron ayudar a que se fijara el proceso en el sistema, pero “está claro que se produjo a la vez ante todo Propio, Apelativo, o Pronombre que se quisiera destacar” (cf. Monedero, op.cit., 292). La oposición a/ø nacería, pues, de una necesidad de enfatizar y pasaría a convertirse después en una oposición expresiva entre un término [+expresivo] marcado con la preposición (“a mí”) y uno [–expresivo] sin preposición (“me”). La regularidad en el uso de la preposición a que se observa con los nombres 18 El ejemplo citado (“do mi misma a Dios”), muy interesante en relación con otros que contienen pronombres personales tónicos siempre precedidos de a, está tomado por Monedero (1983) de los Documentos lingüísticos de España editados por Menéndez Pidal, en 1966, y figura en la página 290 del artículo de esta estudiosa. 24 propios en tiempos del Cantar puede apuntar a que el énfasis inicial diera lugar a un cierto “tratamiento de cortesía”, que alcanzaría también a la máxima dignidad divina: “amar a Dios” (cf. Monedero, op.cit., 293). Este tratamiento de cortesía, o “Acusativo de Dignidad”, sería la forma marcada. La no marcada incluiría los “apelativos especiales o de pertenencia” ‐aquellos que tienen que ver con parentescos familiares‐ que quedan dentro de una esfera de la intimidad a la que no alcanzó esta cortesía. Así pues, en la época del Cantar de Mio Cid, esta construcción no sería enfática con respecto a los nombres propios de persona por ser ya un tratamiento de cortesía más o menos establecido, pero sí lo sería con respecto a los nombres propios de ciudad, en los que hay mucha más vacilación. Las alternancias en el uso de la preposición estarían motivadas por “el enfatismo, la deixis juglaresca, la expresividad y la estética literaria” (cf. Monedero, op.cit., 295). Frente al uso de la preposición a como marca individualizadora que percibe Martín Zorraquino (1976) siguiendo a Lapesa (1964) en el Cantar de Mio Cid, Monedero detecta un valor pragmático de énfasis y expresividad tanto en el Cantar como en otras obras de distinto tipo discursivo. No parecen dos perspectivas que se opongan radicalmente ya que no parecen excluyentes. La presencia de la preposición a ante los pronombres personales tónicos y los nombres propios es bastante regular, aunque haya excepciones en determinados textos, pero no se puede descartar que este uso encontrara su origen en un intento de enfatizar el objeto en cuestión. Además, parece lógico que los elementos que se tratara de destacar fueran concretos e individuales, independientemente de que fueran pronombres personales o nombres propios. García Martín (1992), como Martín Zorraquino (1976) o Lapesa (1964), detecta una tendencia de la preposición a a aparecer ante elementos de carácter humano e individualizado en un análisis que no se limita a una única obra y época. García Martín (1992) estudia la presencia/ausencia de la preposición a ante objeto directo y reconstruye la evolución de esta construcción a lo largo de un periodo de tiempo por medio de la comparación de manuscritos pertenecientes a la 25 misma familia19. De este modo, se pueden observar las tendencias o los cambios que se dieron a lo largo de dicho periodo de tiempo en el uso de la preposición a. Para realizar su análisis, el autor selecciona y compara la frecuencia de casos significativos20 de la construcción (objeto directo con y sin preposición a) en cada una de las obras. Con los resultados obtenidos, llega a la conclusión de que a partir de mediados del siglo XIII se advierte un paulatino descenso en el uso de la construcción del objeto directo con preposición a, que durante el siglo XIV habría una época de crisis del giro preposicional y que en el siglo XV, quizás a finales del XIV, habría una reacción y un aumento del uso de la preposición que seguiría en aumento en adelante (como en los ejemplos del Setenario: “De los que adorauan la tierra” de ca. 1300 frente a “De los que adorauan a la tierra” de ca. 1400, apud cf. García Martín, op.cit, 66). Esta reacción coincidiría con el movimiento cultista y latinizante que caracteriza todo el siglo XV. García Martín (1992) estudia las variables que pudieron influir en la presencia/ ausencia de la construcción y en la evolución de la misma. Aprecia una delimitación cada vez mayor de categorías susceptibles de recibir la preposición, que son (además de la de los nombres propios en general, cf. García Martín, op.cit., 68): 1) los nombres de Dios o de seres únicos, como se observa en “non queríen connosçer nin aorar Dios” (Setenario, ca 1300, apud García Martín, op. cit., 64) frente a “non queríen connosçer nin aorar a Dios” (Setenario, ca. 1400, apud García Martín, op. cit., 64), o en “De cómmo errauan los que orauan el ssol” (Setenario, ca 1300, apud García Martín, op. cit., 67) frente a “De cómmo errauan los que orauan al sol” (Setenario, ca 1400, apud García Martín, op. cit., 67); 2) los sustantivos –incluso los colectivos‐ en singular con artículo determinado, así como el relativo el que o el indefinido todo, entre otras: “Et por esso le llamaron a Dios Ssol de Justicia porque alunbra e escalienta toda cosa ssegunt conuyene” 19 Las obras seleccionadas para su estudio fueron compuestas, según las opiniones más extendidas, desde el segundo cuarto del siglo XIII hasta 1270 y son la Poridat de las poridades, el Setenario y la Historia troyana en prosa y en verso, si bien esta última solo se tiene en cuenta en la primera parte del trabajo de García Martín. Los manuscritos analizados se extienden desde finales del siglo XIII hasta el siglo XV (cf. García Martín, op.cit., 50‐51). 20 García Martín destaca que hay que seleccionar bien los elementos que tenemos que tener en cuenta por relevantes y separarlos de los que son, por ejemplo, mero reflejo de los gustos del copista (cf. García Martín, op.cit., 51, 54). 26 (Setenario, ca 1300, apud García Martín, op. cit., 74) frente a “Et por esso le llamaron a Dios Ssol de Justicia porque alunbra e escalienta a toda cosa ssegunt conuyene” (Setenario, ca 1400, apud García Martín, op. cit., 74). De otro lado, los nombres en plural, animados o inanimados, muestran un comportamiento muy oscilante: “...venían folgados, començaron a maltraer a los troyanos” (Historia troyana, mediados del s. XIV, apud García Martín, op. cit., 68) frente a “estos commo venían folgados, començaron a maltraer los troyanos” (Historia troyana, finales del s. XIV, apud García Martín, op. cit., 68). También se delimitan las categorías que rechazarían la presencia de la preposición a, por ejemplo, los nombres propios de lugar –sobre los que volveremos en seguida‐: “de parte del padre heredó a León, e Gallizia, e Asturias, et aun el rregno de Badaioz” (Setenario, ca 1300, apud García Martín, op. cit., 71) frente a “de parte del padre heredó León, e Gallizia, e Asturias, et aun el rregno de Badaioz” (Setenario, ca 1400, apud García Martín, op. cit., 71); o el pronombre relativo que: “Et ésta sse parte en tres maneras. La primera es a que llaman traiçión mayor...” (Setenario, ca 1300, apud García Martín, op. cit., 71) frente a “Et ésta sse parte en tres maneras. La primeras que llaman traiçión mayor...” (Setenario, ca 1400, apud García Martín, op. cit., 71). Atención especial merecen los nombres geográficos en cuanto objetos directos y su vinculación con la preposición a. Monedero (1983) apunta a que la regularidad en la presencia de a ante nombres propios de persona en función de objeto directo, adquirida por la generalización de su uso, habría dado lugar a que hubiera perdido ya su carácter expresivo en los tiempos del Cantar. Sin embargo, sí serviría como elemento de énfasis con los nombres propios de ciudad (cf. Monedero, op.cit., 295). En el estudio de Martín Zorraquino (1976: 560), los datos referentes a nombres comunes de cosa y animal, y propios de cosa no parecen proporcionar datos concluyentes, pues tanto la presencia como la ausencia de la preposición representan alrededor del 50% de los casos. La autora indica que “los datos permiten deducir que existía, al parecer, una cierta vacilación en el uso de la preposición ante los nombres propios de cosa” (ibídem). En estos nombres propios de cosa se incluye a los nombres propios 27 geográficos, que, precisamente por las vacilaciones que presentan, han sido objeto de diferentes estudios. Carmen Monedero (1978) hace una revisión de los nombres propios geográficos que aparecen en el Cantar de Mio Cid en función de objeto directo con o sin la preposición a e intenta discernir las causas que pueden motivar la presencia de dicha preposición. Monedero realiza un recuento del que resultan 71 nombres propios geográficos en función de objeto directo: 37 aparecen con la preposición a y 34 sin ella. En este estudio, que es previo al que hemos remitido anteriormente (Monedero, 1983), Monedero llega a conclusiones similares a las mencionadas anteriormente. Para la autora, el objeto directo preposicional con nombres geográficos es un elemento innovador dentro del tipo de texto que trata (que se limita en este caso al Cantar de Mio Cid) y funciona, por tanto, como elemento estilístico expresivo y deíctico del juglar para suscitar y mantener la atención del público: “cuanto más innovador fuera un recurso, (...) tanto más eficaz resultaba en la deixis; pero en la medida en que se generalizaba iba perdiendo estas cualidades (...) y obligaba a buscar nuevos procedimientos”. Si bien también apunta que había expresiones, que “no por repetidas y ‘petrificadas’ (...) dejaban de tener su eficacia, pues compensaban su falta de brillantez con la seguridad de una respuesta precisa (...)”. Esta estructura no se limitaba a lo juglaresco, pues aparece en otros tipos de texto, como documentos notariales, “con el mismo afán señalizador y expresivo” (cf. Monedero, 1978: 266‐267). Carmen Monedero apunta a que una de las causas que podría haber dado lugar a la aparición de a ante estos objetos directos podría ser el contagio de la a que ya aparecía con los objetos directos de persona. Los hechos que influirían en este contagio serían la posibilidad de personificar los nombres propios de ciudad y la facilidad de los mismos para aparecer con toda clase de preposiciones. Pero, para la autora, estos dos factores no pueden ser únicos, porque, por un lado, estas condiciones se daban con otros tipos de nombres y, por otro, no sería lógico que hubiera más frecuencia de nombres de ciudad en función de objeto directo introducidos por la preposición a que nombres comunes de persona, como 28 constata la autora (cf. Monedero, op.cit., 271‐273). Los factores verdaderamente determinantes son de tipo estilístico: “Va a aparecer la ‘a’ allí donde se quiera resaltar una ciudad como objeto directo, en las ‘conquistas’ y en los ‘avances’ rápidos, dos acciones plenamente épicas, fundamentales en el argumento del Cantar [Por ejemplo, “Myo Cid gaño a Xerica τ a Onda”, Monedero, op.cit., 273]. No se da con verbos que indiquen lo contrario: dejar, abandonar o vender [“Quitar quiero Casteion”, Monedero, op.cit,: 277]. O indiferentes en cuanto a las conquistas: ver, mirar, tener (=guardar), curiar [“Vos teniendo Valencia, τ yo vençi el campo”, Monedero, op.cit., 282]” (cf. Monedero, op.cit., 273). Carlos Folgar (1988) rechaza rotundamente estos criterios planteados por Carmen Monedero. Su investigación21 se basa en criterios estrictamente gramaticales. Según sus recuentos, si el nombre propio de lugar que funciona como objeto directo aparece precedido de un adjetivo determinativo en función de determinante (artículo determinado, adjetivo demostrativo, etc.), la preposición a no aparece: “troçen las Alcarrias e iban adelant” (Cantar de Mio Cid, apud Folgar, op.cit., 404). Si el nombre propio de lugar no va precedido por adjetivo determinativo, la aparición de a parece opcional: “non conquís a Çaragoça, ont me ferió tal lançada” frente a “Con vos conquís Truquía e Roma a priessa dava” (Roncesvalles, apud Folgar, op.cit., 405). Si el nombre propio de lugar aparece en la construcción “(determinante +)22 nombre común + de + nombre propio de lugar”, la preposición a no aparece: “Poblado ha mio Cid el puerto de Alucat” (Cantar de Mio Cid, apud Folgar, op.cit., 406). Entre los ejemplos encontrados, solo con aquellos que aparecen sin artículo u otro adjetivo determinativo presenta la preposición a variación. Folgar rechaza la justificación del uso de la preposición a ante nombres propios geográficos personificados para explicar esta vacilación. No descarta que los nombres propios geográficos puedan aparecer personificados, pero eso no da cuenta del 21 Folgar estudia la aparición de a con objeto directo toponímico en textos anteriores al año 1230: Cantar de Mío Cid, Razón de amor con los denuestos del agua y el vino, Roncesvalles, La Fazienda de Ultra Mar, El Auto de los Reyes Magos y el Libro de la infancia y muerte de Jesús (cf. Folgar, 1988: 403, n.1). 22 El autor coloca entre paréntesis el elemento “determinante” porque lo considera opcional. 29 comportamiento de la preposición a, pues no explica que los topónimos sin determinante aparezcan tanto con preposición como sin ella y los que tienen determinante siempre sin ella. También rechaza el criterio de la naturaleza semántica de la acción verbal que defiende Monedero (cf. supra, 27‐28 y Monedero, 1978: 273), el criterio rítmico defendido por Reichenkron (cf. supra, 19‐20 y Folgar, op.cit., 410: “La preposición se colocó, a modo de transición átona, entre el verbo y el topónimo cuando el verbo tenía acentuación oxítona”) y el criterio semántico defendido también por Reichenkron que explica la variación presencia/ausencia de la preposición a en virtud de la oposición semántica entre acción dinámica y resultado estático, pues se dan excepciones difíciles de justificar (cf. Folgar, op.cit., 411). Folgar plantea su propio criterio, que también es de naturaleza rítmica23: “Los nombres propios geográficos precedidos de adjetivo determinativo no llevaron nunca (o prácticamente nunca) preposición. (...) Sin embargo, los nombres propios geográficos que no iban introducidos por determinante llevarían siempre la preposición a, probablemente por una tendencia a dotarlos de una estructura rítmica y acentual idéntica a la de los que sí tenían determinante. Es decir, se trataría de conseguir el paralelo rítmico de “troçer las Alcarrias”, que indudablemente era “gañar a Castejón”” (Folgar, 1988: 415‐416). En una segunda etapa de desarrollo de la construcción, se daría un criterio distinto, por el que se eliminaría la preposición a ante los topónimos en función de objeto directo, ya que los nombres de lugar no expresan seres personales, “de modo que no cumplen las condiciones básicas para la anteposición de a ante objeto directo en español (antiguo y moderno)” (Folgar, 1988: 417). Habría una etapa de transición en la que no habría desaparecido todavía la construcción más antigua y tampoco se habría impuesto aún la más reciente, que es la que encontramos en el Cantar de Mio Cid. 23 De este criterio excluye a aquellos nombres propios de lugar que aparecen con la estructura “(determinante+) nombre común + de + nombre propio geográfico”, puesto que en ella la función de núcleo la ejerce un nombre común y el topónimo funciona como complemento del núcleo. 30 A partir del análisis crítico de la bibliografía revisada, constatamos que las fluctuaciones en el uso de la preposición a a lo largo de la historia de nuestra lengua son abundantes y que, mientras que la presencia ante nombre propio de persona y pronombre personal tónico aparece con bastante fijeza ya en los tiempos del Cantar, presenta vacilaciones, por ejemplo, ante nombres comunes o ante los nombres propios geográficos. Las perspectivas de interpretación del fenómeno, los criterios y las diferentes formas de abordarlo también son abundantes y variadas. Encontramos tres enfoques fundamentales: el semántico‐ sintáctico expuesto por Lapesa (1964), Martín Zorraquino (1976) y después por García Martín (1992); otro, que tiene en cuenta aspectos relacionados con la tradición discursiva y con motivaciones pragmáticas y semánticas, como el de Monedero Carrillo de Albornoz, y otro que se centra en aspectos estrictamente gramaticales, como el de Carlos Folgar, que rechaza los criterios estilísticos. Resulta indudable la constatación de la presencia de a ante objetos personales individualizados que muestran Martín Zorraquino (1976) y García Martín (1992). Este hecho no invalida, sin embargo, los criterios pragmáticos de Monedero (1978 y 1983), que parecen relevantes para la aparición del fenómeno, pues resulta convincente que sean las necesidades comunicativas las que motiven la evolución de las construcciones; por ello, el criterio de destacamiento enfático del que habla Monedero es compatible y refuerza la tesis de la aparición de a con nombres que hagan referencia a realidades concretas e individualizadas. En cuanto a la presencia de a ante los nombres propios de lugar, el criterio descrito por Monedero también resulta aceptable, si bien hoy en día la presencia/ausencia de a ante este tipo de nombre parece estar más relacionada con el carácter [±personal] de la realidad a la que se hace referencia (“vencer a Francia” frente a “visitar Francia”). El factor rítmico del que habla Folgar y su rechazo de los criterios estilísticos y pragmáticos parece olvidar, en nuestra opinión, que el texto escrito en cada momento de la historia va acompañado de una lengua hablada a la que no tenemos acceso. Parece aventurado deducir, pues, que dicha lengua hablada no respondiera a motivaciones expresivas o semánticas y que su evolución se debiera 31 exclusivamente a factores rítmicos como los aludidos. Un criterio como el de Folgar limita el fenómeno a un texto y a un tipo de texto concretos y parece claro que el fenómeno del complemento directo preposicional ante nombres propios de lugar no surge únicamente con base en un único tipo de texto. 32 2. LA PRESENCIA DE A ANTE OBJETO DIRECTO EN ESPAÑOL ACTUAL Como hemos expuesto, el objeto directo preposicional parece darse ya en latín y se manifiesta claramente en castellano medieval con muchas vacilaciones que se reflejan a lo largo de la historia de la lengua. El número de estudios sobre esta construcción se multiplica cuando se trata de describir y explicar su funcionamiento en español actual. En este capítulo trataremos de entender en qué consiste el fenómeno y cómo lo explican los siguientes gramáticos: Bello (1847/1988), Seco (1972/1989)24, Alcina y Blecua (1975), Alarcos (1994) Gómez Torrego (1995), Torrego Salcedo (1999), la Nueva gramática de la lengua española de la Real Academia Española (2009) y la versión Nueva gramática de la lengua española Manual (2010)25. 2.1. El objeto directo preposicional: definición y cuestiones problemáticas El objeto directo introducido por la preposición a, denominado también objeto directo preposicional, acompaña a un verbo transitivo como adyacente más interno. Sin embargo, en lugar de aparecer sin mediación alguna, es la preposición a la que lo introduce26. Veamos los siguientes ejemplos: (1) Trae el libro que he dejado ahí 27 (2) Trae al perro que está en el jardín 24 Para realizar este trabajo, en principio, se consultó la obra de Seco de 1972. No obstante, dado que la segunda edición ‐revisada y aumentada‐ de 1989 nos parece más fácil de encontrar, también citamos en nota la ubicación de la referencia en la obra de 1989. 25 En adelante, nos referiremos a la Nueva gramática de la lengua española de la Real Academia Española como NGLE y a la Nueva gramática de la lengua española – Manual como Manual. 26 A este respecto, Torrego Salcedo (1999: 1781) en la Gramática descriptiva de la lengua española dirigida por Bosque y Demonte indica que, en este caso concreto, a no funciona como una preposición y la denomina partícula. Argumenta que los complementos directos que llevan a se pueden sustituir por un pronombre en acusativo (la, lo, las, los) –por tanto sin preposición‐ y pueden realizar la función de sujeto en la oración pasiva, también sin preposición. En este trabajo, no obstante, no se entrará a valorar la certeza de esta consideración y hablaremos, en cualquier caso, de la preposición a. 27 Los ejemplos del capítulo 2 del presente trabajo aparecen numerados consecutivamente desde el (1) hasta el (124). Hemos numerado los ejemplos procedentes de otros autores, a no ser que se encontraran introducidos en una cita del mismo. Todos los ejemplos, tanto ajenos como propios, aparecen entrecomillados. En caso de que no quede claro si se trata de un ejemplo ajeno o propio, lo indicaremos en una nota. 33 (3) Trae a tu hermana a la cena. Mientras que en (1) tenemos un ejemplo de objeto directo sin mediación alguna, (2) y (3) son ejemplos de objeto directo introducido mediante la preposición a. El hecho de que el objeto directo preposicional a menudo acompañe a objetos directos de persona, como ocurre en el ejemplo (3), hace que haya hablantes que lo consideren con frecuencia un objeto indirecto. No obstante, aquel se distingue claramente de este. Lo podemos comprobar sustituyendo cada uno de los objetos directos preposicionales propuestos por el pronombre correspondiente. Veremos que la sustitución solo es posible con los pronombres de objeto directo (excepto en casos de leísmo) y que la preposición se pierde: “Tráelo” para los ejemplos (1) y (2) y “Tráela a la cena”, y no “Tráele a la cena” para el ejemplo (3). En caso de duda, también podemos transformar la oración a la voz pasiva y el objeto directo pasará a ser sujeto sin preposición, como pasaría con un objeto directo típico, mientras que el objeto indirecto no puede pasar a la función de sujeto de la pasiva correspondiente. A continuación, lo ilustramos con un ejemplo extraído de Torrego Salcedo (1999: 1782): (4) “Encarcelaron a un narcotraficante” se transforma en (4a) “Un narcotraficante fue encarcelado”, sin la preposición a. En contraste con los ejemplos precedentes, los ejemplos que proponemos a continuación, acompañados de un objeto directo sin preposición y de un objeto indirecto, admiten una sustitución y una transformación parcialmente diferentes a las mostradas. El elemento que funciona como objeto indirecto es sustituido por le: (5) “Trae un libro a tu hermana” se sustituye por (5a) “Tráele un libro”. Además, el objeto indirecto no puede ser sujeto de una oración pasiva y conserva su estatuto de objeto indirecto, con la preposición a: (6) “Entregaron el libro a Juan” se transforma en (6a) “El libro (le) fue entregado a Juan”. El objeto directo preposicional ha sido asociado habitualmente con los objetos directos de persona, pues muy frecuentemente los introduce. Esta asociación hace que muchos gramáticos consideren que la aparición de la preposición está relacionada con el carácter animado de la referencia del objeto directo. Como 34 indica Seco (1972: 97)28, (...) en general, la explicación está en el significado del nombre que funciona como complemento directo: cuando designa a un ser animado, lleva la preposición a; cuando designa a un ser inanimado, no la lleva: Juan alimenta muy bien a su perro, pero Visitarán Barcelona. Bello, por su parte, habla de personalidad y determinación: “La preposición a se antepone a menudo al acusativo cuando no es formado por un caso complementario; y significa entonces personalidad y determinación” (cf. Bello, 1847/1988: § 889, 567): “Admiro a César, a Napoleón, a Bolívar” (cf. Bello, 1847/1988: § 890, 567). A pesar de la sencillez que puedan sugerir estas explicaciones, este fenómeno resulta muy complejo, principalmente porque dichas explicaciones no se pueden aplicar de manera sistemática a todos los casos ni resuelven todos los casos problemáticos que presenta el uso o la exclusión de la preposición. Existe un dominio delimitado, que veremos más adelante, en el que la presencia de la preposición a ante el objeto directo puede considerarse obligatoria; sin embargo, existen otros ámbitos (aquellos en los que la preposición puede aparecer o no) para los que todavía no hay una explicación satisfactoria que permita delimitar todos sus usos. El fenómeno del objeto directo preposicional se encuentra todavía en un “proceso de gramaticalización” y hasta que este proceso culmine, será complicado delimitar unas condiciones fijas que expliquen las vacilaciones en su uso sincrónicamente. No obstante, esto no es obstáculo para que muchos gramáticos se hayan aproximado al problema y hayan tratado de ofrecer sus propias postulaciones sobre qué factores son determinantes para la presencia o ausencia de la preposición. A continuación, veremos cuáles son los factores gramaticales, 28 La referencia correspondiente a esta cita se encuentra en las páginas 108 y 109 de la edición publicada en 1989 de esta obra de Seco. 35 léxicos, semánticos y discursivo‐pragmáticos que se han planteado en el estudio de este fenómeno. 2.2. Diferentes propuestas para explicar los casos problemáticos del objeto directo preposicional Como hemos indicado antes, el objeto directo preposicional ha despertado el interés de muchos gramáticos, que se han aproximado a él de diferentes maneras. Algunos se han centrado en la caracterización de los factores que tienen que ver con la esencia puramente gramatical del objeto directo preposicional, mientras que otros han estudiado factores que tienen que ver con los elementos que lo componen y lo rodean. A continuación, revisaremos críticamente diferentes propuestas para explicar el comportamiento del objeto directo preposicional. En primer lugar, nos ocuparemos del estudio estrictamente sintáctico del objeto directo preposicional como distinguidor de funciones en el enunciado, destacando la explicación de Alarcos (1994), que otorga a la a una función exclusiva y clara. Después, trataremos otros planteamientos no estrictamente sintácticos: en qué circunstancias (semánticas, referenciales, léxicas, etc.) se da la aparición de a delante del objeto directo. Para ello, analizaremos la naturaleza del elemento o elementos que refleja el objeto directo mismo, la naturaleza del verbo que lo rige y, por último, dedicaremos un pequeño apartado a la naturaleza del sujeto del verbo que rige el objeto directo. 2.2.1. La preposición a como marca diferenciadora de función en el objeto preposicional (frente a otras funciones primarias) Alarcos (1994: § 335, 346‐347) justifica el uso de la preposición a como una marca del sistema para distinguir la función de objeto directo de la del sujeto en algunas oraciones en las que determinadas combinaciones impiden la correcta 36 interpretación de lo comunicado, como en los ejemplos (de Alarcos) que incorporamos a continuación: (7) Dibujaba la niña el niño (8) Mató el elefante el tigre (9) Favorece la codicia la ambición. En estos casos en que la distinción funcional de los elementos nominales en el seno de la oración resulta ambigua, la preposición a se introduce para diferenciar la relación entre elementos: (7a) Dibujaba a la niña el niño (7b) Dibujaba la niña al niño (8a) Mató al elefante el tigre (8b) Mató el elefante al tigre (9a) Favorece a la codicia la ambición (9b) Favorece la codicia a la ambición. Según Alarcos (op.cit., § 336, 348), si bien la preposición a permite distinguir el objeto directo del sujeto, puede producir la indiferenciación del objeto directo frente al indirecto. En estos casos, se suele considerar el directo el primero y el indirecto el segundo: (10) “El maestro presentó a su mujer a Juan”. Sin embargo, la construcción es ambigua y se suele suprimir la preposición a delante del directo (10a) “El maestro presentó su mujer a Juan”, pero no (10b) “El maestro presentó Juan a su mujer”. Aunque los equívocos se resuelven introduciendo el personal átono correspondiente: (10c) “Se la presentó a Juan”, (10d) “Se lo presentó a su mujer”. La explicación de Alarcos, aunque puede parecer convincente, no tiene en cuenta lo siguiente. En primer lugar, en la mayor parte de las ocasiones el contexto suele dejar claras las relaciones internas de los elementos oracionales y el interlocutor distingue sujeto de objeto. A efectos prácticos, los casos de ambigüedad que nos podemos encontrar en la lengua no suelen impedir la correcta comunicación e interpretación de los enunciados. En segundo lugar, normalmente el orden no marcado de la oración o la sustitución por los pronombres correspondientes suele ser suficiente para disipar las posibles dudas. 37 Por otro lado, parece que el factor meramente funcional no podría explicar la diferencia entre estos casos: (11) El profesor favoreció a la chica (11a) Favoreció el profesor a la chica. (12) El buen tiempo favoreció a la chica (12a) Favoreció el buen tiempo a la chica. (13) El buen tiempo favoreció la venta (13a) Favoreció el buen tiempo la venta. (13b) Favoreció la venta el buen tiempo. Mientras que en (11), (11a), (12) y (12a) la presencia de a es obligatoria, no lo es en (13), (13a) y (13b). Si no hubiera preposición en ninguno de los casos, el único ejemplo en el que podría darse una interpretación ambigua de las funciones de los elementos de la oración sería (11) y es probable que el contexto sirviera para disipar cualquier duda. En (12) y (13), el contenido semántico de cada elemento –lo referido en la realidad por el sujeto y el objeto‐ impiden que se dé una interpretación ambigua. Sin embargo (12) requiere la preposición y (13), no. Existe cierto consenso con respecto a otros casos de posible ambigüedad que dan lugar a variaciones entre objeto directo con y sin preposición. Bello lo explica de la siguiente manera: Cuando es necesario distinguir el acusativo de otro complemento formado con la preposición a, podemos y aun debemos omitirla en el acusativo, que en otras circunstancias la exigiría: “Prefiero el discreto al valiente” [frente a “Prefiero al discreto al valiente”] (...). Esto sucede principalmente cuando concurren acusativo y dativo; y nunca se extiende a los nombres propios de persona que carecen de artículo; por lo que no sería permitido: “Presentaron Zenobia al vencedor” [“Presentaron a Zenobia al vencedor”] (...). Cuando es inevitable la repetición del a, suele preceder el acusativo: “El traidor Judas vendió a Jesús a los sacerdotes y fariseos”. Pero si ambos términos fuesen nombres propios de persona, sin artículo, sería preciso adoptar otro giro 38 (...); porque ni “Recomendaron Pedro a Juan”, ni “Recomendaron a Pedro a Juan”, pudieran tolerarse (Bello, 1847/1988: §900, 570)29. De nuevo, y como bien indica Carmen Pensado (1995: 25), “la ambigüedad es más teórica que real” y, en muchas ocasiones, las posibilidades de evitar la ambigüedad no se aprovechan: ‘entregó a sus corchetes al asturiano’” (ibídem). Aparte de lo cuestionable de la ambigüedad, otro punto relevante que tenemos que tener en cuenta a la hora de considerar la propuesta que nos ocupa resulta del hecho de que los ejemplos que nos ofrece Alarcos parecen ejemplos construidos, que pueden coincidir a primera vista con la perspectiva intuitiva del hablante, pero que resultan alejados de los ejemplos que podrían extraerse de un corpus de ejemplos reales, fundamental para comprender el alcance de la cuestión que se plantea: ¿cuándo y por qué aparece –o no aparece‐ a ante el objeto directo? 2.2.2. La presencia del objeto directo preposicional en dependencia de ciertas características del elemento nominal objeto directo Entre los factores que tienen que ver con la naturaleza del nombre o pronombre que funciona como objeto directo, la NGLE (2009), su versión Manual (2010) y otros autores, como Alcina y Blecua (1975) o Bello (1947/1988), destacan algunas tendencias generales. Como ya hemos mencionado, algunos gramáticos atribuyen hasta cierto punto la presencia del objeto directo preposicional al carácter animado del elemento que funciona como tal objeto. Este factor no resulta suficiente si observamos que hay objetos directos animados que carecen de preposición: (14) “Buscamos programadores que conozcan java”. Por eso, al carácter de personal o animado se le añade el de determinado: (15) “Busco al médico de mi prima”, frente a (16) “Busco un médico”. De nuevo, este criterio 29 El giro al que debiera recurrirse sería, en nuestra opinión: “A Pedro le recomendaron a Juan”. En este ejemplo, ambos objetos –directo e indirecto‐ presentan la preposición a, pero el orden de la oración deja claro que el primero de ellos es el objeto indirecto y el segundo, el objeto directo. 39 parece insuficiente si consideramos enunciados como (17) “Busco a un médico especializado en el corazón”. Pareciera que a este carácter de personal también tuviéramos que añadir el de específico. Queda claro que, aparte de grandes tendencias reconocibles, existen numerosos casos de vacilación y alternancia, por ejemplo, en los que un objeto directo de cosa se introduce con a y uno de persona sin ella. Por eso, algunos gramáticos han tratado de recoger tendencias generales que den cuenta de los casos de variación difíciles de explicar. 2.2.2.1. Nombres propios de persona o animal y pronombres personales: ámbito de obligatoriedad de la presencia de la preposición a De acuerdo con lo que indican Alcina y Blecua (1975: 859‐860), está claro que la preposición a se utiliza sistemáticamente con los nombres propios de persona o de animal: (18) “Fui a ver a Platero”. De la misma manera, se emplea la preposición a ante los pronombres personales que aluden a persona (“a él”, “a mí”) y los indefinidos alguien, nadie y el relativo quien. A esta explicación de Alcina y Blecua, la NGLE (2009: 2630‐2637) y el Manual (2010: 658‐659) añaden varias matizaciones. En primer lugar, ambos textos indican que se utiliza la preposición a delante del pronombre personal tónico en función de objeto directo, que siempre aparece en construcciones de duplicación pronominal: (19) “Me viste a mí”. En segundo lugar, además de los pronombres indefinidos que denotan personas (alguien, nadie) y del relativo quien, hay que añadir, como inductores de la presencia de a, a los interrogativos, exclamativos y relativos que denotan a personas: (20) “Matar a cualquiera, no. Acabar con un tirano, sí”, (21) “No obligaremos a nadie”, (22) “Díselo a quien quieras”, (23) “¿A quién buscan?”, etc. En tercer lugar, el relativo que es incompatible con la preposición a: (24) “las personas a que amamos” cuando se refiere a personas y es objeto directo, excepto cuando aparece con artículo: (24a) “las personas a las que amamos” (cf. NGLE: 2632). De acuerdo con lo que indica la NGLE (2632), los relativos el que, la que, los que, las que se asimilan a los que incorporan léxicamente el antecedente (quien, quienes) en que requieren la preposición si se 40 refieren a personas. Por otra parte, tras ciertos verbos de percepción o de sentido prospectivo, uno, alguien y alguno también pueden aparecer sin preposición: (25) “Nunca he visto alguien así” o (26) “Aureliano Segundo quitó el candado buscando alguien con quien conversar” (cf. NGLE: 2623). Además, a pesar de que la preposición es obligatoria cuando nos referimos a la obra de una persona: (27) “Traducir a Platón”; no lo es cuando se trata de una obra concreta: (28) “Subastaron ese Picasso” (cf. NGLE: 2633). A pesar de que no existe una jerarquía totalmente clara de los factores determinantes de la presencia de la preposición a que sea aceptada por todos los gramáticos, los puntos que acabamos de mencionar resultan de los más claros dentro del ámbito de la obligatoriedad de la preposición. 2.2.2.2. Nombres propios de nación o ciudad: algunas vacilaciones en la presencia/ausencia de la preposición a Siguiendo con los nombres propios, Alcina y Blecua (1975: 859‐860) indican que los nombres propios de nación o ciudad, pueblo, etc. conocieron el uso con preposición, como hemos visto en el capítulo precedente, pero ahora dicho uso es muy poco frecuente: (29) “Visitaremos Sevilla”30. Con los nombres de ríos y accidentes geográficos tampoco se usa la preposición. Sin embargo, como matiza Seco (1972: 97), la presencia/ausencia de la preposición con este tipo de nombres propios no es tan simple como parece. Observemos los enunciados (30) “Hemos recorrido Francia” e (31) “Inglaterra venció a Francia”. Para Seco, la diferencia entre las dos oraciones es que, en el primer caso, nos referimos al territorio francés (ser inanimado), mientras que, en el segundo, nos referimos al pueblo francés (ser animado)31. 30 Este ejemplo ha sido inventado por mí misma. 31 En la edición de 1989 de la obra de Seco, esta referencia se puede encontrar en la página 109. 41 2.2.2.3. Nombres comunes de persona: ámbito de vacilación en la presencia/ausencia de la preposición a En cuanto a los nombres comunes de persona, hay cierto consenso entre los gramáticos con respecto a las causas que motivan la variación respecto de la presencia y ausencia de a. Bello (1847/1988: §893, 567‐568) nos explica que los nombres comunes de persona con artículo determinado también llevan la preposición a: (32) “Conozco al gobernador de Gibraltar”. En otros casos, cuando el objeto directo hace referencia a una persona determinada para el sujeto, lleva la preposición a: (33) “Fueron a buscar a un médico experimentado que gozaba de una grande reputación”; cuando no hace referencia a una persona determinada, no: (34) “Fueron a buscar un médico experimentado, que conociera bien las enfermedades del país” (ibídem). Por eso, si el nombre común carece de artículo, no hay preposición, pues hace referencia a personas indeterminadas: (35) “Es preciso que el ejército tenga oficiales inteligentes” (cf. Bello, 1847/1988: §894, 568). Alcina y Blecua (1975: 859‐860) indican que la selección de la preposición o su ausencia suele variar según la determinación del sustantivo del objeto directo. Cuando no lleva artículo, el sustantivo resulta [–determinado] y puede darse la construcción sin preposición a con un alto grado de cohesión significativa entre verbo y sustantivo. Con artículo, el sustantivo aparece [+determinado] y puede llevar la preposición a o no llevarla. En general, y para muchos casos, esto implica una diferencia o matización de significado: (36) “Conoció reyes” (‐ determinado) frente a (37) “Conoció a los reyes” (+determinado). Mientras que, en (36), conocer tiene un significado de ‘tener trato con reyes indeterminados’; en (37), conocer indica que ‘unos reyes particulares fueron presentados al sujeto’. Observamos que, mientras el nombre propio se emplea siempre con preposición, el común vacila según las particularidades de cada verbo, especialmente con los nombres de persona que designan empleos, títulos, dignidades, etc. y en general con los empleos cuando dependen de verbos como necesitar, buscar y semejantes: (38) “Odio los hombres medianos y fríos”, (39) 42 “(...) solía encontrarse (...) paseantes de mal aspecto” (cf. Alcina y Blecua, 1975: 859‐860). Según la NGLE (2009: 2633‐2634) y el Manual (2010: 659), los nombres comunes de persona no llevan la preposición a delante del objeto directo cuando se designa a tipos de individuos: (40) “Mitrione prefería los borrachitos a los presos políticos”. Los verbos elegir, designar, votar, nombrar y otros similares admiten alternancias, por ejemplo, cuando designan a un individuo: (41) “elegir al próximo presidente”, o a un cargo: (41a) “elegir el próximo presidente”. Estas distinciones, sin embargo, plantean ciertas dudas. En primer lugar, la NGLE propone un ejemplo con un verbo que funciona con unas particularidades concretas (preferir) para postular una regla general. Una construcción similar con el mismo verbo y con la preposición a podría interpretarse también en referencia a grupos de individuos: (40a) “Mitrione prefería a los borrachitos”. Lo mismo podemos decir si esta construcción no presentara la preposición: (40b) “Mitrione prefería los borrachitos”. Veamos otro de los ejemplos: (42) “¿Quién elige los jugadores que hay que fichar?”. Si tuviéramos entre manos la alternativa con preposición (42a) “¿Quién elige a los jugadores que hay que fichar?”, podríamos darle la misma interpretación que a la anterior, si el contexto lo propicia; es decir, tanto la presencia como la ausencia de la preposición a resultan gramaticales ante un sustantivo que designe a grupos de individuos. Por otro lado, la NGLE también afirma que algunos verbos, como preferir y elegir, “pertenecen al grupo de los llamados intensionales, que se caracterizan por inducir la interpretación inespecífica de los grupos nominales indefinidos” (cf. NGLE, 2009: 2634). Con ellos se usa el pronombre qué y no quién cuando no sustituyen propiamente a personas, sino a clases o tipos de entidades. Pero si comparamos estos dos ejemplos, (43) “¿A quién prefieres?” y (44) “¿Qué prefieres?”, necesitaríamos un contexto apropiado para valorar si cada uno de ellos se ha utilizado para referirse a una persona concreta o a un tipo de personas respectivamente, o no. A nuestro juicio, los ejemplos que nos 43 proporciona la NGLE resultan insuficientes para valorar la precisión de sus afirmaciones. 2.2.2.4. Nombres comunes de animal y de cosa personificada: ámbito de vacilación en la presencia/ausencia de la preposición a Los nombres comunes de animales llevan preposición cuando se asimilan a los de persona: (45) “(...) y no encuentra a su gato y lo busca (...)” (cf. Manual, 2010: 659). La NGLE (2009: 2635) indica que la preposición ante los nombres comunes de animales puede depender de la cercanía afectiva con el animal; sin embargo, podemos encontrar algunos ejemplos como (46) “Detenido un chino vecino de Zaragoza por matar a palos a un perro”32. En este caso no parece tanto una cuestión de cercanía afectiva sino simplemente de cercanía o familiaridad del perro como animal doméstico y habitual en nuestra cultura. Habría que comprobar si este tipo de construcción se puede ampliar a otros animales más o menos relacionados culturalmente con el hablante. También requieren la presencia de la preposición a las cosas personificadas, como en (47) “(...)como llamando a la muerte” (cf. Manual, 2010: 659), y los nombres colectivos personificados (colegio, compañía, empresa, junta, “consejo”), como en (48) “Si yo defiendo a una empresa tildaré como (...)” (ibídem). No obstante, Bello indica que las cosas que se personifican toman la preposición a ante el objeto directo cuando son determinadas: (49) “Saludar las aves a la aurora”, (50) “Calumniar a la virtud” (Bello, 1847/1988: § 898, 569). Alcina y Blecua (1975: 859) apuntan que (...) no se puede más que sospechar causas entre las cuales puede figurar un cierto grado de personificación. Parece importante el grado de frecuencia con que el verbo aparezca construido con nombres de persona. Se ha señalado el uso con a con verbos como “preceder”, 32 Este es un ejemplo obtenido por mí misma procedente de ABC.es (2013): 201303041903.html (Consultado el 5/03/2013) 44 “calificar”, “determinar”, “seguir” y en los que se implica la idea de antagonismo o dirección como “atacar”, “mirar”, “atinar”, “alcanzar”, “renunciar”, etc.: “En todo caso, no condenes a esta esperanza sin oír antes lo que tengo que decir en su defensa”. Según la NGLE (2010: 2637), los verbos que habitualmente rigen un objeto directo personal favorecen la personificación de objetos directos de cosa: (51) “¿Cómo no iba a odiar al otoño?”, y más si implican la formación de juicios de valor: acusar, culpar, exculpar, inculpar, juzgar, perdonar, etc. Además de la personificación, que también dependerá del significado del sustantivo: (52) “invadir el planeta” (astro) o (53) “(...) que sacude al planeta en la actualidad” (habitantes), también se utiliza la preposición a cuando se da una interpretación metonímica de una parte del cuerpo: (54) “Prefiero morirme de hambre que alimentar a una cabeza que no es la mía” (ibídem). 2.2.2.5. Grupos nominales indefinidos: ámbito de especial vacilación en la presencia/ausencia de la preposición a Una de las áreas de mayor vacilación en el uso o la ausencia de la preposición a ante objeto directo es la de los grupos nominales indefinidos. A pesar de que se trata de sintagmas que se presentan principalmente sin la preposición, existen casos que se introducen con ella. Según la NGLE (2009: 2638), no suelen llevar preposición los nombres comunes de persona sin artículo u otro determinativo: (55) “La universidad debe formar investigadores” o (56) “Nunca vi persona igual”. Pueden llevarla las construcciones que proporcionan información determinativa, asimilable a la que pueden aportar artículos y demostrativos, como ocurre con los grupos nominales coordinados: (57) “(...) la novena cruzada para expulsar a sarracenos y judíos” o (58) “El trópico desgasta a hombres y mujeres”, o con aquellos cuya clase aparece caracterizada por un modificador: (55a) “La universidad debe formar a investigadores cualificados”. También admiten la preposición ciertos objetos sin 45 determinante con complemento predicativo: (59) “Había visto a soldados reponerse de peores heridas”. Estas tres excepciones apuntan a lo mismo: los complementos directos preposicionales reflejan grupos nominales determinados en el sentido de que contienen información suficiente para identificar individuos o grupos de personas o cosas, aunque no se trate necesariamente de grupos nominales con determinante (cf. NGLE, 2009: 2638‐2639). Se dan alternancias de presencia y ausencia de a ante objeto directo con los grupos nominales formados con el artículo indefinido o con cuantificadores: (60) “Buscaba un amigo”, (60a) “Buscaba a un amigo” (cf. NGLE, 2009: 2639). Una explicación tradicional de estas alternancias responde al carácter específico o inespecífico de los grupos nominales, es decir, que se haga referencia a un grupo reconocible o no. Esta propuesta se reconoce a día de hoy como inexacta: (61) “Usted no necesita a un cirujano, sino a un confesor”. En este caso, la preposición no introduce individuos particulares o específicos, sino una referencia de clase: sin embargo, la preposición a está presente (ibídem). Suelen presentar la preposición los objetos con una expresión cuantitativa presuposicional, es decir, que designa un conjunto extraído de otro mayor: (62) “Vi en la calle a muchos de mis estudiantes” (cf. NGLE, 2009: 2640). Los grupos nominales formados por indefinidos y relativas restrictivas en subjuntivo reciben de forma característica la interpretación inespecífica, pero son compatibles con la presencia de la preposición a: (63) “Pero necesitaba a alguien que tuviera los pies en el suelo y ejerciera (...)”, aunque no necesariamente (cf. ibídem). Finalmente, el cuantificador indefinido cualquier/a, que posee interpretación inespecífica, es compatible con la preposición a: (64) “El tiempo y las conveniencias convierten a cualquier persona seria en un miserable coleccionista de cargos y de honores” (ibídem). Este ejemplo, sin embargo, es discutible, ya que presenta una construcción predicativa que podría influir en la presencia de la preposición a; todos los ejemplos ilustrados dejan claro que el carácter específico/inespecífico del objeto directo no está sistemáticamente correlacionado con la presencia/ausencia de la preposición a. 46 2.2.3. El objeto directo preposicional en dependencia de ciertas características del núcleo verbal Aunque algunos autores se centran únicamente en las características del elemento que funciona como objeto directo a la hora de determinar la presencia o ausencia de la preposición a ante objeto directo, otros, como Bello (1847/1988: 567‐570), Torrego Salcedo (1999: 1781‐1793) o la NGLE (2009: 2641‐2649), indican que los factores que determinan que el objeto directo esté o no introducido por una preposición dependen tanto de la naturaleza del nombre o pronombre que funciona como objeto directo como de la naturaleza del núcleo verbal del que este depende. En lo que al verbo respecta, su influencia sobre el objeto directo preposicional es prácticamente regular. Esther Torrego Salcedo (1999) elabora un análisis de los factores verbales que, desde su punto de vista, determinan la presencia o ausencia del objeto directo preposicional. Según la autora (op.cit., 1781), mientras que en el dominio del nombre existe mucha vacilación con respecto a las pautas de comportamiento de la preposición a, en el dominio del verbo la regularidad es casi absoluta. Torrego realiza una clasificación de tres grupos de verbos y explica que, en función de estos, la preposición a del objeto directo puede ser “obligatoria, opcional, o estar simplemente proscrita” (cf. Torrego Salcedo, op.cit., 1781)33. Los verbos que pueden llevarla o no llevarla de manera opcional “ven alterado su sentido según [la preposición a] esté presente o no”. El cambio de sentido del verbo depende en gran medida de la clase semántica verbal de que se trate en cada caso, pero siempre consiste en un incremento de participación del sujeto en la acción (ibídem). La NGLE (2009: 2641) realiza una clasificación muy similar a la de Torrego Salcedo y divide los verbos en tres grupos: verbos transitivos que exigen la preposición a cuando se construyen con objetos directos de persona; verbos 33 Este estudio de Torrego Salcedo se limita a los objetos directos animados y deja de lado las vacilaciones que puedan darse con objetos directos inanimados, pues, “en líneas generales, cuando el nombre que hace de complemento directo es inanimado, la preposición a no aparece (...)” (cf. Torrego Salcedo, 1999: 1782). 47 transitivos que rechazan la preposición a cuando se construyen con objetos directos de persona, y verbos transitivos compatibles con la preposición a cuando se construyen con objetos directos de persona. Sin embargo, como veremos más adelante, su propuesta de clasificación se basa en rasgos diferentes a los que alude Torrego Salcedo. Torrego Salcedo (1999: 1787) estudia el grupo de los verbos transitivos con objeto directo preposicional obligatorio y concluye que los verbos que exigen a con sustantivos animados no presentan irregularidades. En su opinión, es el aspecto léxico verbal el que determina cómo se comportará el verbo con respecto al complemento preposicional. Los verbos “télicos” o “delimitados”34, que son aquellos que expresan eventos con un límite temporal intrínseco (es decir, que implican por sí solos una acción terminada), imponen la preposición a al complemento directo animado: (65) “Marta insultó a un compañero” y no (65a) “Marta insultó un compañero”. Con estos verbos, en la mayoría de los casos, el sujeto se interpreta como agente si es animado (cf. ejemplo (65)) y como causante si es inanimado: (66) “La medicina curó al herido” (cf. Torrego Salcedo, op.cit., 1787). No obstante, esta propuesta de Torrego Salcedo parece no poder aplicarse a todos los casos de manera sistemática. Si tomamos el verbo encontrar, que puede clasificarse como télico, vemos que la preposición a no siempre comparece ante un objeto directo animado: (67) “Marta encontró a un amigo”, pero (67a) “Marta encontró un amigo”35. Aunque estos dos ejemplos puedan interpretarse con significados diferentes, esta diferencia de significado no implica ninguna diferencia en el aspecto léxico del verbo ni en el aspecto final de las expresiones. Si la presencia de la preposición a dependiera exclusivamente del aspecto léxico verbal, el ejemplo (67a) no sería gramatical. 34 Los verbos que Torrego Salcedo (1999) denomina “télicos” se corresponden con los que Elena de Miguel (1999) denomina “delimitados” en su contribución a la Gramática descriptiva de la lengua española dirigida por Bosque y Demonte. Más adelante haremos referencia a dicha contribución, titulada “El aspecto léxico”, puesto que nos hemos servido de ella para entender y comprobar el aspecto léxico de los verbos planteados como objeto de estudio en la segunda parte de este trabajo. 35 Estos ejemplos han sido inventados por mí misma. 48 En el grupo de verbos que exige la presencia de la preposición a, Torrego Salcedo (1999: 1790‐1791) también incluye los objetos directos que son afectados por la acción del verbo, que aparecen siempre con preposición si son animados: (68) “Golpearon a un extranjero”, y no (68a) “Golpearon un extranjero”; (69) “Odia a un vecino y no (69a) “Odia un vecino” (cf. Torrego Salcedo (1999: 1790‐1791). Existe, al margen de esta clasificación, una serie de verbos que exige la presencia de la preposición a, pero no cumple con los criterios indicados: animacidad del objeto, agentividad o causatividad del sujeto, etc. Se trata de los “verbos de transitividad atípica”36. Ejemplos de ellos son: (70) “Un adjetivo califica a un sustantivo”; (71) “Los días siguen a las noches”; (72) “El uno precede al dos”, etc. (cf. Torrego Salcedo, 1999: 1788). Según la NGLE (2009: 2641‐2642), este primer grupo de verbos (aquellos que exigen la presencia de la preposición a ante objeto directo de persona) está compuesto por una serie de verbos que alternan objeto directo con indirecto, como ayudar (“ayudarlo”, “ayudarla”/ “ayudarle”) o servir (“servirlo”, “servirla”/ “servirle”): (73) “Tenía que ayudar a muchas personas”, (74) “Sirvió a varios reyes”. La alternancia entre objeto directo e indirecto se da sobre todo en español europeo en verbos como amenazar, insultar o saludar (“saludarlo” o “saludarla” frente a “saludarle”), que son verbos que pueden caracterizarse con un verbo de apoyo más un sustantivo (“lanzar una amenaza”, “proferir un insulto” o “hacer un saludo”) (cf. NGLE, 2009: 2641‐2642). En nuestra opinión, más que de una alternancia entre objeto directo o indirecto, se trata del uso del pronombre le con función de objeto directo, ya que, el objeto introducido por la preposición a y sustituido por le responde a otras pruebas de transitividad, como la transformación a la voz pasiva: (74a) “Varios reyes fueron servidos (por él)”. La NGLE (2009: 2642), siguiendo a Torrego Salcedo, incluye en este grupo los verbos de afección. Estos verbos también presentan la alternancia entre lo/la y le: asustar, impresionar, aburrir, etc. Por tanto, ejemplos como (75) “Asustó a los 36 Bello indica, con respecto a algunos de estos verbos, que aquellos que implican un orden pueden apartarse de la norma [“los apelativos de cosa no suelen llevar preposición por muy determinados que sean”]: “La primavera precede al estío”. Argumenta que, en realidad, lo que rigen estos verbos es un dativo (Bello, 1847/1988, § 897, 568‐569). 49 niños” o (76) “Saludamos a los lectores” requieren la presencia de la preposición a y su ausencia resulta agramatical: (75a) “Asustó los niños” o (76a) “Saludamos los lectores”. Para Torrego Salcedo (1999), el segundo grupo de verbos lo componen aquellos cuyo objeto directo puede o no llevar la preposición. La presencia o ausencia de la preposición a acarrea alternancias aspectuales37. Este fenómeno se da con los verbos “atélicos” o “no delimitados”, que expresan eventos de carácter no terminativo (es decir, que hacen referencia a un evento que no ha terminado). Según Torrego Salcedo (1999: 1790), el objeto directo preposicional hace que un verbo se entienda como télico incluso si aisladamente designa un evento de aspecto atélico. En el ejemplo (77) “Besaron un niño” tendríamos un evento atélico, no concretado. Sin embargo, si introducimos el objeto directo con la preposición a: (77a) “Besaron a un niño”, el evento se delimita mediante un objeto directo identificable y concreto (ibídem). Según Torrego Salcedo, en el caso sin preposición no cabe identificar al objeto, mientras que en el caso con preposición, el objeto se individualiza y hay una mayor participación por parte del sujeto: podemos identificar al objeto (al niño) que también es un participante del evento. De este modo, la presencia del complemento directo preposicional transforma el carácter atélico de un verbo en télico (cf. Torrego Salcedo, 1999: 1788‐1789). A nuestro juicio, Torrego Salcedo expone un ejemplo en el que se da un cambio de aspecto que nos parece difícil de percibir, sobre todo por la falta de un contexto adecuado en el que interpretar el significado de la oración. En principio, el verbo besar, como hemos dicho, parece tener un aspecto léxico atélico y hace referencia a una actividad durativa, aunque dicha duración sea escasa. Podemos comprobarlo aplicando algunas de las pruebas que explica Elena de Miguel (1999: 2999). Los eventos durativos son compatibles con los modificadores temporales durativos (como los introducidos por durante) e incompatibles con 37 De manera general, la presencia de la preposición es obligatoria cuando se trata de un objeto definido: “Besé al niño” frente a “Besé el niño” o “Besaba al niño” frente a “Besaba el niño”, y opcional cuando es indefinido. Por eso, en todos los ejemplos que aparecen a continuación se muestran objetos directos indefinidos (cf. Torrego Salcedo, 1999: 1782). 50 los modificadores temporales delimitadores (como los introducidos por en). Dado su aspecto no delimitado (y durativo), la oración anterior es compatible con un complemento circunstancial de tiempo introducido por durante, como (77b) “Besaron a un niño durante un segundo”, aunque sea una duración escasa. No obstante, el ejemplo (77c) “Besaron a un niño en un segundo” también parece posible, aunque menos probable, y podría interpretarse como “completaron la tarea de besar a un niño en un segundo”. Hacemos, pues, una interpretación diferente de cada ejemplo: mientras que en el primero haríamos referencia a la duración de la actividad de “besar a un niño”, en el segundo haríamos referencia al tiempo tardado en completar la tarea de “besar a un niño”. En ninguno de los casos parece que sea la presencia de a la que determina el aspecto del conjunto, sino más bien el complemento circunstancial temporal. En nuestra opinión, la presencia de a sería obligatoria en el ejemplo (77b) y su ausencia sería poco probable: (77d) “#Besaron un niño durante un segundo”, y opcional en (77c): “Besaron (a) un niño en un segundo”. Si ampliamos el tiempo del complemento circunstancial: (77d) “Besaron a un niño durante tres minutos” y (77e) “Besaron (a) un niño en tres minutos”, apreciamos más claramente que, mientras que, en (77d), “durante tres minutos” hace referencia a la duración de la actividad de “besar a un niño”; en (77e), “en tres minutos” no hace referencia al tiempo que duró la actividad de “besar”, sino al tiempo que se tardó en conseguir realizar la actividad de “besar (a) un niño”. Si nos fijamos en otros verbos télicos, observamos que las expresiones temporales introducidas por en hacen referencia al tiempo que se tarda en realizar determinada actividad. (78) “Escribí el informe en diez minutos” implica que estuve escribiendo diez minutos hasta que acabé el informe y (79) “Me fumé el cigarrillo en cinco minutos” implica que estuve fumando durante cinco minutos hasta que me acabé el cigarrillo. De modo que parece más probable que sea la locución “en tres minutos” la que aporta su aspecto delimitador a la expresión y no la presencia de la preposición a. Torrego Salcedo (1999: 1790) también habla de los verbos estativos como conocer, con los que la preposición a implica, según ella, un cambio de aspecto, además de un cambio de significado: (80) “Conocieron un músico de jazz”, (80a) “Conocieron a un músico de jazz”. Según la autora, mientras que en el primer 51 ejemplo el sujeto no es agentivo y el evento no es télico, el segundo implica una mayor participación del sujeto y el evento es delimitado. Nos parece que el cambio de aspecto del que habla Torrego Salcedo no se aprecia con claridad en estos ejemplos. En pretérito perfecto simple, ambas oraciones pueden interpretarse como sinónimas con un significado delimitado de ‘enterarse de su existencia’: (81) “Conocimos (a) mucha gente en poco tiempo”, y no con el significado no delimitado de ‘tener trato y comunicación con alguien’: (82) “Conozco (a) un buen podólogo en la ciudad”. Ambos significados pueden expresarse con objetos directos preposicionales si cambiamos el tiempo verbal: (80b) “Conocieron a un músico de jazz durante el verano”, que tiene un sujeto agentivo y el primer significado, frente a (80c) “Conocían a un músico de jazz cuando eran jóvenes”, que tiene un sujeto no agentivo y responde al segundo significado mencionado. La NGLE (2009: 2643) explica que este segundo grupo (aquel formado por verbos que rigen objetos directos compatibles con la preposición a, es decir, que pueden llevarla y no llevarla) es el grupo más conflictivo, del que todavía no se han establecido con nitidez los límites entre factores sintácticos, semánticos y discursivos. La NGLE (ibídem) considera que se deben descartar de este grupo las alternancias que se deban a cambios de régimen: (83) “mirar el cielo” (objeto directo) frente a (83a) “mirar al cielo” (complemento de dirección) o (84) “contestar una pregunta” (objeto directo) frente a (84a) “contestar a una pregunta” (complemento de régimen). A nuestro juicio, tenemos que tener muy claro cuándo nos encontramos ante un cambio de régimen y cuándo no. Por ejemplo, en el caso de “contestar (a) una pregunta”, podemos considerar que se trata de un objeto directo compatible con la preposición a, pues ambas opciones –con a y sin ella‐ parecen poder sustituirse por “contestarla” con más facilidad que por “contestar a ella”. Dado que en este grupo existe una gran vacilación, la NGLE agrupa las distintas tendencias según grupos de verbos. Sin embargo, en algunas ocasiones, los factores expuestos no parecen exclusivamente relacionados con la naturaleza del 52 verbo, sino que se solapan con otros factores ya tratados que dependen de la naturaleza del objeto directo implicado en cada caso. En primer lugar, existen verbos que responden a distintas acepciones y, en función de ellas, van seguidos de objeto directo preposicional o no: (85) “Distinguir un hombre” (‘percibirlo’), (85a) “Distinguir a un hombre” (‘honrarlo’); (86) “Abandonar una ciudad” (‘salir de ella’), (86a) “Abandonar a una ciudad” (‘dejarla en el abandono’), etc. (cf. NGLE, 2009: 2643). No obstante, es posible que el significado que se interprete no siempre dependa de la presencia o ausencia de la preposición; por ejemplo, (85a) “Distinguir a un hombre” también podría interpretarse como ‘percibirlo’ en contexto. En segundo lugar, tenemos los verbos de creación: dibujar, esculpir, pintar, fotografiar, etc. La ausencia de a favorece la interpretación del objeto directo como el resultado del ejercicio ((87) “dibujar una niña”), mientras que su presencia favorece la interpretación de que el objeto es la fuente de inspiración ((87a) “dibujar a una niña”). Se puede considerar que la ausencia de a se utiliza cuando se habla del resultado, por tanto, no se habla tanto de una persona concreta sino del “objeto” resultante de la acción de dibujar: (88) “‐¿Qué dibujaste aquí? –Una niña” (cf. NGLE, 2009: 2643). Si esto es así, nos encontramos ante dos objetos directos distintos desde el punto de vista semántico que se distinguirían formalmente mediante la preposición a. Observamos que esta alternancia se da únicamente con objetos directos de persona con artículo indefinido. Si el objeto directo lleva, por ejemplo, un artículo definido o un determinante posesivo, la presencia de la preposición es obligatoria: (89) “Dibujé a mi prima” o (90) “Dibujé a la niña”. Por lo tanto, podemos pensar también que el matiz semántico que implica la ausencia de a es el de indeterminación, que en nuestro ejemplo consistiría en dibujar una niña desconocida o una niña inexistente. Si tomamos el ejemplo (88), observamos que también se puede formular de la siguiente manera: (88a) “¿Qué dibujaste aquí?” ‐ A mi prima”. A nuestro juicio, el hecho de que la NGLE considere que la respuesta posible es “Una niña” y no “A una niña” se debe a que se plantea un contexto ideal en el que aquel que pregunta sabe que la niña dibujada no es una niña real. 53 Otros factores que plantea la NGLE (2009: 2644) en relación con la naturaleza del verbo parecen estar motivados en realidad por la naturaleza del objeto y ya se han visto con anterioridad, por ejemplo, la distinción entre cuantificación y distribución ((91) “Ayer, en solo quince minutos, conté noventa hombres y treinta y dos mujeres” frente a (92) “El técnico blanquiazul ha concentrado a diecisiete jugadores”) o la distinción entre tipo de individuo o individuo particular de los verbos intensionales ((93) “Necesito el mejor abogado” frente a (93a) “Necesito al mejor abogado”). Otra posibilidad es la invitación a individualizar a personas, animales o cosas: (94) “Vimos tres policías” frente a (94a) “Vimos a tres policías”. En estos casos es difícil afirmar que la presencia o ausencia de la preposición a dependa exclusivamente del núcleo verbal (cf. supra, § 2.2.2.3, 41‐42 y § 2.2.2.5., 44‐45). Otro de los factores que destaca la NGLE (2009: 2644‐2645) sobre la presencia de la preposición a en función de la naturaleza de los verbos hace referencia al carácter más o menos agentivo de los mismos: (95) “Muchos seudónimos ocultan un delator” (‐agentivo) frente a (96) “Durante la guerra había ocultado a un delator” (+agentivo)38. No obstante, a nuestro juicio, el ejemplo (95) también podría presentarse con la preposición a sin que su interpretación variara: (95a) “Muchos seudónimos ocultan a un delator”. En verbos como controlar, vigilar o mirar, la NGLE (2009: 2645) observa una interpretación más activa del verbo cuando la preposición a comparece: (97) “controlar (a) la asociación”, (98) “vigilar (al – el) club”, (99) “mirar (a) la gente que pasa”. En nuestra opinión, la presencia/ausencia de la preposición a en los ejemplos (97) y (98) también podría deberse al carácter más o menos animado del objeto directo, por ejemplo, “vigilar al club” puede referirse a las personas que lo conforman, mientras que “vigilar el club” puede hacer referencia a las instalaciones. Con el verbo recordar, sin embargo, la presencia/ausencia de la preposición no responde al aspecto más 38 Este aspecto que la NGLE clasifica en la sección que dedica a la presencia/ausencia de la preposición a dependiendo de grupos de verbos se retomará en el siguiente apartado de este trabajo, que está dedicado a las características de los sujetos y su relación con la presencia/ausencia de la preposición porque, en última instancia, la agentividad o no agentividad también puede considerarse una característica del sujeto. 54 o menos agentivo del verbo en los términos que acabamos de ver: (100) “Esta ciudad me recuerda a París” (no agentivo), (101) “Elisa me recuerda mis obligaciones” (agentivo), (102) “Elisa me recuerda a mi abuela” (no agentivo). En estos tres ejemplos, la preposición a aparece cuando el verbo no es agentivo. A esto debe añadirse que los verbos que expresan estado o situación y que, por tanto, no exigen un sujeto agentivo, son perfectamente compatibles con el objeto directo preposicional: (103) “Nos merecemos (a) otros políticos” (cf. NGLE, 2009: 2645). En los ejemplos comentados, procedentes de la NGLE, han aparecido algunos que hacen referencia a complementos directos de cosa, aunque no fuera la intención inicial de la NGLE incluir este tipo de objetos directos, sino solo los referidos a personas (cf. NGLE, 2009: 2641). Además, al final de su clasificación, introduce ciertas notas sobre los objetos directos de cosa que presentan vacilación y justifica la presencia de la preposición a (cf. NGLE, 2009: 2645‐2646). Por un lado, remite a la presencia de a como elemento que destaca un proceso de personificación y acentúa el grado de animacidad del objeto directo: (104) “Atacar los aviones” (‐animado) frente a (104a) “Atacar a los aviones” (+animado) (cf. NGLE, 2009: 2645). Por otro lado, también la sintaxis de la construcción predicativa parece favorecer la presencia de la preposición a en el objeto directo: (105) “mirando (a) las nubes pasar”, aunque no siempre: (106) “Dejé sin contestar (a) estas cartas”. En vista de que la personificación y la construcción predicativa no llegan a explicar todos los casos expuestos (en (107) “ver (a) los trenes pasar”, por ejemplo, las dos opciones parecen igualmente aceptables y posibles), se plantea una última explicación, que consiste en el denominado “uso distintivo de la preposición a”. Se denomina así al uso de la preposición a “para diferenciar el objeto directo de otros complementos del verbo, en especial del sujeto, aunque no únicamente de este” (cf. NGLE, 2009: 2646). La a tendría, pues, una función distinguidora del sujeto, como en (108) “acompañar (a) la música”, o del predicativo, como en (109) “Los autores denominaron a esta entidad anisakis gastro‐alérgica”; aunque hay otras ocasiones, en las que ha triunfado la 55 indistinción, como en (110) “Le presentó a mi madre a Nicolás Blanch” (donde a introduce tanto el objeto directo como el indirecto). Por último, tenemos el tercer grupo de verbos: aquellos que rechazan la preposición. Torrego Salcedo no se detiene en ellos, pero indica que aquellos verbos que no admiten un sujeto agente o causante no admiten tampoco un complemento directo con la preposición a: (111) “Hay a unas delegadas en la sala” (cf. Torrego Salcedo, 1999: 1785). La Real Academia Española sí ofrece información sobre este grupo de verbos. Según la NGLE (2009: 2642), suelen ser verbos de causación: (112) “La crisis producirá miles de desocupados”. En este grupo también se incluyen verbos como pedir, demandar o solicitar: (113) “Habían pedido más jueces”. Aunque cuando se habla de un objeto directo particular, estos verbos pueden llevar la preposición: (114) “Habíamos pedido a este juez”. También los verbos haber y tener se incluyen en este grupo: (115) “Hay a muchas personas interesadas” y (116) “Tuvo a un hijo”. Si bien tener puede incluir la preposición a cuando significa ‘dar a luz’: (117) “Tuvo a su tercer hijo en Madrid”; cuando aparece en construcciones presentativas encabezadas por un complemento locativo: (118) “Aquí tenemos a un joven delincuente”; en las enumeraciones: (119) “Teníamos con nosotros a un abogado, a un deportista y a un militar”, y cuando se construye con complementos predicativos: (120) “Tenemos a varios profesionales trabajando en ello”. A pesar de que la NGLE crea tres grupos de verbos siguiendo la clasificación de Torrego Salcedo según requieran la preposición a, sean compatibles con ella o la rechacen, vemos que en la mayoría de los casos no alude a las características de los verbos en cuestión. En realidad, agrupa los verbos según afinidades (verbos causativos, verbos de percepción, verbos intensionales o verbos de creación, por ejemplo) y explica las tendencias que se dan en cada grupo, pero la presencia/ausencia de la preposición a responde, en la mayoría de los casos, a la naturaleza del objeto: si es o no personalizado, si es más o menos definido, si es partitivo o distributivo, etc. Sí coincide con Torrego Salcedo al mencionar que el carácter más o menos agentivo del verbo influye en la presencia de la preposición a (+agentivo) o en su ausencia (‐agentivo), pero esta tesis se limita a 56 ciertos ejemplos, pues hay otros que la contradicen. De manera general, comprobamos que tampoco la clasificación según la naturaleza del verbo nos da respuestas unívocas y sin excepciones con respecto a ninguno de los tres grupos de verbos estudiados. 2.2.4. El objeto directo preposicional en dependencia de ciertas características del sujeto de la oración en la que aparece Como hemos adelantado en el apartado anterior, tanto Torrego Salcedo (1999) como la NGLE (2009) señalan que la presencia o ausencia de la preposición a ante el objeto directo se ve influida por el aspecto más o menos agentivo del verbo en cuestión (cf. supra: 47 y 53‐54). Por un lado, según Torrego Salcedo (1999: 1782, 1784, 1785), en la mayor parte de los verbos cuyo objeto directo presenta la preposición a, el sujeto es o bien agente o bien causa. Según Torrego Salcedo (1999: 1785), “la prohibición de la preposición con el complemento de haber puede verse como una manifestación específica de una propiedad más general. Tal como ya se mencionó, sólo los predicados que llevan un sujeto con valor semántico de agente admiten el complemento directo con a”. Por tanto, para tener un objeto directo preposicional necesitamos un verbo con un sujeto con valor semántico de agente o causante. Torrego Salcedo (ibídem) lo ilustra con el siguiente ejemplo: (121) Este abogado escondió a muchos prisioneros (sujeto agente) (122) Esta montaña escondió a muchos prisioneros (sujeto no agente) (122a) Esta montaña escondió muchos prisioneros (sujeto no agente). Para Torrego Salcedo (ibídem), la a está proscrita en el ejemplo (122), en el que el sujeto “esta montaña” es inanimado y no funciona ni como causa ni como agente. Como hemos visto anteriormente (cf. supra: 53‐54), la NGLE (2009: 2644‐2645) apuntaba en la misma dirección con verbos como ocultar: (95) “Muchos pseudónimos ocultan un delator” (no agentivo) frente a (96) “Durante la guerra había ocultado a un delator”, pero explicaba que con otros verbos, 57 como recordar, ocurría todo lo contrario y el objeto directo introducido por la preposición a era el que requería un sujeto no agentivo: (102) “Elisa me recuerda a mi abuela”. Es decir, esta propuesta, que se da con algunos verbos, no se puede extender de manera sistemática a todos los verbos de acción. Entre los siguientes dos ejemplos con el verbo buscar acompañado de un objeto directo y de un objeto directo preposicional, no encontramos ninguna diferencia en la agentividad del sujeto: (123) Busco un informático que me arregle el ordenador de urgencia (124) Busco a un informático que me arregló el ordenador de urgencia. Por otro lado, los ejemplos que nos brindan Torrego Salcedo y la NGLE no nos parecen acertados en el sentido de que, mientras que ambas consideran proscrita la a en: (122) “Esta montaña escondió a muchos prisioneros” o en (95a) “Muchos pseudónimos ocultan a un delator”, la intuición de muchos hablantes podría considerar aceptables y gramaticales estos ejemplos con objeto directo preposicional. Es decir, la observación de Torrego Salcedo parece acertada si la limitamos a ciertos verbos, pero no es definitiva a la hora de delimitar la presencia/ausencia de a delante del objeto directo. SEGUNDA PARTE. ESTUDIO DEL USO DE LA PREPOSICIÓN A ANTE OBJETO DIRECTO CON VERBOS DE PERCEPCIÓN VISUAL EN ESPAÑOL ACTUAL 59 3. SOBRE LAS CARACTERÍSTICAS DEL CORPUS OBJETO DE ESTUDIO Tal y como hemos señalado en la “Introducción” del presente trabajo, en la segunda parte del mismo, hemos tratado de analizar el comportamiento de la preposición a delante del objeto directo en un corpus de datos que incluye un conjunto de verbos de percepción visual, pues estos presentan características específicas que permiten comprobar la adecuación o inadecuación de las postulaciones teóricas que hemos revisado en la primera parte del trabajo. 3.1. Datos cuantitativos y cualitativos de los ejemplos analizados El corpus que hemos analizado (cf. “Introducción”, 6), que se recoge en el apéndice al final de este trabajo, consta de 194 ejemplos en los que aparecen los verbos de percepción visual seleccionados para la realización de este trabajo (atisbar, columbrar, contemplar, divisar, mirar, observar, otear, ver y vislumbrar), acompañados, en la mayor parte de los casos, de un objeto directo de las características que precisamos a continuación. En primer lugar, contamos 91 ejemplos de objetos directos típicos, es decir, no introducidos por la preposición a ni por ningún otro elemento, como en (4) “(...) si el viajero tenía la fortuna de (...), podía atisbar unos calzones de hilo, amarillentos, (...)”39, en (27) “(...) los niños que le piden un autógrafo para columbrar la condición de su alma deportiva” o en (81) “Desde tu parapeto provisional has divisado un abeto con mucha nieve a su alrededor.”40. 39 En este apartado, los ejemplos están numerados de acuerdo con la numeración que aparece en el apéndice. 40 Enumeramos en esta nota todos los ejemplos de objeto directo sin preposición a: 1, 2, 4, 6, 7, 8, 9, 12, 13, 14, 15, 16, 17, 18, 19, 21, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 58, 61, 62, 63, 65, 68, 70, 73, 74, 75, 76, 77, 78, 81, 83, 85, 95, 107, 108, 109, 111, 112, 113, 114, 115, 116, 117, 132, 133, 134, 135, 136, 137, 138, 139, 141, 142, 143, 144, 145, 147, 148, 152, 153, 156, 157, 170, 171, 172, 173, 174, 176, 180, 181, 183, 186, 187, 188, 189, 190, 191, 193, 194. Algunos de estos ejemplos pueden interpretarse como activas impersonales, pero también como pasivas reflejas. Es el caso de: 7, 12, 15, 65, 68, 70, 143, 174, 180, 187 y 190. Hemos decidido tener en cuenta estos ejemplos que permiten una doble interpretación, aunque somos conscientes de que los resultados obtenidos no serán indiscutibles (cf. infra, 62‐63). 60 En segundo lugar, hemos recogido una serie de 87 ejemplos en los que la preposición a está presente delante del objeto directo, como ocurre en (5) “Nadie les espera. No atisban a persona alguna conocida”, en (91) “En sus conciertos no mira al público porque teme que una mujer guapa sea capaz de hacerle perder la concentración” y en (131) “Por esas razones te recomiendo que observes a Casiopea; una constelación muy visible (...)”41. Si analizamos el corpus según el tipo de objeto directo, nuestros recuentos resultan en lo siguiente. En primer lugar, contamos 76 ejemplos de objeto directo de cosa, como en (17) “(...) comenzó a atisbar la verdadera relación que le unía a su suegro (...)” o en (107) “(...) un paseo singular sobre la actualidad cultural, observando aspectos que no aparecen en otras revistas (...)”42. La mayoría de ellos no presentan la preposición a, pero hay dos excepciones: (49) “Por otro lado, puede servir de introducción la pregunta de Yates "... ¿por qué las neurociencias contemplan a la física para desarrollar sus teorías?” y (50) “En todo caso se hace más evidente que nunca la insuficiencia de contemplar a la ciudad como un espacio cerrado, puesto que cada vez (...)”. No incluimos en este grupo algunos ejemplos de difícil clasificación como: (85) “¿Es tolerante con la gente de ideas políticas contrarias a las suyas? ‐ Sí, porque miro la parte humana de las personas” o (88) “(...) pero ahora miro a lo más profundo del ser y regreso con palabras (...)” y que trataremos en otro apartado. De forma paralela a este gran grupo de objetos directos de cosa, tenemos otro grupo de 8 ejemplos que también contienen objetos directos de cosa pero que hacen referencia a personas. Son aquellos con las palabras “figura” y “silueta”: 41 Enumeramos en esta nota todos los ejemplos de presencia de la preposición a delante del objeto directo: 5, 11, 20, 22, 30, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 64, 66, 69, 71, 72, 79, 80, 84, 86, 87, 88, 90, 91, 92, 93, 94, 98, 99, 100, 101, 102, 103, 104, 105, 106, 110, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 146, 149, 150, 151, 154, 155, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 175, 177, 179, 184, 185. 42 Enumeramos en esta nota todos los ejemplos de oraciones con objeto directo de cosa: 1, 4, 6, 7, 8, 9, 12, 13, 14, 15, 16, 17, 18, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39, 49, 50, 58, 61, 62, 63, 65, 68, 76, 81, 83, 95, 107, 108, 109, 111, 112, 113, 115, 116, 117, 118, 132, 133, 134, 135, 136, 137, 139, 141, 142, 143, 144, 145, 147, 148, 152, 153, 157, 170, 171, 172, 173, 176, 181, 183, 186, 188, 189, 191, 193, 194. 61 (19) “Era un hombre delgado, pero solamente conseguí atisbar su silueta borrosa a través de la mampara de cristal (...)” o (75) “Cuando estaba muy cerca de mi fila divisé mejor su figura, y tuve por vez primera (...)”.43 Otro tipo de objeto directo que abunda es el de persona, que hemos dividido en dos grupos en nuestra clasificación. Por un lado, tenemos aquellos que hacen referencia a personas individuales o individualizables (aunque aparezcan en plural) y, por otro, los colectivos. No hemos hecho distinción entre nombres comunes y propios en este apartado. Los ejemplos en los que aparece un objeto directo de persona individual (o individualizable), como en (11) “Allí conocí al dueño de la casa, el compañero Indalecio, y atisbé brevemente a una chica guapísima, su esposa (...)” o (60) “Apenas enfoqué la calle que desembocaba en ella, divisé al Poeta y a dos fieles amigos de mi tío, (...)”, son 4244. Hemos incluido en este grupo el ejemplo (77) en el que el objeto directo es “unos ángeles” y aparece sin preposición, puesto que nos parece más apropiado incluirlo en este grupo que en el de los objetos directos de cosa, y el ejemplo (90), cuyo objeto directo es “al ser”, que sí que lleva la preposición y no parece hacer referencia a una cosa o a un animal. En cuanto a los objetos directos de persona colectiva, contamos 18 ejemplos, como (106) “(...) muchos ciudadanos de a pie miran a la prensa, o al menos a una parte de ella, como una especie de última trinchera (...)” o (160) “‘(...) cuando paso por los Pirineos, por ejemplo, y veo a toda mi gente, a toda mi afición animando’”45. En algunos ejemplos, es el complemento del núcleo el que aporta el carácter personal a un núcleo colectivo, como en (125): “a un grupo de extremistas” o en (168) “a una serie de personajes y personajillos”. En otros, dicho complemento del núcleo aporta el carácter colectivo del objeto, como en (21) “parte de un corrillo de monjas sentadas al bordado”. 43 Los ejemplos que componen este pequeño grupo son los siguientes: 19, 54, 73, 74, 75, 78, 187, 190. 44 Enumeramos en esta nota todos los ejemplos de objetos directos de persona individual encontrados en el corpus: 5, 11, 22, 30, 40, 42, 43, 46, 47, 51, 52, 53, 57, 59, 60, 64, 66, 71, 72, 77, 79, 80, 87, 90, 93, 99, 110, 119, 122, 123, 146, 149, 150, 154, 166, 167, 175, 177, 179, 180, 184, 185. 45 Los ejemplos que componen este grupo son los siguientes: 20, 21, 41, 48, 55, 56, 91, 106, 125, 151, 158, 159, 160, 161, 162, 163, 165, 168. 62 Los nombres geográficos forman el siguiente grupo de objetos directos, compuesto por 18 ejemplos en total46. Este es un grupo heterogéneo que consta, en primer lugar, de nombres geográficos propios como (92) “a Euskadi”, (98) “a Rusia” o (102) “a Europa”. También incluimos un grupo de nombres geográficos que hace referencia a diferentes astros, como (94) “a la Luna” o (131) “a Casiopea”. Además, identificamos un grupo de objetos directos compuestos por un sintagma nominal consistente en un nombre común complementado por un nombre propio de lugar, como ocurre en (69) “al pueblo de Alpedrete de la Sierra” o en (174) “la sierra de Alcudia”. Por último, encontramos un ejemplo de nombre de lugar común, que hace clara referencia a un país: (120) “a la principal superpotencia”, y un caso dudoso que, a pesar de ser un lugar, resulta tan abstracto que podríamos clasificarlo también como objeto directo de cosa: (104) “a la parte exterior de estas galaxias”. Finalmente, destacamos un grupo de 9 objetos directos de animal, como en (2) “El objetivo de la expedición es zoológico ‐atisbar los corderos azules y el aún más raro leopardo de las nieves‐,(...)” o en (121) “(...) actividades culturales y jornadas de campo para observar a estas aves. El "gru‐gru" de las zancudas ya es audible en (...)”47. En estos grupos no incluimos aquellos ejemplos en los que el objeto directo aparece sustituido por el pronombre de objeto directo correspondiente (114: “Los animales, si tienes tiempo para observarlos, dan a menudo lecciones a las personas.”). En nuestro corpus se dan casos de pasiva refleja que, obviamente, tampoco se incluyen en estos grupos; sin embargo, (como ya hemos indicado en la nota 40) existen algunos casos que pueden interpretarse como una pasiva refleja, pero también como una activa impersonal. A pesar de que las conclusiones obtenidas en virtud de estos ejemplos no pueden considerarse indiscutibles, sí los hemos tenido en cuenta ya que, de acuerdo con la segunda interpretación, tienen un objeto directo. No obstante, cuando nos encontremos 46 Los ejemplos que componen este pequeño grupo son los siguientes: 44, 69, 70, 92, 94, 98, 100, 101, 102, 104, 120, 127, 128, 131, 155, 164, 169, 174. 47 Los ejemplos que componen este pequeño grupo son los siguientes: 2, 45, 103, 105, 121, 126, 129, 130, 156. 63 con uno de estos ejemplos, lo indicaremos de la manera pertinente. Tampoco incluimos en estos grupos algunos casos en los que el verbo aparece utilizado como intransitivo (3: “(...) donde a menudo se concentran unos turistas atisbando sobre el césped”), ni tampoco aquellos en los que resulta difícil determinar si estamos ante un objeto directo o un complemento de otro tipo, como pasa a menudo con el verbo “mirar” cuando aparece seguido de la preposición a (89: “La muchacha lo presentó como el amigo que le surtía de sebos para el jabón y Celestino le miro [sic] a los zapatos para adivinarle las intenciones (...)”). Estos últimos casos los trataremos más detenidamente más adelante. 3.2. Caracterización de los verbos objeto de estudio Como ya hemos indicado en la “Introducción” de este trabajo, nuestro estudio se centra en una serie de verbos transitivos de percepción visual: atisbar, columbrar, contemplar, divisar, mirar, observar, otear, ver y vislumbrar. Nuestro objetivo consiste en comprobar si existen tendencias destacables en el comportamiento de este grupo de verbos con respecto al objeto directo preposicional en virtud de sus puntos en común, es decir, como verbos que comparten una parte de su significado y funcionan como transitivos. También trataremos de comprobar si el aspecto de los eventos a los que se hace referencia o la agentividad o no agentividad de los sujetos de dichos verbos tienen algo que ver con la presencia de la preposición a. En primer lugar, todos estos verbos comparten un aspecto básico de su significado (primario): hacen referencia a una acción que está relacionada con la percepción a través del sentido de la vista. De acuerdo con el Diccionario de la lengua española de la Real Academia Española48 y el Diccionario del uso del español de María Moliner49, todos ellos remiten a ‘percibir’, ‘ver’ o ‘mirar50. 48 En adelante nos referiremos a dicha obra de la Real Academia Española con la abreviatura DRAE. 49 En adelante nos referiremos a dicha obra con la abreviatura DUE. 64 Partiendo de esta base, cada verbo aporta un significado particular y unos matices que lo hacen único o singular. Estos matices hacen que podamos agrupar los verbos según significados más afines. Por ejemplo, los verbos atisbar, observar y otear comparten una parte de su significado: ‘observar con cuidado’. A su vez, atisbar, columbrar, divisar y vislumbrar comparten la acepción de ‘ver o percibir confusamente’ o de ‘ver desde lejos sin distinguir muy bien’. También los verbos observar, contemplar y mirar comparten las acepciones de ‘poner atención’, ‘considerar’, ‘juzgar’ o ‘advertir’ y los verbos atisbar, columbrar y vislumbrar pueden significar ‘conjeturar’. Otra de las características que comparten todos los verbos seleccionados es que pueden funcionar como transitivos. Si los sometemos a las pruebas tradicionales de transitividad (sustitución pronominal y transformación a la voz pasiva), el resultado lo confirma: (1) (...) si el viajero tenía la fortuna de (...), podía atisbar unos calzones de hilo, amarillentos (...)” [Ap. (4)]→ podía atisbarlos / unos calzones de hilo podían ser atisbados51 50 El DRAE nos da las siguientes acepciones de los verbos estudiados, de las que podemos deducir sus puntos en común (la referencia a la percepción visual), así como otros aspectos que los distinguen. En todos los casos, remitimos a la entrada de cada verbo dentro del diccionario citado. El verbo atisbar se define como ‘1. tr. Mirar, observar con cuidado, recatadamente. 2. tr. vislumbrar (‖ ver tenue o confusamente). 3. tr. vislumbrar (‖ conocer por indicios, conjeturar)’. El verbo columbrar presenta las siguientes acepciones: ‘1. tr. Divisar, ver desde lejos algo, sin distinguirlo bien. 2. tr. Rastrear o conjeturar por indicios algo’. Contemplar se define como ‘1. tr. Poner la atención en algo material o espiritual. 2. tr. considerar (‖ juzgar). 4. tr. Rel. Dicho del alma: Ocuparse con intensidad en pensar en Dios y considerar sus atributos divinos o los misterios de la religión’. En el caso de contemplar, mostramos también la definición del DUE (s.v. contemplar) puesto que sí hace referencia a la percepción visual: ‘1. tr. Mirar una cosa o prestar atención a un acontecimiento, con placer, tranquila o pasivamente: (...). Observar. 2. Estar en contemplación de Dios. (...)’. Divisar se define como ‘1. tr. Ver, percibir, aunque confusamente, un objeto’. Mirar se define como ‘1. tr. Dirigir la vista a un objeto. (...) 2. tr. Observar las acciones de alguien. 3. tr. Revisar, registrar. 4. tr. Tener en cuenta, atender. 5. tr. Pensar, juzgar (...)’. Observar se define como ‘1. tr. Examinar atentamente. 3. tr. Advertir, reparar. 4. tr. Mirar con atención y recato, atisbar’. Otear se define como ‘1. tr. Registrar desde un lugar alto lo que está abajo. 2. tr. Escudriñar, registrar o mirar con cuidado’. Ver se define como ‘1. tr. Percibir por los ojos los objetos mediante la acción de la luz. 2. tr. Percibir algo con cualquier sentido o con la inteligencia. 3. tr. Observar, considerar algo. 4. tr. Reconocer con cuidado y atención algo, leyéndolo o examinándolo. 5. tr. Visitar a alguien o estar con él para tratar de algún asunto (...)’. Y, finalmente, vislumbrar significa ‘1. tr. Ver un objeto tenue o confusamente por la distancia o falta de luz. 2. tr. Conocer imperfectamente o conjeturar por leves indicios algo inmaterial’. 51 A partir de este punto, los ejemplos aparecen numerados consecutivamente desde el número (1) hasta el (101). Además, la referencia completa de cada ejemplo aparece en el apéndice, en el lugar que se indica al final de cada ejemplo entre corchetes detrás de la abreviatura “Ap.”. 65 (2) (...) que durante su vida concibió y columbró toda la amplitud de esta hermosa realidad (...) [Ap. (28)] → la columbró / toda la amplitud (...) fue columbrada (3) (...) dice que una de las cosas que más le gratifica al contemplar sus ochenta años es que aún se le encienden los ojos (...) [Ap. (32)] → al contemplarlos / al ser contemplados sus ochenta años (4) Al divisar el mueble que ha quedado abierto y vacío el cajón, da un grito (...) [Ap. (58)] → Al divisarlo / Al ser divisado el mueble (5) ¡Vaya pregunta! Me da corte. Bueno, en primer lugar miro el tórax y los brazos. [Ap. (83)] → los miro / el tórax y los brazos son mirados (6) (...) se alejaron en su día porque observamos en EE una actitud muy agresiva hacia nosotros [Ap. (109)] → la observamos / una actitud muy agresiva fue observada (7) (...) mientras que de espaldas al escenario otea la torreta del pozo que se vislumbra tras el ventanal (...) [Ap. (133)] → la otea / la torreta del pozo es oteada (8) ‐ ¿Cómo ves a tu generación? En mi generación todos quieren (...) [Ap. (151)] → ¿Cómo la ves? / ¿Cómo es vista tu generación? (9) (...) un entramado a través del cual es imposible vislumbrar al filósofo [Ap. (185)]→ es imposible vislumbrarlo / es imposible que el filósofo sea vislumbrado. El aspecto es otro elemento que nos permite caracterizar a estos verbos. El aspecto léxico es la información aspectual inherente a cada predicado, el contenido semántico que nos indica el modo en que tiene lugar el evento que se describe (cf. De Miguel, 1999: 2981). Una de las distinciones aspectuales básicas que ya hemos visto en la primera parte de este trabajo es la que establece la oposición entre eventos “delimitados” y “no delimitados”. Según Elena de Miguel (1999: 2982), un evento tiene un aspecto no delimitado cuando no precisa concluir para tener lugar, como ocurre con trabajar, mientras que los eventos delimitados, como llegar, no están completos hasta que no concluyen. Como indica Elena de Miguel (1999: 3020), “en los eventos no delimitados, no hay distinción entre cesar y terminar: están realizados en cualquier momento del intervalo en que ocurren y, por tanto, lo están en cualquier momento en que cesen”. 66 Si consideramos nuestro grupo de verbos en sus acepciones primarias y fuera de contexto, su aspecto léxico resulta ser no delimitado, o “atélico” de acuerdo con la terminología de Torrego Salcedo (1999); es decir, ninguno de ellos precisa concluir para tener lugar, se trata de eventos continuos cuya acción se ve realizada de igual modo durante todo el tiempo en que tienen lugar. El carácter continuo de estos verbos hace que sean compatibles con los complementos de tiempo introducidos por durante (que se considera un modificador compatible con los eventos continuos), como en (10) “Vio la tele durante tres horas”, y no con aquellos introducidos por en, (compatible con los eventos delimitados) (cf. De Miguel, 1999: 2999‐3001): (10a) “Vio la tele en tres horas”. Si sometemos nuestro grupo de verbos a la prueba sugerida por Elena de Miguel (1999: 3020‐3021), vemos que todos ellos comparten un aspecto léxico no delimitado: (11) Atisbé el barco durante diez minutos52 (12) Columbró toda la amplitud de esta hermosa realidad durante toda su vida [Ap. (28)]53 (13) (...) los Reyes (...) contemplaron durante algunos minutos el gran desfile medieval (...)54 (14) (...) durante tres segundos interminables, cuando divisaron el carruaje que traía de vuelta a la ciudad a Leopoldo Manuel y María del Rosario55 (15) Miramos el cielo durante tres horas, pero no vimos ninguna estrella” (16) La información señala que (...) el fotógrafo localizó un nido, el cual observó durante nueve semanas56 52 Este ejemplo (11), así como los ejemplos (15), (17) y (10) han sido inventados por mí misma para ejemplificar lo que se indica en este apartado. 53 El ejemplo (28) del apéndice ha sido ligeramente modificado para ejemplificar lo que se indica en este apartado. 54 Este ejemplo procede de ABC Electrónico, 29/04/1997: “Los Reyes rinden homenaje al antiguo Reino de Aragón”, Prensa Española, Madrid en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [21/09/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que en este apartado se indica, el 21/09/2013. 55 Este ejemplo procede de Durán, Armando (1978): ¡Viva la revolución! y otros textos banales. Monte Avila Editores, Barcelona, en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [10/10/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que en este apartado se indica, el 10/10/2013. 67 (17) Oteó el horizonte durante una hora (10) Vio la tele durante tres horas (18) Sólo durante el segundo cuarto de hora de la primera mitad, se pudo vislumbrar la calidad de algunos de los hombres del medio campo castellano57. El aspecto léxico de los verbos puede verse modificado por otros elementos, como el contexto, las perífrasis, los complementos o el tiempo verbal. Según Elena de Miguel (1999: 2988‐2989), el tiempo en el que aparece el verbo puede implicar cambios en el aspecto del evento referido. En el caso de ver (y de otros verbos de percepción similares, como atisbar o divisar), esto puede dar lugar a que se den dos interpretaciones posibles cuando se utiliza el perfecto simple, una como evento no delimitado y otra como evento delimitado. Elena de Miguel lo explica a partir del siguiente ejemplo de Bello (1947): (19) “Luego que vimos la costa nos dirigimos a ella”, que se puede interpretar como un evento no delimitado: “se empezó a ver la costa y se siguió viendo, pero sólo el primer momento de verla es el que precede a la acción de dirigirse a ella”, y como evento delimitado: “se vio la costa y se dejó de ver”. Aunque, a nuestro juicio, esta doble interpretación no implica un aspecto delimitado en el evento, sino una duración breve o mínima del evento de “ver la costa”, que no resta para que el evento sea continuo y no delimitado. A pesar de tratarse de eventos no delimitados, los verbos de percepción pueden referirse en mayor o menor grado al comienzo del evento. En el ejemplo (20) “Al divisar el mueble que ha quedado abierto (...)” [Ap. (58)], interpretamos que el evento comienza, pero no hay ninguna señal de que acabe. Lo mismo puede ocurrir con la mayoría de los verbos seleccionados. El verbo atisbar, por ejemplo, hace especial referencia al comienzo del evento: (21) “Allí conocí al dueño de la casa, el compañero Indalecio, y atisbé brevemente a una chica guapísima, su esposa (...)” [Ap. (11)]. En esta oración, “atisbé brevemente” hace referencia al 56 Prensa Libre, 10/07/1996 en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [21/09/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que en este apartado se indica, el 21/09/2013. 57 Este ejemplo ha sido extraído de la edición online del Collins Dictionary en el 9/19/2013. 68 primer momento en que se percibe a la chica y además implica que se trata de un evento que se prolonga en el tiempo, aunque su duración sea breve. No obstante, el hecho de que el aspecto léxico (intrínseco) de los verbos de percepción visual y el aspecto más habitual con el que aparecen da dentro de un contexto sean no delimitados, no implica que estos verbos no puedan encontrarse en contextos en los que presenten un aspecto delimitado. Por ejemplo, el verbo ver en su acepción de ‘examinar’, puede implicar un aspecto delimitado: (22) “Tengo que ir a que me vea el médico” (DUE, s.v. ver); por eso, es compatible con el modificador temporal introducido por en: (23) “El médico me vio en diez minutos”. Además, el aspecto también depende de los complementos que acompañen al verbo y del contexto. Por eso, al siguiente ejemplo que aparece en el DUE: (24) “Estoy viendo los documentos que me trajiste”, le podemos dar dos interpretaciones diferentes: “Estoy examinando los documentos que me trajiste, pero todavía no he acabado”, de aspecto delimitado, y “Tengo ante mis ojos los documentos que me trajiste”, de aspecto no delimitado. Por último, podemos caracterizar estos verbos en función de si requieren o no un sujeto agentivo. En principio, y si nos fijamos en su acepción primaria, estos verbos parecen dividirse en dos grupos. Para comprobar cuáles de estos verbos requieren un sujeto agentivo y cuáles no utilizamos la prueba que propone Elena de Miguel (1999: 3014‐3015) consistente en comprobar si el verbo es compatible con el imperativo. No hemos encontrado ningún ejemplo en la base de datos del CREA en el que los verbos atisbar, columbrar, divisar, otear, ver o vislumbrar aparezcan en imperativo. No obstante, si tomamos los siguientes ejemplos del corpus y creamos una frase en imperativo parece difícil que resulten aceptables para el usuario (en el caso de ver hemos tomado un ejemplo del artículo de Elena de Miguel, 1999: 3015): (25) (...) esto podría permitir atisbar una luz al final del túnel [Ap. (1)]→ ¡Atisba la luz! (26) Al otro extremo del salón, se columbraban nuestros dos bultos (...) [Ap. (29)] → ¡Columbra los dos bultos! 69 (27) Apenas enfoqué la calle que desembocaba en ella, divisé al Poeta y a dos fieles amigos de mi tío (...) [Ap. (60)] → ¡Divisa al Poeta! (7) (...) mientras que de espaldas al escenario otea la torreta del pozo (...) [Ap. (133)] → ¡Otea la torreta! (28) ¡Ve esta montaña!58 (9) (...) un entramado a través del cual es imposible vislumbrar al filósofo [Ap. (185)]→ ¡Vislumbra al filósofo! El resto de verbos sí presenta ejemplos en imperativo: (29) ¡Abrid los ojos y contemplad Egipto (...)59 (30) Ahora puedo decirles a los padres: "Mirad, mirad lo que les gusta a vuestros hijos"60 (31) (...) observad a Bruce Springsteen como [sic] levanta las manos como si ganara una carrera, como [sic] se le marcan las venas en la frente y le caen las gotas de sudor61. Según estos ejemplos, los verbos atisbar, columbrar, divisar, otear, ver y vislumbrar requerirían un sujeto no agentivo, ya que remiten a eventos involuntarios, y los verbos contemplar, mirar y observar requerirían un sujeto agentivo, pues las acciones a las que se refieren son voluntarias. Sin embargo, las cosas no son tan sencillas, pues algunas de las acepciones a las que responden los verbos del primer grupo requieren un sujeto agentivo. En el caso de atisbar, el DRAE le otorga la siguiente acepción: ‘Mirar algo con cuidado y disimulo. Acechar, espiar’, que requiere claramente un sujeto agentivo que actúa de manera voluntaria. Ocurre lo mismo con el verbo otear, pues tiene una acepción 58 Este ejemplo ha sido extraído de De Miguel, 1999: 3015. 59 Este ejemplo procede de Moix, Terenci (1986/1993): No digas que fue un sueño. Planeta, Barcelona, en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [12/08/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que se indica en este apartado, el 12/08/2013. 60 Este ejemplo procede de “Memorias de un pensador: Fernando Savater” en El País. Babelia. Diario El País S.A., Madrid, 22/02/2003 en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [12/08/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que se indica en este apartado, el 12/08/2013. 61 Este ejemplo procede de Joan‐Elies, Adell (1998): La música en la era digital. Milenio, Lérida, en REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [12/08/2013]. Este ejemplo, a pesar de proceder de la base de datos CREA, no aparece en el apéndice, pues se ha extraído exclusivamente para ejemplificar lo que se indica en este apartado, el 12/08/2013. 70 que lo define como verbo agentivo. Según el DUE, otear significa, entre otras cosas, ‘Buscar con la vista para descubrir algo. Escudriñar, observar’. Por último, sucede lo mismo con el verbo ver. Su acepción primaria hace referencia a percibir mediante el sentido de la vista o los ojos y puede considerarse un acto involuntario, no agentivo, como ocurre en este ejemplo que aparece en el DUE: (32) “Desde nuestra ventana vemos la costa de Francia”. De esta acepción primaria derivan otras secundarias. En algunas, el sujeto participa activamente a pesar de ser un verbo no compatible con el imperativo, por ejemplo en la siguiente: ‘Mirar cierta cosa con atención para enterarse de ella o enterarse por ella de algo: Estoy viendo los documentos que me trajiste’ (DUE, s.v. ver)62. En el caso particular de ver, el imperativo en segunda persona del singular “ve” no suele parecer aceptable al hablante, quizá por poco usado; sin embargo, en la segunda persona de formalidad, sí resulta más habitual y aceptable: (33) “Vea usted estos documentos”, y también resulta aceptable una perífrasis de obligación que pueda sustituir al imperativo: (34) “Debes/Tienes que ver estos documentos”. 62 Este ejemplo del DUE (s.v. ver) ha aparecido con anterioridad en esta misma parte de nuestro trabajo con el número (24). 71 4. LOS VERBOS DE PERCEPCIÓN VISUAL Y EL OBJETO DIRECTO DE PERSONA Como indicábamos en la presentación de nuestro corpus, hemos realizado dos grupos con los objetos directos de persona: aquellos que hacen referencia a una o varias personas individuales o individualizables (es decir, designan a una entidad que no es un colectivo) y aquellos que hacen referencia a un colectivo (o varios) (cf. supra, 61). En este apartado procederemos de la misma manera: primero estudiaremos los casos de objetos directos de persona individual o individualizable y después los de persona colectiva. 4.1. El objeto directo de persona no colectiva con preposición a El objeto directo preposicional está estrechamente vinculado con los objetos directos de persona. Por eso, la mayoría de objetos directos de persona que encontramos en nuestro corpus aparecen precedidos por la preposición a. Todos los verbos de nuestro corpus, tanto en su sentido recto como en sentido figurado, presentan ejemplos de objetos directos de persona individual introducidos por la preposición a, excepto el verbo otear: (35) Allí conocí al dueño de la casa, el compañero Indalecio, y atisbé brevemente a una chica guapísima, su esposa (...) [Ap. (11)] (36) Dionisio columbró en la cara interna de sus párpados, tal y como lo había visto dieciocho horas antes entre los tankas y grabados expuestos en la vitrina de un anticuario del zoco, al mismísimo dios Chomolugna cabalgando y guiando un león (...) [Ap. (30)] (37) Se tiene que tener en cuenta (...) que ciertos estereotipos que contemplaban al anciano/a como asexuado/a y ridículos sus posibles sentimientos amorosos, van desapareciendo (...) [Ap. (40)] (27) Apenas enfoqué la calle que desembocaba en ella, divisé al Poeta y a dos fieles amigos de mi tío (...) [Ap. (60)] (38) Lo publica BLANCO Y NEGRO y tengo que emigrar a Filipinas. Y miro a mi marido y es para echarse a llorar [Ap. (87)] (39) (...) ahora estoy en una cafetería sentado y procuro observar a los transeúntes [Ap. (110)] 72 (40) Como voy poco de copas, casi no veo a nadie. Yo tengo mi propio mundo [Ap. (150)] (9) (...) un entramado a través del cual es imposible vislumbrar al filósofo. El autor, actualmente profesor (...) [Ap. (185)]. Entre los objetos directos de persona individual (o individualizable) introducidos por la preposición a, encontramos sintagmas nominales de diverso tipo. En su mayoría, estos objetos directos aparecen en grupos nominales definidos consistentes en un sustantivo precedido de un artículo determinado (17 casos), como en (41): “Ya en esta actividad se vislumbra al hombre que será puente entre dos países, (...)” [Ap. (179)]; o de un determinante posesivo (8 casos), como en (42): “En cuanto divisó a su santo compañero, se allegó a él desalado para (...)”[Ap. (64)]; en un nombre propio (3 casos), como en (43): “Junto al gitano, en el mismo burladero, vislumbré a José Manuel Dominguín, hijo de Pepe y (...)”[Ap. (175)]; o en un pronombre personal (1 caso), como en (44): “‐ La veo a ella como una innovadora” [Ap. (146)]. Por otro lado, también tenemos grupos nominales indefinidos cuyos núcleos aparecen precedidos de artículo indefinido (4 casos) como en (45): “(...) pero lo que resultaba insólito era contemplar a un personaje con esa apariencia sosteniendo un severo discurso (...)” [Ap. (51)], de un determinante indefinido (4 casos), como en (46): “Nadie les espera. No atisban a persona alguna conocida” [Ap. (5)] o consistentes en un pronombre indefinido (2 casos), como en (40): “Como voy poco de copas, casi no veo a nadie (...)”[Ap. (151)]. Vemos que el ámbito de la personalidad, al menos hasta ahora, presenta una gran regularidad en cuanto a la presencia de la preposición a, pues comparece en casi todos los casos encontrados. Únicamente hemos encontrado dos ejemplos en los que un objeto directo de este tipo aparece sin preposición. Se trata de (47): “Iba enfrascado en sus pensamientos ya muy cerca de su cueva cuando, de pronto, divisó en el cielo unos ángeles que llevaban entre las alas el alma de san Pablo (...)” [Ap. (77)]; y (48): “La primera de las alternativas es difícil, puesto que no se vislumbra otro líder socialista con más capacidad que el propio González (...)” [Ap. (180)] (cf. infra, 73‐74, donde comentamos más ampliamente estos 73 ejemplos). En estos ejemplos, el aspecto o la agentividad del sujeto no parece tener influencia en la presencia de la a. Todos ellos presentan un aspecto no delimitado, algunos de ellos también ingresivo, como ocurre con los verbos atisbar, columbrar y divisar, como vemos en los ejemplos (27), (35), (36), (42) o (43). En cuanto a la agentividad, tenemos ejemplos que presentan un sujeto agentivo con los verbos contemplar, mirar y observar, como en (37), (38), (39) o (45); mientras que otros presentan un sujeto no agentivo con verbos como atisbar, columbrar, divisar, ver o vislumbrar, como en (9), (27), (35), (36), (40), (41), (42), (43) o (44). Independientemente del tipo de sujeto, todos los ejemplos presentan la preposición a. 4.2. El objeto directo de persona no colectiva sin preposición a Los objetos directos de persona individual o individualizable (no colectiva) sin preposición a suponen la excepción en nuestro corpus: (47) Iba enfrascado en sus pensamientos ya muy cerca de su cueva cuando, de pronto, divisó en el cielo unos ángeles que llevaban entre las alas el alma de san Pablo (...)[Ap. (77)]63 (48) La primera de las alternativas es difícil, puesto que no se vislumbra otro líder socialista con más capacidad que el propio González (...) [Ap. (180)]. En (47), el objeto directo presenta un núcleo (“ángeles”) precedido por un artículo indeterminado (“unos”), pero determinado por una subordinada de relativo: “que llevaban entre las alas el alma de san Pablo”. Podríamos interpretar que esta es la primera referencia que se hace en el texto de estos ángeles y, por tanto, que hasta este punto eran desconocidos para el lector, pero no indeterminados o inespecíficos. El ejemplo (48), sin embargo, se enmarca en el ámbito de lo inespecífico del que hablábamos en la primera parte de este 63 El ejemplo (47) se engloba en este grupo porque parece difícil enmarcar “unos ángeles” en otro grupo diferente, pues claramente no se trata ni de cosas ni de animales. 74 trabajo (cf. supra, 44‐45): el objeto directo hace referencia a un líder inespecífico que, en principio, no existe y no responde a unas características determinadas. A nuestro juicio, ambos ejemplos serían posibles con la preposición a: (47a) “(...) divisó en el cielo a unos ángeles que llevaban entre las alas el alma de san Pablo (...)” o (48a) “(...) no se vislumbra a otro líder socialista con más capacidad que el propio González”. Si los núcleos de estos objetos estuvieran determinados por un artículo determinado, parece acertado pensar que aparecerían con más probabilidad precedidos por la preposición a: (47b) “(...) divisó en el cielo a los ángeles que llevaban entre las alas el alma de san Pablo (...)” o (48b) “(...) no se vislumbra al líder socialista con más capacidad que el propio González (...)” y con menos probabilidad sin ella: (47c) “(...) divisó en el cielo los ángeles que llevaban entre las alas el alma de san Pablo (...)” o (48c) “(...) no se vislumbra el líder socialista con más capacidad que el propio González (...)”. Este es uno de los ámbitos de aparición de la preposición a con objeto directo de persona en los que más vacilación se ha encontrado en la casuística en general (cf. supra, 44‐45). Aunque la intuición del usuario tienda a preferir la presencia de a con los núcleos determinados y la ausencia de la misma con los indeterminados, vemos que algunos núcleos indeterminados aparecen precedidos de la preposición a (ibídem). No podemos decir que los verbos de percepción escapen a estas vacilaciones, pero sí parece, en vista de los pocos ejemplos encontrados, que existe una tendencia a presentar, cada vez con más frecuencia, los objetos directos de persona siempre precedidos por la preposición a. Es decir, la determinación o indeterminación del objeto directo es cada vez menos determinante a la hora de introducir la preposición a delante de objetos directos que hagan referencia a personas, pues el carácter personal de dichos objetos se impone como condición principal para la presencia de la preposición a. 75 4.3. El objeto directo de persona colectiva con preposición a El objeto directo de persona colectiva sigue la misma tendencia que el de persona individual o individualizable y se presenta en la mayoría de los casos introducido por la preposición a. De hecho, en nuestro corpus, apenas encontramos ejemplos que se aparten de esta tendencia y aparezcan sin la mediación de dicha preposición. El objeto directo de persona colectiva acompaña a los verbos atisbar, contemplar, mirar, observar y ver cuando aparecen tanto en su sentido recto: (49) (...) le tenía cariño al salón del biombo, era la habitación más luminosa, me gustaba atisbar a la gente pasando por la calle (...)[Ap. (20)] (50) Y ya no quise oír la música ni contemplar a las parejas que bailaban a nuestro alrededor (...)[Ap. (56)] (51) En sus conciertos no mira al público porque teme que una mujer guapa sea capaz de hacerle perder la concentración [Ap. (91)] (52) Un habitante de Tel Aviv observa a un grupo de extremistas judíos que bloquean (...)[Ap. (125)] (53) “(...) cuando paso por los Pirineos, por ejemplo, y veo a toda mi gente, a toda mi afición animando” [Ap. (160)], como en sentidos figurados derivados: (54) Primero, no se contempla a los sindicatos, se contempla únicamente las extralimitaciones y las infracciones (...)[Ap. (48)] (55) (...) muchos ciudadanos de a pie miran a la prensa, o al menos a una parte de ella, como una especie de última trinchera (...)[Ap. (106)] (56) Ibarra apoyó esta argumentación diciendo que “basta observar a la Comisión Ejecutiva o al Comité Federal para ver que (...)”[Ap. (124)] (57) (...) el dirigente socialista señaló que “no veo a un PP gobernando con mayoría relativa con el apoyo de IU (...)”[Ap. (158)]. Por ejemplo, en (51), mirar presenta el sentido de ‘dirigir la vista a un objeto’ que recoge el DRAE o ‘aplicar a algo el sentido de la vista, para verlo’ del DUE, frente al ejemplo (55), que responde a la acepción de ‘considerar’ del Diccionario de 76 construcción y régimen de Rufino José Cuervo o al ‘pensar, juzgar’ del DRAE. En algunos casos, el sujeto es agentivo, como ocurre en el ejemplo (49), en el que atisbar puede interpretarse con el sentido de ‘mirar’ o ‘contemplar’, y en otros el sujeto realiza una acción involuntaria, como ocurre en (53), que parece remitir al sentido primario de ver: ‘Percibir algo por el sentido de la vista’ (DUE, s.v. ver). El aspecto, por su parte, es no delimitado en todos los casos. La determinación también parece ser un lugar bastante común en estos objetos directos. La mayoría de nuestros ejemplos presentan un núcleo precedido por un artículo determinado (“a la gente”, “a las parejas”, “al público”); o por un determinante posesivo, como en (8): “¿Cómo ves a tu generación? En mi generación todos quieren (...)” [Ap. (151)]; o por otros medios de determinación como en (58) “(...) es un gozo inefable contemplar a tan nutrido mocerío adelantando a automóviles utilitarios, descapotables, tractores, autobuses y camiones (...)” [Ap. (55)]. También tenemos casos en los que el colectivo es un nombre propio, como en (59): “(...) no puede haber acuerdo ni veo a IU regenerando la democracia con la derecha” [Ap. (159)]. A pesar de contar con ejemplos en los que el objeto directo aparece presentado por un artículo indeterminado, aparecen determinados de otro modo, por ejemplo, en (52): “Un habitante de Tel Aviv observa a un grupo de extremistas judíos que bloquean (...)” [Ap. (125)], donde la subordinada de relativo “que bloquean (...)” permite que interpretemos que se trata de “un grupo de extremistas” concreto. Otros ejemplos con artículo indeterminado son los siguientes: (57) (...) el dirigente socialista señaló que “no veo a un PP gobernando con mayoría relativa con el apoyo de IU (...)” [Ap. (158)]. (60) (...) y no ha tenido nunca un problema con Hacienda, y, sin embargo, vemos a una serie de personajes y personajillos, unos que presumen de (...) [Ap. (168)]. Estos dos ejemplos presentan la preposición a a pesar de incluir el artículo indeterminado y serían perfectamente posibles sin la presencia de la misma: (57a) “(...) no veo un PP gobernando (...)” o (60a) “(...) vemos una serie de 77 personajes y personajillos, (...)”. De modo que parece que la presencia de a se impone también en los objetos directos de persona colectiva. La oposición determinación/indeterminación no nos sirve para explicar la presencia y la ausencia de la preposición a en los ejemplos encontrados.64 4.4. El objeto directo de persona colectiva sin preposición a Los ejemplos de este tipo de objeto directo sin preposición son escasos. En nuestro corpus solo hemos localizado dos: (61) Dirigió la vista al ángulo recóndito del huertecito mixto de flores y hortalizas, semioculto por los tendederos y los matorrales, y atisbó parte de un corrillo de monjas sentadas al bordado [Ap. (21)] (62) (...) a partir de la obra de Marx y Weber, quienes contemplan la sociedad moderna qua sociedad capitalista [Ap. (41)]. En (61) interpretamos el sentido más básico de percepción visual del verbo atisbar (‘ver algo, en sentido propio o figurado, muy débilmente o sólo si se mira muy atentamente’, DUE, s.v. atisbar) y un sujeto no agentivo. En (62) identificamos un sujeto voluntario y una acepción metafórica del verbo contemplar (‘considerar una cosa en cierto modo’, DUE, s.v. contemplar). En 64 Otros ejemplos que podrían incluirse en este grupo se tratan en el apartado dedicado a los objetos directos referidos a nombres geográficos, pero consideramos importante hacer mención de ellos aquí. En primer lugar, destacamos el siguiente ejemplo: “Es inevitable que en los próximos días Estados Unidos haga gala de su conocida ignorancia sobre el fútbol, pero la extraña decisión de celebrar el Mundial en Estados Unidos va a constituir para el resto del mundo una atalaya privilegiada desde la que observar a la principal superpotencia” [Ap. (120)]. En este ejemplo, “a la principal superpotencia” se utiliza como sinónimo de Estados Unidos. Ni “principal superpotencia” ni “Estados Unidos” hacen referencia al espacio físico del país, sino al país considerado como sociedad, como conjunto de individuos, como ente que realiza acciones. Otros ejemplos, como “¿Cómo ve a esta Andalucía que tiene en sus manos?” [Ap. (155)], también se recogen en el apartado dedicado a los nombres geográficos, pero podrían haberse recogido en este apartado, puesto que existen razones que lo justificarían. En este ejemplo, “esta Andalucía” puede percibirse como un colectivo referido a un conjunto de personas, en lugar de como nombre geográfico. El emisor se refiere a “Andalucía” entendida como un conjunto de personas concretas en unas circunstancias concretas. Podemos apreciarlo recuperando el contexto en el que aparece la frase: “Andalucía está en una situación económica y social dinámica. Andalucía ha tenido un desarrollo económico muy importante en los últimos años por la gestión de un gobierno socialista y porque la tierra y la gente de Andalucía son un potencial económico que no se había explotado hasta ahora” (REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [28/08/2013]). 78 ambos casos, nos encontramos ante eventos no delimitados. En (61) se hace referencia a una parte indeterminada de un todo (“parte de un corrillo”), que remite a un colectivo referido a personas (“un corrillo de monjas”). Se ha optado por omitir la preposición a, mientras que en otros ejemplos de nombres colectivos con el mismo verbo, aparece: (49) (...) le tenía cariño al salón del biombo, era la habitación más luminosa, me gustaba atisbar a la gente pasando por la calle (...) [Ap. (20)]. Para intentar comprender la ausencia de la preposición en el ejemplo (61) en lugar de la construcción “y atisbó a parte de un corrillo de monjas” presentamos algo de contexto: Dirigió la vista al ángulo recóndito del huertecito mixto de flores y hortalizas, semioculto por los tendederos y los matorrales, y atisbó parte de un corrillo de monjas sentadas al bordado. Celestino tenía bien observado al grupo homogéneo de las hermanas del vecino convento de las Clarisas Pobres. Buscaban las mujeres el último amable solecito que incidía cada época del año en una parte diferente del huerto. En este extracto encontramos dos ejemplos que contrastan: 1) “parte de un corrillo de monjas” y 2) “al grupo homogéneo de las hermanas”. El primero tiene un sujeto no agentivo y un aspecto no delimitado, frente al sujeto agentivo del segundo y su aspecto limitado, aportado por la perífrasis tener + participio y el adverbio bien. Mientras que en el segundo ejemplo se hace referencia a un grupo concreto de individuos y el uso de la preposición a se ajusta a la norma más extendida, en el primer ejemplo se pretende hacer referencia a una parte indefinida del corrillo, que se considerará consecuentemente un objeto inanimado. El autor no considera esa “parte” como un grupo de personas; es decir, “parte” no hace referencia a un número de monjas del corrillo, sino más bien a una parte no contable de un todo. Así, podría atisbar parte de una sotana, pies, piernas, labores, etc. No creemos que se trate de una oposición 79 determinado/indeterminado, porque los dos ejemplos hablan de unas monjas concretas y conocidas por el sujeto. Consideramos más bien que se trata de una oposición animado/inanimado, porque hay un claro matiz semántico que distingue ambos ejemplos. En el ejemplo (62), el objeto directo “la sociedad” aparece introducido sin mediación de la preposición a y seguido de un predicativo: “qua sociedad capitalista”. Contemplar responde a su acepción de ‘considerar una cosa en cierto modo’ (DUE, s.v. contemplar). El mismo objeto directo de (62) podría aparecer introducido por la preposición a y resultaría aceptable para el usuario, sobre todo, si no apareciera el predicativo: (62a) “(...) a partir de la obra de Marx y Weber, quienes contemplan a la sociedad moderna” (aunque la oración tendría un sentido distinto); o si el predicativo estuviera introducido por como + artículo indefinido: (62b) “(...) a partir de la obra de Marx y Weber, quienes contemplan a la sociedad moderna como una sociedad capitalista”. En cualquiera de estos casos, parece haber una clara –aunque sutil‐ diferencia de significado con respecto a (62). Mientras que en (62a) y en (62b) se hace referencia a una sociedad concreta concebida como un grupo de individuos, no parece que ocurra lo mismo en (62). En realidad, (62) parece hacer referencia a un concepto más bien abstracto de sociedad. No es un objeto identificable con un grupo de individuos concreto, ni siquiera con un grupo de individuos sin concretar, ya que al tratarse de un concepto abstracto se pueden incluir en él otros elementos que no sean personas, por ejemplo, costumbres, corrientes de pensamiento, ideologías, etc. De hecho, tal y como entendemos este objeto directo, se aparta del concepto de personalidad de este apartado y podría haberse clasificado de manera más general como un objeto de cosa. Ambos ejemplos, pues, carecen de la preposición a porque, de acuerdo con la interpretación que hacemos de los mismos, no se presentan como entes animados, sino como elementos inanimados. Esto apoyaría la afirmación que ya hemos realizado con anterioridad: el carácter personal del objeto directo tiene cada vez más peso en la presencia de la preposición. Los objetos directos de persona que no presentan la preposición a son escasos independientemente de 80 que se presenten como determinados o como indeterminados, mientras que la regularidad de aquellos que sí la presentan es casi absoluta. 81 5. LOS VERBOS DE PERCEPCIÓN VISUAL Y EL OBJETO DIRECTO DE LUGAR El objeto directo de lugar ha sido un tema muy tratado en el estudio del objeto directo preposicional por las vacilaciones que ha sufrido a lo largo de su historia en nuestra lengua, como hemos visto en la primera parte de este trabajo (cf. supra, 14‐15, 26‐29). Hoy en día, las gramáticas destacan una tendencia clara: este tipo de objeto directo aparece cada vez con más frecuencia sin la preposición a, a diferencia de lo que ha pasado en etapas anteriores (cf. supra, 40). Esta tendencia se da con los verbos de percepción visual, pero convive con otras tendencias que requieren la presencia de la preposición a. En nuestro corpus hemos localizado un grupo heterogéneo de 18 ejemplos de verbos de percepción visual acompañados por un objeto directo de lugar. De ellos, 16 presentan la preposición a, frente a 2 que no la presentan. A pesar de que la mayoría de los ejemplos aparecen con preposición, distinguimos tres tendencias diferentes que vamos a presentar a continuación. 5.1. El objeto directo de lugar con preposición a En primer lugar, identificamos una serie de ejemplos en los que la preposición a aparece siempre y el objeto directo es el nombre propio de una región, un país, un continente, etc. o un nombre común complementado por un nombre propio geográfico: (63) Este eje mediterráneo debe contemplar a los países del Magreb, que culturalmente se encuentran muy relacionados con los del sur de Europa (...) [Ap. (44)] (64) Ahora Bandrés, desde la Cámara europea, (...) mira de reojo a Euskadi y está apasionado por la construcción de una Europa de los pueblos [Ap. (92)] (65) (...) trasluce la sensación que tiene Aute cada vez que mira a las Américas: “Europa es un museo muerto (...)” [Ap. (101)] (66) “¿Cómo ve a esta Andalucía que tiene en sus manos?” (67) Al igual que Vietnam, su política exterior ya no ve a Estados Unidos 82 como el beligerante de una guerra (...) [Ap. (164)]. Todos estos ejemplos presentan un sujeto agentivo, un verbo con aspecto no delimitado y, además, se trata de verbos de percepción visual que muestran un significado derivado de su acepción primaria. El ejemplo (63) presenta el verbo contemplar con su acepción de ‘considerar’ (DRAE, s.v. contemplar) y no con la acepción de ‘mirar una cosa’ (DUE, s.v. contemplar). En los ejemplos (64) y (65), el verbo mirar no aparece con su significado de ‘dirigir la vista a un objeto’ (DRAE, s.v. mirar) o de ‘aplicar a algo el sentido de la vista, para verlo’ (DUE, s.v. mirar), sino con el de ‘tener en cuenta’, ‘pensar’, ‘considerar’ u ‘observar’. Lo mismo ocurre con los ejemplos (66) y (67): el verbo ver presenta el sentido de ‘entender una cosa’ (DUE, s.v. ver) o de ‘percibir algo (...) con la inteligencia’ (DRAE, s.v. ver) y no de ‘percibir por los ojos los objetos mediante la acción de la luz’ (DRAE, s.v. ver). Los objetos directos de lugar que aparecen en estos ejemplos: “los países del Magreb”, “Euskadi”, “las Américas”, etc., no hacen referencia exclusiva a las áreas geográficas o a los territorios físicos y políticos a los que denominan, sino también a las gentes que los habitan, a las sociedades que constituyen, a su economía, etc. En la primera parte de este trabajo veíamos que Seco matizaba que había una diferencia según si un topónimo se refería a un territorio (ser inanimado) o a las gentes que lo habitan (ser animado)65. En nuestra opinión, algo parecido ocurre con los verbos de percepción visual. Percibir una región o un país puede hacer referencia a percibir a sus gentes o a percibir un ente físico, un territorio. Si nos referimos a un ente físico, a un territorio, probablemente lo ‘percibimos a través del sentido de la vista’. Si hablamos de sus gentes, de su sociedad, probablemente estemos hablando de ‘considerarlas’ o de ‘tenerlas en cuenta’. De modo que se establece una distinción entre topónimos introducidos por a y sin ella, no solo en función de si se trata de entes animados o inanimados, sino también en función del significado del verbo en cada caso concreto. 65 Cf. supra, página 40. 83 Esta distinción queda clara si comparamos los ejemplos anteriores con otros como: (68) “Vimos Roma en dos días” o (69) “Este verano veremos Andalucía”66. Con algunos verbos de percepción es más difícil establecer esta distinción. Por ejemplo, con el verbo mirar resulta más difícil encontrar ejemplos en los que impliquemos que se mira un país en su sentido físico a través del sentido de la vista, si no se mira, por ejemplo, sobre un mapa: (70) “Mira América y mira Europa. ¡Qué diferencia!”67. 5.2. El objeto directo de lugar sin preposición a La preposición a no comparece en aquellos ejemplos en los que el verbo de percepción se interpreta con su acepción primaria en su sentido recto y relacionada con el sentido de la vista –salvo casos excepcionales‐ y el objeto directo hace referencia al objeto percibido por el sentido de la vista: (71) Estupendas vistas desde esta pequeña cumbre (...) donde se divisa el pueblecito del Atazar [sic] y la lámina de agua del embalse (...) [Ap. (70)]68 (72) (...) detrás de la escueta cordillera, parda y malva, se vislumbraba la sierra de Alcudia [Ap. (174)]. En (71), el verbo divisar se interpreta como ‘ver, percibir, aunque confusamente, un objeto’ (DRAE, s.v. divisar) y, en (72), el verbo vislumbrar’ como ‘ver un objeto tenue o confusamente por la distancia o falta de luz’ (DRAE, s.v. vislumbrar). En ambos casos los objetos directos (si se trata de construcciones activas 66 Estos dos ejemplos han sido inventados por mí misma. 67 Ejemplo inventado por mí misma. 68 Este ejemplo, como el siguiente, no constituyen pruebas indiscutibles de la ausencia de la preposición a ante objeto directo, puesto que pueden interpretarse tanto como oraciones en pasiva refleja como en activa impersonal. Si se trata del primer tipo, la transformación en plural contendría al nombre adyacente del verbo como sujeto en plural: “se divisan los pueblecitos” o “se vislumbran las sierras”; mientras que la construcción activa impersonal mantendría el verbo en tercera persona del singular y el nombre en plural (con o sin preposición a): “se divisa los pueblecitos”, “se divisa a los pueblecitos” o “se vislumbra las sierras”, “se vislumbra a las sierras”. Además, en el caso de que se tratara de activas impersonales, el objeto directo se podría sustituir por el pronombre correspondiente, por ejemplo: “se la divisa” o “se las divisa”, mientras que si se tratara de una pasiva refleja esto no sería posible. 84 impersonales) hacen referencia a áreas geográficas, concebidas en su sentido espacial más que humano. También en ambos casos los sujetos implícitos no son agentivos y los eventos a los que se refieren son no delimitados. Como excepción a esta tendencia tenemos el siguiente ejemplo: (73) Un par de curvas más y se llega a la parte alta del cerro, 930 m., desde donde se divisa ya al pueblo de Alpedrete de la Sierra a nuestros pies [Ap. (69)]. Este ejemplo ha sido extraído de la misma fuente que (71)69. El autor ha elegido dos opciones diferentes –con y sin preposición a‐ para dos estructuras prácticamente iguales. Son ejemplos paralelos que comparten el tipo de agentividad del sujeto, el aspecto del evento y el significado con que se interpreta el verbo. Además, en ambos casos se trata de un nombre común que hace referencia a una población en su sentido de área geográfica. A nuestro juicio, sería más habitual que el ejemplo (73) hubiera aparecido sin la preposición a, ya que no hace referencia a una población en su sentido humano y el verbo tiene un significado estrechamente ligado al sentido de la vista. No obstante, el uso de la preposición puede deberse a un contagio del uso habitual de la palabra “pueblo” como referencia a unas gentes, a un grupo de personas. 5.3. Los nombres de astros y constelaciones Los nombres de astros y constelaciones representan un lugar en el espacio, pero resulta difícil interpretarlos como nombres de lugar por el usuario habitual, pues no son lugares a los que habitualmente nos dirijamos o en los que ubiquemos cosas. En todos los ejemplos que hemos encontrado en nuestro corpus con este tipo de objeto directo, la preposición a comparece. Los verbos se interpretan con 69 Pliego, Domingo (1992): 100 Excursiones por la Sierra de Madrid. La librería, Madrid apud REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [17/08/2013] 85 su acepción primera, relacionada directamente con el sentido de la vista. Sin embargo, los nombres de los astros no se pueden interpretar en un sentido [+humano] o [+animado]. (74) Entonces, miráis a la Luna y cada uno donde está, insisto, (...) [Ap. (94)] (75) Una experiencia interesante consiste en observar a Venus cuando es visible por la madrugada (...) [Ap. (127)] (76) Si sustituimos los ojos por dos observatorios diferentes y, en lugar del dedo, observamos a Marte simultáneamente, podemos determinar la distancia a que se encuentra [Ap. (128)] (77) Por esas razones te recomiendo que observes a Casiopea; una constelación muy visible (...) [Ap. (131)] (78) Claro. Fijaros, el Sol, nada más. ¿Qué [sic] cuando vemos a Marte por el telescopio, ¿qué vemos? [Ap. (169)]. En nuestra opinión, estos objetos directos deben interpretarse como objetos de cosa, pues son el objeto de la mirada o de la percepción del sujeto o del emisor en cada caso y el objeto mirado no se percibe como una área o una extensión geográfica, sino como un objeto de estudio. Por eso, en todos los ejemplos puede eliminarse la preposición a sin que el resultado resulte agramatical o inapropiado para el usuario (75a: “Una experiencia interesante consiste en observar Venus cuando es visible por la madrugada (...)”), pero el emisor ha preferido introducir dicha preposición. En todos estos ejemplos, el emisor es conocedor o experto en la materia de la que trata. A nuestro juicio, esa es la razón principal que explica la presencia de la preposición a. Se trata del tema central del discurso del que habla un experto en la materia. La a sirve para dar entidad a dicho tema central en un texto especializado frente a lo habitual en un texto no especializado, que sería la ausencia de la preposición, como podría darse en un ejemplo como (79) “Nos pasamos la noche contemplando la Luna y las estrellas”70. Estaríamos, pues, ante un uso pragmático de la preposición a para dotar de entidad al objeto de estudio. 70 Este es un ejemplo inventado por mí misma. 86 No obstante, no debemos olvidar que se trata de nombres propios que designan a entidades únicas para el ser humano. El carácter único de dichos elementos y el hecho de que se les designe con un nombre propio también es relevante a la hora de considerar la presencia/ausencia de la preposición a. En cualquier caso, la presencia/ausencia de la preposición a ante objetos directos de cosa cuyo núcleo es un nombre propio es un tema lo suficientemente amplio como para dedicarle otro estudio. 87 6. LOS VERBOS DE PERCEPCIÓN VISUAL Y EL OBJETO DIRECTO REFERIDO A UN ANIMAL O A VARIOS ANIMALES Los objetos directos que reflejan nombres de animal son muy interesantes porque, por un lado, son animados y por eso parece que debieran aparecer precedidos de la preposición a, pero, obviamente, no representan a personas y no siempre nos resultan lo suficientemente familiares o cercanos como para presentarlos con la preposición a. En nuestra cultura, ciertos animales son domésticos y están más vinculados a nuestra vida diaria que otros, que nos resultan extraños y lejanos. Una vez repasado nuestro corpus, detectamos que existe una tendencia a presentar los objetos directos de este tipo con la preposición a, de modo que su carácter [+animado] podría considerarse más influyente en la presencia de a que su carácter [–personal]. De los 9 ejemplos encontrados, 7 llevan la preposición a: (80) (...) que acudieron al recinto de la Ciutadella para contemplar a la orca y también visitar la nueva (...) [Ap. (45)] (81) (...) un pájaro mira a otro que ha caído entre las redes. Esta escena (...) [Ap. (103)] (82) Ustedes que nos miran a los monos como si estuviéramos en Segunda División B en la liga de la escala animal, en verdad han descendido mucho (...) [Ap. (105)] (83) (...) actividades culturales y jornadas de campo para observar a estas aves. El "gru‐gru" de las zancudas ya es audible en (...) [Ap. (121)] (84) (...) un paseo en bote desde Telegraph Cave para observar a las orcas en la Reserva Ecológica (...) [Ap. (126)] (85) Como señala Cynthia Moss, que ha pasado 30 años observando a una familia de elefantes salvajes en (...) [Ap. (129)] (86) (...) los puedes también utilizar para estudiar y observar a las especies si eres naturalista aficionado (...) [Ap. (130)] y 2 no la llevan: (87) El objetivo de la expedición es zoológico ‐atisbar los corderos azules y el aún más raro leopardo de las nieves‐,(...) [Ap. (2)] 88 (88) (...)aquí no se ve muy bien, pero ahí veis un pájaro, (...) [Ap. (156)]. Todos los ejemplos presentan un sujeto agentivo, excepto (87) y (88), y los eventos a los que hacen referencia son no delimitados. De todos estos ejemplos, solo uno contiene un verbo de percepción visual que no muestra el sentido recto de percibir a través del sentido de la vista. Se trata de (82) “Ustedes que nos miran a los monos como si estuviéramos en Segunda División B en la liga de la escala animal, en verdad han descendido mucho (...)”, en el que presenta su acepción de ‘pensar, juzgar’ (DRAE, s.v. mirar) en el sentido de ‘considerar’. Se trata de un ejemplo de personificación del objeto directo: el narrador/emisor del texto es un animal, un mono que se dirige a los seres humanos, de modo que aparece personificado, pues realiza una acción propiamente humana. El objeto directo de este ejemplo hace referencia a la especie a la que pertenece el emisor/narrador, de modo que, al igual que este, el objeto directo aparece personificado e introducido por la preposición a. Si nos fijamos en los casos de presencia de la preposición a, a excepción de (82), todos ellos podrían funcionar igualmente sin preposición, por ejemplo: (83a) “(...) actividades culturales y jornadas de campo para observar estas aves”. Según las explicaciones vistas en la primera parte, los nombres comunes de animal aparecen precedidos por a cuando se asimilan a los de persona o cuando hay una cierta cercanía afectiva con el animal (cf. supra, 43). En estos casos, no parece que se trate de animales domésticos o especialmente cercanos culturalmente para el receptor. En realidad, parece que en su mayoría (a excepción de (80)), se trata de textos especializados y la a sirve para dar entidad a un tema que se considera central. En estos textos se presenta una gran “familiaridad” con los animales citados, pero no por tratarse de un animal doméstico, sino porque nos encontramos ante un elemento fundamental del discurso y un tema conocido por el experto. Los dos ejemplos en los que no aparece la preposición a presentan sujetos no agentivos. En ambos ejemplos podría comparecer la preposición a y no resultarían agramaticales para el receptor, dependiendo, siempre, del contexto: 89 (87a) El objetivo de la expedición es zoológico ‐atisbar a los corderos azules y al aún más raro leopardo de las nieves‐,(...)”. (88a) “(...) aquí no se ve muy bien, pero ahí veis a un pájaro, (...)”. La presencia de la preposición a podría explicarse por una interpretación diferente de la que asignaríamos a los ejemplos sin preposición. En (87a) detectaríamos un sujeto agentivo de atisbar, que respondería a la acepción de ‘mirar, observar con cuidado, recatadamente’ (DRAE, s.v. atisbar) frente al sujeto involuntario de (87), en el que interpretamos el sentido de ‘ver algo, en sentido propio o figurado, muy débilmente o sólo si se mira muy atentamente’ (DUE, s.v. atisbar). En (88a) interpretaríamos, a nuestro juicio, que se trata de un tema al que se da importancia en el discurso en cuestión, con un sentido más cercano al de ‘observar’; sin embargo, en el ejemplo (88), parece que el objeto directo no es un elemento importante para el emisor, para el que parece resultar más importante el hecho de ver que el objeto que se ve. Vemos que los objetos directos de animal –no doméstico ni conocido‐ aparecen precedidos con frecuencia de la preposición a tras verbos de percepción cuando estos implican un cierto estudio o focalización en el objeto directo en cuestión. Esto se da, sobre todo, con verbos como observar o contemplar, que están relacionados con acepciones como ‘examinar’, ‘mirar en detalle’, etc. y que implican acciones que se realizan de manera voluntaria e intencionada. A nuestro juicio, existe una diferencia de uso que depende, por un lado, del sentido que se asigne al verbo implicado y, por otro, del emisor, es decir, de si se trata de un experto en la materia o de un usuario común, no experto. Consideramos que es muy posible que el texto especializado presente este tipo de objetos con preposición con más frecuencia que el texto divulgativo o no especializado. 90 7. LOS VERBOS DE PERCEPCIÓN VISUAL Y EL OBJETO DIRECTO DE COSA 7.1. El objeto directo de cosa sin preposición a El objeto directo que con más frecuencia encontramos en nuestro corpus es el objeto directo que designa una cosa y lo más habitual es que aparezca en forma de objeto directo típico, es decir, sin la mediación de la preposición a. Todos los verbos seleccionados para nuestro análisis presentan en alguna ocasión un objeto directo de este tipo. El objeto directo de cosa típico aparece tanto cuando los verbos son utilizados con su sentido recto y su acepción más primaria, es decir, más cercana a la percepción visual a través de la vista: (4) Al divisar el mueble que ha quedado abierto y vacío el cajón, da un grito y se deja caer sobre el asiento más cercano [Ap. (58)] (7) Finalmente lo tira en plena irascibilidad contra el suelo mientras que de espaldas al escenario otea la torreta del pozo que se vislumbra tras el ventanal [Ap. (133)] (89) (...) si el viajero tenía la fortuna de (...), podía atisbar unos calzones de hilo, amarillentos, (...) [Ap. (4)] (90) No me negará que los hombres iban a la revista para contemplar de cerca sus piernas [Ap. (33)]71 como cuando aparecen en sentido figurado y con acepciones derivadas de la primera: (2) Es nuestra mejor respuesta y homenaje a Julio Rey Pastor, (...), que durante su vida concibió y columbró toda la amplitud de esta hermosa realidad (...) [Ap. (28)] 72 (3) (...) dice que una de las cosas que más le gratifica al contemplar sus 71 A pesar de ser las “piernas” una parte del cuerpo humano, no consideramos que presenten un carácter [+humano] en este ejemplo, de modo que las incluimos en los objetos directos de cosa. 72 El verbo columbrar ofrece menos cantidad de ejemplos que la mayoría de verbos que hemos analizado. Su uso está más restringido y parece más limitado a textos literarios, aunque no hay aspectos destacables de su comportamiento, pues se ajusta a lo esperado. 91 ochenta años es que aún se le encienden los ojos ante una mujer bella [Ap. (32)] (6) (...) se alejaron en su día porque observamos en EE una actitud muy agresiva hacia nosotros (...)[Ap. (109)] (25) “(...) esto podría permitir atisbar una luz al final del túnel” [Ap. (1)] (91) Desde que estoy en España hago un negocio cuando veo la posibilidad, directa o indirectamente (...) [Ap. (145)]. Por ejemplo, en (90), el verbo contemplar se utiliza con su acepción de: ‘Mirar una cosa o prestar atención a un acontecimiento, con placer, tranquila o pasivamente’ (DUE, s.v. contemplar), mientras que en (3) nos encontramos con el mismo tipo de objeto directo, pero, en este caso, el verbo presenta un sentido figurado, más cercano a ‘recordar’ o ‘considerar’, que no incluye la intervención del sentido de la vista. La agentividad del verbo parece no tener influencia en la ausencia de la preposición con este tipo de objeto directo. En nuestros ejemplos observamos que hay verbos con un sujeto agentivo, como en (3), (7) o (90), y otros con un sujeto no agentivo, como en (4), (25) o (89), y ninguno de ellos presenta la preposición a. Lo mismo ocurre con el aspecto. Estos objetos directos no parecen tener influencia alguna respecto del aspecto de los eventos en estos ejemplos. En algunos de ellos, además de apreciarse el aspecto no delimitado del evento (“contemplar de cerca sus piernas”, “mientras que de espaldas al escenario otea la torreta del pozo”), se aprecia también un aspecto ingresivo, como en (4), (89) y (91). En ellos, se indica el comienzo del evento, sin mencionar su fin. Los ejemplos con el verbo otear son muy uniformes, pues su significado es siempre el mismo73 y los objetos directos son muy similares. Los objetos directos de persona destacan por su ausencia. Los objetos directos encontrados son 73 Las acepciones que presenta el DUE sobre este verbo son las siguientes: otear (del antig. «oto», alto): 1 tr. Poder ver desde cierto lugar alto una extensión de terreno: ‘Desde aquella altura se otea todo el valle’. Descubrir, divisar, dominar. 2 Buscar con la vista para descubrir algo. Escudriñar, observar. El significado que encontramos en nuestros ejemplos se ajusta, sobre todo, a la acepción número 1 de este diccionario. 92 objetos directos de cosa en los que no aparece la mediación de la preposición a. Casi todos ellos hacen referencia al horizonte y al paisaje que se divisa desde la distancia, suponemos que desde un punto elevado: (7) Finalmente lo tira en plena irascibilidad contra el suelo mientras que de espaldas al escenario otea la torreta del pozo que se vislumbra tras el ventanal. También hay ejemplos en los que el verbo otear presenta un sentido figurado: (92) Habla de las ansias con que un alma entumecida otea la llegada de la primavera, como el avance de un ejército (...). Además, como ocurre con el resto de los verbos analizados, el aspecto léxico de este verbo es no delimitado y en los ejemplos seleccionados, la presencia del objeto directo no modifica el carácter aspectual de los eventos referidos. En cuanto a la agentividad del sujeto, algunos ejemplos con otear permiten interpretar tanto que se trata de un sujeto agentivo como de uno no agentivo. Si nos fijamos en el ejemplo (7), ambas interpretaciones son posibles. Sería necesario un contexto más amplio para poder concretar si se refiere a un acto voluntario o a uno involuntario. 7.2. El objeto directo de cosa precedido por la preposición a Los objetos directos de cosa introducidos por la preposición a son, en general, escasos. En nuestro corpus se limitan a ciertos verbos, por lo que deducimos que la presencia de la preposición a se debe a circunstancias especiales. 7.2.1. Algunos casos con el verbo contemplar El verbo contemplar nos brinda dos ejemplos de objeto directo de cosa 93 introducidos por la preposición a: (93) Por otro lado, puede servir de introducción la pregunta de Yates "... ¿por qué las neurociencias contemplan a la física para desarrollar sus teorías?" [Ap. (49)] (94) En todo caso se hace más evidente que nunca la insuficiencia de contemplar a la ciudad como un espacio cerrado, puesto que cada vez (...) [Ap. (50)]. En estos ejemplos identificamos dos eventos a los que corresponde un sujeto agentivo y cuyo aspecto es no delimitado. En el ejemplo (93), “la física” aparece introducida por la preposición a. El sujeto de esta oración es “las neurociencias”, que son quienes “contemplan”. En este caso, podemos hablar de una personificación de la disciplina, que realiza una acción que se puede considerar humana: contemplar. El objeto directo “la física” se sitúa en el mismo plano humano que las “neurociencias”, de modo que una disciplina contempla a la otra de una manera determinada. El ejemplo podría ser parafraseado con: “¿Por qué los neurocientíficos tienen en cuenta a la física/ a los físicos para desarrollar sus teorías?”. La presencia de a se debe, pues, al carácter animado (por personificado) de sujeto y objeto. En (94) tenemos el objeto directo “la ciudad”. Resulta difícil considerarlo un colectivo que haga referencia a un conjunto de individuos porque el mismo contexto lo define como un “espacio cerrado”, por eso lo clasificamos entre los objetos directos de cosa. Consecuentemente, no esperaríamos la presencia de la preposición a y una oración como “se hace más evidente que nunca la insuficiencia de contemplar la ciudad como un espacio cerrado” resultaría aceptable para el usuario. La presencia o ausencia de la preposición a no implica ningún cambio de significado en la oración ni tampoco de aspecto. Para poder explicar la presencia de a, resulta esencial entender la oración en su contexto. Se trata de un ejemplo procedente del libro titulado Los procesos de urbanización74. En este apartado del libro, “la ciudad” es el tema principal: 74 Vinuesa Angulo, Julio; Vidal Domínguez, María Jesús: Los procesos de urbanización. Síntesis, 94 Esta consideración que nace de la obra de Christaller (1933) y que en buena medida se sintetiza en la idea de "ciudadregión" de Dickinson (1961) o en la concepción del "sistema urbano como protagonista del desarrollo" de P. Gould (1969), termina por ser el soporte teórico de diversas estrategias de política regional, arbitradas por los gobiernos de diferentes tipos de países. Con ellas se persigue mejorar la estructuración del sistema de asentamientos urbanos, con objeto de minorar los desequilibrios regionales y de optimizar la potencialidad de las redes de ciudades para difundir los avances culturales y el desarrollo económico e integrar en ellos a las diferentes comarcas. En todo caso se hace más evidente que nunca la insuficiencia de contemplar a la ciudad como un espacio cerrado, puesto que cada vez pueden ser más lejanos los factores exógenos con capacidad para intervenir en su evolución. La idea es que los problemas de crecimiento de una ciudad no son sólo de carácter endógeno y tampoco pueden ser resueltos sólo con medidas internas (...). En este ejemplo, la presencia de a responde a razones pragmáticas: un intento de dar protagonismo o focalizar la atención en el tema objeto de estudio. Existen factores semánticos que pueden influir en la presencia de a, pues el hecho de ser el tema central convierte a “la ciudad” en una entidad con cierta personalidad, pues aparece como protagonista y se estudia su desarrollo y su crecimiento. 7.2.2. Algunos aspectos relevantes de los giros con el verbo mirar El verbo mirar presenta una casuística particular que lo diferencia del resto de los verbos seleccionados. Al igual que los demás verbos, presenta casos de objeto directo de cosa sin preposición a, que sería lo esperable al tratarse de un verbo transitivo: (5) ¡Vaya pregunta! Me da corte. Bueno, en primer lugar miro el tórax y Madrid. (Tema: Urbanismo). REAL ACADEMIA ESPAÑOLA: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. [12/08/2013]. Este extracto del libro que sirve de contexto para situar el ejemplo (94) lo hemos recuperado de la base de datos CREA exclusivamente para este apartado, de modo que no aparece en el apéndice. 95 los brazos. Me encantan unos brazos musculosos y fuertes (...) [Ap. (83)]. Este tipo de objeto directo no plantea dudas. Sin embargo, no todos los casos son tan claros. El verbo mirar se utiliza a menudo seguido de un complemento circunstancial de lugar con un sentido direccional similar al de la preposición hacia. Este uso puede haberse contagiado a algunos objetos directos, aunque no siempre es fácil distinguir cuándo nos encontramos ante un objeto directo y cuándo ante un complemento circunstancial de dirección: (95) ¡Vaya mesa! Si parece el hotel Ritz. (Las dos mujeres le miran el cuerpo) ¿Puedo ya cenar? [Ap. (95)] (96) La miro a la cara y veo, la identifico entonces como R. (mi novia) [Ap. (95)] (97) “(...) Le miro a los ojos y tiene su mirada pesada producto del alcohol (...)”[Ap. (86)] (98) “La muchacha lo presentó como el amigo que le surtía de sebos para el jabón y Celestino le miro [sic] a los zapatos para adivinarle las intenciones (...)” [Ap. (89)] Mientras que los ejemplos (95) y (96) parecen claros, en los ejemplos (97) y (98) resulta difícil discernir qué tipo de complemento acompaña al verbo mirar. En (95) identificamos un objeto directo de cosa introducido sin preposición a (“el cuerpo”) y un objeto indirecto en forma pronominal (“le”). En el ejemplo (96) tenemos un objeto directo pronominal (“la”) que deja claro que “a la cara” no es un objeto directo, parece más bien un complemento circunstancial de dirección que concreta hacia dónde dirige la mirada el sujeto, sabiendo que el objeto de esa mirada es “la”. Sin embargo, en los ejemplos (97) y (98), tenemos dos complementos introducidos por a (“a los ojos” y “a los zapatos”) que podrían clasificarse como complementos de dirección (“hacia dónde miro”) o como objetos directos preposicionales, correspondientes a “le miro los ojos” o “le miro los zapatos”. En caso de que fueran complementos de dirección, el pronombre “le” constituiría un leísmo aceptado de persona y se trataría de un objeto directo (“lo”) y, en el caso de que fueran objetos directos introducidos por la preposición a, los pronombres realizarían la función de objeto indirecto. A nuestro juicio, 96 ambos casos son ejemplos de objetos directos aunque pueda parecer lo contrario. El sentido de las oraciones nos permite hacer estas interpretaciones de cada uno de ellos: (97) “Le miro los ojos y tiene la mirada (...)” o “Se los miro y (...)”, y (98) “Le miró los zapatos para adivinarle las intenciones” o “Se los miró (...)”75. Cuando el complemento introducido por a es un nombre común de lugar, resulta todavía más difícil saber si nos encontramos ante un objeto directo introducido por a o ante un complemento circunstancial de lugar. A pesar de que ya hemos dedicado un apartado a los objetos directos de lugar (cf. supra, 81‐86), incluimos este ejemplo en este apartado porque, de acuerdo con la interpretación que hacemos de esta oración, se trataría de un objeto de cosa y no de lugar: (99) Si miramos a la parte exterior de estas galaxias, encontramos que el gas se mueve sorprendentemente rápido (...) [Ap. (194)]76 En este ejemplo, la sustitución por el pronombre de acusativo la parece lógica: “Si la miramos, encontramos que (...)”, aunque la sustitución por hacia tampoco parece equivocada: “Si miramos hacia la parte exterior de estas galaxias (...)”. No obstante, por el contexto podemos interpretar que no miramos en dirección a la parte exterior de esa galaxia, sino que la examinamos o la observamos: es el objeto de estudio. La misma construcción sin la preposición parece aceptable: “Si 75 A este respecto, el Diccionario panhispánico de dudas de la Real Academia Española dice lo siguiente (s.v. mirar): 1. Cuando significa ‘dirigir la vista a alguien o algo’, puede ser transitivo o intransitivo: a) En la construcción transitiva, el complemento directo de persona va precedido de la preposición a, no así el de cosa: «Miró a su mujer con los ojos pidiendo socorro» (Jodorowsky Pájaro [Chile 1992]); «Jacob [...] mira las lomas con nostalgia» (Pinto Despertar [C. Rica 1994]). Si este complemento es un pronombre átono de tercera persona, deben usarse las formas lo(s), la(s): «Yo estaba mirando a Carlos. [...] Lo miraba como si él fuera la música» (Mastretta Vida [Méx. 1990]). Cuando el verbo lleva dos complementos, uno de persona y otro de cosa, el de persona funciona como indirecto y exige, por tanto, el empleo de la forma le(s): «Yo estaba mirando a Carlos. [...] Le miraba las piernas» (Mastretta Vida [Méx. 1990]). En expresiones como mirar a los ojos o mirar a la cara el complemento de persona es también directo: «Le agarró el mentón y la miró a los ojos» (Canto Ronda [Arg. 1980]). 76 Dado que este ejemplo resulta complicado, hemos consultado en la base de datos CREA el contexto en el que aparece: “(...) Así, tenemos buenas razones para pensar que existe en el universo una gran cantidad de materia que no observamos, pero que ejerce fuerza gravitacional. La evidencia más inmediata procede de galaxias en forma de disco (como nuestra propia Vía Láctea) que se encuentran en rotación. Si miramos a la parte exterior de estas galaxias, encontramos que el gas se mueve sorprendentemente rápido; mucho más rápidamente de lo que debería debido a la atracción gravitacional producida por las estrellas y gases que detectamos en su interior (...)”. 97 miramos la parte exterior de estas galaxias, encontramos que (...)”. En cualquier caso, sin poder afirmar con rotundidad que estamos ante un objeto directo, podemos afirmar que las dudas al respecto se deben a la influencia de “mirar a” con sentido de dirección. En caso de que fuera un objeto directo, tendríamos que explicar la presencia de a por medio de la hipótesis que ya hemos utilizado con anterioridad: el intento de destacar o dar entidad a un tema dentro de un texto especializado. Otros ejemplos que queremos destacar por presentar vacilación en cuanto a la presencia de la preposición a son los siguientes: (100) ¿Es tolerante con la gente de ideas políticas contrarias a las suyas? ‐ Sí, porque miro la parte humana de las personas [Ap. (85)] (101) (...) pero ahora miro a lo más profundo del ser y regreso con palabras (...) [Ap. (88)]. En (100), “la parte humana de las personas” aparece sin preposición a. Por el contrario, el ejemplo (101) sí que utiliza la preposición a para introducir un objeto directo similar en cuanto a su sentido: “lo más profundo del ser”. En el caso de (100), podemos pensar que el carácter humano del complemento de “parte” no pesa lo suficiente como para modificar un objeto de cosa, pues una “parte” no es un ente animado, sino más bien, en este caso, una área sin definir. No obstante, no nos extrañaría encontrarnos con un ejemplo como “miro a la parte humana de las personas”, que podría deberse tanto a una humanización de esa “parte de las personas”, como a la influencia de los complementos de lugar: “miro en esa dirección para ver qué encuentro”. El ejemplo (101) sí parece influenciado por los complementos de dirección y podría interpretarse como “miro al ser en lo más profundo” o “busco en lo más profundo del ser”. Es un caso difícil, porque podría considerarse un objeto directo: “lo miro” o un complemento de lugar: “miro ahí”. En todos los ejemplos vistos, constatamos que el carácter agentivo del sujeto o el aspecto no delimitado del verbo no tiene relación alguna con la presencia o 98 ausencia de a ente el objeto directo, pues todos los sujetos son agentivos y todos los eventos resultan no delimitados con independencia de la presencia de la misma. 99 CONCLUSIONES La revisión crítica de la bibliografía sobre la presencia de la preposición a delante del objeto directo en español, en la primera parte de nuestro trabajo, nos ha permitido acercarnos al origen de esta construcción, que se encuentra en la desaparición de los casos latinos, y a su expansión a lo largo del tiempo. A pesar de que los estudiosos parecen tener bastante certeza sobre la etimología de la construcción, todavía no existe consenso sobre qué sentido tuvo la aparición del fenómeno y existen diversas postulaciones al respecto: algunas apuntan a la distinción de rasgos referenciales con alcance léxico‐semántico (seres animados/seres inanimados), otras a la distinción de sujeto y objeto personal, otras a factores rítmicos, etc. Por otro lado, si hablamos de su expansión, el uso que se ha hecho de esta construcción a lo largo del tiempo ha dado lugar a que haya ciertos elementos en función de objeto directo que privilegien la presencia de a, como los nombres propios de persona y animal o los pronombres tónicos; mientras que con otros elementos –como los nombres de lugar‐ solo podemos hablar de condiciones favorables o desfavorables a la presencia de la misma, pues el comportamiento de la presencia/ausencia de a todavía no es sistemático. Los estudiosos que han abordado el fenómeno en textos del castellano medieval ofrecen diferentes perspectivas de interpretación tanto semántico‐sintácticas como pragmático‐ discursivas o incluso algunas estrictamente gramaticales. A nuestro juicio y en vista de los datos aportados por los estudiosos, las motivaciones expresivas y semánticas del uso de la construcción siempre deben tenerse en cuenta en el estudio de la evolución del fenómeno, así como otros factores como la oralidad o la función de la construcción en diferentes tipos de texto. En español actual, la revisión de la bibliografía en conjunto permite apreciar que existen diferentes perspectivas de aproximación a la presencia y la ausencia de la preposición a delante del objeto directo y que, a pesar de que hay áreas en las que se dan vacilaciones y fluctuaciones en el uso de la construcción, también existen ámbitos de uso en los que la exigencia de la presencia de la preposición 100 está clara. Los enfoques más productivos a la hora de dar respuesta a la gran casuística que se da en el uso o la ausencia de a en la construcción que nos ocupa son aquellos que tienen en cuenta los factores relacionados con las características del objeto directo mismo y aquellos que se centran en factores que tienen que ver con el verbo que rige el objeto directo. De nuevo, como ocurre en los estudios del fenómeno en castellano medieval, las perspectivas estrictamente gramaticales, que no tienen en cuenta factores semánticos ni pragmáticos, parecen resultar insuficientes. Los enfoques principales de estudio del fenómeno, tanto aquellos que estudian la presencia de a en dependencia del objeto directo mismo, como aquellos que la estudian en relación con el verbo en cuestión, dan lugar a una jerarquía de factores que favorecen la presencia de dicha preposición. Así, el primer enfoque distingue entre un ámbito de presencia de la preposición a obligatoria (nombres propios, pronombres tónicos y otros pronombres) y otros ámbitos de vacilación (de menor a mayor grado de vacilación): nombres comunes determinados referidos a personas, nombres comunes determinados referidos a animales, nombres comunes indeterminados referidos a personas, nombres de lugar, etc. En cuanto a los enfoques que se centran en las características del verbo, distinguen también un grupo de verbos que siempre exigen la preposición a frente a otros grupos de verbos en los que se da una presencia opcional de la misma debida a diferentes factores (por ejemplo, a cambios de significado del verbo o a cambios en el aspecto del evento). En la segunda parte de nuestro trabajo, el análisis de los ejemplos recogidos en nuestro corpus nos ha permitido apreciar cuáles son algunos de los factores que favorecen la presencia de la preposición a ante objeto directo con los verbos de percepción visual seleccionados: atisbar, columbrar, contemplar, divisar, mirar, observar, otear, ver y vislumbrar. A pesar de que el número de ejemplos de nuestro corpus es reducido y se limita a ciertos años (de 1990 a 1995) y exclusivamente a España, hemos podido distinguir, de una parte, varios factores determinantes en la presencia de la preposición a en las construcciones aludidas 101 y, de otra, ciertas vacilaciones que, con todo, revelan tendencias significativas en la selección de la preposición a ante objeto directo. En primer lugar, la presencia de la preposición a queda privilegiada ante los nombres de persona, tanto individual como colectiva, incluso si se trata de nombres comunes. De acuerdo con las gramáticas, los nombres propios de persona siempre presentan la preposición a cuando funcionan como objeto directo. La a también comparece obligatoriamente ante los pronombres tónicos y ante la mayoría de los casos de objeto directo de persona determinada. Sin embargo, el ámbito de la indeterminación ha sido siempre lugar de vacilaciones y fluctuaciones. Pues bien, en nuestro corpus observamos que las vacilaciones relacionadas con la oposición determinado/indeterminado son mínimas y la presencia de a apenas encuentra excepciones. De modo que el carácter [±humano] o [±personal] del objeto directo se impone al carácter [±determinado], hasta el punto de que, incluso ante objetos directos de persona no determinada, lo excepcional es la ausencia de la preposición. Por otra parte, la oposición entre presencia y ausencia de la preposición a presenta tres tendencias fundamentales con los objetos directos que designan un nombre geográfico. La ausencia de la preposición a domina los casos en los que los verbos se utilizan en su sentido recto, que hace referencia a la percepción visual a través del sentido de la vista. Esta tendencia se confirma en las gramáticas y en los trabajos que abordan el tema, que tienden a matizar que la presencia de a ante objeto directo de lugar es cada vez menor. Sin embargo, en este trabajo hemos comprobado que la presencia de la preposición a abunda en aquellos casos en los que los verbos deben interpretarse con un significado derivado o metafórico, en el que el sentido de la vista ya no está presente. Es decir, la presencia de la preposición a ante nombre geográfico depende, en cierta medida, de la interpretación que se haga del significado del verbo en cuestión. Este factor semántico va unido normalmente a otro factor semántico incluido en el significado del elemento que funciona como objeto directo: su carácter [+animado]. Cuando el verbo se interpreta en su sentido primario, el objeto directo de lugar suele ser inanimado y hace referencia al área geográfica en su 102 sentido físico; sin embargo, cuando el verbo se interpreta en sentidos derivados, el nombre geográfico suele hacer referencia a la sociedad o las gentes que habitan y conforman el área geográfica en cuestión y, por tanto, se pueden entender como objetos directos [+animados]. En cuanto a los nombres de animal, la mayoría de los estudiosos indican que es la cercanía afectiva con el animal la que determina en mayor medida la presencia de la preposición a. No obstante, en nuestro estudio hemos detectado otra tendencia interesante. La presencia de la preposición a con este tipo de nombres comunes en nuestro corpus es frecuente. Al margen de casos excepcionales en los que la preposición a aparece porque se da una personificación del objeto directo referido a un animal, el uso de la preposición a en nuestro corpus responde a razones fundamentalmente pragmáticas: se trata de textos en los que el emisor es un experto y utiliza la preposición a para referirse al tema de su discurso. Esta a sirve para dar entidad al objeto directo como tema de estudio del experto, en el que se focaliza la atención en el texto. La cercanía del animal con el emisor de la que hablan los estudios sobre el tema se transforma en la cercanía del experto con su objeto de estudio o de trabajo. Esta tendencia también se aprecia con los nombres de astros. Con ellos se da un uso similar al que hemos visto con los nombres de animal. La presencia de la preposición a es frecuente y, a nuestro juicio, se da una clara diferencia entre el uso que haría un usuario común y el que hace el experto. La presencia de la preposición a aporta, por un lado, entidad al objeto; por otro, deja clara la cercanía o familiaridad del experto con su objeto de estudio. En el caso particular de los nombres de astros, este uso pragmático de a se une al hecho de que se trata de nombres propios de entidades únicas, que tienden también a introducirse con la preposición a por contagio de los nombres propios de persona o de ser animado. A esta tendencia eminentemente pragmática, que hemos detectado en los objetos directos referidos a animales y a astros, se une otra de orden semántico y consideramos que ambas van de la mano. Observamos que la presencia de a 103 también depende del sentido que se asigne al verbo implicado. Los verbos que se utilizan con un sentido de ‘examinar’, ‘mirar detenidamente’ o ‘mirar en detalle’ presentan con más frecuencia la preposición a, mientras que aquellos que hacen referencia a la mera percepción involuntaria del sentido de la vista, no la llevan. La presencia de la preposición a ante objetos directos de cosa es, como se podía esperar, casi inexistente. No obstante, se dan algunos ejemplos. La presencia de a ante este tipo de objetos puede deberse a una personificación del objeto, como han explicado los gramáticos con frecuencia. En otras ocasiones se debe a las razones pragmáticas que ya hemos indicado: la preposición a se utiliza para focalizar un objeto de estudio que adquiere una entidad dentro del texto. En cualquier caso, observamos que los casos de vacilación en el uso de la preposición con objetos directos de cosa con estos verbos son mínimos. De manera contraria a la tendencia que destacan algunos gramáticos, hemos comprobado que ni el aspecto del evento referido ni la agentividad del sujeto tienen peso en la presencia/ausencia de la preposición a ante el objeto directo con este grupo de verbos. En la mayoría de los casos, el aspecto de los eventos resulta no delimitado sin que este afecte a la presencia/ausencia de la preposición a. Igualmente, la preposición a aparece tanto con sujetos agentivos como no agentivos y lo mismo ocurre en los casos en los que la preposición no comparece: los sujetos son tanto agentivos como no agentivos. Por tanto, estos últimos factores son de menor relevancia que las propiedades de la entidad que funciona como objeto directo o el propio significado de percepción de los verbos estudiados a la hora de propiciar la presencia de a ante el objeto directo. Por último, a pesar de que este estudio se centra en los verbos de percepción visual como grupo, hemos apreciado que el verbo mirar refleja un comportamiento ajustado a unas tendencias particulares. El uso de a con este verbo se debe a menudo a la frecuencia de uso de dicho verbo con complementos circunstanciales que presentan una dirección similar a la que indica la preposición hacia y que aparecen introducidos por la preposición a. Por esta razón, existen casos de objeto directo que muestran la preposición a por 104 contagio de los complementos de lugar y existen otros casos en los que no queda claro si nos encontramos ante un objeto directo precedido de la preposición a o ante un complemento circunstancial de lugar. A pesar de que obras de referencia como el Diccionario panhispánico de dudas (2005) explican cómo deben considerarse ciertos complementos de mirar introducidos por la preposición a, los casos de duda siguen siendo, a nuestro juicio, tan frecuentes y difíciles de explicar que solo este asunto requeriría un estudio propio. En síntesis, pues, los estudios que los gramáticos han realizado hasta el momento sobre el objeto directo preposicional en nuestra lengua plantean ciertas tendencias sobre la presencia de la preposición a ante objeto directo, tendencias que hemos podido matizar para el grupo de los verbos de percepción visual seleccionado: en primer lugar, las vacilaciones en el área de los objetos directos de persona indeterminada son mínimas; en segundo lugar, el valor semántico tanto del verbo como del objeto directo en cada caso particular tienen peso en la selección de la presencia de la preposición a y, en tercer lugar, percibimos que existen condiciones pragmáticas que favorecen la presencia de dicha preposición. Asimismo, las explicaciones relativas al aspecto del verbo y a la agentividad del sujeto parecen no ser muy relevantes a la hora de determinar la presencia de la preposición a en nuestro grupo de verbos. 105 BIBLIOGRAFÍA Alarcos Llorach, Emilio, 1994: Gramática de la lengua española, Madrid, R.A.E ‐ Espasa Calpe. Alcina Franch, Juan y José Manuel Blecua, 1975: Gramática española, Barcelona, Ariel. Bello, Andrés, 1847/1988: Gramática de la lengua castellana destinada al uso de los americanos, Madrid, Arco/Libros. Corominas, Juan y José A. Pascual, 1991‐1997: Diccionario crítico etimológico castellano e hispánico, Madrid, Gredos. Cuervo, Rufino José, 1987‐1994: Diccionario de construcción y régimen de la lengua castellana, Santafé de Bogotá, Instituto Caro y Cuervo. De Miguel, Elena, 1999: “El aspecto léxico”, en Ignacio Bosque y Violeta Demonte, dirs., Gramática descriptiva de la lengua española, Madrid, R.A.E. ‐ Espasa Calpe, pp. 2977‐3060. Fernández‐Ordóñez, Inés, 1999: “Leísmo, laísmo y loísmo”, en Ignacio Bosque y Violeta Demonte, dirs., Gramática descriptiva de la lengua española, Madrid, R.A.E. ‐ Espasa Calpe, pp. 1317‐1397. Folgar, Carlos, 1988: “A+ topónimo objeto directo en español arcaico”, en Verba, 15, pp. 403‐420. García Martín, José María, 1992: “Evolución del objeto directo preposicional en la tradición textual de algunas obras castellanas del siglo XIII”, en Cahiers de linguistique hispanique médiévale, 17, pp. 47‐86. García Martín, José María, 1998: “Factores diafásicos en el uso del objeto directo preposicional en las lenguas románicas: estado de la cuestión”, en Estudios de la UCA ofrecidos a la memoria del profesor B. Justel Calabozo, Cádiz, Universidad de Cádiz, pp. 263‐272. 106 Gómez Torrego, Leonardo, 1997: Gramática didáctica del español, Madrid, SM. Horno Chéliz, Mª del Carmen, 2002: Lo que la preposición esconde: estudio sobre la argumentalidad preposicional en el predicado verbal, Zaragoza, Prensas Universitarias de Zaragoza. Laca, Brenda, 1995: “Sobre el uso del acusativo preposicional en español”, en Carmen Pensado, El complemento directo preposicional, Madrid, Visor Libros, pp. 61‐91. Lapesa, Rafael, 1964: “Los casos latinos: restos sintácticos y sustitutos en español”, en el Boletín de la Real Academia Española, tomo 44, cuaderno 171, pp. 57‐106. Lapesa, Rafael, 1981: Historia de la lengua española, Madrid, Gredos, 9ª edición. Marcos Marín, Francisco, 1978: Estudios sobre el pronombre, Madrid, Gredos. Martín Zorraquino, María Antonia, 1976: “A + objeto directo en el Cantar de Mio Cid”, en Mélanges de langues et de littératures romanes offerts à Carl Theodor Gossen, Berna, Francke, pp. 555‐566. Moliner, María, 2007: Diccionario del uso del español [CD‐ROM], Madrid, Gredos, 3ª edición. Monedero Carrillo de Albornoz, Carmen, 1978: “El objeto directo preposicional y la estilística épica. (Nombres geográficos en el Cantar de Mio Cid)”, en Verba, 5, pp. 259‐303. Monedero Carrillo de Albornoz, Carmen, 1983: “El objeto directo preposicional en textos medievales. (Nombres propios de persona y títulos de dignidad)”, en el Boletín de la Real Academia de la Lengua, número 63, pp. 241‐302. Pensado, Carmen (ed.), 1995: El complemento directo preposicional, Madrid, Visor Libros. Real Academia Española, 1960: Diccionario histórico de la lengua española, Madrid, [en línea]. 107 Real Academia Española, 2001: Diccionario de la lengua española (22.a ed.) [en línea]. Real Academia Española, 2005: Diccionario panhispánico de dudas [en línea]. Real Academia Española, 2009: Nueva gramática de la lengua española, Madrid, R.A.E. ‐ Espasa‐Calpe. Real Academia Española, 2010: Manual de la Nueva gramática de la lengua española, Madrid, R.A.E. ‐ Espasa‐Calpe. Real Academia Española: Banco de datos (CREA) [en línea]. Corpus de referencia del español actual. Seco, Manuel, 1972: Gramática esencial del español: introducción al estudio de la lengua, Madrid, Aguilar. (Se ha tenido en cuenta también la edición de 1989.) Seco, Manuel, 1998: Diccionario de dudas y dificultades de la lengua española, Madrid, Espasa Calpe, 10ª edición. Torrego Salcedo, Esther, 1999: “El complemento directo preposicional”, en Ignacio Bosque y Violeta Demonte, dirs., Gramática descriptiva de la lengua española, Madrid, R.A.E. ‐ Espasa Calpe, pp. 1779‐1805.
188362	https://www.petscare.com/en-ca/news/faq/how-rare-is-brindle	How Rare Is Brindle? Coat Patterns in Pets Explained How Rare Is Brindle? Coat Patterns in Pets Explained Blog Home News FAQ How rare is brindle? How rare is brindle? Brindle is relatively uncommon in some breeds but more prevalent in others; overall, it's a moderately rare coat pattern in dogs and other animals. How Rare Is a Brindle Coat Color in Animals? The term brindle refers to a distinctive coat pattern found in various animals, most notably dogs, but also in cats, horses, and even cattle. Characterized by subtle stripes or streaks of color—often brown, black, or red—the brindle pattern adds an eye-catching and unique flair to an animal’s appearance. But exactly how rare is it? What Is a Brindle Coat? The brindle pattern resembles tiger-striping and results from specific genetic expressions. It often appears as dark stripes overlaying a lighter base color. Brindle differs from spots or solid colors in that it creates a layered, almost textured look. The Genetics Behind Brindle The brindle pattern is typically the result of the interaction between various genes, particularly at the K locus in dogs. The K br allele is responsible for brindling when inherited alongside other compatible genes. Importantly, it’s not just a single dominant or recessive trait; the outcome depends on multiple genetic factors and combinations. Brindle in Dogs Brindle is relatively common in certain breeds but rare in others. Here's a look at its occurrence: Common in: Boxers, Greyhounds, Dutch Shepherds, Boston Terriers, Staffordshire Bull Terriers Occasional in: Great Danes, Mastiffs, Dachshunds, Bull Terriers Rare in: Golden Retrievers, Poodles, Beagles In breeds like the Boxer or Dutch Shepherd, brindle is almost expected. In contrast, it’s exceedingly rare—and generally considered a disqualifying trait—in breeds known for solid or specific colors. Brindle in Cats In felines, brindle refers to what’s often called “tortoiseshell” or “torbie” (tortoiseshell-tabby mix) patterns. It’s slightly more common in females due to X-linked inheritance but still not the most prevalent pattern. This makes brindle cats a bit more unique compared to their tabby or solid-colored counterparts. Brindle in Horses and Other Animals In horses, the brindle pattern is considered extremely rare. Most often, it's linked to chimerism—an unusual combination of two different DNA sets in the same organism. Cattle may also display brindle patterns, where it is seen more frequently in some breeds like the Highland or Texas Longhorn. Factors That Influence Brindle Rarity Several factors contribute to the rarity of brindle: Breed standards: Breeding for show purposes may suppress brindle patterns in some breeds. Genetic diversity: Closed gene pools may either suppress or encourage brindle, depending on the gene frequency. Breeder preference: Some breeders may avoid brindle due to market demand or aesthetics. Misconceptions About Brindle Many people confuse brindle with similar patterns like merle or sable. However, each pattern has distinct characteristics and genetic pathways. Brindle tends to be more subtle than merle and does not usually include diluted or blotched colors. Popularity and Appeal Though not the rarest pattern, brindle’s striking appearance often makes animals with this trait more popular among adopters and pet owners. Their unique coat can sometimes increase their chances of being adopted, especially from shelters. Conclusion: Moderately Rare and Highly Distinctive While brindle is not the rarest coat pattern, it is certainly less common than solid, spotted, or tabby coat colors. Its presence in an animal often adds notable visual appeal, making brindle pets highly desirable among certain enthusiasts. Bottom line: The rarity of brindle depends on the species and breed, but overall, it’s a moderately rare and genetically fascinating coat pattern. Share on: brindle brindle dogs brindle cats brindle horses coat color genetics brindle pattern pet coat types rare dog colors dog breed colors tiger stripes dog genetics horse coat colors unique pet coats cat brindle patterns tortoiseshell torbie cats rare pet traits pet breed standards genetic coat traits dog color patterns brindle coat brindle dog breeds brindle cats rarity genetics of brindle brindle vs merle Recommended Guinea Pig Theft in Marana Leads to Swift Law Enforcement Action and Recovery Learn how swift law enforcement action recovered stolen guinea pigs in Marana and discover key security tips for animal rescue centers. Read the article Special Needs Animal Rescue: From the Heart Animal Sanctuary Hosts Annual Fundraiser Join From the Heart Animal Sanctuary's 'Happy Together' fundraiser supporting special needs and medically fragile animals' care in Paso Robles. Read the article Animal Rescue in Gibson County: 11 Pets Saved from Severe Neglect Nashville Humane Association rescues 11 animals from severe neglect in Gibson County, providing urgent care and rehabilitation. Read the article Read All Articles Today is the perfect time to get your Pet Health Report free Limited Time Upload a photo of your pet to receive instant health and care insights. Get Free Pet Report Today is the perfect time to get your Pet Health Report free Limited Time Upload a photo of your pet to receive instant health and care insights. 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188363	https://www.cuemath.com/ncert-solutions/the-product-of-two-perfect-squares-is-a-perfect-square-state-whether-the-statement-is-true-or-false/	The product of two perfect squares is a perfect square. State whether the statement is true or false. Solution: Given, the product of two perfect squares is a perfect square. We have to determine if the given statement is true or false. Number obtained when a number is multiplied by itself is called the square of the number. if m = n², then m is a perfect square where m and n are natural numbers. Example: Consider two perfect squares 4 and 9 Product of two perfect squares = 4 × 9 = 36 Square of 6 = 36 Therefore, the product of two perfect squares is a perfect square. ✦ Try This: The product of two perfect cubes is a perfect cube. State whether the statement is true or false. ☛ Also Check: NCERT Solutions for Class 8 Maths NCERT Exemplar Class 8 Maths Chapter 3 Problem 51 The product of two perfect squares is a perfect square. State whether the statement is true or false Summary: The given statement, ”The product of two perfect squares is a perfect square” is true ☛ Related Questions:
188364	https://www.webassign.net/question_assets/unccolphyseml1/lab_8/manual.html	The Photoelectric Effect and the Quantization of Light Contents> The Photoelectric Effect and the Quantization of Light The Photoelectric Effect and the Quantization of Light Introduction When a light with a sufficiently high frequency shines on a metal plate, electrons are ejected from the plate. This effect is known as the photoelectric effect. The electrons ejected in this process are called photoelectrons. Figure 1 There are several important features of the photoelectric effect that cannot be explained by the classical theory of electromagnetic waves. These are the following. • Electrons are ejected only when the frequency of the incident light is higher than a threshold value, f o, regardless of how intense the light is. The threshold frequency depends on the material being illuminated. • Electrons are emitted from the surface almost instantaneously, even at low light intensities. • When a photoelectric effect is observed, the number of electrons ejected is proportional to the intensity of the incident light. However, the maximum kinetic energy (KE MAX) of the photoelectrons is independent of the light intensity. • The maximum kinetic energy of the photoelectrons increases with higher frequency light. In order to explain the photoelectric effect, Einstein proposed in 1905 that light of frequency f carries energy in discrete packets, each packet containing an amount of energy, E, given by: ( 1 ) E = hf where h is a constant now called Planck's constant. The value of Planck's constant was determined experimentally to be the following. ( 2 ) h = 6.6260755 × 10–34 J · s Today the energy packets are called photons. When light is incident on a metal surface, an electron in the metal can absorb the photon, thus the photon transfers its energy and momentum to the electron. If the electron acquires enough energy from the photon, it can escape from the metal. Since the electron is bound to the metal, an amount of energy is required to escape the metal. The amount of energy needed for the least tightly bound electron to escape is known as the work function, ϕ. The value of ϕ depends on the specific metal surface being illuminated and is typically on the order of a few electron volts (eV). Therefore, according to Einstein's photoelectric theory (which earned him a Nobel prize), an electron can only be ejected if the photon's frequency is larger than a threshold value, f o, given by the equation below. ( 3 ) hf o = ϕ If the incident photon has energy greater than ϕ, then the excess energy becomes the kinetic energy of the electron. The electrons with the largest kinetic energy are those that are the least tightly bound. Since ϕ is the energy required for the least tightly bound electron to escape, the maximum kinetic energy is given below. ( 4 ) KE MAX = hf – ϕ The maximum KE of the ejected electrons, therefore, increases with increasing light frequency and is independent of the light intensity, as observed experimentally. However, the number of photons striking the metal surface increases linearly with the light intensity. Thus, the number of electrons ejected is proportional to the intensity of the incident light. Apparatus Figure 2 The apparatus for this experiment has three essential parts: a high intensity mercury light source that provides photons of different frequencies, a diffraction grating/lens system to spatially separate and focus the light or photons with different frequencies, and the target, which is the anode of a vacuum phototube that is housed in the h/e apparatus together with the associated electronics. The arrangement of the three parts is shown in Fig. 2.When the light from the mercury lamp passes through the lens/grating assembly, it produces a spectrum as shown in Fig. 3. Note that the spectrum is not symmetric on the left and right sides because the diffraction grating is blazed to enhance the first-order pattern on one side. The frequencies and the wavelengths of the spectral lines are shown in the table below. Figure 3 In this experiment, light from each spectral line is focused on a phototube as shown in Fig. 4. The phototube is a vacuum tube with a semi-cylindrical metal electrode (anode) and a thick metal wire at the center (cathode).When a photon with sufficient energy strikes the anode of the phototube, an electron is released and collected at the cathode (see Fig. 4). The process leaves the anode with a positive charge and a positive potential with respect to the cathode (which is grounded). As more electrons are ejected from the anode it becomes more positively charged. The positive charges attract the photoelectrons as they try to leave the anode, thus causing them to lose kinetic energy as they move to the cathode. The amount of kinetic energy lost per electron is equal to eV, where e is the charge of the electron and V is the potential between the anode and the cathode. When V becomes large enough such that eV = KE MAX, even the most energetic electrons (with maximum KE) will be attracted back to the anode and hence, V will have reached a maximum. This maximum value of V is called the stopping potential. Thus, if we can measure this potential and substitute eV for KE MAX in Eq. (3)hf o = ϕ, we get the following. Figure 4 ( 5 ) eV = hf - ϕ Dividing Eq. (5)eV = hf - ϕ by e we get the following. ( 6 ) V = h e f − ϕ e A plot of the measured stopping potential as a function of the frequency of the photon should be a straight line (see Fig. 5) whose slope gives Planck's constant h divided by e. The intercept gives the work function ϕ divided by e, which is 1.602 × 10–19 C. Figure 5 To facilitate measurement of the stopping potential, the anode is connected to a built-in amplifier with an ultra-high input impedance (> 10 12 Ω). The output from this amplifier is connected to the output jacks on the front panel of the apparatus. This high impedance, unity gain ( V OUT/V IN = 1 ) amplifier lets you measure the stopping potential with a digital voltmeter. While the impedance (i.e. resistance) of the amplifier is very high, it is not infinite and some charge leaks off. Thus charging the apparatus is analogous to filling a bathtub with different water flow rates while the drain is partly open. Procedure Part 1: Apparatus Set Up The apparatus should be mostly assembled when you arrive. You may, however, need to check focus and alignment and adjust these as needed. See Fig. 2 for overall assembly view. 1 Before checking the following adjustments, place the hinged bars at about a ninety-degree angle so that you have clearance to work. Note that one light source serves two photoheads, so do not move the light source without consulting with the other students sharing your lamp. 2 The Lens/Grating Assembly mounts on the support bars of the Light Aperture Assembly. The grating is blazed (marked), to produce the brightest spectrum on one side. During your experiment the Lens/Grating Assembly may need to be turned around in order to have the brightest spectrum on a convenient side of your lab table. Consult with your lab instructor if this seems to be the case. 3 If the Light Source is not on, turn it on. Do not turn off the lamp unless told to do so by your instructor. (The lamp is expensive and its lifetime is shortened more by being turned off and on than by running continually.) Check the alignment of the Light Source and the Aperture to ensure that the light illuminates the Lens/Grating assembly. 4 The amplifier in the h/e Apparatus is powered by two 9-volt batteries. You should check the output voltage of the batteries before using the apparatus. Battery test points are located on the side panel of the Apparatus near the ON/OFF switch (see Fig. 6). Batteries functioning below the recommend minimum operating level of 6 volts may cause erroneous results in your experiments. To check the batteries, use a voltmeter and measure the potential difference between the OUTPUT GROUND TERMINAL and each BATTERY TEST TERMINAL. If either battery tests below its minimum rating of +6V or –6V, ask your instructor to replace it before you start taking data. Figure 4 5 Rotate the h/e Apparatus about the pin of the Coupling Bar Assembly until one of the colored maxima in the first order shines directly on the slot in the white reflective mask. It is important that light of only one color enters the slot. There must be no overlap from adjacent spectral maxima. Figure 7 6 See Fig. 7. Rotate the light shield of the h/e Apparatus out of the way to reveal the white photodiode mask inside the Apparatus. Rotate the h/e Apparatus back and forth on the base support rod until the image of the aperture is centered on the window in the phototube mask. (If the light will not fall on the window, you may not have the entire assembly directly in front of the mercury lamp.) Then tighten the thumbscrew on the base support rod to hold the Apparatus in place. 7 Slide the Lens/Grating Assembly back and forth on its support rods, until you achieve the sharpest possible image of the aperture on the window in the phototube mask. The Assembly will probably end up very near to the end of the rod. Tighten the thumbscrew on the Lens/Grating Assembly and replace the light shield to prevent stray light from striking the phototube. 8 Connect the digital voltmeter (DVM) to the OUTPUT TERMINALS of the h/e Apparatus (see Fig. 6). Select the 2V or 20V DC range on the meter. 9 Turn ON the power switch on the h/e apparatus. The output potential difference displayed on your digital voltmeter is a direct measurement of the stopping potential for the photoelectrons. Part 2: Intensity Dependence According to the quantum model of light, the maximum kinetic energy, KE MAX , of photoelectrons depends only on the frequency of the incident light, and is independent of the intensity. Thus, the higher the frequency of the light, the greater its energy. In contrast, the classical wave model of light predicted that KE MAX would depend on light intensity. In other words, the brighter the light, the greater its energy. In this part of the experiment, you will investigate the intensity dependence of the photoelectric effect. You will select two spectral lines from the mercury light source and investigate the maximum energy of the photoelectrons as a function of the intensity. 1 Adjust the h/e Apparatus so that only one of the spectral colors falls upon the opening of the mask of the phototube. If you select the green or yellow spectral line, place the corresponding colored filter over the white reflective mask on the h/e Apparatus. 2 Place the Relative Transmission Filter in front of the white reflective mask (and over the colored filter, if one is used) so that the light passes through the section marked 100%. Press the INSTRUMENT DISCHARGE BUTTON (see Fig. 6), release it, and record the DVM potential difference once the instrument has reached a maximum potential difference. The charging time may take several seconds. Record the approximate charging time. The electrodes of the phototube should be discharged before each measurement by using the discharge button, otherwise the potential difference reading may be inaccurate. 3 Repeat Step 2 until you have tested the sections of the filter marked 100%, 80%, 60%, 40% and 20%. Note: a small drop (a few percent or less) in potential difference with decreasing light intensity is most likely due to charge leaking off the high impedance amplifier in the h/e Apparatus. The charging time should increase with decreasing intensity, but do not be concerned if this is not the case since the charge leaking from the amplifier may alter the results. 4 Repeat the procedure for a different color from the spectrum. Part 3: Stopping Potential Dependence on Frequency In this part of the experiment you will measure the stopping potential for different frequencies of light incident on the phototube. 1 You can see five colors in three orders of the mercury light spectrum. Adjust the h/e Apparatus carefully so that only one color falls on the openings of the mask of the phototube. 2 For each color in the first order, record the stopping potential measured with the DVM. Use the yellow and green colored filters on the Reflective Mask of the h/e Apparatus when you measure the yellow and green spectral lines. 3 Move to the second order and repeat the process. 4 Move the power switch on the h/e Apparatus to the OFF position, but leave the mercury light source ON unless yours is the last lab section of the day (check with your TA to confirm). Analysis 1 Make a scatter plot of V vs. f from the data collected in Part 3. 2 If the plot forms a straight line, use the linear least square fit to determine the slope and the intercept of the line. 3 Calculate the work function in units of electron volts (eV) from the intercept and determine its uncertainty. 4 Calculate the threshold frequency, f o. Discussion Part 2: Intensity Dependence Summarize your experimental results for the intensity dependence of the stopping potential.Describe the effect that passing different amounts of the same colored light through the Variable Transmission Filter has on the stopping potential, and thus, the maximum energy of the photoelectrons. Defend whether this experiment supports a wave or a quantum model of light based on your lab results. Part 3: Stopping Potential Dependence on Frequency Summarize your experimental results for the Planck's constant, the work function and the threshold frequency.Compare your value of h to the accepted value of 6.626 × 10–34 J · s. Discuss possible systematic errors and how they might affect your results. Be sure that both you and your TA each initial your data, and that you hand in a copy of your data before leaving the lab. Remember to pledge your work. Copyright © 2011 Advanced Instructional Systems, Inc. and the University of North Carolina \| Credits
188365	https://stattrek.com/openvideo/playlist/ap-statistics	menu search Log inGet StartedLog out Home My Channel Studio About Terms Privacy Home My Channel Studio About Terms Privacy Toggle dark & light mode AP Statistics play_arrow 4:24 AP Statistics: Introduction to Variables Sep 28, 2022 stattrek.com This video describes the use of variables in AP Statistics. It answers three questions: What is a variable? What is the difference between categorical and quantitative variables? And what is the difference between discrete and continuous variables? View play_arrow 7:56 AP Statistics: Populations and Samples Sep 29, 2022 stattrek.com AP Statistics is all about the study of data sets. This lesson describes two important types of data sets - populations and samples. Along the way, we'll introduce simple random sampling, the main method used in AP Statistics to select samples. View writ play_arrow 16:40 AP Statistics: Mean and Median (measures of central tendency) Oct 1, 2022 stattrek.com The mean and the median are summary measures used in AP Statistics to describe the most "typical" value in a set of values. This video explains how to compute each measure and when to use each measure. (Statisticians sometimes refer to the mean and median play_arrow 17:33 AP Statistics: Measures of Variability Oct 1, 2022 stattrek.com Statisticians use summary measures to describe the amount of variability or spread in a data set. For AP Statistics, students are expected to know about four measures of variability: the range, the interquartile range (IQR), variance, and standard deviati play_arrow 13:39 AP Statistics: Percentiles, Quartiles, z-Scores (measures of position) Oct 6, 2022 stattrek.com Statisticians often talk about the position of a value, relative to other values in a set of data. The most common measures of position are percentiles, quartiles, and standard scores (aka, z-scores). This video shows how to compute each measure of positi play_arrow 11:16 AP Statistics: Patterns in Data Oct 6, 2022 stattrek.com Graphic displays are useful for seeing patterns in data. This video lesson shows how properties of data sets - center, spread, shape, clusters, gaps, and outliers - are revealed in charts and graphs. Features clear, easy-to-understand examples. View writ play_arrow 6:53 AP Statistics: What is a dotplot? Oct 6, 2022 stattrek.com Dotplots are charts that compare frequency counts within groups. Using easy-to-understand examples, this video lesson explains (1) how to read a dotplot and (2) how to construct a dotplot. View written lesson here: play_arrow 9:12 AP Statistics: What are bar charts and histograms? Oct 6, 2022 stattrek.com This video lesson tells you everything you need to know about bar charts and histograms - what they are, when to use them, how they differ, how to construct them, and how to read them. Key points are reinforced with easy-to-understand examples. View writ play_arrow 9:32 AP Statistics: What is a stemplot? Oct 6, 2022 stattrek.com This video lesson tells you everything you need to know about stemplots - what they are, when to use them, how to construct them, and how to read them. Key points are reinforced with easy-to-understand examples. View written lesson here: play_arrow 11:24 AP Statistics: Binomial Probability Distribution Sep 30, 2022 stattrek.com This video tells you everything you need to know about the binomial probability distribution in AP Statistics. By the end of the video, you will understand binomial experiments, you will recognize binomial random variables, you will know how to compute bi play_arrow 8:18 AP Statistics: Cumulative Frequency Plots Oct 11, 2022 stattrek.com A cumulative frequency plot is a way to display cumulative information graphically. It shows the number, percentage, or proportion of observations that are less than or equal to particular values. This video tells you everything you need to know about cum play_arrow 9:35 AP Statistics: What is a scatterplot? Oct 11, 2022 stattrek.com A scatterplot is a graphic tool used to display the relationship between two quantitative variables. This video lesson tells you everything you need to know about scatterplots- what they are, when to use them, how to construct them, and how to read them. play_arrow 11:34 AP Statistics: How to Compare Data Sets Oct 11, 2022 stattrek.com Common graphical displays (e.g., dotplots, boxplots, stemplots, bar charts) can be effective tools for comparing data from two or more data sets. This video shows how to use various graphs to compare data sets in terms of center, spread, shape, and unusua arrow_upward × 2025-09-29 07:05:49
188366	https://www.finra.org/rules-guidance/notices/17-08	FINRA Utility Menu For the Public FINRA DATA FINRA Data provides non-commercial use of data, specifically the ability to save data views and create and manage a Bond Watchlist. For Industry Professionals FINPRO Registered representatives can fulfill Continuing Education requirements, view their industry CRD record and perform other compliance tasks. For Member Firms FINRA GATEWAY Firm compliance professionals can access filings and requests, run reports and submit support tickets. For Case Participants DR PORTAL Arbitration and mediation case participants and FINRA neutrals can view case information and submit documents through this Dispute Resolution Portal. Need Help? \| Check System Status Log In to other FINRA systems FINRA Utility Menu For the Public FINRA DATA FINRA Data provides non-commercial use of data, specifically the ability to save data views and create and manage a Bond Watchlist. For Industry Professionals FINPRO Registered representatives can fulfill Continuing Education requirements, view their industry CRD record and perform other compliance tasks. For Member Firms FINRA GATEWAY Firm compliance professionals can access filings and requests, run reports and submit support tickets. For Case Participants DR PORTAL Arbitration and mediation case participants and FINRA neutrals can view case information and submit documents through this Dispute Resolution Portal. Need Help? \| Check System Status Log In to other FINRA systems FINRA Main Navigation SEC Approves Amendments to Require Mark-Up/Mark-Down Disclosure on Confirmations for Trades With Retail Investors in Corporate and Agency Bonds Pricing Disclosure in the Fixed Income Markets \| \| \| --- \| \| Regulatory Notice \| \| \| Notice Type Rule Amendment \| Referenced Rules & Notices FINRA Rule 2121 FINRA Rule 2232 FINRA Rule 4512 SEA Rule 10b-10 Regulatory Notice 14-52 Regulatory Notice 15-36 \| \| Suggested Routing Compliance Fixed Income Legal Operations Systems Trading Training \| Key Topics Fixed Income Pricing Information TRACE Transaction Confirmations \| Executive Summary The Securities and Exchange Commission (SEC) approved amendments to FINRA Rule 2232 (Customer Confirmations) that require member firms to disclose additional transaction-related information to retail customers for trades in certain fixed income securities. Specifically, amended Rule 2232 requires a member to disclose the amount of mark-up or mark-down it applies to trades with retail customers in corporate or agency debt securities if the member also executes an offsetting principal trade in the same security on the same trading day. The amended rule also requires members to disclose two additional items on all retail customer confirmations for corporate and agency debt security trades: (1) a reference, and a hyperlink if the confirmation is electronic, to a web page hosted by FINRA that contains publicly available trading data for the specific security that was traded, and (2) the execution time of the transaction, expressed to the second. These amendments will become effective on May 14, 2018. The amended rule text is available on FINRA's website. Questions regarding this Notice should be directed to: Background and Discussion Pursuant to SEA Rule 10b-10, firms currently are required to provide transaction cost information when acting as principal with customers for equity trades; however, no comparable requirement had existed for bond trades. As part of an initiative to provide retail fixed income investors with additional information about the costs of their transactions, FINRA filed proposed amendments to Rule 2232 to require enhanced transaction cost and related information on customer confirmations.1 FINRA's amendments to Rule 2232 were developed as the result of a multi-year process during which FINRA twice solicited feedback on related proposals.2 Throughout the process, FINRA worked closely with the Municipal Securities Rulemaking Board (MSRB) to develop similar rules, as appropriate, to ensure consistent disclosures to customers across debt securities and to reduce the operational burdens for firms that trade multiple fixed income securities. The SEC approved FINRA's amendments to Rule 2232, as well as the MSRB's parallel confirmation disclosure proposal, on November 17, 2016.3 Mark-Up Disclosure Requirements When Disclosure is Required New Rule 2232(c) requires members to disclose to a non-institutional customer4 the amount of mark-up or mark-down5 the customer paid for a trade in a corporate or agency debt security,6 if the member also executes one or more offsetting principal trades in the same security on the same trading day which in the aggregate meet or exceed the size of the customer trade. The following example explains how the "offsetting" language that describes when disclosure is triggered under the rule is intended to operate: If a member purchases 100 bonds at 9:30 a.m., and then sells to three customers, who each buy 50 bonds in the same security on the same day, without purchasing any more of the bonds, the rule requires mark-up disclosure on two of the three trades, since one of the trades would need to be satisfied out of the member's prior inventory, or its short position, rather than offset by the member's same-day principal transaction.7 FINRA notes that a disclosure obligation under Rule 2232(c) could be triggered by an offsetting principal trade executed by a member's affiliate. Specifically, if a member's offsetting principal trade is executed with a broker-dealer affiliate and did not occur at arm's length,8 the member is required to "look through" to the time and terms of the affiliate's trade to comply with the rule. New Rule 2232(d) contains two exceptions to the mark-up disclosure requirements of Rule 2232(c). First, mark-up disclosure is not triggered by principal trades that a member executes on a trading desk that is functionally separate from a trading desk that executes customer trades, provided the member maintains policies and procedures reasonably designed to ensure that the functionally separate trading desk has no knowledge of the customer trades.9 Second, mark-up disclosure does not need to be provided for bonds that are acquired by a member in a fixed-price offering and sold to non-institutional customers at the same offering price on the same day the member acquired the bonds. Members may develop reasonable policies and procedures to identify and account for offsetting trades that trigger the disclosure obligations of Rule 2232(c). Members may also choose to provide mark-up disclosure more broadly, for example to all trades with retail customers. Methods to Calculate and Disclose Mark-Ups Members need to calculate the mark-up that is disclosed on a customer confirmation from the prevailing market price (PMP) for the security, consistent with existing FINRA Rule 2121 (Fair Prices and Commissions) and the supplementary material thereunder, particularly Supplementary Material .02 (Additional Mark-Up Policy for Transactions in Debt Securities, Except Municipal Securities). Members may base their mark-up calculations for confirmation disclosure purposes on the information they have available to them as a result of reasonable diligence at the time they input relevant transaction information into systems to generate confirmations. In other words, amended Rule 2232 does not prevent members from maintaining real-time, intra-day confirmation generation processes. Members can engage third-party service providers to facilitate mark-up disclosure consistent with Rule 2232; however, members retain compliance responsibility and are expected to exercise due diligence and oversight over third party relationships. Members may also choose to automate their mark-up disclosure calculation process according to reasonable, consistently applied policies and procedures, if consistent with Rule 2121. Where mark-up disclosure is provided on customer confirmations, Rule 2232(c) requires firms to express the disclosed mark-up as both a dollar amount and a percentage of PMP.10 Members may include accompanying language to provide explanation of mark-up-related concepts, or a member's particular methodology for calculating mark-ups, provided such statements are accurate. However, members may not label mark-ups as "estimated" or "approximate" figures.11 Requirement to Disclose a Reference or Link to Security-Specific Trade Data For all trades with non-institutional customers in corporate and agency debt securities, whether mark-up disclosure is triggered or not, new Rule 2232(e) requires members to provide a reference, and a hyperlink if the confirmation is electronic, to a web page hosted by FINRA that contains TRACE publicly available trading data for the specific security that was traded, along with a brief description of the type of information available on that page.12 FINRA noted during the rulemaking process that it was working to develop a short Uniform Resource Locator (URL) to try to mitigate the operational burdens of this requirement.13 Based on a variety of factors, including logistical concerns such as length of the link and space available on confirmations, discussions with member firms, as well as research into investor preferences, FINRA has established the following URL: org/. Paper confirmations would be required to include this URL in print form; electronic confirmations would be required to include this URL as a hyperlink to the web page. FINRA believes this URL is consistent with the principles it discussed during the rulemaking process, most notably, the need to develop a short, uniform link template that could be subject to automation by members. FINRA also notes that this URL is similar in length and convention to the URL that the MSRB will require for confirmations for transactions in municipal securities with retail customers. Should members wish to provide feedback on the URL that FINRA has chosen, FINRA will consider this input. Time of Execution Disclosure Requirement Rule 2232(e) further requires members to disclose the time of execution, expressed to the second, for all non-institutional customer trades in corporate and agency debt securities. As with the URL requirement, trade time disclosure is required even in cases where mark-up disclosure is not triggered. Providing customers the time of execution will assist them in identifying their individual trade when accessing the TRACE publicly available information. Implementation Period The effective date for the above-described amendments to Rule 2232 is May 14, 2018. This effective date, which is eighteen months from when the amendments were approved by the SEC, is the same as the effective date for the MSRB's parallel confirmation disclosure requirements.14 FINRA recognizes members may still have specific implementation questions and remains committed to working closely with the industry and MSRB during the implementation period to issue further guidance as necessary. See Securities Exchange Act Release No. 78573 (August 15, 2016), 81 FR 55500 (August 19, 2016) (Notice of Filing of SR-FINRA-2016-032) ("Proposal"). See Regulatory Notice 14-52 (November 2014) and Regulatory Notice 15-36 (15-36). The proposals contained in these Regulatory Notices, and how they differ from the amendmentsto Rule 2232 that FINRA ultimately filed with the SEC, are discussed in the proposal, supra note 1. See Securities Exchange Act Release No. 79346 (November 17, 2016), 81 FR 84659 (November 23, 2016) (Order Approving SR-FINRA-2016-032) ("Approval Order"); Securities Exchange Act Release No. 79347 (November 17, 2016), 81 FR 84637 (November 23, 2016) (Order Approving SR-MSRB-2016-12). The term "non-institutional customer" is defined in Rule 2232(f)(4) to mean a customer with an account that is not an institutional account. The term "institutional account" is defined in Rule 4512(c) to mean an account of "(1) a bank, savings and loan association, insurance company or registered investment company; (2) an investment adviser registered either with the SEC under Section 203 of the Investment Advisers Act or with a state securities commission (or any agency or office performing like functions); or (3) any other person (whether a natural person, corporation, partnership, trust or otherwise) with total assets of at least $50 million." For ease of reference, unless otherwise noted, the term "mark-up" refers to mark-ups and mark-downs, collectively. The terms "corporate debt security" and "agency debt security" are defined in Rules 2232(f)(1) and (2), respectively. See Approval Order, supra note 3, at 84663 (discussing FINRA's response to comments submitted to the SEC concerning the proposed amendments.) The term "arms-length transaction" is defined in Rule 2232(f)(3) to mean "a transaction that was conducted through a competitive process in which non-affiliate firms could also participate, and where the affiliate relationship did not influence the price paid or proceeds received by the member." FINRA has noted that as a general matter, it expects the competitive process used in an "arms-length" transaction to be one in which non-affiliates have frequently participated. See Approval Order, supra note 3, at 84662 n. 43. For example, this exception allows an institutional desk within a firm to service an institutional customer without necessarily triggering the disclosure requirement for an unrelated trade performed by a separate retail desk within the firm. However, given the "no knowledge" provision of this exception, a firm could not avoid triggering the mark-up disclosure requirement if trades executed on an institutional desk were used to source transactions at the retail desk. See Proposal, supra note 1, at 55502. Specifically, mark-ups must be disclosed on confirmations as a total dollar amount (i.e., the dollar difference between the customer's price and the security's PMP), and as a percentage amount (i.e., the mark-up's percentage of the security's PMP). See Approval Order, supra note 3, at 84671. Each security-specific web page will include information about the prices of other transactions in the same bond, as well as additional market data and educational material that FINRA believes will be useful to retail investors. See Approval Order, supra note 3, at 84667 (citing to FINRA's response to comments). See MSRB Regulatory Notice 2016-28 (November 2016). Referenced Rules & Notices FINRA Utility Menu FINRA Main Navigation General Inquiries 301-590-6500 Securities Helpline for Seniors® 844-574-3577 (Mon-Fri 9am-5pm ET) File a Regulatory Tip To report on abuse or fraud in the industry Arbitration & Mediation FINRA operates the largest securities dispute resolution forum in the United States File an Investor Complaint File a complaint about fraud or unfair practices. Small Firm Help Line 833-26-FINRA (Mon-Fri 9am-6pm ET) Office of the Ombuds Report a concern about FINRA at 888-700-0028 Footer Legal Links © 2025 FINRA. All Rights Reserved. FINRA is a Registered Trademark of the Financial Industry Regulatory Authority, Inc.
188367	https://math.stackexchange.com/questions/247987/prove-that-2k-k3-for-all-k-ge10	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Prove that $2^k > k^3 $ for all $k\ge10$ Ask Question Asked Modified 12 years, 10 months ago Viewed 1k times 2 $\begingroup$ I've no clue how to go ahead with this, all I know is it will be solved with induction. Proved it's true for $k=10$ Assumed it's true for $k$ Need to prove that $2^{k+1} > ({k+1})^3$ Any pointers? I'm struggling with tough Induction questions so if you have any general tips to solve such questions it'll be great. induction Share edited Nov 30, 2012 at 11:44 Brian M. Scott 633k5757 gold badges824824 silver badges1.4k1.4k bronze badges asked Nov 30, 2012 at 11:43 caughtinalandslidecaughtinalandslide 32511 gold badge55 silver badges1212 bronze badges $\endgroup$ Add a comment \| 3 Answers 3 Reset to default 4 $\begingroup$ If you know $2^k > (k)^3$ and want to prove $2^{k+1} > ({k+1})^3$ the obvious thing to do is multiply the first by two so that you have $2^{k+1} > 2 k^3$ now if we could show that $2k^3 \ge (k+1)^3$ we could put these two inequalities together to complete the proof. By expanding out and collecting like terms we have to prove $k^3 \ge 3k^2 + 3k + 1$ but that's an easy consequence of $k^3 \ge 3k^2 + 3k^2 + k^2$ which holds because $k \ge 10$ (dividing both sides by $k^2$). Share answered Nov 30, 2012 at 11:48 user50336user50336 $\endgroup$ Add a comment \| 4 $\begingroup$ When you pass from $2^k$ to $2^{k+1}$, the number doubles. If you could show that passing from $k^3$ to $(k+1)^3$ increases the number by a factor less than $2$, youd be in business. How big is $\frac{(k+1)^3}{k^3}$? If $k\ge 10$, then $$\frac{(k+1)^3}{k^3}=\left(\frac{k+1}k\right)^3=\left(1+\frac1k\right)^3\le\left(1+\frac1{10}\right)^3=\left(\frac{11}{10}\right)^3=\frac{1331}{1000}<2\;.\tag{1}$$ Thus, if $2^k>k^3$ and $k\ge 10$, we have $$2^{k+1}=2\cdot 2^k>\frac{(k+1)^3}{k^3}\cdot 2^k>\frac{(k+1)^3}{k^3}\cdot k^3=(k+1)^3\;,$$ exactly as desired. Writing it out in one go like that may make it look more mysterious than it really is. All I really did was ask myself what happens to the two sides of the inequality $2^k>k^3$ when $k$ is replaced by $k+1$. Clearly the lefthand side is doubled. What happens to the righthand side (in terms of multiplicative increase) isnt so obvious, but at least we can say that it gets multiplied by $\frac{(k+1)^3}{k^3}$: $$\begin{array}{c} &2^k&>&k^3\ \text{multiply by }2&\downarrow&&\downarrow&\text{multiply by }\frac{(k+1)^3}{k^3}\ &2^{k+1}&\overset{?}>&(k+1)^3 \end{array}$$ Clearly the inequality in the bottom row will be true if $2\ge\frac{(k+1)^3}{k^3}$, so we just have to make sure that this is the case $-$ which is exactly what I did with the calculation $(1)$. Share answered Nov 30, 2012 at 11:59 Brian M. ScottBrian M. Scott 633k5757 gold badges824824 silver badges1.4k1.4k bronze badges $\endgroup$ 0 Add a comment \| 0 $\begingroup$ You have: $(k+1)^3 \stackrel{k\ge 10}{\lt} 2\cdot k^3 \stackrel{hyp.}{\lt} 2\cdot 2^k$ Share answered Nov 30, 2012 at 11:49 user127.0.0.1user127.0.0.1 7,45566 gold badges3535 silver badges4747 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 6 Proving $ \left(\frac{n^2+1}{n^2}\right)^n\ge\frac{n+1}{n}$ by induction. 19 For the Fibonacci numbers, show for all $n$: $F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$ 0 Strong induction definition clarification: is the hypothesis that it is true for all $k$ with $0 \le k < n$ or for all $k$ with $0 \le k \le n$? 4 Prove by induction that for all $n$, $8$ is a factor of $7^{2n+1} +1$ Prove $(1 +\frac{ 1}{n}) ^ {n} \ge 2$ Prove by induction that for the Fibonacci numbers $F(n)$ with $n \ge 6$, $F(n) \ge 2^{n/2}$ Why is Mathematical Induction used to prove solvable inequalities? 2 For all integers $n ¥ 2, n^3 > 2n + 1$ 0 Show that $x_n$ is increasing and $x_n < 2$ for all $n \geq 2$, for the sequence. $x_1 = 1$, $x_{n+1} = \frac{1}{5}(x_n^2 + 6)$ for $n \geq 1$. 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188368	https://pubmed.ncbi.nlm.nih.gov/11395974/	The diverse pathology of post-transplant lymphoproliferative disorders: the importance of a standardized approach - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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The diverse pathology of post-transplant lymphoproliferative disorders: the importance of a standardized approach M A Nalesnik1 Affiliations Expand Affiliation 1 Division of Transplantation Pathology, University of Pittsburgh Medical Center, Pittsburgh, Pennsylvania, USA. nalesnikma@msx.upmc.edu PMID: 11395974 DOI: 10.1034/j.1399-3062.2001.003002088.x Item in Clipboard Review The diverse pathology of post-transplant lymphoproliferative disorders: the importance of a standardized approach M A Nalesnik. Transpl Infect Dis.2001 Jun. Show details Display options Display options Format Transpl Infect Dis Actions Search in PubMed Search in NLM Catalog Add to Search . 2001 Jun;3(2):88-96. doi: 10.1034/j.1399-3062.2001.003002088.x. Author M A Nalesnik1 Affiliation 1 Division of Transplantation Pathology, University of Pittsburgh Medical Center, Pittsburgh, Pennsylvania, USA. nalesnikma@msx.upmc.edu PMID: 11395974 DOI: 10.1034/j.1399-3062.2001.003002088.x Item in Clipboard Full text links Cite Display options Display options Format Abstract Post-transplant lymphoproliferative disorders (PTLD) are a diverse group of abnormal lymphoid growths that include both hyperplasias and neoplasias. They have been divided into several general pathologic categories that have prognostic significance. These include early or hyperplastic PTLD, polymorphic PTLD, and lymphomatous or monomorphic PTLD. The majority of PTLDs are of B-cell origin and contain Epstein-Barr virus (EBV). However, PTLDs of T- or NK-cell origin have been described, and late-arising EBV-negative lymphoid tumors are becoming more frequently reported in this population. Other lymphoid neoplasms, such as those arising from mucosal-associated lymphoid tissue (MALTomas), have recently been recognized in transplant patients, and their relationship to PTLD is uncertain. Multicentric PTLD may represent either advanced-stage disease or multiple independent primary tumors. Likewise, recurrent PTLD may represent true recurrence or the emergence of a second primary tumor. Transplant patients are also at risk for other opportunistic neoplasms, including EBV-associated leiomyosarcomas that may be seen alone or in conjunction with PTLD. This underscores the necessity for pathologic diagnosis of mass lesions in this patient population. The pathologist should strive to categorize the form of post-transplant lymphoproliferation in accordance with currently accepted criteria. The diagnosis should incorporate the histopathologic appearance, cell phenotype, clonal status, and EB viral status. The pathologist may play a special role in guiding therapy by ascertaining the presence of such markers as CD20 on tumor cells. Specialized techniques, such as molecular analysis of oncogenes/tumor suppressor genes and evaluation of host:donor status of PTLD, may play important roles in diagnostic evaluation in the future. PubMed Disclaimer Similar articles Clinicopathologic characteristics of post-transplant lymphoproliferative disorders.Nalesnik MA.Nalesnik MA.Recent Results Cancer Res. 2002;159:9-18. doi: 10.1007/978-3-642-56352-2_2.Recent Results Cancer Res. 2002.PMID: 11785849 Review. Role of Epstein-Barr virus status and immunophenotypic studies in the evaluation of exfoliative cytology specimens from patients with post-transplant lymphoproliferative disorders.Gibson SE, Picarsic J, Swerdlow SH, Pantanowitz L.Gibson SE, et al.Cancer Cytopathol. 2016 Jun;124(6):425-35. doi: 10.1002/cncy.21694. Epub 2016 Mar 17.Cancer Cytopathol. 2016.PMID: 26992116 [Post-transplant lymphoproliferative disease in liver transplant recipients--Merkur University Hospital single center experience].Filipec-Kanizaj T, Budimir J, Colić-Cvrlje V, Kardum-Skelin I, Sustercić D, Naumovski-Mihalić S, Mrzljak A, Kolonić SO, Sobocan N, Bradić T, Dolić ZM, Kocman B, Katicić M, Zidovec-Lepej S, Vince A.Filipec-Kanizaj T, et al.Acta Med Croatica. 2011 Sep;65 Suppl 1:37-43.Acta Med Croatica. 2011.PMID: 23126028 Croatian. Characteristic pattern of chromosomal imbalances in posttransplantation lymphoproliferative disorders: correlation with histopathological subcategories and EBV status.Poirel HA, Bernheim A, Schneider A, Meddeb M, Choquet S, Leblond V, Charlotte F, Davi F, Canioni D, Macintyre E, Mamzer-Bruneel MF, Hirsch I, Hermine O, Martin A, Cornillet-Lefebvre P, Patey M, Toupance O, Kémény JL, Deteix P, Raphaël M.Poirel HA, et al.Transplantation. 2005 Jul 27;80(2):176-84. doi: 10.1097/01.tp.0000163288.98419.0d.Transplantation. 2005.PMID: 16041261 Reexamining post-transplant lymphoproliferative disorders: Newly recognized and enigmatic types.Aguilera N, Gru AA.Aguilera N, et al.Semin Diagn Pathol. 2018 Jul;35(4):236-246. doi: 10.1053/j.semdp.2018.02.001. Epub 2018 Mar 3.Semin Diagn Pathol. 2018.PMID: 29615296 Review. See all similar articles Cited by [Lymph node pathology - an update].Hartmann S, Hansmann ML.Hartmann S, et al.Pathologe. 2013 Feb;34(1):34-44. doi: 10.1007/s00292-012-1706-5.Pathologe. 2013.PMID: 23319007 Review.German. Clinical characteristics and outcomes of posttransplant lymphoproliferative disorders following allogeneic hematopoietic stem cell transplantation in Korea.Park SH, Choi SM, Lee DG, Choi JH, Yoo JH, Kim HJ, Kim DW, Lee JW, Min WS, Shin WS, Kim CC.Park SH, et al.J Korean Med Sci. 2006 Apr;21(2):259-64. doi: 10.3346/jkms.2006.21.2.259.J Korean Med Sci. 2006.PMID: 16614511 Free PMC article. Epstein-Barr virus-associated T/NK cell-type central nervous system lymphoma which manifested as a post-transplantation lymphoproliferative disorder in a renal transplant recipient.Omori N, Narai H, Tanaka T, Tanaka S, Yamadori I, Ichimura K, Yoshino T, Abe K, Manabe Y.Omori N, et al.J Neurooncol. 2008 Apr;87(2):189-91. doi: 10.1007/s11060-007-9504-2. Epub 2007 Dec 6.J Neurooncol. 2008.PMID: 18058068 Post-transplant lymphoproliferative disorders: role of viral infection, genetic lesions and antigen stimulation in the pathogenesis of the disease.Capello D, Gaidano G.Capello D, et al.Mediterr J Hematol Infect Dis. 2009 Dec 14;1(2):e2009018. doi: 10.4084/MJHID.2009.018.Mediterr J Hematol Infect Dis. 2009.PMID: 21416004 Free PMC article. Late-onset fatal Epstein-Barr virus-associated hemophagocytic syndrome following cord blood cell transplantation for adult acute lymphoblastic leukemia.Kawabata Y, Hirokawa M, Saitoh Y, Kosugi S, Yoshioka T, Fujishima M, Fujishima N, Kameoka Y, Saitoh H, Kume M, Takahashi N, Sawada K.Kawabata Y, et al.Int J Hematol. 2006 Dec;84(5):445-8. doi: 10.1532/IJH97.06101.Int J Hematol. 2006.PMID: 17189228 See all "Cited by" articles Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Humans Actions Search in PubMed Search in MeSH Add to Search Lymphoproliferative Disorders / classification Actions Search in PubMed Search in MeSH Add to Search Lymphoproliferative Disorders / pathology Actions Search in PubMed Search in MeSH Add to Search Organ Transplantation / adverse effects Actions Search in PubMed Search in MeSH Add to Search Postoperative Complications Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Wiley Full text links[x] Wiley [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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188369	https://math.stackexchange.com/questions/1710126/harshad-numbers-with-given-sum	computational mathematics - Harshad numbers with given sum - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Harshad numbers with given sum Ask Question Asked 9 years, 6 months ago Modified9 years, 6 months ago Viewed 370 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. By definition Harshad number for base 10 10 is any number divisible by sum of its decimal digits. Wikipedia gives some information on such numbers but i still have some questions and unforunately i wasn't able to answer them myself. So let s s be some positive integer. a) Does there exist Harshad number with digit-sum s s? b) If so, how can we construct such number? c) How can we construct smallest such number? For example if s=11 s=11 then one possible answer for question b) is 1010101010101010101010 1010101010101010101010 (11 11 ones) while the answer for question c) is 209 209. elementary-number-theory computational-mathematics Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Mar 23, 2016 at 12:23 IgorIgor 2,203 18 18 silver badges 22 22 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. This post only provides answers to a) and b). To answer your question a), yes. We'll prove this by answering b), constructing a Harshad number with digit sum n n. First, we define l(n)l(n) to be n n without its factors 2 2 and 5 5. Examples are: l(23)=23 l(23)=23, l(24)=3 l(24)=3, l(25)=1 l(25)=1. If we can construct a number with digit sum n n that is divisible by l(n)l(n), then surely we can make it divisible by n n by just adding enough 0's at the end (equivalent to multiplying it with 10 repeatedly, so this wont change the fact that it is divisible by l(n)). Let A be a set of n non-negative integers, that is, A⊆N 0 and \|A\|=n. Now the number ξ=∑a∈A 10 a has digit sum n, and this doesn't depend on the numbers in a. For it to be divisible by l(n), we'd need ξ=∑a∈A 10 a≡0 mod l(n) and since 10 ϕ(l(n))≡1 mod l(n) (note that we use gcd(10,l(n))=1 here), we can choose A={0⋅ϕ(l(n)),1⋅ϕ(l(n)),2⋅ϕ(l(n)),⋯,(n−1)⋅ϕ(l(n))} so that we get (all equivalences are mod l(n)): ξ≡∑a∈A 10 a≡n−1∑k=0 10 k⋅ϕ(l(n))≡n−1∑k=0(10 ϕ(l(n)))k≡n−1∑k=0 1 k≡n−1∑k=0 1≡n≡0 mod l(n) where the last, n≡0 mod l(n), holds since l(n)∣n, because l(n) has the same factors as n and maybe a little less, but never more. Now we can append enough zeroes at the end so that it is not only divisible by l(n), but also by n. To get back to your example n=11, we get l(11)=11, and ϕ(l(n))=ϕ(11)=10, so we get A={0,10,20,30,⋯,100}, arriving at the number 10000000001000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001 and we don't need to append zeroes because 11 didn't contain any factors 2 or 5. This construction method proves your question b). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 25, 2016 at 12:47 user304329 user304329 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory computational-mathematics See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 7There are infinitely many non-harshad Fibonacci numbers Related 5Finding numbers satisfying a given condition 2Rules for divisibility based on digit sum in general basis. 5Infinite positive integer sequence with distinctive sum of digits 1Let s be the smallest positive integer with the property that its digit sum and the digit sum of s + 1 are both divisible by 19. Number of digits s? 1How many five-digit numbers are divisible by 5, have equal first and last digits, and have a digit-sum divisible by 5? 7There are infinitely many non-harshad Fibonacci numbers 2Sum of all combination of five digit numbers with digits 1,3,5,7,9 (no repetition) Hot Network Questions In Dwarf Fortress, why can't I farm any crops? 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188370	https://unacademy.com/content/cbse-class-11/study-material/mathematics/pure-imaginary-number/	Access free live classes and tests on the app Login Join for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd CBSE Class 11 » CBSE Class 11 Study Materials » Mathematics » Pure Imaginary Number Pure Imaginary Number A pure imaginary number is any complex number with a fundamental part that is 0. Share There are a lot of numbers that are only pure imaginary but not accurate. There are a lot of pure imaginary numbers, and this lesson will show you how to evaluate them, simplify or divide them, multiply them, and solve equations. Learning how to work with imaginary numbers will help us work better with complex numbers. In the past, complex numbers were linked to the idea of solving equations. In the 16th century, mathematicians tried to find algebraic solutions to the cubic equation. When they tried to solve equations, they found that many had real-world solutions. However, the methods used to solve them led to the need to find the square roots of negative numbers. The way of solving cubics that Tartaglia used often made it necessary to figure out the square root of negative numbers even when the solutions were all real, even though they were all real. Definition: Imaginary numbers There are two types of complex numbers: one that is real and one that is not real. An imaginary number has the form bi, where the first number is real and the second is imaginary. People need to know that imaginary numbers are part of another set called complex numbers, and they are not the same as real ones. Let’s also remember what complex numbers are. Imaginary numbers chart There is also an interesting thing about it. The value changes four times when you multiply it. Because √-1= i when you multiply it by itself, for example, In that case, i⨯i = -1. In this example, -iis the answer. That makes it all the way around. This makes it easy to figure out how many exponents of I you need to work out. If: i = √-1 i2 = -1 i3 = -√-1 i4 = 1 i5 = √-1 and so on. This cycle will go on through the exponents, also called the imaginary numbers chart, and will keep going. Knowledge of how imaginary numbers grow is practical when multiplying and dividing imaginary numbers. It is possible to use the rules of exponents to apply to I after grouping the coefficients and fictional terms. The real numbers are then multiplied as usual. The same thing happens when you divide things up, too. By following the same rules for multiplying and dividing real and imaginary numbers, you can make them easier to work with, like variables and coefficients. This isn’t the only time that imaginary numbers have been in pop culture. Pure imaginary numbers The number i is not alone! We can make infinitely many pure imaginary numbers by taking multiples of this imaginary unit and adding them together. When you take the squares of these numbers, you can see how they connect to the real numbers. Let’s find out how this works by squaring the number 3i. The properties of integer exponents stay the same, so we can square 3i, just as we’d think we would be able to do. (3i)2=(3)2⨯i2 We know that i2=-1 hence the answer is -9. Why do we have imaginary numbers anyway? The answer is straightforward. The imaginary unit ‘i’ can help us solve many equations that don’t have real number solutions. Many equations can’t be solved in one number system, but another, more general one can solve them. Here are some examples why we need different types of numbers. We can’t solve x+8=1; we need integers for this! To solve 3x-1=0, which means that 3x minus 1, equals 0, we need rational numbers. We can’t do this with only integers. We can’t solve x2 with only rational numbers hence we developed irrational numbers. The square of 2 is 2×2. It’s time to get into the irrational numbers and the actual number system. As we saw in the above examples that only integers of rational numbers were not sufficient to define maths therefore we developed irrational numbers. Similarly,real numbers can’t help us solve x2=-1, squared, equals, minus, 1, so we can’t do that. It will help us if we use imaginary numbers for this task! Classifying the New Types of Numbers Pure imaginary numbers are a group of numbers made by taking the square roots of negative numbers and then adding them together. In symbols, we could write a pure imaginary number as bi, the product of a natural number b, and an imaginary number i. They are not on the number line, so there is no place for them on the number line. It’s also possible to talk about adding an actual number to a purely imaginary number. When you do this, you get what is called a complex number. A complex number has the form a+bi, where an is a natural number and bi is a number that is only pure imaginary. If these numbers aren’t on the number line, can we make pictures of complex or imaginary numbers by putting them on paper? Yes, that’s true. But a one-dimensional number line won’t be enough for you. It needs to be on a second number line to show the pure imaginary part of the complex number. In other words, we need to show complex numbers with a two-dimensional picture to do that. The number 2+3i is shown as a point in this picture. So the actual numbers are points on the horizontal axis, which is where they are. Imaginary numbers are those numbers that aren’t real and aren’t on the horizontal axis, so they are not real. It’s like this: Pure imaginary numbers are points on the vertical axis (other than the origin). Complex numbers include all the possible issues in the picture, covering the whole thing. There are a lot of different kinds of numbers that are real. Odd numbers are just like them also. Just like a whole number is also a fraction, so is a fraction. Conclusion An actual number multiplied by the imaginary unit I is known as an imaginary number.- b2 is the square of an imaginary number bi, a number that isn’t real, such as 5i, has a square that’s -25. As a general rule, zero is both real and imaginary, so it is both.Complex numbers are used at the places where real numbers fail to give the solution. For example, the solution of equation x 8=-1 cannot be given by real numbers. Frequently asked questions Get answers to the most common queries related to the CBSE Class 11 Examination Preparation. What is an example of a pure imaginary number? Ans : We can make infinitely many pure imaginary numbers by taking multiples of this imagina...Read full What are purely natural and purely imaginary numbers? Ans :Purely natural: If I’m(z)=0, then z is said to be accurate. Purely imaginary: If Re(z)=0...Read full Why is 0 imaginary? Ans : It is both, but not as stated—the square root of a nonpositive real value. Zero is i...Read full Is Infinity an imaginary number? Ans: A real number and an imaginary number are not the same things. Infinity can’t be an imag...Read full Ans : We can make infinitely many pure imaginary numbers by taking multiples of this imaginary unit and adding them together. It’s possible to have pure imaginary numbers with only one actual number and one imaginary number. For example, 3i, -3i, and 4.5i are all pure imaginary numbers with only one imaginary number. Ans :Purely natural: If I’m(z)=0, then z is said to be accurate. Purely imaginary: If Re(z)=0, it is said to be fictional. Ans : It is both, but not as stated—the square root of a nonpositive real value. Zero is imaginary since it is nonpositive, and it’s a square root. Ans: A real number and an imaginary number are not the same things. Infinity can’t be an imaginary number because Infinity is not a number. Share via
188371	https://www.youtube.com/watch?v=AuYRQVmpSaE	Chi-Square Test of Independence \| Contingency Table \| Hypothesis Test Joshua Emmanuel 159000 subscribers 641 likes Description 31439 views Posted: 14 May 2023 Chi-Square Test of Independence - Contingency Table A brief explanation of the chi square test of Independence. 00:00 Contingency Table 00:56 Hypothesis Test 01:08 Null & Alternative Hypothesis 01:47 Test Statistic 02:18 Calculating Expected Values 03:43 Calculating chi-square statistic 04:13 Degrees of freedom 04:31 Critical value 04:51: P-values, Decisions and Conclusions Goodness of fit test: Test of Independence Excel Template: 18 comments Transcript: Contingency Table Chi Square Test of Independence This is a contingency table. Also called a 2-way table, cross classification table, cross tabulation or just crosstab. It shows the observed number of cases or frequency for two categorical variables. By Categorical, we mean nominal or ordinal. These variables are ordinal since their categories can be ranked. The row variable is Age Category consisting of 4 categories. The column variable is AI usage consisting of 3 categories. We refer to a table with this structure as a 4 x 3 contingency table because it has 4 rows and 3 columns. The Chi square test of independence is often employed to determine if a significant association or relationship exists between two categorical variables. Note that we are not testing for a correlation. That term “correlation” is reserved for the relationship between two quantitative Hypothesis Test variables. Let’s begin with this 3x2 contingency table. That is, we have 3 rows and 2 columns. Suppose we want to test if there is a relationship or association between age category and AI usage. Null & Alternative Hypothesis Then the null hypothesis will be that there is no relationship between the two variables. In essence, Age and AI usage are independent. The alternative hypothesis will be that there is a relationship. That is, Age and AI usage are dependent. In the null hypothesis, Independence there means that there is no difference in the distribution of AI usage among the age categories. In the alternative hypothesis, dependence or relationship implies that AI usage distribution will differ among the age categories. Specifically, the question here is, are some age categories likely to use AI more than Test Statistic others? The test statistic for chi square test of independence is the same as the one used for the Goodness of fit test …in that, it compares observed frequencies with expected frequencies. The null hypothesis here basically says that the observed and expected frequencies for each cell are essentially equal while the alternative states that they are different. Here again is our 2-way table showing rows and column totals. The expected frequencies are calculated on the assumption that the two variables are Calculating Expected Values independent. As a result, the formula for calculating the expected values for each cell turns out to be Row total x column total, divided by sample size n (or grand total) which is 1440 in this case. I’m going to place the expected frequencies in this table on the right here. For the “Under 30” and “Yes” cell, the row total is 490 and the column total is 809 so the expected value is calculated as 490 809/ 1440. And that gives 275.28. For the expected value for the “Under 30” and “No” cell, the row total is 490 times 631, the column total, and divided by the grand total of 1440. For row 2 column 1, we have 456809 divided by 1440 which gives 256.18. We continue in that fashion for the row 2 column 2 Row 3 column 1 And row 3 column 2. Now, when we compute the expected totals, note that they must equal the observed totals except for rounding errors. Note that each expected cell must have a value of at least 5 for the test to be valid. However, some authors only require that no more than 20% of the expected cells should be less than 5. Calculating chi-square statistic Let’s now compute the value of the chi-square statistic for the problem. For the first cell, we take the observed (312) minus the expected (275.28), square the result, and divide by the expected. We do the same for the 178, the 273, the 183, the 224, and the 270. On computing these, we have an observed chi square value of 37.26. Now back to the hypothesis test. The degrees of freedom for the chi square test of independence is Degrees of freedom (number of rows – 1) times (number of columns – 1) Since we have 3 rows and 2 columns, the degrees of freedom computes to a value of 2. Now, suppose we’re testing at the significance level alpha of 0.05, Critical value we can obtain the critical chi square value for the test by using software or from the chi square table. With df = 2 and alpha = 0.05, we have a critical value of 5.99. In essence, we will reject the null hypothesis and say that the result is significant if our computed chi square statistic is greater than 5.99. Since our chi square stat is 37.26, which is greater than 5.99, we will reject the null hypothesis and conclude that Age and AI usage are dependent. In other words, there is significant association, or the distribution of AI usage is different among the age categories. We can calculate the P-value of the test using software. or estimate it from the chi square table. Going again to df = 2 in the table, we find that the chi square statistic of 37.26 is larger than all critical values listed there. We therefore say that the P-value is less than .005, or whatever the smallest alpha listed is. Now, suppose the test statistic were 8.33, then the P-value will lie between .01 and .025. Recall that we reject the null hypothesis if P-value is less than alpha. Since this P-value is still less than .05, the result is still significant, and we still reject the null hypothesis. The test statistic is also greater than the critical value corroborating the decision to reject the null hypothesis. However, if the observed chi square were 4.84, the P-value will lie between 0.1 and 0.05, and the association is no longer significant or we fail to reject the null hypothesis), since the P-value is now greater than 0.05, or because the test statistic is no longer greater than the critical value. Therefore, we don’t have enough evidence to conclude that Age and AI usage are dependent in this case. And that’s it for this video. Thanks for watching.
188372	https://www.studocu.com/en-au/document/baulkham-hills-high-school/maths-extension-1/the-auxiliary-angle-method/113159842	Auxiliary Angle Method: Understanding Trigonometric Equations and Waves - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools High School Degrees Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Auxiliary Angle Method: Understanding Trigonometric Equations and Waves The Auxiliary Angle Method Original title: The Auxiliary Angle Method Subject Maths Extension 1 452 documents Degree • Grade HSC • 12 School Baulkham Hills High School Academic year:2024/2025 Uploaded by: doraemon Baulkham Hills High School 0 followers 15 Uploads0 upvotes Follow Recommended for you 26 Y12 Ext 1 Mathematics Study Notes: Key Concepts and Formulas Maths Extension 1 Study notes 100% (5) 9 Maths Extension ONE Summary Notes - Polynomials & Calculus Concepts Maths Extension 1 Study notes 100% (4) 19 WME Ext 1 HSC 2023 Trial Exam Solutions Maths Extension 1 Study notes 100% (2) 13 Trigonometric Integration Lesson Notes: Key Concepts & Examples Maths Extension 1 Study notes 100% (1) 5 IF 2 Inverse Functions: Examples and Detailed Explanations Maths Extension 1 Study notes 100% (1) Comments Please sign in or register to post comments. Report Document Related documents Y12 Ext 1 Mathematics Study Notes: Key Concepts and Formulas Substitution and Harder Integration: EXT2 C1 HSC Exam Prep Trig Integration Solutions for EXT2 C1 2019 HSC Exam Proof and Inequalities: EXT2 P1 Study Guide Parametric Representation of Curves: Key Concepts and Exercises Solving for the variable angle of inclination Preview text The Auxiliary Angle Method When our trigonometric equations have 2 different trigonometric expressions written in the equation, we aim to write it as a single wave function. We are able to do this because any function written in the form can be written as a function of just or. This process is done by expanding the four results below by using the trigonometric identity results . We will just verify this graphically, but the ability to write multiple trigonometric functions as just a single function is remarkably important in all modern electronics and communications that use waves. It is the start of an important mathematical field known as Fourier Series. youtube/watch?v=LznjC4Lo7lE Sidenote: (This argument could be extended to as well since functions of can be rewritten as and further still this argument could then also extend to , and. We will mainly focus though on rewriting into one of four forms: 1. 2. 3. 4. Where and. The coefficient is called the a___, the angle is called the p__ s_. and is calculated the same for all four forms, but the auxiliary angle depends on which form is chosen. Expand using Set the equation as equal to your expansion in part 1. Equate the coefficients of in the function (ie: those coefficients are just 1) with the coefficients of in the expansion (these ‘coefficients’ may need you to shuffle your expanded expression around to see them) Set up simultaneous equations, square them, add them together, factorise and use the Pythagorean identity. This will solve for. Substitute back into your simultaneous equations to solve for. Auxiliary Angle Method: Understanding Trigonometric Equations and Waves Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Auxiliary Angle Method: Understanding Trigonometric Equations and Waves Subject: Maths Extension 1 452 documents Degree • Grade: HSC • 12 Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save The Auxiliary Angle Method When our trigonometric equations have 2 dif ferent trigonometric expressions written in the equation, we aim to write it as a single wave function. W e are able to do this because any function written in the form can be written as a function of just or . This process is done by expanding the four results below by using the trigonometri c identity results . W e will just verify this graphically, but the ability to write multiple trigonometric funct ions as just a single function is remarkably important in all modern electronics and communi cations that use waves. It is the start of an important mathematical field known as Fourier Series. Sidenote: (This ar gument could be extended to as well since functions of can be rewritten as and further still this ar gument could then also extend to , and . W e will mainly focus though on rewriting into one of four forms: 1. 2. 3. 4. Where and . The coef ficient is called the a___, the angle i s called the p__ s_. and is calculated the same for all four forms, but the auxiliary angle depends on which form is chosen. 1.Expand using 2.Set the equation as equal to your expansion in part 1. 3.Equate the coef ficients of in the function (ie: those coef ficients are just 1) with the coef ficients of in the expansion (these ‘coeffici ents’ may need you to shuffl e your expanded expression around to see them) 4.Set up simultaneous equations, square them, add them together, factorise and use the Pythagorean identity . T his will solve for . 5.Substitute back into your simultaneous equations to solve for . Too long to read on your phone? Save to read later on your computer Save to a Studylist Document continues below Discover more from: Maths Extension 1 HSC (New South Wales Higher School Certificate)Grade: 12 452 documents Go to course 16 CSSA Ext 1 2022 HSC Trial Exam Solutions and Marking Guidelines Maths Extension 1 Practice materials 100% (14) 18 Vectors EXT1 V1 2020 HSC: Geometry & Projectile Motion Guide Maths Extension 1 Practice materials 100% (11) 14 Mathematics Extension 1 HSC Study Notes: Trigonometric Identities & More Maths Extension 1 Other 100% (10) 12 2023 HSC Trial Exam - Mathematics Ext 1 Independent E1 Paper Maths Extension 1 Practice materials 100% (9) 16 Polynomials (Yr 11): HSC Exam Questions & MC Practice (EXT1 F2)Maths Extension 1 Practice materials 100% (8) 19 Trigonometry EXT1 HSC Exam Questions Compilation and Analysis Maths Extension 1 Practice materials 100% (7) Discover more from: Maths Extension 1HSC (New South Wales Higher School Certificate)Grade: 12 452 documents Go to course 16 CSSA Ext 1 2022 HSC Trial Exam Solutions and Marking Guidelines Maths Extension 1 100% (14) 18 Vectors EXT1 V1 2020 HSC: Geometry & Projectile Motion Guide Maths Extension 1 100% (11) 14 Mathematics Extension 1 HSC Study Notes: Trigonometric Identities & More Maths Extension 1 100% (10) 12 2023 HSC Trial Exam - Mathematics Ext 1 Independent E1 Paper Maths Extension 1 100% (9) 16 Polynomials (Yr 11): HSC Exam Questions & MC Practice (EXT1 F2) Maths Extension 1 100% (8) 19 Trigonometry EXT1 HSC Exam Questions Compilation and Analysis Maths Extension 1 100% (7) 1 out of 7 Share Download Download More from:Maths Extension 1 More from: Maths Extension 1 HSC (New South Wales Higher School Certificate)Grade: 12 452 documents Go to course 16 CSSA Ext 1 2022 HSC Trial Exam Solutions and Marking Guidelines Maths Extension 1 Practice materials 100% (14) 18 Vectors EXT1 V1 2020 HSC: Geometry & Projectile Motion Guide Maths Extension 1 Practice materials 100% (11) 14 Mathematics Extension 1 HSC Study Notes: Trigonometric Identities & More Maths Extension 1 Other 100% (10) 12 2023 HSC Trial Exam - Mathematics Ext 1 Independent E1 Paper Maths Extension 1 Practice materials 100% (9) More from: Maths Extension 1HSC (New South Wales Higher School Certificate)Grade: 12 452 documents Go to course 16 CSSA Ext 1 2022 HSC Trial Exam Solutions and Marking Guidelines Maths Extension 1 100% (14) 18 Vectors EXT1 V1 2020 HSC: Geometry & Projectile Motion Guide Maths Extension 1 100% (11) 14 Mathematics Extension 1 HSC Study Notes: Trigonometric Identities & More Maths Extension 1 100% (10) 12 2023 HSC Trial Exam - Mathematics Ext 1 Independent E1 Paper Maths Extension 1 100% (9) 16 Polynomials (Yr 11): HSC Exam Questions & MC Practice (EXT1 F2) Maths Extension 1 100% (8) 19 Trigonometry EXT1 HSC Exam Questions Compilation and Analysis Maths Extension 1 100% (7) More from:doraemon More from: doraemon impact 4 impact 4 Baulkham Hills High SchoolHSC • Grade: 12 Discover more 2 Taking the Sum of a Sequence: Understanding Partial Sums Maths Advanced Study notes None 2 Integration by Substitution Techniques for Calculus Success Mathematics Extension 1 Study notes None 2 Integration by Substitution - Solutions & Examples Mathematics Extension 1 Study notes None 2 Internal Sources of Finance: Retained Profits Analysis (2.2)Business Studies Practice materials None More from: doraemonimpact 4 impact 4 Baulkham Hills High SchoolHSC • Grade: 12 Discover more 2 Taking the Sum of a Sequence: Understanding Partial Sums Maths Advanced None 2 Integration by Substitution Techniques for Calculus Success Mathematics Extension 1 None 2 Integration by Substitution - Solutions & Examples Mathematics Extension 1 None 2 Internal Sources of Finance: Retained Profits Analysis (2.2) Business Studies None 2 Solving for the variable angle of inclination Maths Extension 1 None 2 Function Notation: Understanding f(x) and Its Applications Mathematics Extension 1 None Recommended for you 26 Y12 Ext 1 Mathematics Study Notes: Key Concepts and Formulas Maths Extension 1 Study notes 100% (5) 9 Maths Extension ONE Summary Notes - Polynomials & Calculus Concepts Maths Extension 1 Study notes 100% (4) 19 WME Ext 1 HSC 2023 Trial Exam Solutions Maths Extension 1 Study notes 100% (2) 13 Trigonometric Integration Lesson Notes: Key Concepts & Examples Maths Extension 1 Study notes 100% (1) 26 Y12 Ext 1 Mathematics Study Notes: Key Concepts and Formulas Maths Extension 1 100% (5) 9 Maths Extension ONE Summary Notes - Polynomials & Calculus Concepts Maths Extension 1 100% (4) 19 WME Ext 1 HSC 2023 Trial Exam Solutions Maths Extension 1 100% (2) 13 Trigonometric Integration Lesson Notes: Key Concepts & Examples Maths Extension 1 100% (1) 5 IF 2 Inverse Functions: Examples and Detailed Explanations Maths Extension 1 100% (1) 7 Applications of Arithmetic & Geometric Series in Real Life Maths Extension 1 100% (1) Related documents Y12 Ext 1 Mathematics Study Notes: Key Concepts and Formulas Substitution and Harder Integration: EXT2 C1 HSC Exam Prep Trig Integration Solutions for EXT2 C1 2019 HSC Exam Proof and Inequalities: EXT2 P1 Study Guide Parametric Representation of Curves: Key Concepts and Exercises Solving for the variable angle of inclination Get homework AI help with the Studocu App Open the App English Australia Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help Frequently asked questions Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English Australia Studocu is not affiliated to or endorsed by any school, college or university. 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188373	https://www.youtube.com/watch?v=rdqsKpGpErc	Find value of k given constant term for expansion of (x^3/3+k/x)^12 Ms Shaws Math Class 51100 subscribers 152 likes Description 15027 views Posted: 20 Oct 2019 Binomial Theorem 18 comments Transcript: everyone and we're going to find the value of K for the expansion of X cubed divided by 3 plus K divided by X all raised to the 12th power and they give you the constant term so this means this is really times x to the 0 power remember X to the 0 power equals 1 so just remember that equals your constant term so we get started we're going to use this formula here and instead of that and of course your calculator would be like that but we're going to use n choose R like that all right so let's get started and rewrite this our n is 12 our a is X cubed or one third of X cubed and our B is K X to the negative one I just rewrote that I put the X up in the numerator so that's X to the negative one so let's fill this out we have n is 12 so n choose R our a is X cubed divided by 3 to the 12 minus our following the formula and our B is K X to the negative 1 to the R power now we want to first solve for R so the first thing I'm going to do is just look at my X's here so if you just look at your X's this is X cubed divided by 3 is really one third times X cubed so I'm just looking at the x's so I'm going to take X cubed times 12 minus R and then we have times X and negative 1 R and we want this to equal X to this zero because this is our constant term therefore I'm going to combine my exes so have X this is 36 minus 3r times X to the negative R equals x to the zero power so we have X you use your product property you have 36 minus 4r I'm just combining those two equals x to the 0 now that my bases are the same they're exes I can just solve for R so 36 minus 4 R equals 0 36 equals 4 R or you could just do this in your head so we know that r is 9 now the next one I'm going to plug in my R and then solve for K so basically my R is 9 so I'm going to plug this in I have 12 choose 9x cubed divide it by 3 to the 12 minus 9 power and then I have K X to the negative 1 to the 9th power so this is just 3 so this gives you 12 choose 9 it's just 220 and then this is going to be X to the 9th and then 3 cubed is 27 and then we have K to the 9th and X to the negative ninth so putting all this together we have 220 divided it by 27 times X to the 9th K to the 9th and X to the negative ninth well X to the ninth times X to the negative ninth its X to the 9 minus 9 which is X to the 0 and we set it up that way so we can just disregard that so now we have to 20/20 7 K to the 9th power equals our constant which is one one two six four zero and put that over one if you want now solving for K you just multiply both sides of the equation by 27 divided by 220 and you get K to the 9th power equals 512 now if you take the ninth root of both sides you get K equals 2 that's because 2 to the 9th equals 512 so that's our final answer thank you have a nice day bye bye [Music]
188374	https://www.sciencedirect.com/science/article/abs/pii/S0030402615018525	Analysis influence of fiber alignment error on laser–diode fiber coupling efficiency - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (10) Cited by (14) Optik Volume 127, Issue 6, March 2016, Pages 3276-3280 Analysis influence of fiber alignment error on laser–diode fiber coupling efficiency Author links open overlay panel Yu Junhong a b, Guo Linhui a b, Wu Hualing a b, Meng Huicheng a b, Tan Hao a b, Gao Songxin a b, Wu Deyong a b Show more Add to Mendeley Share Cite rights and content Abstract The fiber alignment error has restricted the stability and efficiency of laser–diode fiber-coupled module. The theoretical calculation model is established by optical ray tracing based on software ZEMAX, and the fiber coupling efficiency has been calculated by software MATLAB. The calculated efficiency has been compared with experimental results, and the comparative results show that the calculated efficiency is entirely valid. Introduction Laser–diodes coupled into a fiber have many applications in fiber laser pumping, material processing, and biology medicine fields , , due to its compactness, high electro-optical efficiency and reliability etc. The fiber alignment error will cause the optical-optical efficiency declination , , , and the fiber can even be destroyed with the heat accumulation , . So the mismatch between fiber and focus lens is one of the main restrictions on stability and reliability of laser–diode fiber-coupled module. In this work, the theoretical calculation model is established by optical ray tracing based on software ZEMAX, and the fiber coupling efficiency has been calculated by software MATLAB in three situations: lateral alignment error, longitudinal alignment error and rotational angle alignment error. Comparison with other laser–diode fiber coupling efficiency calculation papers , , , our works have several different points. First, the calculation model of laser–diode is established more correct, former papers considered the laser–diode beam in fast and slow axes are both TEM 00 Gaussian beam (the super-Gaussian factor G=1) , , but actually the slow axis beam are Hermite–Gaussian beam (the super-Gaussian factor G=2) , so in order to obtain more precise results, we have modified the calculation model. Second, in the lateral alignment error and rotational angle alignment error situations we have calculated the efficiency versus two directions rather than one direction in former paper. Third, in the rotational angle alignment error situation, we have considered the longitudinal offset caused by rotational angle to make the calculated results more precise. Finally, the calculation efficiency has been compared with the experimental results, this has validated the correction of the calculation model. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Laser–diode fiber coupling simulation model The laser–diode beam after beam shaping goes through the aspherical lens to focus on the multiple-mode fiber core tip is the last and the most important step of laser–diode fiber coupling process. As shown in Fig. 1, whether the position of fiber core tip is matching the focused beam spot can strongly determine the efficiency and stability of laser–diode fiber coupling module. In order to analyze the influence of fiber alignment error on laser–diode fiber coupling efficiency, we have established Coupling efficiency analysis The alignment error of fiber core tip in the position space and angle space includes three situations: lateral offset alignment error, longitudinal offset alignment error and rotational angle alignment error. We will calculate the influence of fiber coupling efficiency in these three situations, respectively. And in this part we do not consider the Fresnel reflection loss of the fiber in all the calculation results. Conclusions Based on the ZEMAX and MATLAB software, laser–diode fiber coupling efficiency calculation model considering the fiber alignment error has been established, and the calculation efficiency model has been optimized in longitudinal offset and rotational angle error situations. The comparison results between experimental efficiency and calculation efficiency have validated the correction of the efficiency calculation model. Acknowledgment The authors would like to acknowledge GuoLinhuiand Wu Hualing, for their helpful discussions and insights. Recommended articles References (10) H. Injeyan et al. High-power diode laser arrays High-Power Laser Handbook (2011) R. Diehl High-Power Diode Lasers: Fundamentals, Technology, Applications (2000) Yong Zhao et al. Analysis on the coupling efficiency between semiconductor laser diode and single-mode fibers Opt. Technol. (1999) Bin Lin et al. Analysis on the laser diode multimode fiber coupling efficiency Acta Opt. Sin. (2003) Haitao Chen et al. Influence of the fiber displacement on coupling light beam into single-mode fiber Laser&Infrared (2011) There are more references available in the full text version of this article. Cited by (14) Research on the influence of laser thermal characteristics on misalignment of optical components 2021, Optik Citation Excerpt : In fact, the laser's thermal stress will not only cause the deformation of the laser chip but also cause the misalignment of other optical devices . The misalignment of optical components will cause the laser to fail to achieve the expected output performance [14–17]. In the research of laser optical device misalignment, scholars mostly use numerical analysis and simulation analysis to obtain the influence of each optical device's offset on the output performance of the laser [18–26]. Show abstract With the improvement of application requirements, the demand for output performance and reliability of high-power semiconductor lasers have become more stringent. The great heat flux generated during the laser operation will seriously affect the performance indicators of the laser, and will also cause the misalignment of the optical components. This article first explores the thermal characteristics of the optical components during laser operation through experiment and then combines the simulation and experiment to reveal the generation law of the optical component misalignments. Furthermore, the factors that affect the magnitude of the misalignment are further studied. Finally, feasibility solutions to reduce the misalignment of the optical component is discussed. It provides an important basis for guaranteeing the high output performance and reliability of the laser. ### Review of the technology of a single mode fiber coupling to a laser diode 2020, Optical Fiber Technology Citation Excerpt : The fundamental mode of single mode elliptic core fibers was analyzed by Gaussian approximation, and the longitudinal, transverse and azimuthal expressions were obtained . The optical simulation software Zemax [33,34] has been used to optimize the microlens [35,36]. All in all, to increase the coupling efficiency and raise the process tolerance of an external bulk lens’ coupling scheme, a better way could be to reduce the coupling loss caused by physical effects, choose a suitable material and explore new structures for the bulk lens. Show abstract In this paper, the technology of a single mode fiber coupling to a semiconductor laser diode has been summarized and the latest developments in the bulk optics coupling scheme and the microlens fiber coupling scheme have been reviewed. The review has focused on optimizing the optical structure and the coupling parameters to improve the coupling efficiency and packaging performance. The advanced manufacturing technology as well as common modeling methods and applications of coupling systems have also been reviewed. Finally, the paper has summarized the key technologies of a single mode fiber coupling to a laser diode and its direction of development in the future. ### Influence of positioning errors on the coupling efficiency of a single emitter laser array 2020, Optik Citation Excerpt : There have been only a few attempts at reporting the influence of the alignment error of the fiber through experimental approaches. Yu and He both obtained results on the effect of fiber alignment error on coupling efficiency by using both simulation and experimental analysis methods [27,28]. However, studies that merely focus on fiber alignment errors are one-sided; the collimated laser array beam must be coupled into the fiber through combining and focusing . Show abstract A high coupling efficiency is required to guarantee high-output power for lasers. However, positioning errors in optical devices can lead to different degrees of attenuation for the coupling efficiency. This paper, investigates the effects of positioning errors on the coupling efficiency of optical devices. Specifically, the combining, focusing, and coupling processes of a three-wire single emitter laser array are studied through experiments and are supported by simulations. The results show that a 50 μ m position error in the Y-axis of the fiber will reduce the coupling efficiency by 44 %, and a rotation error of 0.5° in the X-axis of the reflector will reduce the coupling efficiency by 100 %. Moreover, the sensitivity of the coupling efficiency to each error is revealed. ### Wide-Angle Optical Metasurface for Vortex Beam Generation 2023, Nanomaterials ### Measurement of carbonization region on leather cutting in CO2 and diode laser-based laser beam process 2022, Proceedings of the Institution of Mechanical Engineers Part E Journal of Process Mechanical Engineering ### Review of Issues and Solutions in High-Power Semiconductor Laser Packaging Technology 2021, Frontiers in Physics View all citing articles on Scopus View full text Copyright © 2015 Elsevier GmbH. All rights reserved. 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188375	https://eng.libretexts.org/Bookshelves/Civil_Engineering/Fluid_Mechanics_(Bar-Meir)/10%3A_Inviscid_Flow_or_Potential_Flow/10.3_Potential_Flow_Functions_Inventory	Skip to main content 10.3 Potential Flow Functions Inventory Last updated : Mar 5, 2021 Save as PDF 10.2.3.1: Existences of Stream Functions 10.3.1: Flow Around a Circular Cylinder Donate Page ID : 781 Genick Bar-Meir Potto Project ( \newcommand{\kernel}{\mathrm{null}\,}) This section describes several simple scenarios of the flow field. These flow fields will be described and exhibits utilizition of the potential and stream functions. These flow fields can be combined by utilizing superimposing principle. Uniform Flow The trivial flow is the uniform flow in which the fluid field moves directly and uniformly from one side to another side. This flow is further simplified, that is the coordinates system aligned with to flow so the –coordinate in the direction of the flow. In this case the velocity is given by and according to definitions in this chapter Hence, it can be obtained that where is arbitrary function of the and is arbitrary function of . In the same time these function have to satisfy the condition These conditions dictate that Hence These lines can be exhibits for various constants as shown in Figure below. Fig. 10.6 Uniform Flow Streamlines and Potential Lines. Line Source and Sink Flow Another typical flow is a flow from a point or a line in a two dimensional field. This flow is only an idealization of the flow into a single point. Clearly this kind of flow cannot exist because the velocity approaches infinity at the singular point of the source. Yet this idea has its usefulness and is commonly used by many engineers. This idea can be combined with other flow fields and provide a more realistic situation. Fig. 10.7 Streamlines and Potential lines due to Source or sink. The volumetric flow rate (two dimensional) denotes the flow rate out or in to control volume into the source or sink. The flow rate is shown in Figure 10.7 is constant for every potential line. The flow rate can be determined by The volumetric flow rate (two dimensional) denotes the flow rate out or in to control volume into the source or sink. The flow rate is shown in Figure 10.7 is constant for every potential line. The flow rate can be determined by Where is the volumetric flow rate, is distance from the origin and is the velocity pointing out or into the origin depending whether origin has source or sink. The relationship between the potential function to velocity dictates that Explicitly writing the gradient in cylindrical coordinate results as Equation (10) the gradient components must satisfy the following The integration of equation results in where is the radius at a known point and is the potential at that point. The stream function can be obtained by similar equations that were used or Cartesian coordinates. In the same fashion it can be written that Where in this case (the shortest distance between two adjoining stream lines is perpendicular to both lines) and hence equation (13) is Note that the direction of and is identical. The integration of equation (14) yields It traditionally chosen that the stream function is zero at . This operation is possible because the integration constant and the arbitrary reference. In the case of the sink rather than the source, the velocity is in the opposite direction. Hence the flow rate is negative and the same equations obtained. Free Vortex Flow Fig. 10.8 Two dimensional Vortex free flow. In the diagram exhibits part the circle to explain the stream lines and potential lines. As opposed to the radial flow direction (which was discussed under the source and sink) the flow in the tangential direction is referred to as the free vortex flow. Another typical name for this kind of flow is the potential vortex flow. The flow is circulating the origin or another point. The velocity is only a function of the distance from the radius as And in vector notation the flow is The fundamental aspect of the potential flow is that this flow must be irrotational flow. The gradient of the potential in cylindrical coordinates is Hence, equation (20) dictates that From these equations it can be seem that and Equation (23) states that the potential function depends on the angle, while it also a function of the radius. The only what the above requirement is obtained when the derivative of and the equation are equal to a constant. Thus, It can be observed from equation (23) that the velocity varies inversely with the radius. This variation is referred in the literature as the natural vortex as oppose to forced vortex where the velocity varies in any different functionality. It has to be noted that forced vortex flow is not potential flow. The stream function can be found in the "standard'' way as It can observed, in this case, from Figure 10.8 that hence Thus, The source point or the origin of the source is a singular point of the stream function and there it cannot be properly defined. Equation (24) dictates that velocity at the origin is infinity. This similar to natural situation such as tornadoes, hurricanes, and whirlpools where the velocity approaches a very large value near the core. In these situation the pressure became very low as the velocity increase. Since the pressure cannot attain negative value or even approach zero value, the physical situation changes. At the core of these phenomenon a relative zone calm zone is obtained. The Circulation Concept In the construction of the potential flow or the inviscid flow researchers discover important concept of circulation. This term mathematically defined as a close path integral around area (in two dimensional flow) of the velocity along the path. The circulation is denoted as and defined as Where the velocity represents the velocity component in the direction of the path. The symbol indicating that the integral in over a close path. Fig. 10.9 Circulation path to illustrate varies calculations. Mathematically to obtain the integral the velocity component in the direction of the path has to be chosen and it can be defined as Substituting the definition potential function into equation (29) provides And using some mathematical manipulations yields [ \label{if:eq:cpManipulation} \Gamma = \oint_C \overbrace{\dfrac{d\phi}{ds} }^ ParseError: invalid DekiScript (click for details) ``` Callstack: at (Bookshelves/Civil_Engineering/Fluid_Mechanics_(Bar-Meir)/10:_Inviscid_Flow_or_Potential_Flow/10.3_Potential_Flow_Functions_Inventory), /content/body/div/p/span, line 1, column 1 ``` \,ds = \oint_C d\phi ] The integration of equation (31) results in Unless the potential function is dual or multi value, the difference between the two points is zero. In fact this what is expected from the close path integral. However, in a free vortex situation the situation is different. The integral in that case is the integral around a circular path which is In this case the circulation, is not vanishing. In this example, the potential function is a multiple value as potential function the potential function with a single value. Example 10.5 Calculate the circulation of the source on the path of the circle around the origin with radius for a source of a given strength. Solution 10.5 The circulation can be carried by the integration Since the velocity is perpendicular to the path at every point on the path, the integral identically is zero. Thus, there are two kinds of potential functions one where there are single value and those with multi value. The free vortex is the cases where the circulation add the value of the potential function every rotation. Hence, it can be concluded that the potential function of vortex is multi value which increases by the same amount every time, . In this case value at is different because the potential function did not circulate or encompass a singular point. In the other cases, every additional enclosing adds to the value of potential function a value. It was found that the circulation, is zero when there is no singular point within the region inside the path. For the free vortex the integration constant can be found if the circulation is known as In the literature, the term is, some times, referred to as the "strength'' of the vortex. The common form of the stream function and potential function is in the form of Superposition of Flows For incompressible flow and two dimensional the continuity equation reads The potential function must satisfy the Laplace's equation which is a linear partial differential equation. The velocity perpendicular to a solid boundary must be zero (boundary must be solid) and hence it dictates the boundary conditions on the potential equation. From mathematical point of view this boundary condition as In this case, represents the unit vector normal to the surface. A solution to certain boundary condition with certain configuration geometry and shape is a velocity flow field which can be described by the potential function, . If such function exist it can be denoted as . If another velocity flow field exists which describes, or is, the solutions to a different boundary condition(s) it is denoted as . The Laplacian of first potential is zero, and the same is true for the second one . Hence, it can be written that Since the Laplace mathematical operator is linear the two potential can be combined as The boundary conditions can be also treated in the same fashion. On a solid boundary condition for both functions is zero hence and the normal derivative is linear operator and thus It can be observed that the combined new potential function create a new velocity field. In fact it can be written that The velocities and are obtained from and respectively. Hence, the superposition of the solutions is the characteristic of the potential flow. Source and Sink Flow or Doublet Flow In the potential flow, there is a special case where the source and sink are combined since it represents a special and useful shape. A source is located at point B which is from the origin on the positive coordinate. The flow rate from the source is and the potential function is Fig. 10.10 Combination of the Source and Sink located at a distance from the origin on the coordinate. The source is on the right. The sink is at the same distance but at the negative side of the coordinate and hence it can be represented by the potential function The description is depicted on Figure 10.10. The distances, and are defined from the points and respectively. The potential of the source and the sink is In this case, it is more convenient to represent the situation utilizing the cylindrical coordinates. The Law of Cosines for the right triangle () this cases reads In the same manner it applied to the left triangle as Therefore, equation (47) can be written as It can be shown that the following the identity exist Caution: mathematical details which can be skipped where is a dummy variable. Hence, substituting into equation (50) the identity of equation (51) results in The several following stages are more of a mathematical nature which provide minimal contribution to physical understanding but are provide to interested reader. The manipulations are easier with an implicit solution and thus Equation (53), when noticing that the , can be written as In Cartesian coordinates equation (54) can be written as Equation (55) can be rearranged by the left hand side to right as and moving to left side result in Add to both sides transfers equation (??) The hyperbolic identity can be written as End Caution: mathematical details It can be noticed that first three term on the right hand side are actually quadratic and can be written as equation (59) represents a circle with a radius and a center at . The potential lines depicted on Figure 10.11. For the drawing purposes equation (59) is transformed into a dimensionless form as Notice that the stream function has the same dimensions as the source/sink flow rate. Fig. 10.11 Stream and Potential line for a source and sink. It can be noticed that stream line (in blue to green) and the potential line are in orange to crimson. This figure is relative distances of and . The parameter that change is and . Notice that for give larger of the circles are smaller. The stream lines can be obtained by utilizing similar procedure. The double stream function is made from the combination of the source and sink because stream functions can be added up. Hence, The angle and shown in Figure 10.11 related other geometrical parameters as and The stream function becomes Caution: mathematical details which can be skipped Rearranging equation (64) yields Utilizing the identity Equation (65) transfers to As in the potential function cases, Several manipulations to convert the equation (66) form so it can be represented in a "standard'' geometrical shapes are done before to potential function. Reversing and finding the common denominator provide or End Caution: mathematical details Equation (68) can be rearranged, into a typical circular representation Equation (69) describes circles with center on the coordinates at . It can be noticed that these circles are orthogonal to the the circle that represents the the potential lines. For the drawing it is convenient to write equation (69) in dimensionless form as Dipole Flow It was found that when the distance between the sink and source shrinks to zero a new possibility is created which provides benefits to new understanding. The new combination is referred to as the dipole. Even though, the construction of source/sink to a single location (as the radius is reduced to zero) the new "creature'' has direction as opposed to the scalar characteristics of source and sink. First the potential function and stream function will be presented. The potential function is To determine the value of the quantity in equation (71) the L'Hôpital's rule will be used. First the appropriate form will be derived so the technique can be used. Caution: mathematical details which can be skipped Multiplying and dividing equation (71) by yields Equation (72) has two parts. The first part, , which is a function of and and the second part which is a function of . While reducing to zero, the flow increases in such way that the combination of is constant. Hence, the second part has to be examined and arranged for this purpose. It can be noticed that the ratio in the natural logarithm approach one . The L'Hopital's rule can be applied because the situation of nature of . The numerator can be found using a short cut End Caution: mathematical details at [ \label{if:eq:doubleImit2Lhopital} \lim_{r_0\rightarrow 0} \dfrac{ \dfrac{\cancelto{0}{2\,r_0} - 2\,r\,\cos\theta} {r^2+\cancelto{0} ParseError: EOF expected (click for details) ``` Callstack: at (Bookshelves/Civil_Engineering/Fluid_Mechanics_(Bar-Meir)/10:_Inviscid_Flow_or_Potential_Flow/10.3_Potential_Flow_Functions_Inventory), /content/body/div/p/span, line 1, column 4 ``` } {4} = - \dfrac{\cos\theta}{r} ] Combining the first and part with the second part results in After the potential function was established the attention can be turned into the stream function. To establish the stream function, the continuity equation in cylindrical is used which is The transformation of equations (??) and (??) to cylindrical coordinates results in The relationship for the potential function of the cylindrical coordinates was determined before an appear the relationship (??) and (??) Thus the relationships that were obtained before for Cartesian coordinates is written in cylindrical coordinates as In the case of the dipole, the knowledge of the potential function is used to obtain the stream function. The derivative of the potential function as respect to the radius is And From equation (83) after integration with respect to one can obtain and from equation (84) one can obtain that The only way that these conditions co–exist is to be constant and thus is zero. The general solution of the stream function is then Caution: mathematical details which can be skipped The potential function and stream function describe the circles as following: In equation (87) it can be recognized that Thus, multiply equation (87) by and some rearrangement yield Further rearranging equation (88) provides and converting to the standard equation of circles as End Caution: mathematical details The equation (87) (or (90) represents a circle with a radius of with location at and . The identical derivations can be done for the potential function. It can be noticed that the difference between the functions results from difference of the instead of the term is . Thus, the potential functions are made from circles that the centers are at same distance as their radius from origin on the coordinate. It can be noticed that the stream function and the potential function can have positive and negative values and hence there are family on both sides of coordinates. Figure 10.12 displays the stream functions (cyan to green color) and potential functions (gold to crimson color). Notice the larger the value of the stream function the smaller the circle and the same for the potential functions. Fig. 10.12 Stream lines and Potential lines for Doublet. The potential lines are in gold color to crimson while the stream lines are cyan to green color. Notice the smaller value of the stream function translates the smaller circle. The drawing were made for the constant to be one (1) and direct value can be obtained by simply multiplying. It must be noted that in the derivations above it was assumed that the sink is on the left and source is on the right. Clear similar results will obtained if the sink and source were oriented differently. Hence the dipole (even though) potential and stream functions are scalar functions have a direction. In this stage this topic will not be treated but must be kept in question form. Example 10.6 This academic example is provided mostly for practice of the mathematics. Built the stream function of dipole with angle. Start with a source and a sink distance from origin on the line with a angle from coordinates. Let the distance shrink to zero. Write the stream function. Template:HideTOC Contributors and Attributions Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. 2πψQ0=tan−1y x/ 10.2.3.1: Existences of Stream Functions 10.3.1: Flow Around a Circular Cylinder
188376	https://www.sketchy.com/mcat-lessons/genetic-recombination	Opens in a new window Opens an external website Opens an external website in a new window We use cookies to provide necessary functionality and improve your experience. By remaining on this website you indicate your consent. Cookie Policy Try for FreeLogin GET 20% OFF SKETCHY MCAT WITH CODE REG20 \| REGISTRATION DAY SALE MCAT Curriculum / Cell Biology / Genetics & Growth / Genetic Recombination Genetic Recombination Tags: No items found. Cell Biology Bacterial recombination is the transfer of DNA from one bacterium to another to create genetic diversity. It occurs in three main ways: transformation, conjugation, and transduction. Transformation occurs when foreign DNA from the environment is integrated into the genome of the bacterial host cell, typically originating from other lysed bacteria. Conjugation is the transfer of genetic material from one bacterium to another and is considered the prokaryotic form of sexual reproduction. In conjugation, the bacterium sharing genetic material is called the donor male, and the bacterium receiving it is the recipient female. The sharing of genetic material occurs via direct contact between the cells, facilitated by an appendage called a sex pilus. However, only cells with a special chunk of DNA called the fertility factor (F-factor) can synthesize the proteins needed to make up the sex pilus. Lastly, transduction occurs when a virus that infects bacteria, called a bacteriophage, transfers genetic material from one bacterium to another. Bacteriophages can accidentally incorporate a piece of bacterial DNA into their own viral genome when reproducing inside the host. When one of these phages infects another bacterium, it releases the mix of viral and bacterial genes into the new host cell, resulting in new genes for the host. Lesson Outline Introduction to Bacterial Recombination Need for genetic variation in bacteria Three main ways of bacterial recombination: transformation, conjugation, and transduction Transformation Foreign DNA from the environment integrated into the bacterial genome Typical source of genetic material: lysed bacteria Conjugation Transfer of genetic material from one bacterium to another Donor male and recipient female bacteria Direct cell contact and the use of a sex pilus for DNA transfer Fertility factor (F-factor) required for sex pilus formation and DNA transfer Recipient female can become a donor male if F-factor is transferred Transduction Transfer of genetic material by bacteriophages (viruses) Bacteriophages reproduce inside host and can accidentally incorporate bacterial DNA Transfer of mixed viral and bacterial genetic material to a new bacterial host Don't stop here! Get access to 35 more Cell Biology lessons & 8 more full MCAT courses with one subscription! Try 7 Days Free FAQs What is genetic recombination and how does it differ from binary fission in bacteria? Genetic recombination is a process by which genetic material from two different sources is combined to create a new genetic makeup, leading to genetic variation. In bacteria, this can occur through transformation, conjugation, and transduction. On the other hand, binary fission is a type of asexual reproduction in bacteria, where the parent cell divides into two equal daughter cells, each with an identical copy of the parent cell's DNA. Binary fission does not typically involve genetic variation, while genetic recombination promotes genetic diversity in bacterial populations. How do transformation, conjugation, and transduction contribute to genetic recombination in bacteria? Transformation, conjugation, and transduction are three mechanisms by which bacteria exchange genetic material, leading to genetic recombination. In transformation, bacteria take up free DNA molecules present in the environment, which can then be integrated into their own genome. Conjugation involves direct cell-to-cell contact, where DNA transfer takes place through a sex pilus. Transduction is a process in which a bacteriophage (a virus that infects bacteria) transfers DNA from one bacterium to another unintentionally. All three mechanisms lead to the incorporation of new genetic material into the recipient bacteria, increasing the genetic variation within the bacterial population. What are the roles of the sex pilus and the fertility factor in conjugation? The sex pilus is a tube-like structure present on the surface of the donor bacterial cell in conjugation. It connects the donor and recipient cells, allowing for the transfer of genetic material between them. The fertility factor (F factor) is a specific plasmid (small, circular DNA molecule) that allows the bacterial cell to form a sex pilus and transfer genetic material. The donor cell contains the F factor, while the recipient cell typically does not (though the F factor plasmid can be transferred from the donor cell to the recipient during conjugation). How does transduction lead to genetic recombination in bacteria? Transduction occurs when a bacteriophage infects a bacterial cell and accidentally packages a fragment of the host bacterium's DNA into its viral capsid instead of its own genetic material. When the newly assembled bacteriophage infects another bacterium, it injects the previously packaged bacterial DNA into the new host. This foreign DNA can be integrated into the new host bacterium's genome, resulting in genetic recombination and an increase in genetic variation within the bacterial population. Programs MedicalMCATPANP CompanyBlogCareersPrivacy PolicyTerms of UseCookie Policy
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Burst Balloons 313. Super Ugly Number 314. Binary Tree Vertical Order Traversal 315. Count of Smaller Numbers After Self 316. Remove Duplicate Letters 317. Shortest Distance from All Buildings 318. Maximum Product of Word Lengths 319. Bulb Switcher 320. Generalized Abbreviation 321. Create Maximum Number 322. Coin Change 323. Number of Connected Components in an Undirected Graph 324. Wiggle Sort II 325. Maximum Size Subarray Sum Equals k 326. Power of Three 327. Count of Range Sum 328. Odd Even Linked List 329. Longest Increasing Path in a Matrix 330. Patching Array 331. Verify Preorder Serialization of a Binary Tree 332. Reconstruct Itinerary 333. Largest BST Subtree 334. Increasing Triplet Subsequence 335. Self Crossing 336. Palindrome Pairs 337. House Robber III 338. Counting Bits 339. Nested List Weight Sum 340. Longest Substring with At Most K Distinct Characters 341. Flatten Nested List Iterator 342. Power of Four 343. Integer Break 344. Reverse String 345. Reverse Vowels of a String 346. Moving Average from Data Stream 347. Top K Frequent Elements 348. Design Tic-Tac-Toe 349. Intersection of Two Arrays 350. Intersection of Two Arrays II 351. Android Unlock Patterns 352. Data Stream as Disjoint Intervals 353. Design Snake Game 354. Russian Doll Envelopes 355. Design Twitter 356. Line Reflection 357. Count Numbers with Unique Digits 358. Rearrange String k Distance Apart 359. Logger Rate Limiter 360. Sort Transformed Array 361. Bomb Enemy 362. Design Hit Counter 363. Max Sum of Rectangle No Larger Than K 364. Nested List Weight Sum II 365. Water and Jug Problem 366. Find Leaves of Binary Tree 367. Valid Perfect Square 368. Largest Divisible Subset 369. Plus One Linked List 370. Range Addition 371. Sum of Two Integers 372. Super Pow 373. Find K Pairs with Smallest Sums 374. Guess Number Higher or Lower 375. Guess Number Higher or Lower II 376. Wiggle Subsequence 377. Combination Sum IV 378. Kth Smallest Element in a Sorted Matrix 379. Design Phone Directory 380. Insert Delete GetRandom O(1) 381. Insert Delete GetRandom O(1) - Duplicates allowed 382. Linked List Random Node 383. Ransom Note 384. Shuffle an Array 385. Mini Parser 386. Lexicographical Numbers 387. First Unique Character in a String 388. Longest Absolute File Path 389. Find the Difference 390. Elimination Game 391. Perfect Rectangle 392. Is Subsequence 393. UTF-8 Validation 394. Decode String 395. Longest Substring with At Least K Repeating Characters 396. Rotate Function 397. Integer Replacement 398. Random Pick Index 399. Evaluate Division 400. Nth Digit 401. Binary Watch 402. Remove K Digits 403. Frog Jump 404. Sum of Left Leaves 405. Convert a Number to Hexadecimal 406. Queue Reconstruction by Height 407. Trapping Rain Water II 408. Valid Word Abbreviation 409. Longest Palindrome 410. Split Array Largest Sum 411. Minimum Unique Word Abbreviation 412. Fizz Buzz 413. Arithmetic Slices 414. Third Maximum Number 415. Add Strings 416. Partition Equal Subset Sum 417. Pacific Atlantic Water Flow 418. Sentence Screen Fitting 419. Battleships in a Board 420. Strong Password Checker 421. Maximum XOR of Two Numbers in an Array 422. Valid Word Square 423. Reconstruct Original Digits from English 424. Longest Repeating Character Replacement 425. Word Squares 426. Convert Binary Search Tree to Sorted Doubly Linked List 427. Construct Quad Tree 428. Serialize and Deserialize N-ary Tree 429. N-ary Tree Level Order Traversal 430. Flatten a Multilevel Doubly Linked List 431. Encode N-ary Tree to Binary Tree 432. All O`one Data Structure 433. Minimum Genetic Mutation 434. Number of Segments in a String 435. Non-overlapping Intervals 436. Find Right Interval 437. Path Sum III 438. Find All Anagrams in a String 439. Ternary Expression Parser 440. K-th Smallest in Lexicographical Order 441. Arranging Coins 442. Find All Duplicates in an Array 443. String Compression 444. Sequence Reconstruction 445. Add Two Numbers II 446. Arithmetic Slices II - Subsequence 447. Number of Boomerangs 448. Find All Numbers Disappeared in an Array 449. Serialize and Deserialize BST 450. Delete Node in a BST 451. Sort Characters By Frequency 452. Minimum Number of Arrows to Burst Balloons 453. Minimum Moves to Equal Array Elements 454. 4Sum II 455. Assign Cookies 456. 132 Pattern 457. Circular Array Loop 458. Poor Pigs 459. Repeated Substring Pattern 460. LFU Cache 461. Hamming Distance 462. Minimum Moves to Equal Array Elements II 463. Island Perimeter 464. Can I Win 465. Optimal Account Balancing 466. Count The Repetitions 467. Unique Substrings in Wraparound String 468. Validate IP Address 469. Convex Polygon 470. Implement Rand10() Using Rand7() 471. Encode String with Shortest Length 472. Concatenated Words 473. Matchsticks to Square 474. Ones and Zeroes 475. Heaters 476. Number Complement 477. Total Hamming Distance 478. Generate Random Point in a Circle 479. Largest Palindrome Product 480. Sliding Window Median 481. Magical String 482. License Key Formatting 483. Smallest Good Base 484. Find Permutation 485. Max Consecutive Ones 486. Predict the Winner 487. Max Consecutive Ones II 488. Zuma Game 489. Robot Room Cleaner 490. The Maze 491. Non-decreasing Subsequences 492. Construct the Rectangle 493. Reverse Pairs 494. Target Sum 495. Teemo Attacking 496. Next Greater Element I 497. Random Point in Non-overlapping Rectangles 498. Diagonal Traverse 499. The Maze III 500. Keyboard Row 501. Find Mode in Binary Search Tree 502. IPO 503. Next Greater Element II 504. Base 7 505. The Maze II 506. Relative Ranks 507. Perfect Number 508. Most Frequent Subtree Sum 509. Fibonacci Number 510. Inorder Successor in BST II 513. Find Bottom Left Tree Value 514. Freedom Trail 515. Find Largest Value in Each Tree Row 516. Longest Palindromic Subsequence 517. Super Washing Machines 518. Coin Change II 519. Random Flip Matrix 520. Detect Capital 521. Longest Uncommon Subsequence I 522. Longest Uncommon Subsequence II 523. Continuous Subarray Sum 524. Longest Word in Dictionary through Deleting 525. Contiguous Array 526. Beautiful Arrangement 527. Word Abbreviation 528. Random Pick with Weight 529. Minesweeper 530. Minimum Absolute Difference in BST 531. Lonely Pixel I 532. K-diff Pairs in an Array 533. Lonely Pixel II 535. Encode and Decode TinyURL 536. Construct Binary Tree from String 537. Complex Number Multiplication 538. Convert BST to Greater Tree 539. Minimum Time Difference 540. Single Element in a Sorted Array 541. Reverse String II 542. 01 Matrix 543. Diameter of Binary Tree 544. Output Contest Matches 545. Boundary of Binary Tree 546. Remove Boxes 547. Number of Provinces 548. Split Array with Equal Sum 549. Binary Tree Longest Consecutive Sequence II 551. Student Attendance Record I 552. Student Attendance Record II 553. Optimal Division 554. Brick Wall 555. Split Concatenated Strings 556. Next Greater Element III 557. Reverse Words in a String III 558. Logical OR of Two Binary Grids Represented as Quad-Trees 559. Maximum Depth of N-ary Tree 560. Subarray Sum Equals K 561. Array Partition 562. Longest Line of Consecutive One in Matrix 563. Binary Tree Tilt 564. Find the Closest Palindrome 565. Array Nesting 566. Reshape the Matrix 567. Permutation in String 568. Maximum Vacation Days 572. Subtree of Another Tree 573. Squirrel Simulation 575. Distribute Candies 576. Out of Boundary Paths 581. Shortest Unsorted Continuous Subarray 582. Kill Process 583. Delete Operation for Two Strings 587. Erect the Fence 588. Design In-Memory File System 589. N-ary Tree Preorder Traversal 590. N-ary Tree Postorder Traversal 591. Tag Validator 592. Fraction Addition and Subtraction 593. Valid Square 594. Longest Harmonious Subsequence 598. Range Addition II 599. Minimum Index Sum of Two Lists 600. Non-negative Integers without Consecutive Ones 604. Design Compressed String Iterator 605. Can Place Flowers 606. Construct String from Binary Tree 609. Find Duplicate File in System 611. Valid Triangle Number 616. Add Bold Tag in String 617. Merge Two Binary Trees 621. Task Scheduler 622. Design Circular Queue 623. Add One Row to Tree 624. Maximum Distance in Arrays 625. Minimum Factorization 628. Maximum Product of Three Numbers 629. K Inverse Pairs Array 630. Course Schedule III 631. Design Excel Sum Formula 632. Smallest Range Covering Elements from K Lists 633. Sum of Square Numbers 634. Find the Derangement of An Array 635. Design Log Storage System 636. Exclusive Time of Functions 637. Average of Levels in Binary Tree 638. Shopping Offers 639. Decode Ways II 640. Solve the Equation 641. Design Circular Deque 642. Design Search Autocomplete System 643. Maximum Average Subarray I 644. Maximum Average Subarray II 645. Set Mismatch 646. Maximum Length of Pair Chain 647. Palindromic Substrings 648. Replace Words 649. Dota2 Senate 650. 2 Keys Keyboard 651. 4 Keys Keyboard 652. Find Duplicate Subtrees 653. Two Sum IV - Input is a BST 654. Maximum Binary Tree 655. Print Binary Tree 656. Coin Path 657. Robot Return to Origin 658. Find K Closest Elements 659. Split Array into Consecutive Subsequences 660. Remove 9 661. Image Smoother 662. Maximum Width of Binary Tree 663. Equal Tree Partition 664. Strange Printer 665. Non-decreasing Array 666. Path Sum IV 667. Beautiful Arrangement II 668. Kth Smallest Number in Multiplication Table 669. Trim a Binary Search Tree 670. Maximum Swap 671. Second Minimum Node In a Binary Tree 672. Bulb Switcher II 673. Number of Longest Increasing Subsequence 674. Longest Continuous Increasing Subsequence 675. Cut Off Trees for Golf Event 676. Implement Magic Dictionary 677. Map Sum Pairs 678. Valid Parenthesis String 679. 24 Game 680. Valid Palindrome II 681. Next Closest Time 682. Baseball Game 683. K Empty Slots 684. Redundant Connection 685. Redundant Connection II 686. Repeated String Match 687. Longest Univalue Path 688. Knight Probability in Chessboard 689. Maximum Sum of 3 Non-Overlapping Subarrays 690. Employee Importance 691. Stickers to Spell Word 692. Top K Frequent Words 693. Binary Number with Alternating Bits 694. Number of Distinct Islands 695. Max Area of Island 696. Count Binary Substrings 697. Degree of an Array 698. Partition to K Equal Sum Subsets 699. Falling Squares 700. Search in a Binary Search Tree 701. Insert into a Binary Search Tree 702. Search in a Sorted Array of Unknown Size 703. Kth Largest Element in a Stream 704. Binary Search 705. Design HashSet 706. Design HashMap 707. Design Linked List 708. Insert into a Sorted Circular Linked List 709. To Lower Case 710. Random Pick with Blacklist 711. Number of Distinct Islands II 712. Minimum ASCII Delete Sum for Two Strings 713. Subarray Product Less Than K 714. Best Time to Buy and Sell Stock with Transaction Fee 715. Range Module 716. Max Stack 717. 1-bit and 2-bit Characters 718. Maximum Length of Repeated Subarray 719. Find K-th Smallest Pair Distance 720. Longest Word in Dictionary 721. Accounts Merge 722. Remove Comments 723. Candy Crush 724. Find Pivot Index 725. Split Linked List in Parts 726. Number of Atoms 727. Minimum Window Subsequence 728. Self Dividing Numbers 729. My Calendar I 730. Count Different Palindromic Subsequences 731. My Calendar II 732. My Calendar III 733. Flood Fill 734. Sentence Similarity 735. Asteroid Collision 736. Parse Lisp Expression 737. Sentence Similarity II 738. Monotone Increasing Digits 739. Daily Temperatures 740. Delete and Earn 741. Cherry Pickup 742. Closest Leaf in a Binary Tree 743. Network Delay Time 744. Find Smallest Letter Greater Than Target 745. Prefix and Suffix Search 746. Min Cost Climbing Stairs 747. Largest Number At Least Twice of Others 748. Shortest Completing Word 749. Contain Virus 750. Number Of Corner Rectangles 751. IP to CIDR 752. Open the Lock 753. Cracking the Safe 754. Reach a Number 755. Pour Water 756. Pyramid Transition Matrix 757. Set Intersection Size At Least Two 758. Bold Words in String 759. Employee Free Time 760. Find Anagram Mappings 761. Special Binary String 762. Prime Number of Set Bits in Binary Representation 763. Partition Labels 764. Largest Plus Sign 765. Couples Holding Hands 766. Toeplitz Matrix 767. Reorganize String 768. Max Chunks To Make Sorted II 769. Max Chunks To Make Sorted 770. Basic Calculator IV 771. Jewels and Stones 772. Basic Calculator III 773. Sliding Puzzle 774. Minimize Max Distance to Gas Station 775. Global and Local Inversions 776. Split BST 777. Swap Adjacent in LR String 778. Swim in Rising Water 779. K-th Symbol in Grammar 780. Reaching Points 781. Rabbits in Forest 782. Transform to Chessboard 783. Minimum Distance Between BST Nodes 784. Letter Case Permutation 785. Is Graph Bipartite 786. K-th Smallest Prime Fraction 787. Cheapest Flights Within K Stops 788. Rotated Digits 789. Escape The Ghosts 790. Domino and Tromino Tiling 791. Custom Sort String 792. Number of Matching Subsequences 793. Preimage Size of Factorial Zeroes Function 794. Valid Tic-Tac-Toe State 795. Number of Subarrays with Bounded Maximum 796. Rotate String 797. All Paths From Source to Target 798. Smallest Rotation with Highest Score 799. Champagne Tower 800. Similar RGB Color 801. Minimum Swaps To Make Sequences Increasing 802. Find Eventual Safe States 803. Bricks Falling When Hit 804. Unique Morse Code Words 805. Split Array With Same Average 806. Number of Lines To Write String 807. Max Increase to Keep City Skyline 808. Soup Servings 809. Expressive Words 810. Chalkboard XOR Game 811. Subdomain Visit Count 812. Largest Triangle Area 813. Largest Sum of Averages 814. Binary Tree Pruning 815. Bus Routes 816. Ambiguous Coordinates 817. Linked List Components 818. Race Car 819. Most Common Word 820. Short Encoding of Words 821. Shortest Distance to a Character 822. Card Flipping Game 823. Binary Trees With Factors 824. Goat Latin 825. Friends Of Appropriate Ages 826. Most Profit Assigning Work 827. Making A Large Island 828. Count Unique Characters of All Substrings of a Given String 829. Consecutive Numbers Sum 830. Positions of Large Groups 831. Masking Personal Information 832. Flipping an Image 833. Find And Replace in String 834. Sum of Distances in Tree 835. Image Overlap 836. Rectangle Overlap 837. New 21 Game 838. Push Dominoes 839. Similar String Groups 840. Magic Squares In Grid 841. Keys and Rooms 842. Split Array into Fibonacci Sequence 843. Guess the Word 844. Backspace String Compare 845. Longest Mountain in Array 846. Hand of Straights 847. Shortest Path Visiting All Nodes 848. Shifting Letters 849. Maximize Distance to Closest Person 850. Rectangle Area II 851. Loud and Rich 852. Peak Index in a Mountain Array 853. Car Fleet 854. K-Similar Strings 855. Exam Room 856. Score of Parentheses 857. Minimum Cost to Hire K Workers 858. Mirror Reflection 859. Buddy Strings 860. Lemonade Change 861. Score After Flipping Matrix 862. Shortest Subarray with Sum at Least K 863. All Nodes Distance K in Binary Tree 864. Shortest Path to Get All Keys 865. Smallest Subtree with all the Deepest Nodes 866. Prime Palindrome 867. Transpose Matrix 868. Binary Gap 869. Reordered Power of 2 870. Advantage Shuffle 871. Minimum Number of Refueling Stops 872. Leaf-Similar Trees 873. Length of Longest Fibonacci Subsequence 874. Walking Robot Simulation 875. Koko Eating Bananas 876. Middle of the Linked List 877. Stone Game 878. Nth Magical Number 879. Profitable Schemes 880. Decoded String at Index 881. Boats to Save People 882. Reachable Nodes In Subdivided Graph 883. Projection Area of 3D Shapes 884. Uncommon Words from Two Sentences 885. Spiral Matrix III 886. Possible Bipartition 887. Super Egg Drop 888. Fair Candy Swap 889. Construct Binary Tree from Preorder and Postorder Traversal 890. Find and Replace Pattern 891. Sum of Subsequence Widths 892. Surface Area of 3D Shapes 893. Groups of Special-Equivalent Strings 894. All Possible Full Binary Trees 895. Maximum Frequency Stack 896. Monotonic Array 897. Increasing Order Search Tree 898. Bitwise ORs of Subarrays 899. Orderly Queue 900. RLE Iterator 901. Online Stock Span 902. Numbers At Most N Given Digit Set 903. Valid Permutations for DI Sequence 904. Fruit Into Baskets 905. Sort Array By Parity 906. Super Palindromes 907. Sum of Subarray Minimums 908. Smallest Range I 909. Snakes and Ladders 910. Smallest Range II 911. Online Election 912. Sort an Array 913. Cat and Mouse 914. X of a Kind in a Deck of Cards 915. Partition Array into Disjoint Intervals 916. Word Subsets 917. Reverse Only Letters 918. Maximum Sum Circular Subarray 919. Complete Binary Tree Inserter 920. Number of Music Playlists 921. Minimum Add to Make Parentheses Valid 922. Sort Array By Parity II 923. 3Sum With Multiplicity 924. Minimize Malware Spread 925. Long Pressed Name 926. Flip String to Monotone Increasing 927. Three Equal Parts 928. Minimize Malware Spread II 929. Unique Email Addresses 930. Binary Subarrays With Sum 931. Minimum Falling Path Sum 932. Beautiful Array 933. Number of Recent Calls 934. Shortest Bridge 935. Knight Dialer 936. Stamping The Sequence 937. Reorder Data in Log Files 938. Range Sum of BST 939. Minimum Area Rectangle 940. Distinct Subsequences II 941. Valid Mountain Array 942. DI String Match 943. Find the Shortest Superstring 944. Delete Columns to Make Sorted 945. Minimum Increment to Make Array Unique 946. Validate Stack Sequences 947. Most Stones Removed with Same Row or Column 948. Bag of Tokens 949. Largest Time for Given Digits 950. Reveal Cards In Increasing Order 951. Flip Equivalent Binary Trees 952. Largest Component Size by Common Factor 953. Verifying an Alien Dictionary 954. Array of Doubled Pairs 955. Delete Columns to Make Sorted II 956. Tallest Billboard 957. Prison Cells After N Days 958. Check Completeness of a Binary Tree 959. Regions Cut By Slashes 960. Delete Columns to Make Sorted III 961. N-Repeated Element in Size 2N Array 962. Maximum Width Ramp 963. Minimum Area Rectangle II 964. Least Operators to Express Number 965. Univalued Binary Tree 966. Vowel Spellchecker 967. Numbers With Same Consecutive Differences 968. Binary Tree Cameras 969. Pancake Sorting 970. Powerful Integers 971. Flip Binary Tree To Match Preorder Traversal 972. Equal Rational Numbers 973. K Closest Points to Origin 974. Subarray Sums Divisible by K 975. Odd Even Jump 976. Largest Perimeter Triangle 977. Squares of a Sorted Array 978. Longest Turbulent Subarray 979. Distribute Coins in Binary Tree 980. Unique Paths III 981. Time Based Key-Value Store 982. Triples with Bitwise AND Equal To Zero 983. Minimum Cost For Tickets 984. String Without AAA or BBB 985. Sum of Even Numbers After Queries 986. Interval List Intersections 987. Vertical Order Traversal of a Binary Tree 988. Smallest String Starting From Leaf 989. Add to Array-Form of Integer 990. Satisfiability of Equality Equations 991. Broken Calculator 992. Subarrays with K Different Integers 993. Cousins in Binary Tree 994. Rotting Oranges 995. Minimum Number of K Consecutive Bit Flips 996. Number of Squareful Arrays 997. Find the Town Judge 998. Maximum Binary Tree II 999. Available Captures for Rook 1000. Minimum Cost to Merge Stones 1001. Grid Illumination 1002. Find Common Characters 1003. Check If Word Is Valid After Substitutions 1004. Max Consecutive Ones III 1005. Maximize Sum Of Array After K Negations 1006. Clumsy Factorial 1007. Minimum Domino Rotations For Equal Row 1008. Construct Binary Search Tree from Preorder Traversal 1009. Complement of Base 10 Integer 1010. Pairs of Songs With Total Durations Divisible by 60 1011. Capacity To Ship Packages Within D Days 1012. Numbers With Repeated Digits 1013. Partition Array Into Three Parts With Equal Sum 1014. Best Sightseeing Pair 1015. Smallest Integer Divisible by K 1016. Binary String With Substrings Representing 1 To N 1017. Convert to Base -2 1018. Binary Prefix Divisible By 5 1019. Next Greater Node In Linked List 1020. Number of Enclaves 1021. Remove Outermost Parentheses 1022. Sum of Root To Leaf Binary Numbers 1023. Camelcase Matching 1024. Video Stitching 1025. Divisor Game 1026. Maximum Difference Between Node and Ancestor 1027. Longest Arithmetic Subsequence 1028. Recover a Tree From Preorder Traversal 1029. Two City Scheduling 1030. Matrix Cells in Distance Order 1031. Maximum Sum of Two Non-Overlapping Subarrays 1032. Stream of Characters 1033. Moving Stones Until Consecutive 1034. Coloring A Border 1035. Uncrossed Lines 1036. Escape a Large Maze 1037. Valid Boomerang 1038. Binary Search Tree to Greater Sum Tree 1039. Minimum Score Triangulation of Polygon 1040. Moving Stones Until Consecutive II 1041. Robot Bounded In Circle 1042. Flower Planting With No Adjacent 1043. Partition Array for Maximum Sum 1044. Longest Duplicate Substring 1046. Last Stone Weight 1047. Remove All Adjacent Duplicates In String 1048. Longest String Chain 1049. Last Stone Weight II 1051. Height Checker 1052. Grumpy Bookstore Owner 1053. Previous Permutation With One Swap 1054. Distant Barcodes 1055. Shortest Way to Form String 1056. Confusing Number 1057. Campus Bikes 1058. Minimize Rounding Error to Meet Target 1059. All Paths from Source Lead to Destination 1060. Missing Element in Sorted Array 1061. Lexicographically Smallest Equivalent String 1062. Longest Repeating Substring 1063. Number of Valid Subarrays 1064. Fixed Point 1065. Index Pairs of a String 1066. Campus Bikes II 1067. Digit Count in Range 1071. Greatest Common Divisor of Strings 1072. Flip Columns For Maximum Number of Equal Rows 1073. Adding Two Negabinary Numbers 1074. Number of Submatrices That Sum to Target 1078. Occurrences After Bigram 1079. Letter Tile Possibilities 1080. Insufficient Nodes in Root to Leaf Paths 1081. Smallest Subsequence of Distinct Characters 1085. Sum of Digits in the Minimum Number 1086. High Five 1087. Brace Expansion 1088. Confusing Number II 1089. Duplicate Zeros 1090. Largest Values From Labels 1091. Shortest Path in Binary Matrix 1092. Shortest Common Supersequence 1093. Statistics from a Large Sample 1094. Car Pooling 1095. Find in Mountain Array 1096. Brace Expansion II 1099. Two Sum Less Than K 1100. Find K-Length Substrings With No Repeated Characters 1101. The Earliest Moment When Everyone Become Friends 1102. Path With Maximum Minimum Value 1103. Distribute Candies to People 1104. Path In Zigzag Labelled Binary Tree 1105. Filling Bookcase Shelves 1106. Parsing A Boolean Expression 1108. Defanging an IP Address 1109. Corporate Flight Bookings 1110. Delete Nodes And Return Forest 1111. Maximum Nesting Depth of Two Valid Parentheses Strings 1114. Print in Order 1115. Print FooBar Alternately 1116. Print Zero Even Odd 1117. Building H2O 1118. Number of Days in a Month 1119. Remove Vowels from a String 1120. Maximum Average Subtree 1121. Divide Array Into Increasing Sequences 1122. Relative Sort Array 1123. Lowest Common Ancestor of Deepest Leaves 1124. Longest Well-Performing Interval 1125. Smallest Sufficient Team 1128. Number of Equivalent Domino Pairs 1129. Shortest Path with Alternating Colors 1130. Minimum Cost Tree From Leaf Values 1131. Maximum of Absolute Value Expression 1133. Largest Unique Number 1134. Armstrong Number 1135. Connecting Cities With Minimum Cost 1136. Parallel Courses 1137. N-th Tribonacci Number 1138. Alphabet Board Path 1139. Largest 1-Bordered Square 1140. Stone Game II 1143. Longest Common Subsequence 1144. Decrease Elements To Make Array Zigzag 1145. Binary Tree Coloring Game 1146. Snapshot Array 1147. Longest Chunked Palindrome Decomposition 1150. Check If a Number Is Majority Element in a Sorted Array 1151. Minimum Swaps to Group All 1's Together 1152. Analyze User Website Visit Pattern 1153. String Transforms Into Another String 1154. Day of the Year 1155. Number of Dice Rolls With Target Sum 1156. Swap For Longest Repeated Character Substring 1157. Online Majority Element In Subarray 1160. Find Words That Can Be Formed by Characters 1161. Maximum Level Sum of a Binary Tree 1162. As Far from Land as Possible 1163. Last Substring in Lexicographical Order 1165. Single-Row Keyboard 1166. Design File System 1167. Minimum Cost to Connect Sticks 1168. Optimize Water Distribution in a Village 1169. Invalid Transactions 1170. Compare Strings by Frequency of the Smallest Character 1171. Remove Zero Sum Consecutive Nodes from Linked List 1172. Dinner Plate Stacks 1175. Prime Arrangements 1176. Diet Plan Performance 1177. Can Make Palindrome from Substring 1178. Number of Valid Words for Each Puzzle 1180. Count Substrings with Only One Distinct Letter 1181. Before and After Puzzle 1182. Shortest Distance to Target Color 1183. Maximum Number of Ones 1184. Distance Between Bus Stops 1185. Day of the Week 1186. Maximum Subarray Sum with One Deletion 1187. Make Array Strictly Increasing 1188. Design Bounded Blocking Queue 1189. Maximum Number of Balloons 1190. Reverse Substrings Between Each Pair of Parentheses 1191. K-Concatenation Maximum Sum 1192. Critical Connections in a Network 1195. Fizz Buzz Multithreaded 1196. How Many Apples Can You Put into the Basket 1197. Minimum Knight Moves 1198. Find Smallest Common Element in All Rows 1199. Minimum Time to Build Blocks 1200. Minimum Absolute Difference 1201. Ugly Number III 1202. Smallest String With Swaps 1203. Sort Items by Groups Respecting Dependencies 1206. Design Skiplist 1207. Unique Number of Occurrences 1208. Get Equal Substrings Within Budget 1209. Remove All Adjacent Duplicates in String II 1210. Minimum Moves to Reach Target with Rotations 1213. Intersection of Three Sorted Arrays 1214. Two Sum BSTs 1215. Stepping Numbers 1216. Valid Palindrome III 1217. Minimum Cost to Move Chips to The Same Position 1218. Longest Arithmetic Subsequence of Given Difference 1219. Path with Maximum Gold 1220. Count Vowels Permutation 1221. Split a String in Balanced Strings 1222. Queens That Can Attack the King 1223. Dice Roll Simulation 1224. Maximum Equal Frequency 1226. The Dining Philosophers 1227. Airplane Seat Assignment Probability 1228. Missing Number In Arithmetic Progression 1229. Meeting Scheduler 1230. Toss Strange Coins 1231. Divide Chocolate 1232. Check If It Is a Straight Line 1233. Remove Sub-Folders from the Filesystem 1234. Replace the Substring for Balanced String 1235. Maximum Profit in Job Scheduling 1236. Web Crawler 1237. Find Positive Integer Solution for a Given Equation 1238. Circular Permutation in Binary Representation 1239. Maximum Length of a Concatenated String with Unique Characters 1240. Tiling a Rectangle with the Fewest Squares 1242. Web Crawler Multithreaded 1243. Array Transformation 1244. Design A Leaderboard 1245. Tree Diameter 1246. Palindrome Removal 1247. Minimum Swaps to Make Strings Equal 1248. Count Number of Nice Subarrays 1249. Minimum Remove to Make Valid Parentheses 1250. Check If It Is a Good Array 1252. Cells with Odd Values in a Matrix 1253. Reconstruct a 2-Row Binary Matrix 1254. Number of Closed Islands 1255. Maximum Score Words Formed by Letters 1256. Encode Number 1257. Smallest Common Region 1258. Synonymous Sentences 1259. Handshakes That Don't Cross 1260. Shift 2D Grid 1261. Find Elements in a Contaminated Binary Tree 1262. Greatest Sum Divisible by Three 1263. Minimum Moves to Move a Box to Their Target Location 1265. Print Immutable Linked List in Reverse 1266. Minimum Time Visiting All Points 1267. Count Servers that Communicate 1268. Search Suggestions System 1269. Number of Ways to Stay in the Same Place After Some Steps 1271. Hexspeak 1272. Remove Interval 1273. Delete Tree Nodes 1274. Number of Ships in a Rectangle 1275. Find Winner on a Tic Tac Toe Game 1276. Number of Burgers with No Waste of Ingredients 1277. Count Square Submatrices with All Ones 1278. Palindrome Partitioning III 1279. Traffic Light Controlled Intersection 1281. Subtract the Product and Sum of Digits of an Integer 1282. Group the People Given the Group Size They Belong To 1283. Find the Smallest Divisor Given a Threshold 1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix 1286. Iterator for Combination 1287. Element Appearing More Than 25% In Sorted Array 1288. Remove Covered Intervals 1289. Minimum Falling Path Sum II 1290. Convert Binary Number in a Linked List to Integer 1291. Sequential Digits 1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold 1293. Shortest Path in a Grid with Obstacles Elimination 1295. Find Numbers with Even Number of Digits 1296. Divide Array in Sets of K Consecutive Numbers 1297. Maximum Number of Occurrences of a Substring 1298. Maximum Candies You Can Get from Boxes 1299. Replace Elements with Greatest Element on Right Side 1300. Sum of Mutated Array Closest to Target 1301. Number of Paths with Max Score 1302. Deepest Leaves Sum 1304. Find N Unique Integers Sum up to Zero 1305. All Elements in Two Binary Search Trees 1306. Jump Game III 1307. Verbal Arithmetic Puzzle 1309. Decrypt String from Alphabet to Integer Mapping 1310. XOR Queries of a Subarray 1311. Get Watched Videos by Your Friends 1312. Minimum Insertion Steps to Make a String Palindrome 1313. Decompress Run-Length Encoded List 1314. Matrix Block Sum 1315. Sum of Nodes with Even-Valued Grandparent 1316. Distinct Echo Substrings 1317. Convert Integer to the Sum of Two No-Zero Integers 1318. Minimum Flips to Make a OR b Equal to c 1319. Number of Operations to Make Network Connected 1320. Minimum Distance to Type a Word Using Two Fingers 1323. Maximum 69 Number 1324. Print Words Vertically 1325. Delete Leaves With a Given Value 1326. Minimum Number of Taps to Open to Water a Garden 1328. Break a Palindrome 1329. Sort the Matrix Diagonally 1330. Reverse Subarray To Maximize Array Value 1331. Rank Transform of an Array 1332. Remove Palindromic Subsequences 1333. Filter Restaurants by Vegan-Friendly, Price and Distance 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance 1335. Minimum Difficulty of a Job Schedule 1337. The K Weakest Rows in a Matrix 1338. Reduce Array Size to The Half 1339. Maximum Product of Splitted Binary Tree 1340. Jump Game V 1342. Number of Steps to Reduce a Number to Zero 1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold 1344. Angle Between Hands of a Clock 1345. Jump Game IV 1346. Check If N and Its Double Exist 1347. Minimum Number of Steps to Make Two Strings Anagram 1348. Tweet Counts Per Frequency 1349. Maximum Students Taking Exam 1351. Count Negative Numbers in a Sorted Matrix 1352. Product of the Last K Numbers 1353. Maximum Number of Events That Can Be Attended 1354. Construct Target Array With Multiple Sums 1356. Sort Integers by The Number of 1 Bits 1357. Apply Discount Every n Orders 1358. Number of Substrings Containing All Three Characters 1359. Count All Valid Pickup and Delivery Options 1360. Number of Days Between Two Dates 1361. Validate Binary Tree Nodes 1362. Closest Divisors 1363. Largest Multiple of Three 1365. How Many Numbers Are Smaller Than the Current Number 1366. Rank Teams by Votes 1367. Linked List in Binary Tree 1368. Minimum Cost to Make at Least One Valid Path in a Grid 1370. Increasing Decreasing String 1371. Find the Longest Substring Containing Vowels in Even Counts 1372. Longest ZigZag Path in a Binary Tree 1373. Maximum Sum BST in Binary Tree 1374. Generate a String With Characters That Have Odd Counts 1375. Number of Times Binary String Is Prefix-Aligned 1376. Time Needed to Inform All Employees 1377. Frog Position After T Seconds 1379. Find a Corresponding Node of a Binary Tree in a Clone of That Tree 1380. Lucky Numbers in a Matrix 1381. Design a Stack With Increment Operation 1382. Balance a Binary Search Tree 1383. Maximum Performance of a Team 1385. Find the Distance Value Between Two Arrays 1386. Cinema Seat Allocation 1387. Sort Integers by The Power Value 1388. Pizza With 3n Slices 1389. Create Target Array in the Given Order 1390. Four Divisors 1391. Check if There is a Valid Path in a Grid 1392. Longest Happy Prefix 1394. Find Lucky Integer in an Array 1395. Count Number of Teams 1396. Design Underground System 1397. Find All Good Strings 1399. Count Largest Group 1400. Construct K Palindrome Strings 1401. Circle and Rectangle Overlapping 1402. Reducing Dishes 1403. Minimum Subsequence in Non-Increasing Order 1404. Number of Steps to Reduce a Number in Binary Representation to One 1405. Longest Happy String 1406. Stone Game III 1408. String Matching in an Array 1409. Queries on a Permutation With Key 1410. HTML Entity Parser 1411. Number of Ways to Paint N × 3 Grid 1413. Minimum Value to Get Positive Step by Step Sum 1414. Find the Minimum Number of Fibonacci Numbers Whose Sum Is K 1415. The k-th Lexicographical String of All Happy Strings of Length n 1416. Restore The Array 1417. Reformat The String 1418. Display Table of Food Orders in a Restaurant 1419. Minimum Number of Frogs Croaking 1420. Build Array Where You Can Find The Maximum Exactly K Comparisons 1422. Maximum Score After Splitting a String 1423. Maximum Points You Can Obtain from Cards 1424. Diagonal Traverse II 1425. Constrained Subsequence Sum 1426. Counting Elements 1427. Perform String Shifts 1428. Leftmost Column with at Least a One 1429. First Unique Number 1430. Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree 1431. Kids With the Greatest Number of Candies 1432. Max Difference You Can Get From Changing an Integer 1433. Check If a String Can Break Another String 1434. Number of Ways to Wear Different Hats to Each Other 1436. Destination City 1437. Check If All 1's Are at Least Length K Places Away 1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit 1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows 1441. Build an Array With Stack Operations 1442. Count Triplets That Can Form Two Arrays of Equal XOR 1443. Minimum Time to Collect All Apples in a Tree 1444. Number of Ways of Cutting a Pizza 1446. Consecutive Characters 1447. Simplified Fractions 1448. Count Good Nodes in Binary Tree 1449. Form Largest Integer With Digits That Add up to Target 1450. Number of Students Doing Homework at a Given Time 1451. Rearrange Words in a Sentence 1452. People Whose List of Favorite Companies Is Not a Subset of Another List 1453. Maximum Number of Darts Inside of a Circular Dartboard 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence 1456. Maximum Number of Vowels in a Substring of Given Length 1457. Pseudo-Palindromic Paths in a Binary Tree 1458. Max Dot Product of Two Subsequences 1460. Make Two Arrays Equal by Reversing Subarrays 1461. Check If a String Contains All Binary Codes of Size K 1462. Course Schedule IV 1463. Cherry Pickup II 1464. Maximum Product of Two Elements in an Array 1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts 1466. Reorder Routes to Make All Paths Lead to the City Zero 1467. Probability of a Two Boxes Having The Same Number of Distinct Balls 1469. Find All The Lonely Nodes 1470. Shuffle the Array 1471. The k Strongest Values in an Array 1472. Design Browser History 1473. Paint House III 1474. Delete N Nodes After M Nodes of a Linked List 1475. Final Prices With a Special Discount in a Shop 1476. Subrectangle Queries 1477. Find Two Non-overlapping Sub-arrays Each With Target Sum 1478. Allocate Mailboxes 1480. Running Sum of 1d Array 1481. Least Number of Unique Integers after K Removals 1482. Minimum Number of Days to Make m Bouquets 1483. Kth Ancestor of a Tree Node 1485. Clone Binary Tree With Random Pointer 1486. XOR Operation in an Array 1487. Making File Names Unique 1488. Avoid Flood in The City 1489. Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree 1490. Clone N-ary Tree 1491. Average Salary Excluding the Minimum and Maximum Salary 1492. The kth Factor of n 1493. Longest Subarray of 1's After Deleting One Element 1494. Parallel Courses II 1496. Path Crossing 1497. Check If Array Pairs Are Divisible by k 1498. Number of Subsequences That Satisfy the Given Sum Condition 1499. Max Value of Equation 1500. Design a File Sharing System 1502. Can Make Arithmetic Progression From Sequence 1503. Last Moment Before All Ants Fall Out of a Plank 1504. Count Submatrices With All Ones 1505. Minimum Possible Integer After at Most K Adjacent Swaps On Digits 1506. Find Root of N-Ary Tree 1507. Reformat Date 1508. Range Sum of Sorted Subarray Sums 1509. Minimum Difference Between Largest and Smallest Value in Three Moves 1510. Stone Game IV 1512. Number of Good Pairs 1513. Number of Substrings With Only 1s 1514. Path with Maximum Probability 1515. Best Position for a Service Centre 1516. Move Sub-Tree of N-Ary Tree 1518. Water Bottles 1519. Number of Nodes in the Sub-Tree With the Same Label 1520. Maximum Number of Non-Overlapping Substrings 1521. Find a Value of a Mysterious Function Closest to Target 1522. Diameter of N-Ary Tree 1523. Count Odd Numbers in an Interval Range 1524. Number of Sub-arrays With Odd Sum 1525. Number of Good Ways to Split a String 1526. Minimum Number of Increments on Subarrays to Form a Target Array 1528. Shuffle String 1529. Minimum Suffix Flips 1530. Number of Good Leaf Nodes Pairs 1531. String Compression II 1533. Find the Index of the Large Integer 1534. Count Good Triplets 1535. Find the Winner of an Array Game 1536. Minimum Swaps to Arrange a Binary Grid 1537. Get the Maximum Score 1538. Guess the Majority in a Hidden Array 1539. Kth Missing Positive Number 1540. Can Convert String in K Moves 1541. Minimum Insertions to Balance a Parentheses String 1542. Find Longest Awesome Substring 1544. Make The String Great 1545. Find Kth Bit in Nth Binary String 1546. Maximum Number of Non-Overlapping Subarrays With Sum Equals Target 1547. Minimum Cost to Cut a Stick 1548. The Most Similar Path in a Graph 1550. Three Consecutive Odds 1551. Minimum Operations to Make Array Equal 1552. Magnetic Force Between Two Balls 1553. Minimum Number of Days to Eat N Oranges 1554. Strings Differ by One Character 1556. Thousand Separator 1557. Minimum Number of Vertices to Reach All Nodes 1558. Minimum Numbers of Function Calls to Make Target Array 1559. Detect Cycles in 2D Grid 1560. Most Visited Sector in a Circular Track 1561. Maximum Number of Coins You Can Get 1562. Find Latest Group of Size M 1563. Stone Game V 1564. Put Boxes Into the Warehouse I 1566. Detect Pattern of Length M Repeated K or More Times 1567. Maximum Length of Subarray With Positive Product 1568. Minimum Number of Days to Disconnect Island 1569. Number of Ways to Reorder Array to Get Same BST 1570. Dot Product of Two Sparse Vectors 1572. Matrix Diagonal Sum 1573. Number of Ways to Split a String 1574. Shortest Subarray to be Removed to Make Array Sorted 1575. Count All Possible Routes 1576. Replace All 's to Avoid Consecutive Repeating Characters 1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers 1578. Minimum Time to Make Rope Colorful 1579. Remove Max Number of Edges to Keep Graph Fully Traversable 1580. Put Boxes Into the Warehouse II 1582. Special Positions in a Binary Matrix 1583. Count Unhappy Friends 1584. Min Cost to Connect All Points 1585. Check If String Is Transformable With Substring Sort Operations 1588. Sum of All Odd Length Subarrays 1589. Maximum Sum Obtained of Any Permutation 1590. Make Sum Divisible by P 1591. Strange Printer II 1592. Rearrange Spaces Between Words 1593. Split a String Into the Max Number of Unique Substrings 1594. Maximum Non Negative Product in a Matrix 1595. Minimum Cost to Connect Two Groups of Points 1597. Build Binary Expression Tree From Infix Expression 1598. Crawler Log Folder 1599. Maximum Profit of Operating a Centennial Wheel 1600. Throne Inheritance 1601. Maximum Number of Achievable Transfer Requests 1602. Find Nearest Right Node in Binary Tree 1603. Design Parking System 1604. Alert Using Same Key-Card Three or More Times in a One Hour Period 1605. Find Valid Matrix Given Row and Column Sums 1606. Find Servers That Handled Most Number of Requests 1608. Special Array With X Elements Greater Than or Equal X 1609. Even Odd Tree 1610. Maximum Number of Visible Points 1611. Minimum One Bit Operations to Make Integers Zero 1612. Check If Two Expression Trees are Equivalent 1614. Maximum Nesting Depth of the Parentheses 1615. Maximal Network Rank 1616. Split Two Strings to Make Palindrome 1617. Count Subtrees With Max Distance Between Cities 1618. Maximum Font to Fit a Sentence in a Screen 1619. Mean of Array After Removing Some Elements 1620. Coordinate With Maximum Network Quality 1621. Number of Sets of K Non-Overlapping Line Segments 1622. Fancy Sequence 1624. Largest Substring Between Two Equal Characters 1625. Lexicographically Smallest String After Applying Operations 1626. Best Team With No Conflicts 1627. Graph Connectivity With Threshold 1628. Design an Expression Tree With Evaluate Function 1629. Slowest Key 1630. Arithmetic Subarrays 1631. Path With Minimum Effort 1632. Rank Transform of a Matrix 1634. Add Two Polynomials Represented as Linked Lists 1636. Sort Array by Increasing Frequency 1637. Widest Vertical Area Between Two Points Containing No Points 1638. Count Substrings That Differ by One Character 1639. Number of Ways to Form a Target String Given a Dictionary 1640. Check Array Formation Through Concatenation 1641. Count Sorted Vowel Strings 1642. Furthest Building You Can Reach 1643. Kth Smallest Instructions 1644. Lowest Common Ancestor of a Binary Tree II 1646. Get Maximum in Generated Array 1647. Minimum Deletions to Make Character Frequencies Unique 1648. Sell Diminishing-Valued Colored Balls 1649. Create Sorted Array through Instructions 1650. Lowest Common Ancestor of a Binary Tree III 1652. Defuse the Bomb 1653. Minimum Deletions to Make String Balanced 1654. Minimum Jumps to Reach Home 1655. Distribute Repeating Integers 1656. Design an Ordered Stream 1657. Determine if Two Strings Are Close 1658. Minimum Operations to Reduce X to Zero 1659. Maximize Grid Happiness 1660. Correct a Binary Tree 1662. Check If Two String Arrays are Equivalent 1663. Smallest String With A Given Numeric Value 1664. Ways to Make a Fair Array 1665. Minimum Initial Energy to Finish Tasks 1666. Change the Root of a Binary Tree 1668. Maximum Repeating Substring 1669. Merge In Between Linked Lists 1670. Design Front Middle Back Queue 1671. Minimum Number of Removals to Make Mountain Array 1672. Richest Customer Wealth 1673. Find the Most Competitive Subsequence 1674. Minimum Moves to Make Array Complementary 1675. Minimize Deviation in Array 1676. Lowest Common Ancestor of a Binary Tree IV 1678. Goal Parser Interpretation 1679. Max Number of K-Sum Pairs 1680. Concatenation of Consecutive Binary Numbers 1681. Minimum Incompatibility 1682. Longest Palindromic Subsequence II 1684. Count the Number of Consistent Strings 1685. Sum of Absolute Differences in a Sorted Array 1686. Stone Game VI 1687. Delivering Boxes from Storage to Ports 1688. Count of Matches in Tournament 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers 1690. Stone Game VII 1691. Maximum Height by Stacking Cuboids 1692. Count Ways to Distribute Candies 1694. Reformat Phone Number 1695. Maximum Erasure Value 1696. Jump Game VI 1697. Checking Existence of Edge Length Limited Paths 1698. Number of Distinct Substrings in a String 1700. Number of Students Unable to Eat Lunch 1701. Average Waiting Time 1702. Maximum Binary String After Change 1703. Minimum Adjacent Swaps for K Consecutive Ones 1704. Determine if String Halves Are Alike 1705. Maximum Number of Eaten Apples 1706. Where Will the Ball Fall 1707. Maximum XOR With an Element From Array 1708. Largest Subarray Length K 1709. Biggest Window Between Visits 1710. Maximum Units on a Truck 1711. Count Good Meals 1712. Ways to Split Array Into Three Subarrays 1713. Minimum Operations to Make a Subsequence 1714. Sum Of Special Evenly-Spaced Elements In Array 1716. Calculate Money in Leetcode Bank 1717. Maximum Score From Removing Substrings 1718. Construct the Lexicographically Largest Valid Sequence 1719. Number Of Ways To Reconstruct A Tree 1720. Decode XORed Array 1721. Swapping Nodes in a Linked List 1722. Minimize Hamming Distance After Swap Operations 1723. Find Minimum Time to Finish All Jobs 1724. Checking Existence of Edge Length Limited Paths II 1725. Number Of Rectangles That Can Form The Largest Square 1726. Tuple with Same Product 1727. Largest Submatrix With Rearrangements 1728. Cat and Mouse II 1730. Shortest Path to Get Food 1732. Find the Highest Altitude 1733. Minimum Number of People to Teach 1734. Decode XORed Permutation 1735. Count Ways to Make Array With Product 1736. Latest Time by Replacing Hidden Digits 1737. Change Minimum Characters to Satisfy One of Three Conditions 1738. Find Kth Largest XOR Coordinate Value 1739. Building Boxes 1740. Find Distance in a Binary Tree 1742. Maximum Number of Balls in a Box 1743. Restore the Array From Adjacent Pairs 1744. Can You Eat Your Favorite Candy on Your Favorite Day 1745. Palindrome Partitioning IV 1746. Maximum Subarray Sum After One Operation 1748. Sum of Unique Elements 1749. Maximum Absolute Sum of Any Subarray 1750. Minimum Length of String After Deleting Similar Ends 1751. Maximum Number of Events That Can Be Attended II 1752. Check if Array Is Sorted and Rotated 1753. Maximum Score From Removing Stones 1754. Largest Merge Of Two Strings 1755. Closest Subsequence Sum 1756. Design Most Recently Used Queue 1758. Minimum Changes To Make Alternating Binary String 1759. Count Number of Homogenous Substrings 1760. Minimum Limit of Balls in a Bag 1761. Minimum Degree of a Connected Trio in a Graph 1762. Buildings With an Ocean View 1763. Longest Nice Substring 1764. Form Array by Concatenating Subarrays of Another Array 1765. Map of Highest Peak 1766. Tree of Coprimes 1768. Merge Strings Alternately 1769. Minimum Number of Operations to Move All Balls to Each Box 1770. Maximum Score from Performing Multiplication Operations 1771. Maximize Palindrome Length From Subsequences 1772. Sort Features by Popularity 1773. Count Items Matching a Rule 1774. Closest Dessert Cost 1775. Equal Sum Arrays With Minimum Number of Operations 1776. Car Fleet II 1778. Shortest Path in a Hidden Grid 1779. Find Nearest Point That Has the Same X or Y Coordinate 1780. Check if Number is a Sum of Powers of Three 1781. Sum of Beauty of All Substrings 1782. Count Pairs Of Nodes 1784. Check if Binary String Has at Most One Segment of Ones 1785. Minimum Elements to Add to Form a Given Sum 1786. Number of Restricted Paths From First to Last Node 1787. Make the XOR of All Segments Equal to Zero 1788. Maximize the Beauty of the Garden 1790. Check if One String Swap Can Make Strings Equal 1791. Find Center of Star Graph 1792. Maximum Average Pass Ratio 1793. Maximum Score of a Good Subarray 1794. Count Pairs of Equal Substrings With Minimum Difference 1796. Second Largest Digit in a String 1797. Design Authentication Manager 1798. Maximum Number of Consecutive Values You Can Make 1799. Maximize Score After N Operations 1800. Maximum Ascending Subarray Sum 1801. Number of Orders in the Backlog 1802. Maximum Value at a Given Index in a Bounded Array 1803. Count Pairs With XOR in a Range 1804. Implement Trie II (Prefix Tree) 1805. Number of Different Integers in a String 1806. Minimum Number of Operations to Reinitialize a Permutation 1807. Evaluate the Bracket Pairs of a String 1808. Maximize Number of Nice Divisors 1810. Minimum Path Cost in a Hidden Grid 1811. Find Interview Candidates 1812. Determine Color of a Chessboard Square 1813. Sentence Similarity III 1814. Count Nice Pairs in an Array 1815. Maximum Number of Groups Getting Fresh Donuts 1816. Truncate Sentence 1817. Finding the Users Active Minutes 1818. Minimum Absolute Sum Difference 1819. Number of Different Subsequences GCDs 1820. Maximum Number of Accepted Invitations 1821. Find Customers With Positive Revenue this Year 1822. Sign of the Product of an Array 1823. Find the Winner of the Circular Game 1824. Minimum Sideway Jumps 1825. Finding MK Average 1826. Faulty Sensor 1827. Minimum Operations to Make the Array Increasing 1828. Queries on Number of Points Inside a Circle 1829. Maximum XOR for Each Query 1830. Minimum Number of Operations to Make String Sorted 1831. Maximum Transaction Each Day 1832. Check if the Sentence Is Pangram 1833. Maximum Ice Cream Bars 1834. Single-Threaded CPU 1835. Find XOR Sum of All Pairs Bitwise AND 1836. Remove Duplicates From an Unsorted Linked List 1837. Sum of Digits in Base K 1838. Frequency of the Most Frequent Element 1839. Longest Substring Of All Vowels in Order 1840. Maximum Building Height 1841. League Statistics 1842. Next Palindrome Using Same Digits 1844. Replace All Digits with Characters 1845. Seat Reservation Manager 1846. Maximum Element After Decreasing and Rearranging 1847. Closest Room 1848. Minimum Distance to the Target Element 1849. Splitting a String Into Descending Consecutive Values 1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number 1851. Minimum Interval to Include Each Query 1852. Distinct Numbers in Each Subarray 1853. Convert Date Format 1854. Maximum Population Year 1855. Maximum Distance Between a Pair of Values 1856. Maximum Subarray Min-Product 1857. Largest Color Value in a Directed Graph 1858. Longest Word With All Prefixes 1859. Sorting the Sentence 1860. Incremental Memory Leak 1861. Rotating the Box 1862. Sum of Floored Pairs 1863. Sum of All Subset XOR Totals 1864. Minimum Number of Swaps to Make the Binary String Alternating 1865. Finding Pairs With a Certain Sum 1866. Number of Ways to Rearrange Sticks With K Sticks Visible 1868. Product of Two Run-Length Encoded Arrays 1869. Longer Contiguous Segments of Ones than Zeros 1870. Minimum Speed to Arrive on Time 1871. Jump Game VII 1872. Stone Game VIII 1874. Minimize Product Sum of Two Arrays 1876. Substrings of Size Three with Distinct Characters 1877. Minimize Maximum Pair Sum in Array 1878. Get Biggest Three Rhombus Sums in a Grid 1879. Minimum XOR Sum of Two Arrays 1880. Check if Word Equals Summation of Two Words 1881. Maximum Value after Insertion 1882. Process Tasks Using Servers 1883. Minimum Skips to Arrive at Meeting On Time 1884. Egg Drop With 2 Eggs and N Floors 1885. Count Pairs in Two Arrays 1886. Determine Whether Matrix Can Be Obtained By Rotation 1887. Reduction Operations to Make the Array Elements Equal 1888. Minimum Number of Flips to Make the Binary String Alternating 1889. Minimum Space Wasted From Packaging 1891. Cutting Ribbons 1893. Check if All the Integers in a Range Are Covered 1894. Find the Student that Will Replace the Chalk 1895. Largest Magic Square 1896. Minimum Cost to Change the Final Value of Expression 1897. Redistribute Characters to Make All Strings Equal 1898. Maximum Number of Removable Characters 1899. Merge Triplets to Form Target Triplet 1900. The Earliest and Latest Rounds Where Players Compete 1901. Find a Peak Element II 1902. Depth of BST Given Insertion Order 1903. Largest Odd Number in String 1904. The Number of Full Rounds You Have Played 1905. Count Sub Islands 1906. Minimum Absolute Difference Queries 1908. Game of Nim 1909. Remove One Element to Make the Array Strictly Increasing 1910. Remove All Occurrences of a Substring 1911. Maximum Alternating Subsequence Sum 1912. Design Movie Rental System 1913. Maximum Product Difference Between Two Pairs 1914. Cyclically Rotating a Grid 1915. Number of Wonderful Substrings 1918. Kth Smallest Subarray Sum 1920. Build Array from Permutation 1921. Eliminate Maximum Number of Monsters 1922. Count Good Numbers 1923. Longest Common Subpath 1924. Erect the Fence II 1925. Count Square Sum Triples 1926. Nearest Exit from Entrance in Maze 1927. Sum Game 1928. Minimum Cost to Reach Destination in Time 1929. Concatenation of Array 1930. Unique Length-3 Palindromic Subsequences 1931. Painting a Grid With Three Different Colors 1932. Merge BSTs to Create Single BST 1933. Check if String Is Decomposable Into Value-Equal Substrings 1935. Maximum Number of Words You Can Type 1936. Add Minimum Number of Rungs 1937. Maximum Number of Points with Cost 1938. Maximum Genetic Difference Query 1940. Longest Common Subsequence Between Sorted Arrays 1941. Check if All Characters Have Equal Number of Occurrences 1942. The Number of the Smallest Unoccupied Chair 1943. Describe the Painting 1944. Number of Visible People in a Queue 1945. Sum of Digits of String After Convert 1946. Largest Number After Mutating Substring 1947. Maximum Compatibility Score Sum 1948. Delete Duplicate Folders in System 1950. Maximum of Minimum Values in All Subarrays 1952. Three Divisors 1953. Maximum Number of Weeks for Which You Can Work 1954. Minimum Garden Perimeter to Collect Enough Apples 1955. Count Number of Special Subsequences 1956. Minimum Time For K Virus Variants to Spread 1957. Delete Characters to Make Fancy String 1958. Check if Move is Legal 1959. Minimum Total Space Wasted With K Resizing Operations 1960. Maximum Product of the Length of Two Palindromic Substrings 1961. Check If String Is a Prefix of Array 1962. Remove Stones to Minimize the Total 1963. Minimum Number of Swaps to Make the String Balanced 1964. Find the Longest Valid Obstacle Course at Each Position 1966. Binary Searchable Numbers in an Unsorted Array 1967. Number of Strings That Appear as Substrings in Word 1968. Array With Elements Not Equal to Average of Neighbors 1969. Minimum Non-Zero Product of the Array Elements 1970. Last Day Where You Can Still Cross 1971. Find if Path Exists in Graph 1973. Count Nodes Equal to Sum of Descendants 1974. Minimum Time to Type Word Using Special Typewriter 1975. Maximum Matrix Sum 1976. Number of Ways to Arrive at Destination 1977. Number of Ways to Separate Numbers 1979. Find Greatest Common Divisor of Array 1980. Find Unique Binary String 1981. Minimize the Difference Between Target and Chosen Elements 1982. Find Array Given Subset Sums 1983. Widest Pair of Indices With Equal Range Sum 1984. Minimum Difference Between Highest and Lowest of K Scores 1985. Find the Kth Largest Integer in the Array 1986. Minimum Number of Work Sessions to Finish the Tasks 1987. Number of Unique Good Subsequences 1989. Maximum Number of People That Can Be Caught in Tag 1991. Find the Middle Index in Array 1992. Find All Groups of Farmland 1993. Operations on Tree 1994. The Number of Good Subsets 1995. Count Special Quadruplets 1996. The Number of Weak Characters in the Game 1997. First Day Where You Have Been in All the Rooms 1998. GCD Sort of an Array 1999. Smallest Greater Multiple Made of Two Digits 2000. Reverse Prefix of Word 2001. Number of Pairs of Interchangeable Rectangles 2002. Maximum Product of the Length of Two Palindromic Subsequences 2003. Smallest Missing Genetic Value in Each Subtree 2005. Subtree Removal Game with Fibonacci Tree 2006. Count Number of Pairs With Absolute Difference K 2007. Find Original Array From Doubled Array 2008. Maximum Earnings From Taxi 2009. Minimum Number of Operations to Make Array Continuous 2011. Final Value of Variable After Performing Operations 2012. Sum of Beauty in the Array 2013. Detect Squares 2014. Longest Subsequence Repeated k Times 2015. Average Height of Buildings in Each Segment 2016. Maximum Difference Between Increasing Elements 2017. Grid Game 2018. Check if Word Can Be Placed In Crossword 2019. The Score of Students Solving Math Expression 2021. Brightest Position on Street 2022. Convert 1D Array Into 2D Array 2023. Number of Pairs of Strings With Concatenation Equal to Target 2024. Maximize the Confusion of an Exam 2025. Maximum Number of Ways to Partition an Array 2027. Minimum Moves to Convert String 2028. Find Missing Observations 2029. Stone Game IX 2030. Smallest K Length Subsequence With Occurrences of a Letter 2031. Count Subarrays With More Ones Than Zeros 2032. Two Out of Three 2033. Minimum Operations to Make a Uni-Value Grid 2034. Stock Price Fluctuation 2035. Partition Array Into Two Arrays to Minimize Sum Difference 2036. Maximum Alternating Subarray Sum 2037. Minimum Number of Moves to Seat Everyone 2038. Remove Colored Pieces if Both Neighbors are the Same Color 2039. The Time When the Network Becomes Idle 2040. Kth Smallest Product of Two Sorted Arrays 2042. Check if Numbers Are Ascending in a Sentence 2043. Simple Bank System 2044. Count Number of Maximum Bitwise-OR Subsets 2045. Second Minimum Time to Reach Destination 2046. Sort Linked List Already Sorted Using Absolute Values 2047. Number of Valid Words in a Sentence 2048. Next Greater Numerically Balanced Number 2049. Count Nodes With the Highest Score 2050. Parallel Courses III 2052. Minimum Cost to Separate Sentence Into Rows 2053. Kth Distinct String in an Array 2054. Two Best Non-Overlapping Events 2055. Plates Between Candles 2056. Number of Valid Move Combinations On Chessboard 2057. Smallest Index With Equal Value 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points 2059. Minimum Operations to Convert Number 2060. Check if an Original String Exists Given Two Encoded Strings 2061. Number of Spaces Cleaning Robot Cleaned 2062. Count Vowel Substrings of a String 2063. Vowels of All Substrings 2064. Minimized Maximum of Products Distributed to Any Store 2065. Maximum Path Quality of a Graph 2067. Number of Equal Count Substrings 2068. Check Whether Two Strings are Almost Equivalent 2069. Walking Robot Simulation II 2070. Most Beautiful Item for Each Query 2071. Maximum Number of Tasks You Can Assign 2073. Time Needed to Buy Tickets 2074. Reverse Nodes in Even Length Groups 2075. Decode the Slanted Ciphertext 2076. Process Restricted Friend Requests 2077. Paths in Maze That Lead to Same Room 2078. Two Furthest Houses With Different Colors 2079. Watering Plants 2080. Range Frequency Queries 2081. Sum of k-Mirror Numbers 2083. Substrings That Begin and End With the Same Letter 2085. Count Common Words With One Occurrence 2086. Minimum Number of Food Buckets to Feed the Hamsters 2087. Minimum Cost Homecoming of a Robot in a Grid 2088. Count Fertile Pyramids in a Land 2089. Find Target Indices After Sorting Array 2090. K Radius Subarray Averages 2091. Removing Minimum and Maximum From Array 2092. Find All People With Secret 2093. Minimum Cost to Reach City With Discounts 2094. Finding 3-Digit Even Numbers 2095. Delete the Middle Node of a Linked List 2096. Step-By-Step Directions From a Binary Tree Node to Another 2097. Valid Arrangement of Pairs 2098. Subsequence of Size K With the Largest Even Sum 2099. Find Subsequence of Length K With the Largest Sum 2100. Find Good Days to Rob the Bank 2101. Detonate the Maximum Bombs 2102. Sequentially Ordinal Rank Tracker 2103. Rings and Rods 2104. Sum of Subarray Ranges 2105. Watering Plants II 2106. Maximum Fruits Harvested After at Most K Steps 2107. Number of Unique Flavors After Sharing K Candies 2108. Find First Palindromic String in the Array 2109. Adding Spaces to a String 2110. Number of Smooth Descent Periods of a Stock 2111. Minimum Operations to Make the Array K-Increasing 2113. Elements in Array After Removing and Replacing Elements 2114. Maximum Number of Words Found in Sentences 2115. Find All Possible Recipes from Given Supplies 2116. Check if a Parentheses String Can Be Valid 2117. Abbreviating the Product of a Range 2119. A Number After a Double Reversal 2120. Execution of All Suffix Instructions Staying in a Grid 2121. Intervals Between Identical Elements 2122. Recover the Original Array 2123. Minimum Operations to Remove Adjacent Ones in Matrix 2124. Check if All A's Appears Before All B's 2125. Number of Laser Beams in a Bank 2126. Destroying Asteroids 2127. Maximum Employees to Be Invited to a Meeting 2128. Remove All Ones With Row and Column Flips 2129. Capitalize the Title 2130. Maximum Twin Sum of a Linked List 2131. Longest Palindrome by Concatenating Two Letter Words 2132. Stamping the Grid 2133. Check if Every Row and Column Contains All Numbers 2134. Minimum Swaps to Group All 1's Together II 2135. Count Words Obtained After Adding a Letter 2136. Earliest Possible Day of Full Bloom 2137. Pour Water Between Buckets to Make Water Levels Equal 2138. Divide a String Into Groups of Size k 2139. Minimum Moves to Reach Target Score 2140. Solving Questions With Brainpower 2141. Maximum Running Time of N Computers 2143. Choose Numbers From Two Arrays in Range 2144. Minimum Cost of Buying Candies With Discount 2145. Count the Hidden Sequences 2146. K Highest Ranked Items Within a Price Range 2147. Number of Ways to Divide a Long Corridor 2148. Count Elements With Strictly Smaller and Greater Elements 2149. Rearrange Array Elements by Sign 2150. Find All Lonely Numbers in the Array 2151. Maximum Good People Based on Statements 2152. Minimum Number of Lines to Cover Points 2154. Keep Multiplying Found Values by Two 2155. All Divisions With the Highest Score of a Binary Array 2156. Find Substring With Given Hash Value 2157. Groups of Strings 2158. Amount of New Area Painted Each Day 2160. Minimum Sum of Four Digit Number After Splitting Digits 2161. Partition Array According to Given Pivot 2162. Minimum Cost to Set Cooking Time 2163. Minimum Difference in Sums After Removal of Elements 2164. Sort Even and Odd Indices Independently 2165. Smallest Value of the Rearranged Number 2166. Design Bitset 2167. Minimum Time to Remove All Cars Containing Illegal Goods 2168. Unique Substrings With Equal Digit Frequency 2169. Count Operations to Obtain Zero 2170. Minimum Operations to Make the Array Alternating 2171. Removing Minimum Number of Magic Beans 2172. Maximum AND Sum of Array 2174. Remove All Ones With Row and Column Flips II 2176. Count Equal and Divisible Pairs in an Array 2177. Find Three Consecutive Integers That Sum to a Given Number 2178. Maximum Split of Positive Even Integers 2179. Count Good Triplets in an Array 2180. Count Integers With Even Digit Sum 2181. Merge Nodes in Between Zeros 2182. Construct String With Repeat Limit 2183. Count Array Pairs Divisible by K 2184. Number of Ways to Build Sturdy Brick Wall 2185. Counting Words With a Given Prefix 2186. Minimum Number of Steps to Make Two Strings Anagram II 2187. Minimum Time to Complete Trips 2188. Minimum Time to Finish the Race 2189. Number of Ways to Build House of Cards 2190. Most Frequent Number Following Key In an Array 2191. Sort the Jumbled Numbers 2192. All Ancestors of a Node in a Directed Acyclic Graph 2193. Minimum Number of Moves to Make Palindrome 2194. Cells in a Range on an Excel Sheet 2195. Append K Integers With Minimal Sum 2196. Create Binary Tree From Descriptions 2197. Replace Non Coprime Numbers in Array 2198. Number of Single Divisor Triplets 2200. Find All K-Distant Indices in an Array 2201. Count Artifacts That Can Be Extracted 2202. Maximize the Topmost Element After K Moves 2203. Minimum Weighted Subgraph With the Required Paths 2204. Distance to a Cycle in Undirected Graph 2206. Divide Array Into Equal Pairs 2207. Maximize Number of Subsequences in a String 2208. Minimum Operations to Halve Array Sum 2209. Minimum White Tiles After Covering With Carpets 2210. Count Hills and Valleys in an Array 2211. Count Collisions on a Road 2212. Maximum Points in an Archery Competition 2213. Longest Substring of One Repeating Character 2214. Minimum Health to Beat Game 2215. Find the Difference of Two Arrays 2216. Minimum Deletions to Make Array Beautiful 2217. Find Palindrome With Fixed Length 2218. Maximum Value of K Coins From Piles 2219. Maximum Sum Score of Array 2220. Minimum Bit Flips to Convert Number 2221. Find Triangular Sum of an Array 2222. Number of Ways to Select Buildings 2223. Sum of Scores of Built Strings 2224. Minimum Number of Operations to Convert Time 2225. Find Players With Zero or One Losses 2226. Maximum Candies Allocated to K Children 2227. Encrypt and Decrypt Strings 2229. Check if an Array Is Consecutive 2231. Largest Number After Digit Swaps by Parity 2232. Minimize Result by Adding Parentheses to Expression 2233. Maximum Product After K Increments 2234. Maximum Total Beauty of the Gardens 2235. Add Two Integers 2236. Root Equals Sum of Children 2237. Count Positions on Street With Required Brightness 2239. Find Closest Number to Zero 2240. Number of Ways to Buy Pens and Pencils 2241. Design an ATM Machine 2242. Maximum Score of a Node Sequence 2243. Calculate Digit Sum of a String 2244. Minimum Rounds to Complete All Tasks 2245. Maximum Trailing Zeros in a Cornered Path 2246. Longest Path With Different Adjacent Characters 2247. Maximum Cost of Trip With K Highways 2248. Intersection of Multiple Arrays 2249. Count Lattice Points Inside a Circle 2250. Count Number of Rectangles Containing Each Point 2251. Number of Flowers in Full Bloom 2254. Design Video Sharing Platform 2255. Count Prefixes of a Given String 2256. Minimum Average Difference 2257. Count Unguarded Cells in the Grid 2258. Escape the Spreading Fire 2259. Remove Digit From Number to Maximize Result 2260. Minimum Consecutive Cards to Pick Up 2261. K Divisible Elements Subarrays 2262. Total Appeal of A String 2263. Make Array Non-decreasing or Non-increasing 2264. Largest 3-Same-Digit Number in String 2265. Count Nodes Equal to Average of Subtree 2266. Count Number of Texts 2267. Check if There Is a Valid Parentheses String Path 2268. Minimum Number of Keypresses 2269. Find the K-Beauty of a Number 2270. Number of Ways to Split Array 2271. Maximum White Tiles Covered by a Carpet 2272. Substring With Largest Variance 2273. Find Resultant Array After Removing Anagrams 2274. Maximum Consecutive Floors Without Special Floors 2275. Largest Combination With Bitwise AND Greater Than Zero 2276. Count Integers in Intervals 2277. Closest Node to Path in Tree 2278. Percentage of Letter in String 2279. Maximum Bags With Full Capacity of Rocks 2280. Minimum Lines to Represent a Line Chart 2281. Sum of Total Strength of Wizards 2282. Number of People That Can Be Seen in a Grid 2283. Check if Number Has Equal Digit Count and Digit Value 2284. Sender With Largest Word Count 2285. Maximum Total Importance of Roads 2286. Booking Concert Tickets in Groups 2287. Rearrange Characters to Make Target String 2288. Apply Discount to Prices 2289. Steps to Make Array Non-decreasing 2290. Minimum Obstacle Removal to Reach Corner 2291. Maximum Profit From Trading Stocks 2293. Min Max Game 2294. Partition Array Such That Maximum Difference Is K 2295. Replace Elements in an Array 2296. Design a Text Editor 2297. Jump Game VIII 2299. Strong Password Checker II 2300. Successful Pairs of Spells and Potions 2301. Match Substring After Replacement 2302. Count Subarrays With Score Less Than K 2303. Calculate Amount Paid in Taxes 2304. Minimum Path Cost in a Grid 2305. Fair Distribution of Cookies 2306. Naming a Company 2307. Check for Contradictions in Equations 2309. Greatest English Letter in Upper and Lower Case 2310. Sum of Numbers With Units Digit K 2311. Longest Binary Subsequence Less Than or Equal to K 2312. Selling Pieces of Wood 2313. Minimum Flips in Binary Tree to Get Result 2315. Count Asterisks 2316. Count Unreachable Pairs of Nodes in an Undirected Graph 2317. Maximum XOR After Operations 2318. Number of Distinct Roll Sequences 2319. Check if Matrix Is X-Matrix 2320. Count Number of Ways to Place Houses 2321. Maximum Score Of Spliced Array 2322. Minimum Score After Removals on a Tree 2323. Find Minimum Time to Finish All Jobs II 2325. Decode the Message 2326. Spiral Matrix IV 2327. Number of People Aware of a Secret 2328. Number of Increasing Paths in a Grid 2330. Valid Palindrome IV 2331. Evaluate Boolean Binary Tree 2332. The Latest Time to Catch a Bus 2333. Minimum Sum of Squared Difference 2334. Subarray With Elements Greater Than Varying Threshold 2335. Minimum Amount of Time to Fill Cups 2336. Smallest Number in Infinite Set 2337. Move Pieces to Obtain a String 2338. Count the Number of Ideal Arrays 2340. Minimum Adjacent Swaps to Make a Valid Array 2341. Maximum Number of Pairs in Array 2342. Max Sum of a Pair With Equal Sum of Digits 2343. Query Kth Smallest Trimmed Number 2344. Minimum Deletions to Make Array Divisible 2345. Finding the Number of Visible Mountains 2347. Best Poker Hand 2348. Number of Zero-Filled Subarrays 2349. Design a Number Container System 2350. Shortest Impossible Sequence of Rolls 2351. First Letter to Appear Twice 2352. Equal Row and Column Pairs 2353. Design a Food Rating System 2354. Number of Excellent Pairs 2355. Maximum Number of Books You Can Take 2357. Make Array Zero by Subtracting Equal Amounts 2358. Maximum Number of Groups Entering a Competition 2359. Find Closest Node to Given Two Nodes 2360. Longest Cycle in a Graph 2361. Minimum Costs Using the Train Line 2363. Merge Similar Items 2364. Count Number of Bad Pairs 2365. Task Scheduler II 2366. Minimum Replacements to Sort the Array 2367. Number of Arithmetic Triplets 2368. Reachable Nodes With Restrictions 2369. Check if There is a Valid Partition For The Array 2370. Longest Ideal Subsequence 2371. Minimize Maximum Value in a Grid 2373. Largest Local Values in a Matrix 2374. Node With Highest Edge Score 2375. Construct Smallest Number From DI String 2376. Count Special Integers 2378. Choose Edges to Maximize Score in a Tree 2379. Minimum Recolors to Get K Consecutive Black Blocks 2380. Time Needed to Rearrange a Binary String 2381. Shifting Letters II 2382. Maximum Segment Sum After Removals 2383. Minimum Hours of Training to Win a Competition 2384. Largest Palindromic Number 2385. Amount of Time for Binary Tree to Be Infected 2386. Find the K-Sum of an Array 2387. Median of a Row Wise Sorted Matrix 2389. Longest Subsequence With Limited Sum 2390. Removing Stars From a String 2391. Minimum Amount of Time to Collect Garbage 2392. Build a Matrix With Conditions 2393. Count Strictly Increasing Subarrays 2395. Find Subarrays With Equal Sum 2396. Strictly Palindromic Number 2397. Maximum Rows Covered by Columns 2398. Maximum Number of Robots Within Budget 2399. Check Distances Between Same Letters 2400. Number of Ways to Reach a Position After Exactly k Steps 2401. Longest Nice Subarray 2402. Meeting Rooms III 2403. Minimum Time to Kill All Monsters 2404. Most Frequent Even Element 2405. Optimal Partition of String 2406. Divide Intervals Into Minimum Number of Groups 2407. Longest Increasing Subsequence II 2408. Design SQL 2409. Count Days Spent Together 2410. Maximum Matching of Players With Trainers 2411. Smallest Subarrays With Maximum Bitwise OR 2412. Minimum Money Required Before Transactions 2413. Smallest Even Multiple 2414. Length of the Longest Alphabetical Continuous Substring 2415. Reverse Odd Levels of Binary Tree 2416. Sum of Prefix Scores of Strings 2417. Closest Fair Integer 2418. Sort the People 2419. Longest Subarray With Maximum Bitwise AND 2420. Find All Good Indices 2421. Number of Good Paths 2422. Merge Operations to Turn Array Into a Palindrome 2423. Remove Letter To Equalize Frequency 2424. Longest Uploaded Prefix 2425. Bitwise XOR of All Pairings 2426. Number of Pairs Satisfying Inequality 2427. Number of Common Factors 2428. Maximum Sum of an Hourglass 2429. Minimize XOR 2430. Maximum Deletions on a String 2431. Maximize Total Tastiness of Purchased Fruits 2432. The Employee That Worked on the Longest Task 2433. Find The Original Array of Prefix Xor 2434. Using a Robot to Print the Lexicographically Smallest String 2435. Paths in Matrix Whose Sum Is Divisible by K 2436. Minimum Split Into Subarrays With GCD Greater Than One 2437. Number of Valid Clock Times 2438. Range Product Queries of Powers 2439. Minimize Maximum of Array 2440. Create Components With Same Value 2441. Largest Positive Integer That Exists With Its Negative 2442. Count Number of Distinct Integers After Reverse Operations 2443. Sum of Number and Its Reverse 2444. Count Subarrays With Fixed Bounds 2445. Number of Nodes With Value One 2446. Determine if Two Events Have Conflict 2447. Number of Subarrays With GCD Equal to K 2448. Minimum Cost to Make Array Equal 2449. Minimum Number of Operations to Make Arrays Similar 2450. Number of Distinct Binary Strings After Applying Operations 2451. Odd String Difference 2452. Words Within Two Edits of Dictionary 2453. Destroy Sequential Targets 2454. Next Greater Element IV 2455. Average Value of Even Numbers That Are Divisible by Three 2456. Most Popular Video Creator 2457. Minimum Addition to Make Integer Beautiful 2458. Height of Binary Tree After Subtree Removal Queries 2459. Sort Array by Moving Items to Empty Space 2460. Apply Operations to an Array 2461. Maximum Sum of Distinct Subarrays With Length K 2462. Total Cost to Hire K Workers 2463. Minimum Total Distance Traveled 2464. Minimum Subarrays in a Valid Split 2465. Number of Distinct Averages 2466. Count Ways To Build Good Strings 2467. Most Profitable Path in a Tree 2468. Split Message Based on Limit 2469. Convert the Temperature 2470. Number of Subarrays With LCM Equal to K 2471. Minimum Number of Operations to Sort a Binary Tree by Level 2472. Maximum Number of Non-overlapping Palindrome Substrings 2473. Minimum Cost to Buy Apples 2475. Number of Unequal Triplets in Array 2476. Closest Nodes Queries in a Binary Search Tree 2477. Minimum Fuel Cost to Report to the Capital 2478. Number of Beautiful Partitions 2479. Maximum XOR of Two Non-Overlapping Subtrees 2481. Minimum Cuts to Divide a Circle 2482. Difference Between Ones and Zeros in Row and Column 2483. Minimum Penalty for a Shop 2484. Count Palindromic Subsequences 2485. Find the Pivot Integer 2486. Append Characters to String to Make Subsequence 2487. Remove Nodes From Linked List 2488. Count Subarrays With Median K 2489. Number of Substrings With Fixed Ratio 2490. Circular Sentence 2491. Divide Players Into Teams of Equal Skill 2492. Minimum Score of a Path Between Two Cities 2493. Divide Nodes Into the Maximum Number of Groups 2495. Number of Subarrays Having Even Product 2496. Maximum Value of a String in an Array 2497. Maximum Star Sum of a Graph 2498. Frog Jump II 2499. Minimum Total Cost to Make Arrays Unequal 2500. Delete Greatest Value in Each Row 2501. Longest Square Streak in an Array 2502. Design Memory Allocator 2503. Maximum Number of Points From Grid Queries 2505. Bitwise OR of All Subsequence Sums 2506. Count Pairs Of Similar Strings 2507. Smallest Value After Replacing With Sum of Prime Factors 2508. Add Edges to Make Degrees of All Nodes Even 2509. Cycle Length Queries in a Tree 2510. Check if There is a Path With Equal Number of 0's And 1's 2511. Maximum Enemy Forts That Can Be Captured 2512. Reward Top K Students 2513. Minimize the Maximum of Two Arrays 2514. Count Anagrams 2515. Shortest Distance to Target String in a Circular Array 2516. Take K of Each Character From Left and Right 2517. Maximum Tastiness of Candy Basket 2518. Number of Great Partitions 2519. Count the Number of K-Big Indices 2520. Count the Digits That Divide a Number 2521. Distinct Prime Factors of Product of Array 2522. Partition String Into Substrings With Values at Most K 2523. Closest Prime Numbers in Range 2524. Maximum Frequency Score of a Subarray 2525. Categorize Box According to Criteria 2526. Find Consecutive Integers from a Data Stream 2527. Find Xor-Beauty of Array 2528. Maximize the Minimum Powered City 2529. Maximum Count of Positive Integer and Negative Integer 2530. Maximal Score After Applying K Operations 2531. Make Number of Distinct Characters Equal 2532. Time to Cross a Bridge 2533. Number of Good Binary Strings 2534. Time Taken to Cross the Door 2535. Difference Between Element Sum and Digit Sum of an Array 2536. Increment Submatrices by One 2537. Count the Number of Good Subarrays 2538. Difference Between Maximum and Minimum Price Sum 2539. Count the Number of Good Subsequences 2540. Minimum Common Value 2541. Minimum Operations to Make Array Equal II 2542. Maximum Subsequence Score 2543. Check if Point Is Reachable 2544. Alternating Digit Sum 2545. Sort the Students by Their Kth Score 2546. Apply Bitwise Operations to Make Strings Equal 2547. Minimum Cost to Split an Array 2548. Maximum Price to Fill a Bag 2549. Count Distinct Numbers on Board 2550. Count Collisions of Monkeys on a Polygon 2551. Put Marbles in Bags 2552. Count Increasing Quadruplets 2553. Separate the Digits in an Array 2554. Maximum Number of Integers to Choose From a Range I 2555. Maximize Win From Two Segments 2556. Disconnect Path in a Binary Matrix by at Most One Flip 2557. Maximum Number of Integers to Choose From a Range II 2558. Take Gifts From the Richest Pile 2559. Count Vowel Strings in Ranges 2560. House Robber IV 2561. Rearranging Fruits 2562. Find the Array Concatenation Value 2563. Count the Number of Fair Pairs 2564. Substring XOR Queries 2565. Subsequence With the Minimum Score 2566. Maximum Difference by Remapping a Digit 2567. Minimum Score by Changing Two Elements 2568. Minimum Impossible OR 2569. Handling Sum Queries After Update 2570. Merge Two 2D Arrays by Summing Values 2571. Minimum Operations to Reduce an Integer to 0 2572. Count the Number of Square-Free Subsets 2573. Find the String with LCP 2574. Left and Right Sum Differences 2575. Find the Divisibility Array of a String 2576. Find the Maximum Number of Marked Indices 2577. Minimum Time to Visit a Cell In a Grid 2578. Split With Minimum Sum 2579. Count Total Number of Colored Cells 2580. Count Ways to Group Overlapping Ranges 2581. Count Number of Possible Root Nodes 2582. Pass the Pillow 2583. Kth Largest Sum in a Binary Tree 2584. Split the Array to Make Coprime Products 2585. Number of Ways to Earn Points 2586. Count the Number of Vowel Strings in Range 2587. Rearrange Array to Maximize Prefix Score 2588. Count the Number of Beautiful Subarrays 2589. Minimum Time to Complete All Tasks 2590. Design a Todo List 2591. Distribute Money to Maximum Children 2592. Maximize Greatness of an Array 2593. Find Score of an Array After Marking All Elements 2594. Minimum Time to Repair Cars 2595. Number of Even and Odd Bits 2596. Check Knight Tour Configuration 2597. The Number of Beautiful Subsets 2598. Smallest Missing Non-negative Integer After Operations 2599. Make the Prefix Sum Non-negative 2600. K Items With the Maximum Sum 2601. Prime Subtraction Operation 2602. Minimum Operations to Make All Array Elements Equal 2603. Collect Coins in a Tree 2604. Minimum Time to Eat All Grains 2605. Form Smallest Number From Two Digit Arrays 2606. Find the Substring With Maximum Cost 2607. Make K-Subarray Sums Equal 2608. Shortest Cycle in a Graph 2609. Find the Longest Balanced Substring of a Binary String 2610. Convert an Array Into a 2D Array With Conditions 2611. Mice and Cheese 2612. Minimum Reverse Operations 2613. Beautiful Pairs 2614. Prime In Diagonal 2615. Sum of Distances 2616. Minimize the Maximum Difference of Pairs 2617. Minimum Number of Visited Cells in a Grid 2618. Check if Object Instance of Class 2619. Array Prototype Last 2620. Counter 2621. Sleep 2622. Cache With Time Limit 2623. Memoize 2624. Snail Traversal 2625. Flatten Deeply Nested Array 2626. Array Reduce Transformation 2627. Debounce 2628. JSON Deep Equal 2629. Function Composition 2630. Memoize II 2631. Group By 2632. Curry 2633. Convert Object to JSON String 2634. Filter Elements from Array 2635. Apply Transform Over Each Element in Array 2636. Promise Pool 2637. Promise Time Limit 2638. Count the Number of K-Free Subsets 2639. Find the Width of Columns of a Grid 2640. Find the Score of All Prefixes of an Array 2641. Cousins in Binary Tree II 2642. Design Graph With Shortest Path Calculator 2643. Row With Maximum Ones 2644. Find the Maximum Divisibility Score 2645. Minimum Additions to Make Valid String 2646. Minimize the Total Price of the Trips 2647. Color the Triangle Red 2648. Generate Fibonacci Sequence 2649. Nested Array Generator 2650. Design Cancellable Function 2651. Calculate Delayed Arrival Time 2652. Sum Multiples 2653. Sliding Subarray Beauty 2654. Minimum Number of Operations to Make All Array Elements Equal to 1 2655. Find Maximal Uncovered Ranges 2656. Maximum Sum With Exactly K Elements 2657. Find the Prefix Common Array of Two Arrays 2658. Maximum Number of Fish in a Grid 2659. Make Array Empty 2660. Determine the Winner of a Bowling Game 2661. First Completely Painted Row or Column 2662. Minimum Cost of a Path With Special Roads 2663. Lexicographically Smallest Beautiful String 2664. The Knight’s Tour 2665. Counter II 2666. Allow One Function Call 2667. Create Hello World Function 2670. Find the Distinct Difference Array 2671. Frequency Tracker 2672. Number of Adjacent Elements With the Same Color 2673. Make Costs of Paths Equal in a Binary Tree 2674. Split a Circular Linked List 2675. Array of Objects to Matrix 2676. Throttle 2677. Chunk Array 2678. Number of Senior Citizens 2679. Sum in a Matrix 2680. Maximum OR 2681. Power of Heroes 2682. Find the Losers of the Circular Game 2683. Neighboring Bitwise XOR 2684. Maximum Number of Moves in a Grid 2685. Count the Number of Complete Components 2689. Extract Kth Character From The Rope Tree 2690. Infinite Method Object 2692. Make Object Immutable 2693. Call Function with Custom Context 2694. Event Emitter 2695. Array Wrapper 2696. Minimum String Length After Removing Substrings 2697. Lexicographically Smallest Palindrome 2698. Find the Punishment Number of an Integer 2699. Modify Graph Edge Weights 2700. Differences Between Two Objects 2702. Minimum Operations to Make Numbers Non-positive 2703. Return Length of Arguments Passed 2704. To Be Or Not To Be 2705. Compact Object 2706. Buy Two Chocolates 2707. Extra Characters in a String 2708. Maximum Strength of a Group 2709. Greatest Common Divisor Traversal 2710. Remove Trailing Zeros From a String 2711. Difference of Number of Distinct Values on Diagonals 2712. Minimum Cost to Make All Characters Equal 2713. Maximum Strictly Increasing Cells in a Matrix 2714. Find Shortest Path with K Hops 2715. Timeout Cancellation 2716. Minimize String Length 2717. Semi-Ordered Permutation 2718. Sum of Matrix After Queries 2719. Count of Integers 2721. Execute Asynchronous Functions in Parallel 2722. Join Two Arrays by ID 2723. Add Two Promises 2724. Sort By 2725. Interval Cancellation 2726. Calculator with Method Chaining 2727. Is Object Empty 2728. Count Houses in a Circular Street 2729. Check if The Number is Fascinating 2730. Find the Longest Semi-Repetitive Substring 2731. Movement of Robots 2732. Find a Good Subset of the Matrix 2733. Neither Minimum nor Maximum 2734. Lexicographically Smallest String After Substring Operation 2735. Collecting Chocolates 2736. Maximum Sum Queries 2737. Find the Closest Marked Node 2739. Total Distance Traveled 2740. Find the Value of the Partition 2741. Special Permutations 2742. Painting the Walls 2743. Count Substrings Without Repeating Character 2744. Find Maximum Number of String Pairs 2745. Construct the Longest New String 2746. Decremental String Concatenation 2747. Count Zero Request Servers 2748. Number of Beautiful Pairs 2749. Minimum Operations to Make the Integer Zero 2750. Ways to Split Array Into Good Subarrays 2753. Count Houses in a Circular Street II 2754. Bind Function to Context 2755. Deep Merge of Two Objects 2757. Generate Circular Array Values 2758. Next Day 2759. Convert JSON String to Object 2760. Longest Even Odd Subarray With Threshold 2761. Prime Pairs With Target Sum 2762. Continuous Subarrays 2763. Sum of Imbalance Numbers of All Subarrays 2764. is Array a Preorder of Some ‌Binary Tree 2765. Longest Alternating Subarray 2766. Relocate Marbles 2767. Partition String Into Minimum Beautiful Substrings 2768. Number of Black Blocks 2769. Find the Maximum Achievable Number 2770. Maximum Number of Jumps to Reach the Last Index 2771. Longest Non-decreasing Subarray From Two Arrays 2772. Apply Operations to Make All Array Elements Equal to Zero 2773. Height of Special Binary Tree 2774. Array Upper Bound 2775. Undefined to Null 2776. Convert Callback Based Function to Promise Based Function 2777. Date Range Generator 2778. Sum of Squares of Special Elements 2779. Maximum Beauty of an Array After Applying Operation 2780. Minimum Index of a Valid Split 2781. Length of the Longest Valid Substring 2782. Number of Unique Categories 2784. Check if Array is Good 2785. Sort Vowels in a String 2786. Visit Array Positions to Maximize Score 2787. Ways to Express an Integer as Sum of Powers 2788. Split Strings by Separator 2789. Largest Element in an Array after Merge Operations 2790. Maximum Number of Groups With Increasing Length 2791. Count Paths That Can Form a Palindrome in a Tree 2792. Count Nodes That Are Great Enough 2794. Create Object from Two Arrays 2795. Parallel Execution of Promises for Individual Results Retrieval 2796. Repeat String 2797. Partial Function with Placeholders 2798. Number of Employees Who Met the Target 2799. Count Complete Subarrays in an Array 2800. Shortest String That Contains Three Strings 2801. Count Stepping Numbers in Range 2802. Find The K-th Lucky Number 2803. Factorial Generator 2804. Array Prototype ForEach 2805. Custom Interval 2806. Account Balance After Rounded Purchase 2807. Insert Greatest Common Divisors in Linked List 2808. Minimum Seconds to Equalize a Circular Array 2809. Minimum Time to Make Array Sum At Most x 2810. Faulty Keyboard 2811. Check if it is Possible to Split Array 2812. Find the Safest Path in a Grid 2813. Maximum Elegance of a K-Length Subsequence 2814. Minimum Time Takes to Reach Destination Without Drowning 2815. Max Pair Sum in an Array 2816. Double a Number Represented as a Linked List 2817. Minimum Absolute Difference Between Elements With Constraint 2818. Apply Operations to Maximize Score 2819. Minimum Relative Loss After Buying Chocolates 2821. Delay the Resolution of Each Promise 2822. Inversion of Object 2823. Deep Object Filter 2824. Count Pairs Whose Sum is Less than Target 2825. Make String a Subsequence Using Cyclic Increments 2826. Sorting Three Groups 2827. Number of Beautiful Integers in the Range 2828. Check if a String Is an Acronym of Words 2829. Determine the Minimum Sum of a k-avoiding Array 2830. Maximize the Profit as the Salesman 2831. Find the Longest Equal Subarray 2832. Maximal Range That Each Element Is Maximum in It 2833. Furthest Point From Origin 2834. Find the Minimum Possible Sum of a Beautiful Array 2835. Minimum Operations to Form Subsequence With Target Sum 2836. Maximize Value of Function in a Ball Passing Game 2838. Maximum Coins Heroes Can Collect 2839. Check if Strings Can be Made Equal With Operations I 2840. Check if Strings Can be Made Equal With Operations II 2841. Maximum Sum of Almost Unique Subarray 2842. Count K-Subsequences of a String With Maximum Beauty 2843. Count Symmetric Integers 2844. Minimum Operations to Make a Special Number 2845. Count of Interesting Subarrays 2846. Minimum Edge Weight Equilibrium Queries in a Tree 2847. Smallest Number With Given Digit Product 2848. Points That Intersect With Cars 2849. Determine if a Cell Is Reachable at a Given Time 2850. Minimum Moves to Spread Stones Over Grid 2851. String Transformation 2852. Sum of Remoteness of All Cells 2855. Minimum Right Shifts to Sort the Array 2856. Minimum Array Length After Pair Removals 2857. Count Pairs of Points With Distance k 2858. Minimum Edge Reversals So Every Node Is Reachable 2859. Sum of Values at Indices With K Set Bits 2860. Happy Students 2861. Maximum Number of Alloys 2862. Maximum Element-Sum of a Complete Subset of Indices 2863. Maximum Length of Semi-Decreasing Subarrays 2864. Maximum Odd Binary Number 2865. Beautiful Towers I 2866. Beautiful Towers II 2867. Count Valid Paths in a Tree 2868. The Wording Game 2869. Minimum Operations to Collect Elements 2870. Minimum Number of Operations to Make Array Empty 2871. Split Array Into Maximum Number of Subarrays 2872. Maximum Number of K-Divisible Components 2873. Maximum Value of an Ordered Triplet I 2874. Maximum Value of an Ordered Triplet II 2875. Minimum Size Subarray in Infinite Array 2876. Count Visited Nodes in a Directed Graph 2877. Create a DataFrame from List 2878. Get the Size of a DataFrame 2879. Display the First Three Rows 2880. Select Data 2881. Create a New Column 2882. Drop Duplicate Rows 2883. Drop Missing Data 2884. Modify Columns 2885. Rename Columns 2886. Change Data Type 2887. Fill Missing Data 2888. Reshape Data Concatenate 2889. Reshape Data Pivot 2890. Reshape Data Melt 2891. Method Chaining 2892. Minimizing Array After Replacing Pairs With Their Product 2894. Divisible and Non-divisible Sums Difference 2895. Minimum Processing Time 2896. Apply Operations to Make Two Strings Equal 2897. Apply Operations on Array to Maximize Sum of Squares 2898. Maximum Linear Stock Score 2899. Last Visited Integers 2900. Longest Unequal Adjacent Groups Subsequence I 2901. Longest Unequal Adjacent Groups Subsequence II 2902. Count of Sub-Multisets With Bounded Sum 2903. Find Indices With Index and Value Difference I 2904. Shortest and Lexicographically Smallest Beautiful String 2905. Find Indices With Index and Value Difference II 2906. Construct Product Matrix 2907. Maximum Profitable Triplets With Increasing Prices I 2908. Minimum Sum of Mountain Triplets I 2909. Minimum Sum of Mountain Triplets II 2910. Minimum Number of Groups to Create a Valid Assignment 2911. Minimum Changes to Make K Semi-palindromes 2912. Number of Ways to Reach Destination in the Grid 2913. Subarrays Distinct Element Sum of Squares I 2914. Minimum Number of Changes to Make Binary String Beautiful 2917. Find the K-or of an Array 2918. Minimum Equal Sum of Two Arrays After Replacing Zeros 2919. Minimum Increment Operations to Make Array Beautiful 2920. Maximum Points After Collecting Coins From All Nodes 2921. Maximum Profitable Triplets With Increasing Prices II 2923. Find Champion I 2924. Find Champion II 2925. Maximum Score After Applying Operations on a Tree 2926. Maximum Balanced Subsequence Sum 2927. Distribute Candies Among Children III 2928. Distribute Candies Among Children I 2929. Distribute Candies Among Children II 2930. Number of Strings Which Can Be Rearranged to Contain Substring 2931. Maximum Spending After Buying Items 2932. Maximum Strong Pair XOR I 2933. High-Access Employees 2934. Minimum Operations to Maximize Last Elements in Arrays 2935. Maximum Strong Pair XOR II 2936. Number of Equal Numbers Blocks 2937. Make Three Strings Equal 2938. Separate Black and White Balls 2939. Maximum Xor Product 2940. Find Building Where Alice and Bob Can Meet 2941. Maximum GCD-Sum of a Subarray 2942. Find Words Containing Character 2943. Maximize Area of Square Hole in Grid 2944. Minimum Number of Coins for Fruits 2945. Find Maximum Non-decreasing Array Length 2946. Matrix Similarity After Cyclic Shifts 2947. Count Beautiful Substrings I 2948. Make Lexicographically Smallest Array by Swapping Elements 2949. Count Beautiful Substrings II 2950. Number of Divisible Substrings 2951. Find the Peaks 2952. Minimum Number of Coins to be Added 2953. Count Complete Substrings 2954. Count the Number of Infection Sequences 2955. Number of Same-End Substrings 2956. Find Common Elements Between Two Arrays 2957. Remove Adjacent Almost-Equal Characters 2958. Length of Longest Subarray With at Most K Frequency 2959. Number of Possible Sets of Closing Branches 2960. Count Tested Devices After Test Operations 2961. Double Modular Exponentiation 2962. Count Subarrays Where Max Element Appears at Least K Times 2963. Count the Number of Good Partitions 2964. Number of Divisible Triplet Sums 2965. Find Missing and Repeated Values 2966. Divide Array Into Arrays With Max Difference 2967. Minimum Cost to Make Array Equalindromic 2968. Apply Operations to Maximize Frequency Score 2969. Minimum Number of Coins for Fruits II 2970. Count the Number of Incremovable Subarrays I 2971. Find Polygon With the Largest Perimeter 2972. Count the Number of Incremovable Subarrays II 2973. Find Number of Coins to Place in Tree Nodes 2974. Minimum Number Game 2975. Maximum Square Area by Removing Fences From a Field 2976. Minimum Cost to Convert String I 2977. Minimum Cost to Convert String II 2979. Most Expensive Item That Can Not Be Bought 2980. Check if Bitwise OR Has Trailing Zeros 2981. Find Longest Special Substring That Occurs Thrice I 2982. Find Longest Special Substring That Occurs Thrice II 2983. Palindrome Rearrangement Queries 2992. Number of Self-Divisible Permutations 2996. Smallest Missing Integer Greater Than Sequential Prefix Sum 2997. Minimum Number of Operations to Make Array XOR Equal to K 2998. Minimum Number of Operations to Make X and Y Equal 2999. Count the Number of Powerful Integers 3000. Maximum Area of Longest Diagonal Rectangle 3001. Minimum Moves to Capture The Queen 3002. Maximum Size of a Set After Removals 3003. Maximize the Number of Partitions After Operations 3004. Maximum Subtree of the Same Color 3005. Count Elements With Maximum Frequency 3006. Find Beautiful Indices in the Given Array I 3007. Maximum Number That Sum of the Prices Is Less Than or Equal to K 3008. Find Beautiful Indices in the Given Array II 3009. Maximum Number of Intersections on the Chart 3010. Divide an Array Into Subarrays With Minimum Cost I 3011. Find if Array Can Be Sorted 3012. Minimize Length of Array Using Operations 3013. Divide an Array Into Subarrays With Minimum Cost II 3014. Minimum Number of Pushes to Type Word I 3015. Count the Number of Houses at a Certain Distance I 3016. Minimum Number of Pushes to Type Word II 3017. Count the Number of Houses at a Certain Distance II 3018. Maximum Number of Removal Queries That Can Be Processed I 3019. Number of Changing Keys 3020. Find the Maximum Number of Elements in Subset 3021. Alice and Bob Playing Flower Game 3022. Minimize OR of Remaining Elements Using Operations 3023. Find Pattern in Infinite Stream I 3024. Type of Triangle 3025. Find the Number of Ways to Place People I 3026. Maximum Good Subarray Sum 3027. Find the Number of Ways to Place People II 3028. Ant on the Boundary 3029. Minimum Time to Revert Word to Initial State I 3030. Find the Grid of Region Average 3031. Minimum Time to Revert Word to Initial State II 3032. Count Numbers With Unique Digits II 3033. Modify the Matrix 3034. Number of Subarrays That Match a Pattern I 3035. Maximum Palindromes After Operations 3036. Number of Subarrays That Match a Pattern II 3038. Maximum Number of Operations With the Same Score I 3039. Apply Operations to Make String Empty 3040. Maximum Number of Operations With the Same Score II 3042. Count Prefix and Suffix Pairs I 3043. Find the Length of the Longest Common Prefix 3044. Most Frequent Prime 3045. Count Prefix and Suffix Pairs II 3046. Split the Array 3047. Find the Largest Area of Square Inside Two Rectangles 3048. Earliest Second to Mark Indices I 3062. Winner of the Linked List Game 3063. Linked List Frequency 3064. Guess the Number Using Bitwise Questions I 3065. Minimum Operations to Exceed Threshold Value I 3066. Minimum Operations to Exceed Threshold Value II 3067. Count Pairs of Connectable Servers in a Weighted Tree Network 3068. Find the Maximum Sum of Node Values 3069. Distribute Elements Into Two Arrays I 3070. Count Submatrices with Top-Left Element and Sum Less Than k 3071. Minimum Operations to Write the Letter Y on a Grid 3072. Distribute Elements Into Two Arrays II 3073. Maximum Increasing Triplet Value 3074. Apple Redistribution into Boxes 3075. Maximize Happiness of Selected Children 3076. Shortest Uncommon Substring in an Array 3077. Maximum Strength of K Disjoint Subarrays 3078. Match Alphanumerical Pattern in Matrix I 3079. Find the Sum of Encrypted Integers 3080. Mark Elements on Array by Performing Queries 3081. Replace Question Marks in String to Minimize Its Value 3082. Find the Sum of the Power of All Subsequences 3083. Existence of a Substring in a String and Its Reverse 3084. Count Substrings Starting and Ending with Given Character 3085. Minimum Deletions to Make String K-Special 3086. Minimum Moves to Pick K Ones 3088. Make String Anti-palindrome 🔒 3090. Maximum Length Substring With Two Occurrences 3091. Apply Operations to Make Sum of Array Greater Than or Equal to k 3092. Most Frequent IDs 3093. Longest Common Suffix Queries 3094. Guess the Number Using Bitwise Questions II 🔒 3095. Shortest Subarray With OR at Least K I 3096. Minimum Levels to Gain More Points 3097. Shortest Subarray With OR at Least K II 3098. Find the Sum of Subsequence Powers 3099. Harshad Number 3100. Water Bottles II 3101. Count Alternating Subarrays 3102. Minimize Manhattan Distances 3103. Find Trending Hashtags II 🔒 3104. Find Longest Self-Contained Substring 🔒 3105. Longest Strictly Increasing or Strictly Decreasing Subarray 3106. Lexicographically Smallest String After Operations With Constraint 3107. Minimum Operations to Make Median of Array Equal to K 3108. Minimum Cost Walk in Weighted Graph 3109. Find the Index of Permutation 🔒 3110. Score of a String 3111. Minimum Rectangles to Cover Points 3112. Minimum Time to Visit Disappearing Nodes 3113. Find the Number of Subarrays Where Boundary Elements Are Maximum 3114. Latest Time You Can Obtain After Replacing Characters 3115. Maximum Prime Difference 3116. Kth Smallest Amount With Single Denomination Combination 3117. Minimum Sum of Values by Dividing Array 3119. Maximum Number of Potholes That Can Be Fixed 🔒 3120. Count the Number of Special Characters I 3121. Count the Number of Special Characters II 3122. Minimum Number of Operations to Satisfy Conditions 3123. Find Edges in Shortest Paths 3125. Maximum Number That Makes Result of Bitwise AND Zero 🔒 3127. Make a Square with the Same Color 3128. Right Triangles 3129. Find All Possible Stable Binary Arrays I 3130. Find All Possible Stable Binary Arrays II 3131. Find the Integer Added to Array I 3132. Find the Integer Added to Array II 3133. Minimum Array End 3134. Find the Median of the Uniqueness Array 3135. Equalize Strings by Adding or Removing Characters at Ends 🔒 3136. Valid Word 3137. Minimum Number of Operations to Make Word K-Periodic 3138. Minimum Length of Anagram Concatenation 3141. Maximum Hamming Distances 🔒 3142. Check if Grid Satisfies Conditions 3143. Maximum Points Inside the Square 3144. Minimum Substring Partition of Equal Character Frequency 3145. Find Products of Elements of Big Array 3146. Permutation Difference between Two Strings 3147. Taking Maximum Energy From the Mystic Dungeon 3148. Maximum Difference Score in a Grid 3149. Find the Minimum Cost Array Permutation 3151. Special Array I 3152. Special Array II 3153. Sum of Digit Differences of All Pairs 3154. Find Number of Ways to Reach the K-th Stair 3155. Maximum Number of Upgradable Servers 🔒 3157. Find the Level of Tree with Minimum Sum 🔒 3158. Find the XOR of Numbers Which Appear Twice 3159. Find Occurrences of an Element in an Array 3160. Find the Number of Distinct Colors Among the Balls 3162. Find the Number of Good Pairs I 3163. String Compression III 3164. Find the Number of Good Pairs II 3165. Maximum Sum of Subsequence With Non-adjacent Elements 3167. Better Compression of String 🔒 3168. Minimum Number of Chairs in a Waiting Room 3169. Count Days Without Meetings 3170. Lexicographically Minimum String After Removing Stars 3171. Find Subarray With Bitwise OR Closest to K 3173. Bitwise OR of Adjacent Elements 🔒 3174. Clear Digits 3175. Find The First Player to win K Games in a Row 3176. Find the Maximum Length of a Good Subsequence I 3177. Find the Maximum Length of a Good Subsequence II 3178. Find the Child Who Has the Ball After K Seconds 3179. Find the N-th Value After K Seconds 3180. Maximum Total Reward Using Operations I 3181. Maximum Total Reward Using Operations II 3183. The Number of Ways to Make the Sum 🔒 3184. Count Pairs That Form a Complete Day I 3185. Count Pairs That Form a Complete Day II 3186. Maximum Total Damage With Spell Casting 3187. Peaks in Array 3189. Minimum Moves to Get a Peaceful Board 🔒 3190. Find Minimum Operations to Make All Elements Divisible by Three 3191. Minimum Operations to Make Binary Array Elements Equal to One I 3192. Minimum Operations to Make Binary Array Elements Equal to One II 3193. Count the Number of Inversions 3194. Minimum Average of Smallest and Largest Elements 3195. Find the Minimum Area to Cover All Ones I 3196. Maximize Total Cost of Alternating Subarrays 3197. Find the Minimum Area to Cover All Ones II 3199. Count Triplets with Even XOR Set Bits I 🔒 3200. Maximum Height of a Triangle 3201. Find the Maximum Length of Valid Subsequence I 3202. Find the Maximum Length of Valid Subsequence II 3203. Find Minimum Diameter After Merging Two Trees 3205. Maximum Array Hopping Score I 🔒 3206. Alternating Groups I 3207. Maximum Points After Enemy Battles 3208. Alternating Groups II 3209. Number of Subarrays With AND Value of K 3210. Find the Encrypted String 3211. Generate Binary Strings Without Adjacent Zeros 3212. Count Submatrices With Equal Frequency of X and Y 3213. Construct String with Minimum Cost 3215. Count Triplets with Even XOR Set Bits II 🔒 3216. Lexicographically Smallest String After a Swap 3217. Delete Nodes From Linked List Present in Array 3218. Minimum Cost for Cutting Cake I 3219. Minimum Cost for Cutting Cake II 3221. Maximum Array Hopping Score II 🔒 3222. Find the Winning Player in Coin Game 3223. Minimum Length of String After Operations 3224. Minimum Array Changes to Make Differences Equal 3226. Number of Bit Changes to Make Two Integers Equal 3227. Vowels Game in a String 3228. Maximum Number of Operations to Move Ones to the End 3229. Minimum Operations to Make Array Equal to Target 3231. Minimum Number of Increasing Subsequence to Be Removed 🔒 3232. Find if Digit Game Can Be Won 3233. Find the Count of Numbers Which Are Not Special 3235. Check if the Rectangle Corner Is Reachable 3237. Alt and Tab Simulation 🔒 3238. Find the Number of Winning Players 3239. Minimum Number of Flips to Make Binary Grid Palindromic I 3240. Minimum Number of Flips to Make Binary Grid Palindromic II 3242. Design Neighbor Sum Service 3243. Shortest Distance After Road Addition Queries I 3244. Shortest Distance After Road Addition Queries II 3247. Number of Subsequences with Odd Sum 🔒 3248. Snake in Matrix 3249. Count the Number of Good Nodes 3250. Find the Count of Monotonic Pairs I 3251. Find the Count of Monotonic Pairs II 3253. Construct String with Minimum Cost (Easy) 🔒 3254. Find the Power of K-Size Subarrays I 3255. Find the Power of K-Size Subarrays II 3258. Count Substrings That Satisfy K-Constraint I 3259. Maximum Energy Boost From Two Drinks 3261. Count Substrings That Satisfy K-Constraint II 3263. Convert Doubly Linked List to Array I 🔒 3264. Final Array State After K Multiplication Operations I 3265. Count Almost Equal Pairs I 3266. Final Array State After K Multiplication Operations II 3267. Count Almost Equal Pairs II 3269. Constructing Two Increasing Arrays 🔒 3270. Find the Key of the Numbers 3271. Hash Divided String 3272. Find the Count of Good Integers 3274. Check if Two Chessboard Squares Have the Same Color 3275. K-th Nearest Obstacle Queries 3276. Select Cells in Grid With Maximum Score 3277. Maximum XOR Score Subarray Queries 3279. Maximum Total Area Occupied by Pistons 🔒 3280. Convert Date to Binary 3281. Maximize Score of Numbers in Ranges 3282. Reach End of Array With Max Score 3283. Maximum Number of Moves to Kill All Pawns 3284. Sum of Consecutive Subarrays 🔒 3285. Find Indices of Stable Mountains 3286. Find a Safe Walk Through a Grid 3287. Find the Maximum Sequence Value of Array 3289. The Two Sneaky Numbers of Digitville 3290. Maximum Multiplication Score 3291. Minimum Number of Valid Strings to Form Target I 3292. Minimum Number of Valid Strings to Form Target II 3294. Convert Doubly Linked List to Array II 🔒 3295. Report Spam Message 3296. Minimum Number of Seconds to Make Mountain Height Zero 3297. Count Substrings That Can Be Rearranged to Contain a String I 3298. Count Substrings That Can Be Rearranged to Contain a String II 3299. Sum of Consecutive Subsequences 🔒 3300. Minimum Element After Replacement With Digit Sum 3301. Maximize the Total Height of Unique Towers 3304. Find the K-th Character in String Game I 3305. Count of Substrings Containing Every Vowel and K Consonants I 3306. Count of Substrings Containing Every Vowel and K Consonants II 3307. Find the K-th Character in String Game II 3309. Maximum Possible Number by Binary Concatenation 3310. Remove Methods From Project 3311. Construct 2D Grid Matching Graph Layout 3312. Sorted GCD Pair Queries 3313. Find the Last Marked Nodes in Tree 🔒 3314. Construct the Minimum Bitwise Array I 3315. Construct the Minimum Bitwise Array II 3316. Find Maximum Removals From Source String 3317. Find the Number of Possible Ways for an Event 3318. Find X-Sum of All K-Long Subarrays I 3319. K-th Largest Perfect Subtree Size in Binary Tree 3320. Count The Number of Winning Sequences 3321. Find X-Sum of All K-Long Subarrays II 3323. Minimize Connected Groups by Inserting Interval 🔒 3324. Find the Sequence of Strings Appeared on the Screen 3325. Count Substrings With K-Frequency Characters I 3326. Minimum Division Operations to Make Array Non Decreasing 3327. Check if DFS Strings Are Palindromes 3329. Count Substrings With K-Frequency Characters II 🔒 3330. Find the Original Typed String I 3331. Find Subtree Sizes After Changes 3332. Maximum Points Tourist Can Earn 3334. Find the Maximum Factor Score of Array 3339. Find the Number of K-Even Arrays 🔒 3340. Check Balanced String 3341. Find Minimum Time to Reach Last Room I 3342. Find Minimum Time to Reach Last Room II 3343. Count Number of Balanced Permutations 3344. Maximum Sized Array 🔒 3345. Smallest Divisible Digit Product I 3346. Maximum Frequency of an Element After Performing Operations I 3347. Maximum Frequency of an Element After Performing Operations II 3349. Adjacent Increasing Subarrays Detection I 3350. Adjacent Increasing Subarrays Detection II 3351. Sum of Good Subsequences 3353. Minimum Total Operations 🔒 3354. Make Array Elements Equal to Zero 3355. Zero Array Transformation I 3356. Zero Array Transformation II 3360. Stone Removal Game 3361. Shift Distance Between Two Strings 3364. Minimum Positive Sum Subarray 3365. Rearrange K Substrings to Form Target String 3366. Minimum Array Sum 3367. Maximize Sum of Weights after Edge Removals 3369. Design an Array Statistics Tracker 🔒 3370. Smallest Number With All Set Bits 3371. Identify the Largest Outlier in an Array 3372. Maximize the Number of Target Nodes After Connecting Trees I 3373. Maximize the Number of Target Nodes After Connecting Trees II 3374. First Letter Capitalization II 3375. Minimum Operations to Make Array Values Equal to K 3376. Minimum Time to Break Locks I 3377. Digit Operations to Make Two Integers Equal 3378. Count Connected Components in LCM Graph 3379. Transformed Array 3380. Maximum Area Rectangle With Point Constraints I 3381. Maximum Subarray Sum With Length Divisible by K 3383. Minimum Runes to Add to Cast Spell 🔒 3385. Minimum Time to Break Locks II 🔒 3386. Button with Longest Push Time 3387. Maximize Amount After Two Days of Conversions 3388. Count Beautiful Splits in an Array 3391. Design a 3D Binary Matrix with Efficient Layer Tracking 🔒 3392. Count Subarrays of Length Three With a Condition 3394. Check if Grid can be Cut into Sections 3396. Minimum Number of Operations to Make Elements in Array Distinct 3397. Maximum Number of Distinct Elements After Operations 3398. Smallest Substring With Identical Characters I 3399. Smallest Substring With Identical Characters II 3400. Maximum Number of Matching Indices After Right Shifts 🔒 3402. Minimum Operations to Make Columns Strictly Increasing 3403. Find the Lexicographically Largest String From the Box I 3404. Count Special Subsequences 3407. Substring Matching Pattern 3408. Design Task Manager 3411. Maximum Subarray With Equal Products 3412. Find Mirror Score of a String 3417. Zigzag Grid Traversal With Skip 3418. Maximum Amount of Money Robot Can Earn 3422. Minimum Operations to Make Subarray Elements Equal 🔒 3423. Maximum Difference Between Adjacent Elements in a Circular Array 3424. Minimum Cost to Make Arrays Identical 3427. Sum of Variable Length Subarrays 3430. Maximum and Minimum Sums of at Most Size K Subarrays 3431. Minimum Unlocked Indices to Sort Nums 🔒 3432. Count Partitions with Even Sum Difference 3433. Count Mentions Per User 3437. Permutations III 🔒 3438. Find Valid Pair of Adjacent Digits in String 3440. Reschedule Meetings for Maximum Free Time II 3442. Maximum Difference Between Even and Odd Frequency I 3443. Maximum Manhattan Distance After K Changes 3446. Sort Matrix by Diagonals 3447. Assign Elements to Groups with Constraints 3450. Maximum Students on a Single Bench 🔒 3452. Sum of Good Numbers 3456. Find Special Substring of Length K 3457. Eat Pizzas! 3459. Length of Longest V-Shaped Diagonal Segment 3460. Longest Common Prefix After at Most One Removal 🔒 3461. Check If Digits Are Equal in String After Operations I 3462. Maximum Sum With at Most K Elements 3466. Maximum Coin Collection 🔒 3467. Transform Array by Parity 3471. Find the Largest Almost Missing Integer 3472. Longest Palindromic Subsequence After at Most K Operations 3476. Maximize Profit from Task Assignment 🔒 3477. Fruits Into Baskets II 3478. Choose K Elements With Maximum Sum 3481. Apply Substitutions 🔒 3483. Unique 3-Digit Even Numbers 3484. Design Spreadsheet 3485. Longest Common Prefix of K Strings After Removal 3487. Maximum Unique Subarray Sum After Deletion 3488. Closest Equal Element Queries 3491. Phone Number Prefix 🔒 3492. Maximum Containers on a Ship 3493. Properties Graph 3496. Maximize Score After Pair Deletions 🔒 3498. Reverse Degree of a String 3499. Maximize Active Section with Trade I 3502. Minimum Cost to Reach Every Position 3503. Longest Palindrome After Substring Concatenation I 3504. Longest Palindrome After Substring Concatenation II 3511. Make a Positive Array 🔒 3512. Minimum Operations to Make Array Sum Divisible by K 3516. Find Closest Person 3522. Calculate Score After Performing Instructions 3523. Make Array Non-decreasing 3527. Find the Most Common Response 3528. Unit Conversion I 3531. Count Covered Buildings 3532. Path Existence Queries in a Graph I 3536. Maximum Product of Two Digits 3541. Find Most Frequent Vowel and Consonant 3545. Minimum Deletions for At Most K Distinct Characters 3546. Equal Sum Grid Partition I 3549. Multiply Two Polynomials 🔒 3550. Smallest Index With Digit Sum Equal to Index 3551. Minimum Swaps to Sort by Digit Sum 3552. Grid Teleportation Traversal 3555. Smallest Subarray to Sort in Every Sliding Window 🔒 3556. Sum of Largest Prime Substrings 3560. Find Minimum Log Transportation Cost 3561. Resulting String After Adjacent Removals 3565. Sequential Grid Path Cover 🔒 3566. Partition Array into Two Equal Product Subsets 3567. Minimum Absolute Difference in Sliding Submatrix 3568. Minimum Moves to Clean the Classroom 3571. Find the Shortest Superstring II 🔒 3572. Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values 3573. Best Time to Buy and Sell Stock V 3574. Maximize Subarray GCD Score 3576. Transform Array to All Equal Elements 3577. Count the Number of Computer Unlocking Permutations 3578. Count Partitions With Max-Min Difference at Most K 3579. Minimum Steps to Convert String with Operations 3581. Count Odd Letters from Number 🔒 3582. Generate Tag for Video Caption 3583. Count Special Triplets 3584. Maximum Product of First and Last Elements of a Subsequence 3587. Minimum Adjacent Swaps to Alternate Parity 3590. Kth Smallest Path XOR Sum 3591. Check if Any Element Has Prime Frequency 3596. Minimum Cost Path with Alternating Directions I 🔒 3597. Partition String 3598. Longest Common Prefix Between Adjacent Strings After Removals 3599. Partition Array to Minimize XOR 3602. Hexadecimal and Hexatrigesimal Conversion 3606. Coupon Code Validator 3610. Minimum Number of Primes to Sum to Target 🔒 3612. Process String with Special Operations I 3613. Minimize Maximum Component Cost 3616. Number of Student Replacements 🔒 3618. Split Array by Prime Indices 3619. Count Islands With Total Value Divisible by K 3622. Check Divisibility by Digit Sum and Product 3627. Maximum Median Sum of Subsequences of Size 3 3628. Maximum Number of Subsequences After One Inserting 3631. Sort Threats by Severity and Exploitability 🔒 3633. Earliest Finish Time for Land and Water Rides I 3634. Minimum Removals to Balance Array 3635. Earliest Finish Time for Land and Water Rides II 3637. Trionic Array I 3638. Maximum Balanced Shipments 2413. Smallest Even Multiple Easy Math Number Theory Leetcode Link Problem Description The given problem requires determining the smallest positive integer that can be divided evenly (without leaving a remainder) by both 2 and a given positive integer n. This integer is essentially the least common multiple (LCM) of 2 and n. Since the question only involves 2 as one of the numbers, it simplifies the conditions for the LCM. If n is already an even number, then it itself is the least multiple of both 2 and n (as any even number is divisible by 2). However, if n is odd, then the smallest even multiple is simply twice the value of n, as multiplying an odd number by 2 yields the smallest even number which is a multiple of both 2 and the odd number. Quick Interview Experience Help others by sharing your interview experience Have you seen this problem before? Intuition To solve this problem, we check if n is already an even number. An even number is characterized by having no remainder when divided by 2. We can check this by performing n % 2 and see if the result is 0. If n is even, then n is our answer because n would be the smallest number that is a multiple of both itself and 2. If n is not even (implying it's odd), then the smallest even multiple that is also a multiple of n would be twice n. This is because the only way to make an odd number even (and therefore a multiple of 2) without changing its divisibility by n is to multiply it by 2. Thus, the solution can be succinctly implemented with a single line of code that returns n if it's even, or n 2 if it's odd. Learn more about Math patterns. Solution Implementation ``` 1 class Solution : 2 def smallestEvenMultiple ( self, n: int ) -> int : 3 # Check if n is even by using the modulo operator. 4 # If n is even, it is already the smallest even multiple, so we return it. 5 if n % 2 == 0 : 6 return n 7 else : 8 # If n is odd, the smallest even multiple is n multiplied by 2. 9 return n 2 10 ``` ``` 1class Solution { 2 // Method to calculate the smallest even multiple of a given number n 3 public int smallestEvenMultiple(int n) { 4 // Check if n is already even by checking remainder when divided by 2 5 if (n % 2 == 0) { 6 // If n is even, then it is the smallest even multiple of itself 7 return n; 8 } else { 9 // If n is odd, then the smallest even multiple is n times 2 10 return n 2; 11 } 12 } 13} 14 ``` ``` 1class Solution { 2public: 3 // Function to find the smallest even multiple of a given number 'n' 4 int smallestEvenMultiple(int n) { 5 // If 'n' is already even, it is its own smallest even multiple 6 if (n % 2 == 0) { 7 return n; 8 } 9 // If 'n' is odd, double 'n' to get the smallest even multiple 10 return n 2; 11 } 12}; 13 ``` ``` 1// This function finds the smallest even multiple of a given number. 2// If 'n' is already even, it returns 'n'; else it returns 'n' multiplied by 2. 3function smallestEvenMultiple(n: number): number { 4 // Check if 'n' is even using modulo operator 5 if (n % 2 === 0) { 6 // 'n' is even, so return 'n' itself 7 return n; 8 } else { 9 // 'n' is odd, so return 'n' doubled to make it even 10 return n 2; 11 } 12} 13 ``` Solution Approach The provided solution is straightforward and does not require complicated algorithms or data structures. It leverages a straightforward mathematical fact that every multiple of an even number is even, and the smallest even multiple of an odd number is the number itself multiplied by 2. Here is a step-by-step breakdown of the implementation process: Check if the given integer n is even by using the modulo operator %. In Python, n % 2 == 0 is True if n is divisible by 2 without any remainder. It is a common way to determine if a number is even. If n is even, the function returns n itself since n is the smallest even number that satisfies the condition of being a multiple of both n and 2. If n is odd (the check resulted in False), multiply n by 2 to find the smallest even multiple of n. Odd numbers can only yield even numbers when multiplied by an even number, and multiplying by 2 ensures the smallest possible outcome. Please note that there is no explicit loop or conditional structures needed; the solution uses a ternary-like expression in Python that is a concise way to write an if-else statement on a single line. The equation: return n if n % 2 == 0 else n 2 covers the solution approach completely. In this specific problem, the solution's time complexity is constant, O(1), because the calculation requires a maximum of two operations regardless of the size of n. The space complexity is also constant, O(1), as it does not require any additional space that depends on the input size. Ready to land your dream job? Unlock your dream job with a 2-minute evaluator for a personalized learning plan! Start Evaluator Example Walkthrough Let's consider an example to illustrate the solution approach. Suppose we are given n = 7, which is an odd number. First, we check if n is even by performing n % 2. For our example, 7 % 2 equals 1, which is not zero. So, n is not even. Since n is odd, our solution should return n 2. Hence, we multiply 7 by 2, which gives us 14. The number 14 is the smallest positive integer that is even and can be divided evenly by both 2 and 7 (since the other divisor is 2, and 14 is already an even number). Therefore, the function will return 14 as the answer in this case. Now let's take an even number for n, for instance, n = 6. We perform the modulo operation again: 6 % 2. This time it equals 0, which means n is even. Since n is even, the smallest even multiple of 2 that can also be divided by n is n itself. There is no need to perform any multiplication. The function will return 6 as the least common multiple that meets the problem's conditions. The key takeaway here is the simple check for evenness, which dictates whether we can return n directly or need to return n 2. This example walkthrough demonstrates the directness and efficiency of the solution approach for both odd and even input scenarios. Time and Space Complexity The given code consists of a single function smallestEvenMultiple which takes an integer n and returns the smallest even multiple of n. Time Complexity The time complexity of this function is O(1) which means that it runs in constant time. This is because no matter the size of the input n, the function performs a maximum of one comparison (n % 2 == 0) and possibly one multiplication (n 2). Both operations are basic arithmetic operations that take constant time to complete. Space Complexity The space complexity of this function is also O(1). The function uses a fixed amount of space; it allocates space for just one integer (the return value), and the amount of memory used does not scale with the input size n. The overall performance of the code is thus highly efficient, with both time and space complexities being constant. Learn more about how to find time and space complexity quickly using problem constraints. Unlock Full Access Sign in to get your free access quota and explore: In-Depth Strategy Explanations Step-by-Step Example Walkthroughs Time & Space Complexity Analysis Personalized Progress Tracking Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan: Recommended Readings Math Basics Math for Technical Interviews How much math do I need to know for technical interviews The short answer is about high school level math Computer science is often associated with math and some universities even place their computer science department under the math faculty However the reality is that you Coding Interview Patterns: The Smart, Data-Driven Way to Prepare for Tech Interviews Coding Interview Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way Recursion Review Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https assets algo monster recursion jpg You first call Ben and ask Want a Structured Path to Master System Design Too? 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188378	https://www.youtube.com/watch?v=mdDu3nqSy4I	The sum of n terms of a series is An^2+Bn, then the n th term is Doubtnut 3940000 subscribers 10 likes Description 918 views Posted: 11 Oct 2018 To ask Unlimited Maths doubts download Doubtnut from - The sum of n terms of a series is An^2+Bn, then the n th term is 1 comments Transcript: [Music] in discussion the sum of n terms of the series is a n square plus V n then n term is okay so media SN is equal to a n square plus BN so first time getting s1 sum of first term is you can collude one duckling yoga a plus B okay yes to get sum of two terms is a into two Qashqai is four plus B into 2 so this is equal to 4a plus 2b so you can see that first term here we and say a plus B second term clogging second demagoguing sum of two terms minus 1 terms two years I get 4a plus 2b minus of 4a plus 2b minus of sum of first term means first equipment subtract up the next second term second term is equal to 3a plus B he can sum of K is the first time of gavi's scheme which comes out to be a plus B he's a picky person taking common difference yoga common difference is equal to 3a plus B minus of a plus B so this is equal to 2 a common difference guy gets a figure to it so I put the nth term of that type it Tian Tian chaotic a plus n minus 1 into D so aka first time of the trip is a plus B plus n minus 1 in two days to a ok so nth term of the technique comes out to be a plus B plus 2 a n minus 2 a so TN thunk am i plus B plus 2 a common lately so a common linking to a sahaja yoga here's ever a minus 2 is minus a TI ck a combination to n yeah yeah for Katherine a - a - yoga one so our first term of the type a is equal equal to B plus a into 2 n minus 1 okay what anthem you say piggy B plus a into 2 n minus 1 okay so they say the answer for this question so it's option will be the correct option option B will be the correct option and term of that FP equal to B plus a 2 min to n minus 1 okay [Music]
188379	https://artofproblemsolving.com/wiki/index.php/Fermat%27s_Little_Theorem?srsltid=AfmBOor9lST5KKFs_9ya_AJdw23eSdh6DT9WXjJAeIf6-ieR0RbSnUWB	Art of Problem Solving Fermat's Little Theorem - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Fermat's Little Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Fermat's Little Theorem Fermat's Little Theorem is highly useful in number theory for simplifying the computation of exponents in modular arithmetic (which students should study more at the introductory level if they have a hard time following the rest of this article). This theorem is credited to Pierre de Fermat. Contents [hide] 1 Statement 2 Proof 2.1 Proof 1 (Induction) 2.2 Proof 2 (Inverses) 2.3 Proof 3 (Combinatorics) 2.4 Proof 4 (Geometry) 2.5 Proof 5 (Burnside's Lemma) 2.6 Proof 6 (Lagrange's Theorem) 2.7 Proof 7 (Field Theory) 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Advanced 4 Hints/Solutions 5 Extensions 6 See Also Statement If is an integer, is a prime number and is not divisible by , then . A frequently used corollary of Fermat's Little Theorem is . As you can see, it is derived by multiplying both sides of the theorem by . This form is useful because we no longer need to restrict ourselves to integers not divisible by . This theorem is a special case of Euler's Totient Theorem, which states that if and are relatively prime integers, then , where denotes Euler's totient function. In particular, for prime numbers . In turn, this is a special case of Lagrange's Theorem. In contest problems, Fermat's Little Theorem is often used in conjunction with the Chinese Remainder Theorem to simplify tedious calculations. Proof We offer several proofs using different techniques to prove the statement . If , then we can cancel a factor of from both sides and retrieve the first version of the theorem. Proof 1 (Induction) The most straightforward way to prove this theorem is by applying the induction principle. We fix as a prime number. The base case, , is obviously true. Suppose the statement is true. Then, by the binomial theorem, Note that divides into any binomial coefficient of the form for . This follows by the definition of the binomial coefficient as ; since is prime, then divides the numerator, but not the denominator. Taken , all of the middle terms disappear, and we end up with . Since we also know that , then , as desired. Proof 2 (Inverses) Let . Then, we claim that the set , consisting of the product of the elements of with , taken modulo , is simply a permutation of . In other words, Clearly none of the for are divisible by , so it suffices to show that all of the elements in are distinct. Suppose that . Since , by the cancellation rule, that reduces to which means as Thus, , we have that the product of the elements of is Cancelling the factors from both sides, we are left with the statement . A similar version can be used to prove Euler's Totient Theorem, if we let Proof 3 (Combinatorics) An illustration of the case . Consider a necklace with beads, each bead of which can be colored in different ways. There are ways to pick the colors of the beads. of these are necklaces that consists of beads of the same color. Of the remaining necklaces, for each necklace, there are exactly more necklaces that are rotationally equivalent to this necklace. It follows that must be divisible by . Written in another way, Video explanation Proof 4 (Geometry) For and , respectively. We imbed a hypercube of side length in (the -th dimensional Euclidean space), such that the vertices of the hypercube are at . A hypercube is essentially a cube, generalized to higher dimensions. This hypercube consists of separate unit hypercubes, with centers consisting of the points where each is an integer from to . Besides the centers of the unit hypercubes in the main diagonal (from to ), the transformation carrying maps one unit hypercube to a distinct hypercube. Much like the combinatorial proof, this splits the non-main diagonal unit hypercubes into groups of size , from which it follows that . Thus, we have another way to visualize the above combinatorial proof, by imagining the described transformation to be, in a sense, a rotation about the main diagonal of the hypercube Proof 5 (Burnside's Lemma) Consider the number of ways to color a -beaded oriented necklace in colors up to symmetry where is prime. The group , or the cyclic group of order , acts on the colorings of an oriented necklace by rotation. The identity fixes all of the colorings by definition. If where then permutes the necklace in a single orbit which we can denote as (since the size of the orbit is a factor of ). Hence, if then fixes only the monochromatic paintings. By Burnside's Lemma the number of ways to paint the necklace (up to symmetry) is This simplifies to and since must be an integer we must have that or that which finishes. Proof 6 (Lagrange's Theorem) The key to this proof is to recognize that for some prime where is actually a group. Notice that the order of is . Suppose there exists some such that for some sufficient , . By Lagrange's Theorem we must have that so for some . Therefore we have which yields as desired Proof 7 (Field Theory) Define a field such that is its group of units written as . If we can prove the cyclicity of then the claim follows. We first prove the following lemma: Lemma: For any integer and any finite field , is not a subgroup of . Proof: Our aim is to show that . It is evident that any element in has to satisfy . However, at most elements satisfy this equation (which can be proven inductively), and which means that it can not be a subgroup of since . This completes the proof. Since is abelian, we can use the Fundamental Theorem of Finitely Generated Abelian Groups and we can write as a product of cyclic groups of prime order where the set of prime power orders is unique. We can do this because if any two prime powers are not coprime then contains and consequently , contradicting our lemma. We can therefore write as where . By the Chinese Remainder Theorem we can then write which means is cyclic. Our proof of Fermat's Little Theorem, however, comes as a corollary of this theorem. If (which is always a field for prime ) then must be a cyclic group of order . Hence for any nonzero , or that for prime which completes our proof. Problems Introductory Compute some examples, for example find , and , and check your answers by calculator where possible. Let . What is the units digit of ? (Source) Find mod . (Discussion). An integer is selected at random in the range . What is the probability that the remainder when is divided by is ? (Try to solve this with FLT) (AMC 10A 2017 Q14) Intermediate One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that . Find the value of . (Source) If , find the last two digits of . (Source) Advanced Is it true that if is a prime number, and is an integer , then the sum of the products of each -element subset of will be divisible by ? Hints/Solutions Introductory: Hint: For the first example, we have by FLT (Fermat's Little Theorem). It follows that . Intermediate: Solution (1989 AIME, 9) To solve this problem, it would be nice to know some information about the remainders can have after division by certain numbers. By Fermat's Little Theorem, we know is congruent to modulo 5. Hence, Continuing, we examine the equation modulo , Thus, is divisible by three and leaves a remainder of four when divided by 5. It's obvious that , so the only possibilities are or . It quickly becomes apparent that 174 is much too large, so must be 144. Advanced: Hint: try to establish the identity , and then apply Vieta's formulas. Extensions If is an integer, is a prime number and is not divisible by , then . The above follows from the exponent rule An extension of the Corollary given above is that: Immediately by normal exponent rules, it follows that if: Then, Which means, by repeating the process, we have that we can reduce the exponent to its digital root base See Also Number theory Modular arithmetic Euler's Totient Theorem Order (group theory) This article is a stub. Help us out by expanding it. Retrieved from " Categories: Number theory Theorems Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188380	https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6?srsltid=AfmBOoqAF5J1_AZtA2F6QDjAUVrpyJOW1o1GvP_QXElj6ED_oviCJ0PH	Art of Problem Solving 2004 AIME I Problems/Problem 6 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2004 AIME I Problems/Problem 6 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2004 AIME I Problems/Problem 6 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 5 See also Problem An integer is called snakelike if its decimal representation satisfies if is odd and if is even. How many snakelike integers between 1000 and 9999 have four distinct digits? Solution 1 We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the former case, suppose we pick digits such that . There are five arrangements of these digits that satisfy the condition of being snakelike: , , , , . Thus there are snakelike numbers which do not contain the digit zero. In the second case we choose zero and three other digits such that . There are three arrangements of these digits that satisfy the condition of being snakelike: , , . Because we know that zero is a digit, there are snakelike numbers which contain the digit zero. Thus there are snakelike numbers. Solution 2 Let's create the snakelike number from digits , and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits . There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is . Thus our answer is Solution 3 We will first decide the order of the 4 digits, greatest to least. To do this, we will pretend that we have selected the digits 1,2,3,4, and we need to arrange them to create a snakelike number. By testing all permutations, there are only 5 ways to make a snakelike number: (1,3,2,4),(1,4,2,3),(2,3,1,4),(2,4,1,3),(3,4,1,2). Now we select 4 digits to replace the 1,2,3,4. In first 2 of cases: (1,3,2,4),(1,4,2,3), the leading digit is a 1, which means it is the smallest of our 4 digits. If we select a 0, the leading digit will be a zero, which is bad because all numbers between 1000 and 9999 have nonzero leading digits. So, we need to select our 4 digits only from the pool of 1-9. There are 9 choose 4 ways and there are 2 cases: Thus, there are 252 ways for those 2 cases. For the next 3 cases, selecting a 0 is okay, so we can select from the pool of 0-9. There are 10 choose 4 ways to select our 4 digits and there are 3 cases: For those 3 cases there are 630 ways. Thus, our answer is 630+252 = . -Alexlikemath See also 2004 AIME I (Problems • Answer Key • Resources) Preceded by Problem 5Followed by Problem 7 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188381	https://pressbooks.bccampus.ca/basicelectricity/chapter/fuses-and-circuit-breakers/	Electric Circuits Three Laws for Series Circuits There are three fundamental relationships concerning resistance, current, and voltage for all series circuits. It is important that you learn the three fundamental laws for series circuits. Resistance Whenever individual resistances are connected in series, they have the same effect as one large combined resistance. Since there is only one path for current flow in a series circuit, and since each of the resistors is in line to act as an opposition to this current flow, the overall resistance is the combined opposition of all the in-line resistors. _The total resistance of a series circuit is equal to the sum of all the individual resistances in the circuit__._ Rt = R1 + R2 + R3… Using this formula, you find that the total resistance of the circuit is: RT = 15 Ω + 5 Ω + 20 Ω = 40 Ω Figure 16. Series circuit Current Since there is only one path for electron flow in a series circuit, the current is the same magnitude at any point in the circuit. _The total current in a series circuit is the same as the current through any resistance of the circuit._ IT = I1 = I2 = I3… Given 120 V as the total voltage, and having determined the total resistance of the circuit as 40 Ω, you can now apply Ohms law to determine the total current in this circuit: IT = 120 V/ 40 Ω = 3 A This total circuit current would remain the same through all the individual circuit resistors. Voltage Before any current will flow through a resistance, a potential difference, or voltage, must be available. When resistors are connected in series, they must “share” the total voltage of the source. _The total voltage in a series circuit is equal to the sum of all the individual voltage drops in the circuit._ As current passes through each resistor in a series circuit, it establishes a difference in potential across each individual resistance. This is commonly called voltage drop, and its magnitude is in direct proportion to the value of resistance. The greater the value of resistance, the higher the voltage drop across that resistor. ET = E1 + E2 + E3… Using Ohms law you can determine the voltage across each resistor. 3 A× 15 Ω = 45 V 3 A× 5 Ω = 15 V 3 A× 20 Ω = 60 V The total source voltage is equal to the sum of the individual voltage drops: 45 V + 15 V + 60 V = 120 V An Open in a Series Circuit If an open is introduced, current through the circuit is interrupted. If there is no current flow, the voltage drop across each of the resistive elements is zero. However, the potential difference of the source appears across the open. If a voltmeter is connected across the open, the reading is the same as if it were connected directly across the terminals of the supply source. Figure 17. Open circuit Effects of Line Drop and Line Loss Copper and aluminum are used as conductors because they offer little opposition to the flow of current. Although the resistance is often neglected in simple circuit analysis, it may be necessary to consider the resistance of lines in practical applications. Line Drop Figure 18. Volt drops As the 10 A current flows through each line resistance of 0.15Ω, a small voltage drop appears across each line. This voltage drop across the line conductors is commonly referred to as a line drop. Since there are two lines, the total drop is 2 × 1.5 V = 3 V. The net voltage across the load (117 V) is less than the source voltage. In some situations, it may be necessary to used larger conductors, which have lower resistance, so that the line drop does not reduce the load voltage too significantly. Line Loss Another term associated with conductors is line loss. This is a power loss expressed in watts and is related to heat energy dissipation as current flows through the resistance of the line conductors. Line loss is calculated by using one of the power equations. Using the previous example: P = I 2 × R P = (10A)2 × 0.3Ω P = 30 watts Remember: Line drop is expressed in volts. Line loss is expressed in watts. Attribution DC series circuit video byThe Electric Academyis under aCreative Commons Attribution License.
188382	https://www.zhihu.com/question/1940426838049522709/answer/1940511043701938134	蹲一个大神？ - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册蹲一个大神？关注问题写回答登录/注册蹲一个大神？从数字1到10中选一些数，要求：选出的这些数中，任意两个数的和都不相同。如何证明：最多只能选择5个数？比如选了{1,2,3}：1+2=31+3=42…显示全部关注者 2 被浏览 202 关注问题写回答邀请回答好问题添加评论分享登录后你可以不限量看优质回答私信答主深度交流精彩内容一键收藏登录查看全部 2 个回答金鱼马浙江大学计算机科学技术硕士在读关注谢邀 @不明飞行金鱼考虑从 {1,2,…,n}{1,2,\ldots,n} 中选一个子集，使得该子集中任意两个数的和都不相同，求该子集最大可能的元素个数。个人认为这个问题应该是 NP-难的。比如，记 x i∈{0,1}x_i\in{0,1} 为是否选择数字 i i ，若 x i+x l=x j+x k x_i+x_l=x_j+x_k ，那么这四个肯定不能同时选，这其实等价于一个合取式的否： ¬(x i∧x j∧x k∧x l)\neg (x_i\wedge x_j\wedge x_k\wedge x_l) 。把所有这样的子句合在一起，最终应该能转化为一个 MaxSAT 问题，而后者是 NP-难的。再比如，我们审视一下这个问题：如果把 E:={1,…,n}E:={1,\ldots,n} 的所有满足“任意两个数的和都不相同”的子集拿出来，把这个集族记作 F\mathcal{F} ，则 (E,F)(E,\mathcal{F}) 显然是一个独立系统。而找 F\mathcal{F} 中元素最多的问题相当于找这个独立系统中的“最大独立集”。如果谈到“最大独立集”问题，可能有些人会想到图论里那个，找最大的顶点集使得任意两个点没有边相连。其实这里也是一个图上的最大独立及，不过是一个 “4-超图”，与一般的图结构（每条边连两个点）不同的是，我们允许一条“边”同时连接 4 个点。考虑 n n 个顶点的 4-超图 G={V,E}G={V,E} ，其中 V={1,…,n}V={1,\ldots,n} 。若 i+l=j+k i+l=j+k ，那么我们将 (i,j,k,l)(i,j,k,l) 作为超边加入 E E 。于是问题就变成了超图 G G 的最大独立集问题。这里给出一个非常直接的状态压缩 DP 的代码： ```mathematica n = 10; (生成 elementToHyperMasks 映射表) elementToHyperMasks = Table[{}, {n}]; Do[If[i + l == j + k, mask4 = BitOr[2^(i - 1), 2^(j - 1), 2^(k - 1), 2^(l - 1)]; (将 mask4 添加到元素 i,j,k,l 对应的列表中) AppendTo[elementToHyperMasks, mask4]; AppendTo[elementToHyperMasks, mask4]; AppendTo[elementToHyperMasks, mask4]; AppendTo[elementToHyperMasks, mask4];], {i, 1, n - 3}, {j, i + 1, n - 2}, {k, j + 1, n - 1}, {l, k + 1, n}]; dp = ConstantArray[False, 2^n]; dp = True; bestSize = 0; bestMask = 0; Do[ lowBit = BitAnd[mask, -mask]; prev = mask - lowBit; x = IntegerExponent[lowBit, 2] + 1; (新添加的元素) If[dp && AllTrue[elementToHyperMasks, BitAnd[mask, #] != # &], dp = True; size = DigitCount[mask, 2, 1]; If[size > bestSize, bestSize, bestMask = {size, mask} ]; ], {mask, 1, 2^n - 1} ]; elements = Flatten@Position[Reverse@IntegerDigits[bestMask, 2, n], 1]; bestSize (输出：5) elements (输出：{1,2,3,5,8}) ``` 好在，我们可以证明一个上界。命题: 设 1\le a_1<a_2<\cdots2 ，集合 A={a_1,\ldots,a_k} 满足“任意两个数的和都不相同”，那么 n\ge \binom {k-1}{2}+2 。证明: 其实思路很简单，既然任意两个数的和不同，那么我们可以限制任意两个数的差。首先： A 中的 k 个数字可以构成 k(k-1)/2 个正的差值，这些差值落在 {1,2,\ldots,n-1} 中。但我们还不能说 n-1\ge k(k-1)/2 ，因为这些差值有可能相同。设想一下 a_1-a_2=a_3-a_4>0 ，且 {a_1,a_2}\ne {a_3,a_4} ，那么 a_1,a_2,a_3,a_4 一定不可以互不相同，否则 a_1+a_4=a_2+a_3 ，与“和互不相同”矛盾。所以一定有两个数相同，那就只能 a_1=a_4 或者 a_2=a_3 。若 a_1=a_4 ，则 a_2,a_1,a_3 构成三项等差数列，中项是 a_1 ；若 a_2=a_3 ，则 a_4,a_2,a_1 构成三项等差数列，中项是 a_2 ；得出结论：对每个 a_1-a_2=a_3-a_4 ，都能找出一个三项等差数列，以及一个中项 a 。接下来有个关键：对于每个 a\in A ， a 只能是至多一个三项等差数列的中项，否则， a_1+a_2=b_1+b_2=2a 与 “和互不相同”矛盾。此外，最小以及最大的 a ，肯定不是中项。所以，至多有 k-2 个三项等差数列。总结一下结论： A 中元素构成 k(k-1)/2 个正的差值，这些差值落在 {1,2,\ldots,n-1} 中。至多有 k-2 个额外的差值重复。所以 k(k-1)/2-k+2\le n-1 ，即， n\ge 3+\frac {k(k-1)}{2}-k= \frac12(k-1)(k-2)+2=\binom {k-1}2+2\证毕！\tag{$\rule{1ex}{1.5ex}$}\ 在题主的问题里， n=10 ，所以 \binom {k-1}{2}+2\le 10 ，算出来 k\le \frac12(3+\sqrt{65})\approx 5.53 ，所以 k\le 5 ！证出来了 ✿✿ヽ(°▽°)ノ✿ 对于一般的这种“任意两个数的和都不相同”的集合其实有个名字，叫 “弱 Sidon 集”，也叫“well-spread 集” 。为啥叫“弱”Sidon 集呢？Sidon 集的定义要强一些：一个 A={a_1,a_2,\ldots}\subset\mathbb{Z} 若满足：对任意 a_i,a_j\in A ， i\le j ， a_i+a_j 的值互不相同，则 A 称作 Sidon 集或者 B_2 sequence。这里的定义里允许重复选两遍相同的元素，但“弱”的 Sidon 集里不允许。设 1\le a_1<a_2<\cdots<a_k\le n ，且 A={a_1,\ldots,a_k} 是所谓的“弱 Sidon 集”，并设最大可能的 k 为 g(n) ，可以证明g(n)\le \sqrt n + O(\sqrt n)\不过我感觉没啥用…… 论文里算出来精确的上界是 \frac{m+1}{2}+\sqrt n \left(1+\frac 5 m\right) ，其中 m=\lceil \sqrt n \rceil ，感觉这个上界在 n 不大时有点松了。参考 ^参考了 Well-spread (weak Sidon) sets ^Kayll, P. M. (2005). A note on weak Sidon sequences. Discrete mathematics, 299(1-3), 141-144. 展开阅读全文赞同 3添加评论分享收藏喜欢更多回答肆零肆 Java开发初学者关注数学不会但用代码表示还是挺简单的，你就用Java的冒泡排序算法稍微改改就行，比如设一个变量来存放相加的结果然后在循环里来回比较就行发布于 2025-08-24 02:21 赞同 1添加评论分享收藏喜欢查看全部 2 个回答下载知乎客户端与世界分享知识、经验和见解关于作者金鱼马学生, 1/67656 浙江大学计算机科学技术硕士在读回答 461文章 131关注者 5,055 关注他发私信相关问题最近知乎评论中常出现“蹲一个”这种说法，到底是什么意思？ 1 个回答广告帮助中心知乎隐私保护指引申请开通机构号联系我们举报中心涉未成年举报网络谣言举报涉企侵权举报更多关于知乎下载知乎知乎招聘知乎指南知乎协议更多京 ICP 证 110745 号 · 京 ICP 备 13052560 号 - 1 · 京公网安备 11010802020088 号 · 互联网新闻信息服务许可证：11220250001 · 京网文2674-081 号 · 药品医疗器械网络信息服务备案（京）网药械信息备字（2022）第00334号 · 广播电视节目制作经营许可证:（京）字第06591号 · 互联网宗教信息服务许可证：京（2022）0000078 · 服务热线：400-919-0001 · Investor Relations · © 2025 知乎北京智者天下科技有限公司版权所有 · 违法和不良信息举报：010-82716601 · 举报邮箱：jubao@zhihu.com 想来知乎工作？请发送邮件到 jobs@zhihu.com 登录知乎，问答干货一键收藏打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
188383	https://brainly.com/question/34608331	[FREE] Simplify the expression: A = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +61,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +17,3k Ace exams faster, with practice that adapts to you Practice Worksheets +5,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Simplify the expression: A=(x 2+2 x y+y 2)−(x 2−2 x y+y 2) 1 See answer Explain with Learning Companion NEW Asked by bigshark2213 • 07/09/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 405092913 people 405M 0.0 0 Upload your school material for a more relevant answer 4xy Explanation assuming you require to simplify the expression A = (x² + 2xy + y²) - (x² - 2xy + y²) remove the parenthesis on the first part and distribute the second by - 1 A = x² + 2xy + y² - x² + 2xy - y² ← collect like terms = (x² - x²) + (2xy + 2xy) + (y² - y²) = 0 + 4xy + 0 = 4xy Answered by jimrgrant1 •52.3K answers•405.1M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 405092913 people 405M 0.0 0 Upload your school material for a more relevant answer The simplified expression of A=(x 2+2 x y+y 2)−(x 2−2 x y+y 2) is 4 x y. This is achieved by removing parentheses, distributing the negative sign, and combining like terms. The terms x 2 and y 2 cancel out, leading to the result. Explanation To simplify the expression A=(x 2+2 x y+y 2)−(x 2−2 x y+y 2), we follow these steps: Start with the original expression: A=(x 2+2 x y+y 2)−(x 2−2 x y+y 2) Remove the parentheses in the first part as it’s positive: A=x 2+2 x y+y 2 Distribute the negative sign in the second part: A=x 2+2 x y+y 2−x 2+2 x y−y 2 Now, combine like terms. Notice that x 2 and −x 2 cancel out: A=(x 2−x 2)+(2 x y+2 x y)+(y 2−y 2) A=0+4 x y+0 Thus, the simplified expression is: A=4 x y Examples & Evidence For example, if you let x=1 and y=2, then A=4(1)(2)=8. This shows how the simplified expression behaves with specific values of x and y. The steps followed in the simplification process are based on algebraic principles, demonstrating that the use of distributive property and combining like terms is accurate. Thanks 0 0.0 (0 votes) Advertisement bigshark2213 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 5.0 5 What is the form of the Pythagorean triple generator? O A. (x² + y²)² = (x² - y²)² + (2xy)² O B. (x² + y²)² = (x² + y²)² + (2xy)² O C. (x² + y²)² = (x² - y²)² - (2xy)² O D. (x² - y²)² = (x² + y²)²-(2xy)² Community Answer 4.2 12 Question 4 of 40Darlene wrote this proof of the identity (x+y)2-(x-y)2 = 4xy Which of thefollowing is a justification for Step 5 of her proof?Step 1: (x+y)2-(x-y)² = (x+y)(x+y) - (x-y)(x-y)Step 2: (x+y)(x+y)-(x-y)(x- y) = (x² + xy + xy + y²) - (x²-xy- xy + y²)Step 3: (x²+xy+ xy + y)-(x²-xy- xy + y) = (x² + 2xy + y) - (x²-2xy + y²)Step 4: (x²+2xy + y) - (x²-2xy + y^²) = x² + 2xy + y²-x² + 2xy-j²Step 5: x²+2xy + y²-x² + 2xy-y²= 4xySUBMITA. Definition of squaring a binomialB. Distributive propertyC. Combining like termsOD. Reflexive property Community Answer Choose the correct answer for the following function: Select one: O None of the Others = =< ○ < fz¹ fy >=< 0 < fz₁ fy >=< ○ < fzr fy > = < 2x+2y³ 6zy² x²+2xy³¹ x² +2zy³ 2x+2y³ 6xy² 2x+y³ ¹2x+y³ 2x-2y³ 2x+6xy² x²+2xy³¹x²+2xy³ 2x 6y² x²+2xy³¹ x²+2xy³ f(x, y) = ln(x² + 2xy³) Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Solve for x. 6(x−1)=9(x+2) A. x=−8 B. x=−3 C. x=3 D. x=8 What is the y-intercept of the quadratic function f(x)=(x−8)(x+3)? A. (0,3) B. (−5,0) C. (8,0) D. (0,−24) Karissa begins to solve the equation 2 1(x−14)+11=2 1x−(x−4). Her work is correct and is shown below. 2 1(x−14)+11=2 1x−(x−4)2 1x−7+11=2 1x−x+4 2 1x+4=−2 1x+4 When she subtracts 4 from both sides, 2 1x=−2 1x results. What is the value of x? Which is the graph of f(x)=−(x+3)(x+1)? Which equation has no solution? A. 4(x+3)+2 x=6(x+2) B. 5+2(3+2 x)=x+3(x+1) C. 5(x+3)+x=4(x+3)+3 D. 4+6(2+x)=2(3 x+8) Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
188384	https://www.quora.com/How-do-I-derive-the-formula-for-the-surface-area-of-the-sphere	Something went wrong. Wait a moment and try again. Surface Area of Sphere Solid 3D Geometry Formulas of Mathematics Curved Surface Area Surface Area and Volume 5 How do I derive the formula for the surface area of the sphere? · Deriving the formula for the surface area of a sphere involves understanding the geometry of the sphere and using calculus. Here’s a step-by-step outline of the derivation: Understanding the Sphere A sphere is defined as the set of all points in three-dimensional space that are at a constant distance (the radius, r) from a central point (the center of the sphere). Using Spherical Coordinates To derive the surface area, we can use spherical coordinates, where a point on the surface of the sphere can be represented as: - x=rsinθcosϕ - y=rsinθsinϕ - z=rcos\t Deriving the formula for the surface area of a sphere involves understanding the geometry of the sphere and using calculus. Here’s a step-by-step outline of the derivation: Understanding the Sphere A sphere is defined as the set of all points in three-dimensional space that are at a constant distance (the radius, r) from a central point (the center of the sphere). Using Spherical Coordinates To derive the surface area, we can use spherical coordinates, where a point on the surface of the sphere can be represented as: - x=rsinθcosϕ - y=rsinθsinϕ - z=rcosθ Here, θ is the polar angle (from the positive z-axis), and ϕ is the azimuthal angle (in the xy-plane from the positive x-axis). Surface Area Element The surface area element dS on a sphere can be found using the Jacobian determinant when transforming from Cartesian coordinates to spherical coordinates. The area element on a sphere is given by: dS=r2sinθdθdϕ Setting Up the Integral To find the total surface area S of the sphere, integrate dS over the appropriate ranges for θ and ϕ: - θ varies from 0 to π - ϕ varies from 0 to 2π The integral becomes: S=∫2π0∫π0r2sinθdθdϕ Computing the Integral First, compute the inner integral with respect to θ: ∫π0sinθdθ=[−cosθ]π0=−cos(π)−(−cos(0))=2 Next, compute the outer integral with respect to ϕ: ∫2π0dϕ=2π Combining the Results Now, combine the results of the two integrals: S=r2⋅2⋅2π=4πr2 Conclusion Thus, the formula for the surface area S of a sphere with radius r is: S=4πr2 This formula represents the total surface area of a sphere in three-dimensional space. Related questions How do I derive the formula for the surface area of a sphere without calculus using nets? Can you derive the formula for the surface area of a sphere without using calculus? How do I derive the surface area and volume of a sphere with calculus? How does one derive the equation of volume of a sphere? How do I derive the formulae for finding the volumes, total surface areas and lateral surface areas for spheres and cones? Vivek Shaw I love Proofs · 8y Consider the diagram below of a hemisphere of a sphere Let us first try to find out the surface area of this hemisphere whose Radius is R. Let the center of the sphere be O as shown above in the diagram So we have OA = R, (radius of the sphere) . Let us consider 2 circular strips BE and CD as shown in the diagram above in blue color. Let the height of this circular strip i.e CB or DE be infinitesimally small. Let us first find the surface area of this circular strip. Suppose we bring out this circular strip. it will be in the form of a cylinder with no top and bottom.Since CB is very small we ca Consider the diagram below of a hemisphere of a sphere Let us first try to find out the surface area of this hemisphere whose Radius is R. Let the center of the sphere be O as shown above in the diagram So we have OA = R, (radius of the sphere) . Let us consider 2 circular strips BE and CD as shown in the diagram above in blue color. Let the height of this circular strip i.e CB or DE be infinitesimally small. Let us first find the surface area of this circular strip. Suppose we bring out this circular strip. it will be in the form of a cylinder with no top and bottom.Since CB is very small we can consider CB to be perpendicular to BF as shown in the diagram above. Let us now find the radius of this cylinder or circular strip BCDE. From the right angle triangle OBF, we have angle OBF = Θ So BF = RCosΘ. We know that the surface area of a cylinder is 2∗pi∗r∗hHere r = BF = RCosΘ Let us find h = CB. Consider the triangle OBC. Let the angle BOC = dΘ. Since CB is very small we can consider CB as perpendicular to OB. So OBC is a right angled triangle. Also OC = R. So BC = RSin(dΘ) So surface area of the cylinder or circular strip BCDE is 2∗pi∗r∗h=2∗pi∗RCosΘ∗RSin(dΘ) We know that when angle(dΘ) is very small , we have Sin(dΘ) = dΘ So this smallArea=dA = 2∗pi∗RCosΘ∗RSin(dΘ)=2∗pi∗RCosΘ∗RdΘ=2∗pi∗(R2)∗CosΘ∗dΘ. Now to cover the full surface area of the half of the sphere we have to vary Θ from 0 to pi/2. So, Now for both halves of the sphere, the area is Edit 1: Sorry for my poor drawing. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. 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If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Gaurav Kumar Former Mathematics Learner · Author has 536 answers and 1.7M answer views · 5y Originally Answered: How do you derive the surface area of a sphere? · Shubhankar Datta Master of Science in Mathematics, Jadavpur University (Graduated 2022) · Author has 289 answers and 637.9K answer views · 7y Originally Answered: How can we derive a formula for the volume and the surface area of a sphere? · It's easy to derive the formula of the volume and surface area of a sphere if you have knowledge of calculus, mainly integration. I'll leave you with some hint assuming you are clear with integration. When you revolve a circle with respect to a straight line, you get a sphere. Volume of a solid obtained by revolving a curve y=f(x) about the x-axis is given by integral of πy²dx within the limits. Here you will take limits as -r to r, where (r,0) is the point where the curve intersects the x-axis. Surface area of the solid obtained by revolution of a curve y=f(x) about the x-axis is given by integr It's easy to derive the formula of the volume and surface area of a sphere if you have knowledge of calculus, mainly integration. I'll leave you with some hint assuming you are clear with integration. When you revolve a circle with respect to a straight line, you get a sphere. Volume of a solid obtained by revolving a curve y=f(x) about the x-axis is given by integral of πy²dx within the limits. Here you will take limits as -r to r, where (r,0) is the point where the curve intersects the x-axis. Surface area of the solid obtained by revolution of a curve y=f(x) about the x-axis is given by integral of 2πy×√{1+(dy/dx)²}dx within the limits -r to r. Now come to the sphere. Assume that the sphere is of radius r. Let it is obtained by revolving a circle of radius r and centre at the origin about the x-axis. Then the equation of the circle is x²+y²=r². So you get y in terms of x as y=√(r²-x²). Then apply the formulas given in 2 and 3 to get the volume and surface area of the sphere. Related questions How can you derive the equation for the surface area of a sphere? Can we derive volume and area formulas for a sphere using its equation? What is the curved surface area of a sphere? What is the surface area of a cone frustum formula (simplified)? Can you derive the curved surface area of a frustum of sphere without use of integration? Jacob Palasek B.S. in Information Technology & Mathematics, Grand Valley State University (Graduated 2014) · 1y Originally Answered: How can you derive the equation for the surface area of a sphere? · My first thought was that if you know the formula for the volume of the sphere then you could take two spheres where both radii are very close to each other and subtract the volume of one from the other. And keep doing this until those to radii are so close there is almost zero difference between them. Then I realized I was doing calculus by doing the above. So, when the difference between the two radii is zero then you have the surface area. And that is also the derivative of the formula for a sphere. Because the surface area is the rate of change of the volume. My first thought was that if you know the formula for the volume of the sphere then you could take two spheres where both radii are very close to each other and subtract the volume of one from the other. And keep doing this until those to radii are so close there is almost zero difference between them. Then I realized I was doing calculus by doing the above. So, when the difference between the two radii is zero then you have the surface area. And that is also the derivative of the formula for a sphere. Because the surface area is the rate of change of the volume. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. 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To derive it I would look at an element of area using polar coordinates and take the double integral. I will leave it for someone else to fill in the details on that one! Julian mathematics lover · Author has 346 answers and 54.5K answer views · 1y I suppose that you’ve already know that the formula of the volume of a sphere is V=43r3 We know that the derivative is often described as the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.For the function of the volume of a sphere,f(r)=43r3π,derivative is the change in volume when the diametre makes a very tiny change in length(approach zero),which is the surface area of the sphere. Therefore, we can directly find the derivative of the volume formula of the sphere to get the surface area f I suppose that you’ve already know that the formula of the volume of a sphere is V=43r3 We know that the derivative is often described as the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.For the function of the volume of a sphere,f(r)=43r3π,derivative is the change in volume when the diametre makes a very tiny change in length(approach zero),which is the surface area of the sphere. Therefore, we can directly find the derivative of the volume formula of the sphere to get the surface area formula. limr0→043(r+r0)3π−43r3πr0 =limr0→043r3π+4r2r0π+4r02rπ+43r03π−43r3πr0 =limr0→04r2π+4r0rπ+43r02π =4πr2 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. David Seed MSc numerical maths · Author has 2.5K answers and 1.5M answer views · 5y Originally Answered: How do you derive the surface area of a sphere? · the derivation is as follows. consider an open cylinder with a vertical axis that wraps a sphere with the centre of the sphere at the origin. the cylinder has a circumference of 2πr and a height of 2r so the area is 4πr^2 Now we show that a horizontal slice of small thickness δr has an area 2πrδr for both the cylinder and the sphere. it should be self evident for the cylinder. , the reduced radius of the sphere at angle θ from the horizontal is r cosθ. so the reduced perimeter for that strip is 2πrcosθ but the increased strip width on the sphere is δr/cosθ. Thus the strip area on the cylinder is th the derivation is as follows. consider an open cylinder with a vertical axis that wraps a sphere with the centre of the sphere at the origin. the cylinder has a circumference of 2πr and a height of 2r so the area is 4πr^2 Now we show that a horizontal slice of small thickness δr has an area 2πrδr for both the cylinder and the sphere. it should be self evident for the cylinder. , the reduced radius of the sphere at angle θ from the horizontal is r cosθ. so the reduced perimeter for that strip is 2πrcosθ but the increased strip width on the sphere is δr/cosθ. Thus the strip area on the cylinder is the same as the strip area on the sphere. it helps to draw a diagram. Gopal Menon B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010) · Author has 10.2K answers and 15.2M answer views · 4y Originally Answered: The surface area of a sphere is 4πr^2. How is it derived? · The surface area of a sphere is 4πr^2. How is it derived? Let r be the radius of the sphere. Consider a radius vector at an ∠θ to the horizontal. If this radius vector goes around the vertical axis maintaining the angle θ with the horizontal, it would trace a circle on the surface of the sphere of radius rcosθ. The circumference of the circle would then be 2πrcosθ. Let the radius vector move up slightly through an angle dθ. The arc formed due to this movement would be of length rdθ. If this arc moves around the sphere, it would give us a ring of surface area The surface area of a sphere is 4πr^2. How is it derived? Let r be the radius of the sphere. Consider a radius vector at an ∠θ to the horizontal. If this radius vector goes around the vertical axis maintaining the angle θ with the horizontal, it would trace a circle on the surface of the sphere of radius rcosθ. The circumference of the circle would then be 2πrcosθ. Let the radius vector move up slightly through an angle dθ. The arc formed due to this movement would be of length rdθ. If this arc moves around the sphere, it would give us a ring of surface area 2πrcosθ×rdθ=2πr2cosθdθ. The surface area of the sphere can be obtained by integrating this. ⇒S=2π/2∫02πr2cosθdθ=4πr2π/2∫0cosθdθ. =4πr2[sinθ]π/20 ⇒S=4πr2[1−0]=4πr2. Roger Pickering Spent 6 years at 2 universities doing maths · Author has 14.9K answers and 5.9M answer views · 5y The formula for the surface area of a sphere is Area=4πr2. EDIT: This answer was written for a question about how to calculate the surface area of a sphere. It has been merged with the question of how to derive the formula which is clearly a different order if difficulty. Answers to that question are unlikely to be helpful to the person who submitted the original one. Apparently it is no longer possible to request that a merge should be undone. Muriel Lange Teacher at Huntington Learning Center (2007–present) · Author has 247 answers and 462.7K answer views · 5y Originally Answered: How do you find the total surface area of a sphere? · The formula for the surface area of a sphere is 4(Pi)r^2. That is interesting, because, if you slice the sphere (think of an orange) in half, the area of the cross-section would be (Pi)r^2. the orange peel (infinitesimally thin) would make exactly 4 layers on the cross-section of the orange. Joshua Gross Associate Professor of Computer Science at CSUMB · Author has 34.6K answers and 560.1M answer views · 5y If you are asking how the formula was derived, there are several answers, but this article covers one proof and a thought process leading in that direction: Surface Area of a Sphere \| Brilliant Math & Science Wiki . Hilmar Zonneveld Translator (1985–present) · Author has 58.5K answers and 19.4M answer views · 1y Here is the derivation I saw in my algebra textbook: The surface of the circle is split up, by drawing lots of radii, close together. The resulting small circle areas are approximately the shape of triangles - and the closer together you draw the radii, the more accurate the approximation is. Finally, the areas of all the triangles are added up, and this is, of course, 1/2 times the circumference times the radius. Related questions How do I derive the formula for the surface area of a sphere without calculus using nets? Can you derive the formula for the surface area of a sphere without using calculus? How do I derive the surface area and volume of a sphere with calculus? How does one derive the equation of volume of a sphere? How do I derive the formulae for finding the volumes, total surface areas and lateral surface areas for spheres and cones? How can you derive the equation for the surface area of a sphere? Can we derive volume and area formulas for a sphere using its equation? What is the curved surface area of a sphere? What is the surface area of a cone frustum formula (simplified)? Can you derive the curved surface area of a frustum of sphere without use of integration? How do we get the formula of a sphere? What is the formula for the surface area of a sphere? What is the formula for finding the surface area of a sphere without integrating? What is the formula for finding the area of a sphere with a minimum surface area? How do you calculate the surface area of a sphere? Related questions How do I derive the formula for the surface area of a sphere without calculus using nets? Can you derive the formula for the surface area of a sphere without using calculus? How do I derive the surface area and volume of a sphere with calculus? How does one derive the equation of volume of a sphere? How do I derive the formulae for finding the volumes, total surface areas and lateral surface areas for spheres and cones? How can you derive the equation for the surface area of a sphere? Can we derive volume and area formulas for a sphere using its equation? What is the curved surface area of a sphere? What is the surface area of a cone frustum formula (simplified)? Can you derive the curved surface area of a frustum of sphere without use of integration? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
188385	https://artofproblemsolving.com/community/sentersoft?srsltid=AfmBOoopktrzl1m1z72gYvwATGIhOXvNSa1_19rcGirdOd5-_pwKg57b	AoPS Community Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 1-12. 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Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H A weird congruent case of triangles RopuToran 1 Given two acute triangles and satisfying ,,. Prove that 1 reply RopuToran 3 hours ago MathMaxGreat 14 minutes ago J H Mathcounts prep as I qualed into my school MATHCOUNTS team HoneyHap 14 Hey guys, With somewhat great difficulty, I was able to make it into the school MATHCOUNTS team. Now, my next step is to strengthen my rookie problem-solving skills and level it up. I am thinking because I don’t I have a lot of time, to take Po Shen Loh’s self paced course “Competition Booster Pack: Module 2+4+5” - , so I can be ready to perform well on the Mathcounts chapter round. Do you guys recommend taking this course with not that much time left? I am saying as I am in 8th grade and have never qualed for MATHCOUNTS team until now. So, I am thinking I might be way behind than the rest of my MATHCOUNTS peers in PA. Also, since I live in Pennsylvania, I want to know how competitive is PA in Mathcounts so I can see where I am at and how much practice I need to put in. If you guys have any helpful info, do let me know. 14 replies HoneyHap Yesterday at 9:41 PM HoneyHap 38 minutes ago J H 9 Pythagorean Triples ZMB038 146 Please put some of the ones you know, and try not to troll/start flame wars! Thank you :D 146 replies ZMB038 May 19, 2025 melloncandy Today at 3:21 AM J H Percent & Saline Mix Kushagra2012 2 How many liters of a % saline solution must be added to liters of a % saline solution to produce a % saline solution? %? %. Any pattern? 2 replies Kushagra2012 Saturday at 9:32 PM evt917 Today at 2:35 AM J High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked t 122,703 topics w 3 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H angle wanted , right triangle 2023 OMERJ L2.7 Brazil parmenides51 0 Let be a right triangle with such that . Let and be points on sides and , respectively, such that is parallel to . Finally, let be the intersection point of and . Calculate the angle . 0 replies parmenides51 12 minutes ago 0 replies J H area = 3 perimeter, integer sides, right triangle 2022 OMERJ L2.8 Brazil parmenides51 0 How many non-congruent right triangles have sides with integer lengths and areas numerically equal to three times their perimeters? 0 replies parmenides51 16 minutes ago 0 replies J H Inequalities sqing 25 Let and .Prove that Let and .Prove that Let and .Prove that Let and .Prove that Let and .Prove that Let and .Prove that 25 replies sqing Sep 22, 2025 NamelyOrange 41 minutes ago J H plane geometry chunchun.math.2010 0 Give the pointed triangle ABC inscribed (O). The line through O // BC cut AB AC at x y. Prove that the circle of diameter by, CX and the circle (O) goes through a point. 0 replies chunchun.math.2010 an hour ago 0 replies J Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked t 35,030 topics w 5 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H 9 What do you guys find difficult about the amc 10. littleduckysteve 1 Just for fun, let's see what people find difficult on the Amc 10. I will make another poll for strengths too. 1 reply littleduckysteve Today at 4:58 AM giratina3 11 minutes ago J H 9 What do you guys most strongest in, for Amc 10 littleduckysteve 1 Just for fun, what are you guys good at for amc 10? 1 reply littleduckysteve Today at 5:02 AM giratina3 12 minutes ago J H force overlay inversion vibes v4913 68 Source: USAMO 2023/6 Let be a triangle with incenter and excenters ,, and opposite ,, and , respectively. Let be an arbitrary point on the circumcircle of that does not lie on any of the lines ,, or . Suppose the circumcircles of and intersect at two distinct points and . If is the intersection of lines and , prove that . Proposed by Zach Chroman 68 replies v4913 Mar 23, 2023 bin_sherlo 3 hours ago J H States Question xHypotenuse 3 Hello everyone, I have a bit of confusion regarding states. Here is the question: when should I recursively define states, and when should I "forwardly" define states? For example, a "forwardly" defined states problem would be 2017 AMC 12A Problem 22. In this case, the states would be "forwardly" defined because we calculate the states based on how to go to the "next" state, not how we arrived upon it. An example of a "recursively" defined state would be 2021 aime ii, problem 8. This is because our states depend on how we arrived upon it, not how we go to the next one. From my understanding, we use recursively defined states if there is a set time/#steps involved but we use "forwardly" defined states in any expected value or "eventually reach" problems (any states problem with an absorbing state). I'm sorry if this question is unclear; I really have no idea how to phrase it. That being said, if anyone can understand/help, that would be really appreciate it! Thanks :) 3 replies xHypotenuse Sep 26, 2025 nsking_1209 6 hours ago J High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked t 317,099 topics w 13 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H angle wanted, equal 458o angles, equal segments given parmenides51 0 Source: 2023 L3 p5 OMERJ / Rio de Janeiro Mathematical Olympiad Let be a non-isosceles acute triangle. Let ,, and be points on sides ,, and such that and . Similarly, let ,, and be points on sides ,, and such that (typo in this relation) and . Knowing that , calculate . 0 replies parmenides51 2 minutes ago 0 replies J H Find all real functions yunxiu 70 Source: 2012 European Girls’ Mathematical Olympiad P3 Find all functions such that for all . Netherlands (Birgit van Dalen) 70 replies yunxiu Apr 12, 2012 math-olympiad-clown 2 minutes ago J H asymmetric inequality Noname23 1 Let :. Find min (and prove): 1 reply Noname23 34 minutes ago Noname23 4 minutes ago J H Constructing orthocenter using ruler with width Quantum-Phantom 23 Source: Canada MO 2024/5 Initially, three non-collinear points, ,, and , are marked on the plane. You have a pencil and a double-edged ruler of width . Using them, you may perform the following operations: [list] []Mark an arbitrary point in the plane. []Mark an arbitrary point on an already drawn line. []If two points and are marked, draw the line connecting and . []If two non-parallel lines and are drawn, mark the intersection of and . []If a line is drawn, draw a line parallel to that is at distance away from (note that two such lines may be drawn). [/list] Prove that it is possible to mark the orthocenter of using these operations. 23 replies Quantum-Phantom Mar 8, 2024 ray66 6 minutes ago J College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked t 114,929 topics w 1 user TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H integral of arctangent CyanSwan27 1 some of my friends saw i was bored at math class and just googled some random extra difficult elon musk level math and got this integral question i need integral of arctangent for the question, like how do you find integral of arctangent im not sure if this was the exact problem google it and you probably will find it or the question something like it 1 reply CyanSwan27 Yesterday at 12:18 AM Figaro an hour ago J H $n-a_n$ exists? SatisfiedMagma 4 Source: MU(theta) Let and . Show that exists and find its value. 4 replies SatisfiedMagma Saturday at 10:39 PM Alphaamss 2 hours ago J H Disjoint segments ZETA_in_olympiad 2 Source: 2022 Miklós Schweitzer Problem 5 Is it possible to select a non-degenerate segment from each line of the plane such that any two selected segments are disjoint? 2 replies 1 viewing ZETA_in_olympiad Nov 12, 2022 alinazarboland 3 hours ago J H Question on Goldbach Conjecture Verification MDB001 6 Hello, I was looking at the AoPS page on the Goldbach Conjecture, and something came to mind. When people say the conjecture has been verified up to 10^{18}, what exactly is the verification method? Is it done by generating all primes up to that bound and checking their sums to cover all even numbers in the interval? Or is it done the other way around, starting with an even number and directly decomposing it into two primes? If the method is the first, then the second seems more intriguing. Because if one could start from an even number, especially a large one, say 100 digits or more, and directly retrieve its two primes, what would that imply for the problem itself? Thank you for clarifying. I would be very curious about your perspectives. 6 replies MDB001 Yesterday at 9:23 AM MDB001 4 hours ago J Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked t 18,906 topics w 1 user TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H [RESOLVED] 8char limit wrongfully imposed jkim0656 13 Hi SS, I noticed that if you write a message and surround it in brackets [], it says that the message is shorter than eight characters even when it is definitely longer. Chrome Browser HP laptop I did notice that if you write at least 8 characters after the text in [], it lets the message through. 13 replies jkim0656 Yesterday at 1:05 AM Bangde 37 minutes ago J H Emoji rule O33ochan 19 Why does every post have a rule where there can only be 5 emojis? Can't it have more? I don't know why this rule exists, but if Idid, I would understand why this rule is not meant to be annoying. Thank you! 19 replies O33ochan Saturday at 3:13 PM UniqueScorpion89 5 hours ago J H Topics turning white RollingPanda4616 12 Summary of the problem: When rapidly clicking down topics, some of the topics turn white. Page URL: (found it here but might work elsewhere) Steps to reproduce: Go onto Site Support. View the top ~10 threads, turning them gray. Click the first topic "Suggestion Form". Go down the list of topics as fast as you can, using the things on the left. Click "Read me first/ How to...", then immediately click "Community Safety", and so on. Some topics will turn white randomly. Expected behavior: The topics will stay gray, unless someone else posts. Frequency: ~25% Operating system(s): macOS Sequoia 15.4 Browser(s), including version: Google Chrome 140 Additional information: It only happens on some threads. Refreshing fixes it, but it is still reproduceable. 12 replies RollingPanda4616 Sep 20, 2025 Staragon Today at 3:50 AM J H Is this a Glitch? ZuzabKoit 5 While I was looking at a topic it suddenly started scrolling. I was doing nothing, but I had to close the tab or hit the return to top button for it to stop. Even when I tried scrolling, it overrided my mouse and kept scrolling. Is my account getting hacked or is this a glitch? 5 replies ZuzabKoit Sep 26, 2025 KangarooPrecise Today at 1:37 AM J Other Forums and Collections ? AoPS Forums Forums for AoPS books, courses, and other resources. Forums for AoPS books, courses, and other resources. 3 V New Topic k Locked AoPS Forums Forums for AoPS books, courses, and other resources. Forums for AoPS books, courses, and other resources. 3 V New Topic k Locked Forums for AoPS books, courses, and other resources. AoPS Online Classes Questions and information about AoPS classesAoPS Books Questions and information about AoPS booksAlcumus The free AoPS online learning systemFor the Win! 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188386	https://math.stackexchange.com/questions/1443110/multiplicative-inverse-of-modulo-n-is-unique	Skip to main content Multiplicative Inverse of Modulo n is UNIQUE Ask Question Asked Modified 2 years, 5 months ago Viewed 6k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am trying to prove this: x,b,n∈Z, xb(modn)=1⟹ b is unique. I have tried by using proof by contradiction. That is, assume there are integers b,c where the above first condition is satisfied. Then we know that xb≡xc≡1(modn) meaning n\|(x(b−c)) and xb=xc+k1n,k1∈Z and xb+k2n=1,k2∈Z and xc+k3n=1,k3∈Z. That's all I got. I don't know how to proceed. Can anybody help? Thanks. modular-arithmetic Share CC BY-SA 3.0 Follow this question to receive notifications asked Sep 20, 2015 at 5:19 Jack NicholsonJack Nicholson 3311 silver badge33 bronze badges Add a comment \| 2 Answers 2 Reset to default This answer is useful 11 Save this answer. Show activity on this post. Assume xb≡xc≡1(modn). Then: b≡1b≡xcb≡xbc≡1c≡c(modn) Share CC BY-SA 3.0 Follow this answer to receive notifications answered Sep 20, 2015 at 6:39 Akiva WeinbergerAkiva Weinberger 25.5k22 gold badges5252 silver badges126126 bronze badges Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. Because gcd(x,n)=1 and n\|x(b−c)⟹n\|(b−c)⟺b≡c(modn) Share CC BY-SA 4.0 Follow this answer to receive notifications edited Mar 17, 2023 at 1:51 BakedPotato66 522 bronze badges answered Sep 20, 2015 at 5:20 lab bhattacharjeelab bhattacharjee 279k2020 gold badges213213 silver badges337337 bronze badges 5 @JackNicholson, 4≡1,7≡1(mod3) Congruent values are considered as same in modular arithmetic – lab bhattacharjee Commented Sep 20, 2015 at 5:37 Oh my goodness I finally get it now thanks! – Jack Nicholson Commented Sep 20, 2015 at 5:38 I'm new to finite field arithmetic, can you please explain to me why n divides b−c ? – Aravind A Commented Apr 23, 2020 at 4:29 1 @VivekanandV, en.wikipedia.org/wiki/Euclid%27s_lemma – lab bhattacharjee Commented Apr 23, 2020 at 4:51 @labbhattacharjee Thankyou very much !!! – Aravind A Commented Apr 23, 2020 at 4:55 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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188387	https://mathinsight.org/critical_points_monotone_increase_decrease_refresher	Skip to navigation (Press Enter) Math Insight To create your own interactive content like this, check out our new web site doenet.org! Critical points, monotone increase and decrease A function is called increasing if it increases as the input $x$ moves from left to right, and is called decreasing if it decreases as $x$ moves from left to right. Of course, a function can be increasing in some places and decreasing in others: that's the complication. We can notice that a function is increasing if the slope of its tangent is positive, and decreasing if the slope of its tangent is negative. Continuing with the idea that the slope of the tangent is the derivative: a function is increasing where its derivative is positive, and is decreasing where its derivative is negative. This is a great principle, because we don't have to graph the function or otherwise list lots of values to figure out where it's increasing and decreasing. If anything, it should be a big help in graphing to know in advance where the graph goes up and where it goes down. And the points where the tangent line is horizontal, that is, where the derivative is zero, are critical points. The points where the graph has a peak or a trough will certainly lie among the critical points, although there are other possibilities for critical points, as well. Further, for the kind of functions we'll deal with here, there is a fairly systematic way to get all this information: to find the intervals of increase and decrease of a function $f$: Compute the derivative $f'$ of $f$, and solve the equation $f'(x)=0$ for $x$ to find all the critical points, which we list in order as $x_1 < x_2 < \ldots < x_n$. (If there are points of discontinuity or non-differentiability, these points should be added to the list! But points of discontinuity or non-differentiability are not called critical points.) We need some auxiliary points: To the left of the leftmost critical point $x_1$ pick any convenient point $t_o$, between each pair of consecutive critical points $x_i,x_{i+1}$ choose any convenient point $t_i$, and to the right of the rightmost critical point $x_n$ choose a convenient point $t_n$. Evaluate the derivative $f'$ at all the auxiliary points $t_i$. Conclusion: if $f'(t_{i+1})>0$, then $f$ is increasing on $(x_i,x_{i+1})$, while if $f'(t_{i+1}) <0$, then $f$ is decreasing on that interval. Conclusion: on the ‘outside’ interval $(-\infty,x_o)$, the function $f$ is increasing if $f'(t_o)>0$ and is decreasing if $f'(t_o) <0$. Similarly, on $(x_n,\infty)$, the function $f$ is increasing if $f'(t_n)>0$ and is decreasing if $f'(t_n) <0$. It is certainly true that there are many possible shortcuts to this procedure, especially for polynomials of low degree or other rather special functions. However, if you are able to quickly compute values of (derivatives of!) functions on your calculator, you may as well use this procedure as any other. Exactly which auxiliary points we choose does not matter, as long as they fall in the correct intervals, since we just need a single sample on each interval to find out whether $f'$ is positive or negative there. Usually we pick integers or some other kind of number to make computation of the derivative there as easy as possible. It's important to realize that even if a question does not directly ask for critical points, and maybe does not ask about intervals either, still it is implicit that we have to find the critical points and see whether the functions is increasing or decreasing on the intervals between critical points. Examples Find the critical points and intervals on which $f(x)=x^2+2x+9$ is increasing and decreasing: Compute $f'(x)=2x+2$. Solve $2x+2=0$ to find only one critical point $-1$. To the left of $-1$ let's use the auxiliary point $t_o=-2$ and to the right use $t_1=0$. Then $f'(-2)=-2 <0$, so $f$ is decreasing on the interval $(-\infty,-1)$. And $f'(0)=2>0$, so $f$ is increasing on the interval $(-1,\infty)$. Find the critical points and intervals on which $f(x)=x^3-12x+3$ is increasing, decreasing. Compute $f'(x)=3x^2-12$. Solve $3x^2-12=0$: this simplifies to $x^2-4=0$, so the critical points are $\pm 2$. To the left of $-2$ choose auxiliary point $t_o=-3$, between $-2$ and $+2$ choose auxiliary point $t_1=0$, and to the right of $+2$ choose $t_2=3$. Plugging in the auxiliary points to the derivative, we find that $f'(-3)=27-12>0$, so $f$ is increasing on $(-\infty,-2)$. Since $f'(0)=-12 <0$, $f$ is decreasing on $(-2,+2)$, and since $f'(3)=27-12>0$, $f$ is increasing on $(2,\infty)$. Notice too that we don't really need to know the exact value of the derivative at the auxiliary points: all we care about is whether the derivative is positive or negative. The point is that sometimes some tedious computation can be avoided by stopping as soon as it becomes clear whether the derivative is positive or negative. Exercises Find the critical points and intervals on which $f(x)=x^2+2x+9$ is increasing, decreasing. Find the critical points and intervals on which $f(x)=3x^2-6x+7$ is increasing, decreasing. Find the critical points and intervals on which $f(x)=x^3-12x+3$ is increasing, decreasing. Thread navigation Calculus Refresher Previous: Tangent and normal lines Next: Minimization and maximization refresher Similar pages Introduction to local extrema of functions of two variables Two variable local extrema examples Minimization and maximization refresher Local minima and maxima (First Derivative Test) An algebra trick for finding critical points Minimization and maximization problems Solutions to minimization and maximization problems More similar pages
188388	https://math.stackexchange.com/questions/4737931/prime-solutions-to-p2p3-1-qq1	number theory - Prime solutions to $p^2(p^3-1) = q(q+1)$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prime solutions to p 2(p 3−1)=q(q+1)p 2(p 3−1)=q(q+1) Ask Question Asked 2 years, 2 months ago Modified2 years, 2 months ago Viewed 316 times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Solve: p 2(p 3−1)=q(q+1)p 2(p 3−1)=q(q+1) where p p and q q are prime numbers. Could someone help me with this question? I tried several things but I couldn't get it. I can show what I already did about that question, but I didn't go too far. I don't have the answers. number-theory elementary-number-theory prime-numbers diophantine-equations Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 30, 2023 at 0:45 Servaes 68k 8 8 gold badges 83 83 silver badges 174 174 bronze badges asked Jul 17, 2023 at 23:28 Eric MaireneEric Mairene 97 5 5 bronze badges 6 Hi! Yes, please show whatever you've done, and also give some context for the question, e.g., is it homework; if so, what course and topic; what you think might be relevant but you're not sure how to apply it; etc.Brian Tung –Brian Tung 2023-07-17 23:33:27 +00:00 Commented Jul 17, 2023 at 23:33 It occurs to me to take both primes in Z 6 Z 6, and consider the combinations knowing that the primes are of the form 6 n±1 6 n±1 Luis Alexandher –Luis Alexandher 2023-07-17 23:48:49 +00:00 Commented Jul 17, 2023 at 23:48 Hi! It's not homework, I'm studying number theory by myself because it's fun; this is a question from a brazilian book "Methods for solving Diophantine equations: from Pythagoras to Fermat" of Diego Marques. So I wrote down: p 2(p−1)(p 2+p+1)=q(q+1)p 2(p−1)(p 2+p+1)=q(q+1) then p\|q(q+1)p\|q(q+1) and since g c d(q,q+1)=1 g c d(q,q+1)=1. p p can't divide q q because they are prime and the only case would be p=q p=q, but it doesn't have solution. So p\|q+1 p\|q+1. Back to the equation q\|p 2(p−1)(p 2+p+1)q\|p 2(p−1)(p 2+p+1) but q q doesn't divide "p 2 p 2. So q\|(p−1)(p 2+p+1)q\|(p−1)(p 2+p+1) but q q can't divide p−1 p−1 either, because if so p−−1>q p−−1>q and q+1>p q+1>p, adding give 0>0 0>0 Eric Mairene –Eric Mairene 2023-07-17 23:51:55 +00:00 Commented Jul 17, 2023 at 23:51 so q\|p 2+p+1 q\|p 2+p+1 and p\|q+1 p\|q+1 but from here i dont know what to do Eric Mairene –Eric Mairene 2023-07-18 00:01:43 +00:00 Commented Jul 18, 2023 at 0:01 1 @JohnOmielan Tyvm!Eric Mairene –Eric Mairene 2023-07-18 00:14:32 +00:00 Commented Jul 18, 2023 at 0:14 \|Show 1 more comment 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. Suppose that p p and q q are prime numbers such that q(q+1)=p 2(p 3−1).q(q+1)=p 2(p 3−1). Clearly q>p q>p so q q divides p 3−1 p 3−1, but does not divide p−1 p−1. This implies q≡1(mod 3)q≡1(mod 3) and so p 2(p 3−1)≡q(q+1)≡2(mod 3),p 2(p 3−1)≡q(q+1)≡2(mod 3), which is impossible. Hence no such primes p p and q q exist. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 21, 2023 at 16:01 answered Jul 21, 2023 at 15:53 ServaesServaes 68k 8 8 gold badges 83 83 silver badges 174 174 bronze badges 2 1 Can you explain the deduction that q q is 1 mod 3 3 a bit more?hunter –hunter 2023-07-30 17:56:47 +00:00 Commented Jul 30, 2023 at 17:56 1 @hunter In the group (Z/q Z)×(Z/q Z)×, the element p p has order 3 3 because p 3≡1(mod q)p 3≡1(mod q) but p≢1(mod q)p≢1(mod q). This means q−1≡0(mod 3)q−1≡0(mod 3).Servaes –Servaes 2023-07-31 14:42:48 +00:00 Commented Jul 31, 2023 at 14:42 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. p=q p=q and p 2=q p 2=q are impossible, so we must have p 2∣q+1 p 2∣q+1. Let q+1=k p 2 q+1=k p 2 for some k>0 k>0. Note that q(q+1)=k p 2(k p 2−1)=k 2 p 4−k p 2≡−k p 2(mod p 4)q(q+1)=k p 2(k p 2−1)=k 2 p 4−k p 2≡−k p 2(mod p 4). However, p 2(p 3−1)≡−p 2(mod p 4)p 2(p 3−1)≡−p 2(mod p 4), so k≡1(mod p 2)k≡1(mod p 2). k=1 k=1 has no solutions, so k≥p 2+1 k≥p 2+1. Then, q=k p 2−1≥p 4+p 2−1 q=k p 2−1≥p 4+p 2−1. However, because q≠p,p 2 q≠p,p 2 we must have q∣p 3−1 q∣p 3−1, so q≤p 3−1 q≤p 3−1, contradiction because p≥1 p≥1. Thus, there are no solutions. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 18, 2023 at 3:58 answered Jul 17, 2023 at 23:50 Joshua WangJoshua Wang 6,253 12 12 silver badges 19 19 bronze badges 7 3 k p 2≡p 2(mod p 4)k p 2≡p 2(mod p 4) does not necessarily imply k≡1(mod p 4)k≡1(mod p 4). But it does imply that k=1+p s k=1+p s, for some s≥2 s≥2. Also, you wrote q+1=k p 2 q+1=k p 2 at the beginning and q=k p 2 q=k p 2 at the end.user1134696 –user1134696 2023-07-18 00:02:30 +00:00 Commented Jul 18, 2023 at 0:02 Thank you very much! I was missing the final conclusion Eric Mairene –Eric Mairene 2023-07-18 00:02:49 +00:00 Commented Jul 18, 2023 at 0:02 Thanks for catching that. I believe the proof still works.Joshua Wang –Joshua Wang 2023-07-18 00:06:50 +00:00 Commented Jul 18, 2023 at 0:06 ShyamalSayak, you're right but p\|q+1 p\|q+1 thus p 2(p 3−1)=k p(k p−1)p 2(p 3−1)=k p(k p−1) and p\|k p\|k which implies that k=p k=p but q=(p−1)(p+1)q=(p−1)(p+1) and it's composite for p>2 p>2 Eric Mairene –Eric Mairene 2023-07-18 00:08:01 +00:00 Commented Jul 18, 2023 at 0:08 Essentially, yes. I think it would be nice if you could edit your answer.user1134696 –user1134696 2023-07-18 00:08:15 +00:00 Commented Jul 18, 2023 at 0:08 \|Show 2 more comments This answer is useful 2 Save this answer. Show activity on this post. We are given that p p and q q are prime, with p 2(p 3−1)=q(q+1).(1)p 2(p 3−1)=q(q+1).(1) First note that 2 2(2 3−1)=28 2 2(2 3−1)=28 cannot be factorized in the form q(q+1)q(q+1). Hence p⩾3 p⩾3. Also observe that q≠p q≠p; otherwise eqn 1 1 becomes p 2(p 3−1)=p(p+1)p 2(p 3−1)=p(p+1) and so p(p 3−2)=1 p(p 3−2)=1: a falsehood since p>2 p>2. It follows that p 2 p 2 and q q are coprime. Therefore q q divides p 3−1=(p−1)(p 2+p+1)p 3−1=(p−1)(p 2+p+1). Now q q cannot divide p−1 p−1, because then q⩽p−1 q⩽p−1, and this would give p 2(p 3−1)⩽(p−1)p p 2(p 3−1)⩽(p−1)p and so p(p 2+p+1)⩽1 p(p 2+p+1)⩽1: an impossibility. Consequently q q divides p 2+p+1 p 2+p+1. It follows that q⩽p 2+p+1.q⩽p 2+p+1. From this inequality and eqn 1 1, dividing out the factor p 2+p+1 p 2+p+1 gives p 2(p−1)⩽p 2+p+2 p 2(p−1)⩽p 2+p+2, which may be written (p−3)3+7(p−3)2+14(p−3)+4⩽0.(p−3)3+7(p−3)2+14(p−3)+4⩽0. But this must be false, since p⩾3 p⩾3. Hence the given equation has no solution. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 30, 2023 at 21:03 answered Jul 30, 2023 at 17:03 John BentinJohn Bentin 20.3k 3 3 gold badges 51 51 silver badges 75 75 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions number-theory elementary-number-theory prime-numbers diophantine-equations See similar questions with these tags. 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188389	https://pm.szczecin.pl/uploads/imfich/zaklad-fizyki/11-Capacitance-and-Dielectrics.pdf	Storyline Chapter 25: Capacitance and Dielectrics Physics for Scientists and Engineers, 10e Raymond A. Serway John W . Jewett, Jr. Capacitor Capacitor: combination of two conductors, called plates Definition of Capacitance The capacitance C of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors: Q C V  1 F 1 C/V = Q C V =  Definition of Capacitance Consider capacitor formed from pair of parallel plates: • Each plate connected to one terminal of battery • Acts as source of potential difference • If capacitor initially uncharged: • Battery establishes electric field in connecting wires For plate connected to negative terminal of battery: • Electric field in wire applies force on electrons in wire immediately outside this plate: • Force causes electrons to move onto plate • Movement continues until plate, wire, and terminal all at same electric potential • Once equilibrium attained potential difference no longer exists between terminal and plate: • Result: no electric field present in wire and electrons stop moving • Plate now carries negative charge • Similar process occurs at other capacitor plate: • Electrons move from plate to wire, leaving plate positively charged Calculating Capacitance 0 0 Q E A    = = 0 Qd V Ed A   = = 0 0 / Q Q C V Qd A A C d   = =  = Capacitance 0A C d  = The parallel-plate capacitor: • Proportional to area of plates • Inversely proportional to plate separation ( ) 0 4π 2ln / L C b a  = The cylindrical capacitor: • depends on the radii a and b • is proportional to the length of the cylinders An example of this type of geometric arrangement is a coaxial cable, which consists of two concentric cylindrical conductors separated by an insulator. You probably have a coaxial cable attached to your television set if you are a subscriber to cable television. The coaxial cable is especially useful for shielding electrical signals from any possible external influences. The spherical capacitor: • depends on the radii a and b ( ) 0 4π ab C b a  = − 4π b C a  →  = Combinations of Capacitors Parallel Combination 1 2 V V V  =  =  tot 1 2 1 1 2 2 Q Q Q Q C V C V C V = +   =      =  +  Parallel Combination tot eq Q C V =  ( ) eq 1 2 parallel combination C C C = + ( ) eq 1 2 3 + + parallel combination C C C C = + tot 1 1 2 2 Q C V C V =  +  Series Combination 1 2 Q Q Q = = tot 1 2 1 2 1 2 V V V Q Q C C  =  +  = + Series Combination 1 2 eq 1 2 Q Q Q C C C = + ( ) eq 1 2 1 1 1 series combination C C C = + tot eq Q V C  = ( ) eq 1 2 3 1 1 1 1 + + series combination C C C C = + 1 2 tot 1 2 Q Q V C C  = + Energy Stored in a Charged Capacitor • Because positive and negative charges separated in system of two conductors in charged capacitor: Electric potential energy stored in system • If plates of charged capacitor connected by conductor such as wire Charge moves between each plate and connecting wire until capacitor uncharged. Discharge often observed as visible spark • If you accidentally touch opposite plates of charged capacitor your fingers act as pathway for discharge. Result: electric shock – Could be dangerous if high voltages present (i.e., power supply of home theater system) • Because charges can be stored in capacitor even when system turned off  Unplugging system does not make it safe to open case and touch components inside Energy Stored in a Charged Capacitor Energy Stored in a Charged Capacitor d d d q W V q q C =  = 2 0 0 1 d d 2 Q Q q Q W q q q C C C = = =   ( ) 2 2 1 1 2 2 2 E Q U Q V C V C = =  =  Work necessary to transfer increment of charge dq from plate carrying charge –q to plate carrying charge q (at higher electric potential) is: Total energy: Energy Stored in a Charged Capacitor ( ) 2 1 2 E U C V =  ( ) ( ) 2 2 0 0 1 1 2 2 E A U Ed Ad E d     = =     For parallel-plate capacitor: • potential difference related to electric field through V = Ed • capacitance: C = 0A/d Energy per unit volume (energy density) is: 2 0 1 2 E E U u E Ad  = = Expression generally valid regardless of source of electric field: • Energy density in any electric field square of magnitude of electric field at given point Portable Defibrillator • Stored energy released through te heart by conducting electrodes (paddles) which are placed on both sides of victim’s chest • Defibrillator can deliver energy to patient in 2 ms • Paramedics must wait between applications of energy because of time interval necessary for capacitors to become fully charged • Capacitors serve as energy reservoirs that can be slowly charged and then quickly discharged to provide large amounts of energy in a short pulse • When cardiac fibrillation (random contractions) occurs  heart produces rapid, irregular pattern of beats • Fast discharge of energy through heart can return the organ to normal beat pattern • Emergency medical teams use portable defibrillators that contain batteries capable of charging capacitor to high voltage • Circuitry actually permits capacitor to be charged to much higher voltage than that of battery • Up to 360 J stored in electric field of large capacitor in defibrillator when fully charged Capacitors with Dielectrics 0 V V    = 0 0 0 0 0 0 / Q Q C V V Q V C C    = =   =  = 0A C d  = Dielectric: nonconducting material (i.e., rubber, glass, or waxed paper) 𝜀= dielectric constant of the material Capacitors with Dielectrics 0A C d  = 1) Dielectric provides: • Increase in capacitance • Increase in maximum operating voltage 2) If magnitude of electric field 𝐸= 𝑈/𝑑in dielectric exceeds dielectric strength Insulating properties break down and dielectric begins to conduct, causing discharge. 3) When selecting capacitor for given application, must consider capacitance as well as expected voltage across capacitor in circuit. ( ) 2 2 1 1 2 2 2 E Q U Q V C V C = =  =  Electric Dipole in an Electric Field 2 p aq  Electric dipole: two charges of equal magnitude and opposite sign separated by distance a Electric dipole moment: vector Ԧ 𝑝directed from q toward +q along line joining the charges, with magnitude Electric Dipole in an Electric Field sin Fa   =  sin sin aqE pE    =  = =  τ p E sin Fa    + − = = sin r F rF    =  = Energy of Dipole cos E U pE  = − dW d  = sin pE   =   ( ) sin sin cos cos cos f f f i i i f i f i i f U U d pE d pE d pE pE               − = = = = − = −    E U = − p E Polar Molecules Microwave cooking Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, dispersed as molecular rotations, vibrations and/or translations in solids and liquids raises the temperature of the food, in a process similar to heat transfer by contact with a hotter body Soap and the Dipole Structure of Water • Grease and oil made up of nonpolar molecules • Generally not attracted to water • Plain water not very useful for removing this type of grime • Soap contains long molecules called surfactants • In long molecule  • Polarity characteristics of one end of molecule can be different from those at other end • In surfactant molecule: • One end acts like nonpolar molecule and other acts like polar molecule • Nonpolar end can attach to grease or oil molecule • Polar end can attach to water molecule • Soap serves as chain linking dirt and water molecules together • When water rinsed away: • Grease and oil go with it Induced Polarization • Symmetric molecule (figure): no permanent polarization • Polarization can be induced by placing molecule in electric field.
188390	https://www.thesaurus.com/browse/intricate	Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement intricate Definition for intricate adjective as in complicated, elaborate Strongest matches baroque, complex, convoluted, difficult, labyrinthine, sophisticated, tangled, tortuous, tricky Weak matches abstruse, byzantine, can of worms, daedal, entangled, fancy, hard, high-tech, involved, obscure, perplexing, rococo Discover More Example Sentences Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com. Both Arsenal and City have been sides that have dominated the ball, playing in an intricate fashion. FromBBC At a weekly ballet class in Wokingham, dancers glide elegantly through an intricate series of poses, making it look easy despite ranging in age from 50 to over 80. FromBBC His novels move with kinetic energy, his plots are intricate puzzles shrouded in religious iconography, ancient cryptography and other obscure arcana. FromLos Angeles Times Swift's greatest gift as a lyricist is the way she weaves her own story into her songs, balancing intricate, specific detail with universal themes of love, hope, heartbreak and betrayal. FromBBC They feature an intricate bee design on the tails side, representing his love of nature. FromBBC Advertisement Discover More Related Words Words related to intricate are not direct synonyms, but are associated with the word intricate. Browse related words to learn more about word associations. abstruse adjectiveas in difficult to understand Greek to me abstract clear as dishwater complex complicated deep enigmatic esoteric heavy hidden incomprehensible intricate involved muddy obscure perplexing profound puzzling recondite subtle unfathomable vague byzantine adjectiveas in complex complicated convoluted daedal detailed devious difficult elaborate intricate involved labyrinthine tortuous complex adjectiveas in difficult to understand Daedalean Gordian abstruse bewildering byzantine circuitous complicated confused convoluted crabbed cryptic discursive disordered disturbing enigmatic entangled excursive hidden impenetrable inscrutable interwoven intricate involved jumbled knotted knotty labyrinthine mazy meandering mingled mixed muddled obscure paradoxical perplexing puzzling rambling recondite round-about sinuous snarled sophisticated tangled tortuous undecipherable unfathomable winding complicated adjectiveas in difficult, complex Daedalean Gordian abstruse arduous byzantine can of worms convoluted difficult elaborate entangled fancy gasser hard high-tech interlaced intricate involved knotty labyrinthine mega factor mixed perplexing problematic puzzling recondite sophisticated troublesome various wheels within wheels convolute verbas in make complicated cloud coil contort intricate loop snaking spiral tangle twist wind Viewing 5/40related words From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement
188391	https://www.amazon.com/Computer-Graphics-Principles-Practice-2nd/dp/0201848406	Books › Computers & Technology › Programming › Graphics & Multimedia Enjoy fast, free delivery, exclusive deals, and award-winning movies & TV shows. Join Prime Hardcover $27.64 - $58.04) Paperback $28.99 Other Used and New from $3.09 Hardcover from $3.09 Paperback from $3.25 Buy new: -42% $58.04$58.04 FREE delivery Thursday, September 4 Ships from: Amazon Sold by: Hope's Ark Save with Used - Acceptable $27.64$27.64 FREE delivery Thursday, September 4 on orders shipped by Amazon over $35 Ships from: Amazon Sold by: QCompany Other sellers on Amazon New & Used (36) from $3.09$3.09 + $3.98 shipping Download the free Kindle app and start reading Kindle books instantly on your smartphone, tablet, or computer - no Kindle device required. Read instantly on your browser with Kindle for Web. Using your mobile phone camera - scan the code below and download the Kindle app. Image Unavailable Image not available for Color: To view this video download Flash Player VIDEOS 360° VIEW IMAGES Follow the authors John F. HughesJohn F. Hughes Follow James D. FoleyJames D. Foley Follow Something went wrong. Please try your request again later. OK Computer Graphics: Principles and Practice 2nd Edition by Andries Van Dam (Author), Steven K. Feiner (Author), John F. Hughes (Author), James D. Foley (Editor) & 1 more 4.6 4.6 out of 5 stars 71 ratings Sorry, there was a problem loading this page. Try again. See all formats and editions {"desktop_buybox_group_1":[{"displayPrice":"$58.04","priceAmount":58.04,"currencySymbol":"$","integerValue":"58","decimalSeparator":".","fractionalValue":"04","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"4M%2FtjsttDNpRb4b3urehkM1G8N2VYWbIOYubwiCKEwW2xS6RLXVMeAFNd0D%2FyNgsS40VIKGg8uuX%2BJmtSrVbh6LlXezsbPtVFBBMRKN88dbeDtrZJ8tAYtdpbv3XY5wQOh%2BKk%2BdiAmNKyN4CM3dBUo%2FgVawzcARqIQXUFtppg3lEMXnIFacCN%2BMQvADVfrmn","locale":"en-US","buyingOptionType":"NEW","aapiBuyingOptionIndex":0}, {"displayPrice":"$27.64","priceAmount":27.64,"currencySymbol":"$","integerValue":"27","decimalSeparator":".","fractionalValue":"64","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"4M%2FtjsttDNpRb4b3urehkM1G8N2VYWbItAnyANo8pt7JxQtIfiDc%2FHUCYXUk6quPD1WF79G19AjcK4hOaFDVV1bb%2BIkLO6LjwQnthvRKLbzTmy6obWr%2B95WD8kyT9x1xpOpZRBjMfjfHieEuyHwhkZagBwWZ14ZEM6buM6WyNlaMza9D4Q2HDwMeRrnjxvgT","locale":"en-US","buyingOptionType":"USED","aapiBuyingOptionIndex":1}]} Purchase options and add-ons A guide to the concepts and applications of computer graphics covers such topics as interaction techniques, dialogue design, and user interface software. Read more Report an issue with this product or seller Previous slide of product details ISBN-10 9780201848403 2. ISBN-13 978-0201848403 3. Edition 2nd 4. Publisher Addison-Wesley Professional 5. Publication date January 1, 1995 6. Language English 7. Dimensions 6.75 x 1.75 x 9.75 inches 8. Print length 1175 pages Next slide of product details See all details There is a newer edition of this item: Computer Graphics: Principles and Practice $133.97 (47) Only 2 left in stock - order soon. Frequently bought together This item: Computer Graphics: Principles and Practice $54.02$54.02 Get it Sep 8 - 9 Only 1 left in stock - order soon. Ships from and sold by MMC Store LLc. + Computer Graphics, C Version (2nd Edition) $57.02$57.02 Only 1 left in stock - order soon. Ships from and sold by SozBooks. Total price: $111.04$111.04 To see our price, add these items to your cart. Try again! Details Added to Cart These items are shipped from and sold by different sellers. Show details Hide details Choose items to buy together. Popular titles by this author Page 1 of 2 Start over Previous set of slides Computer Graphics: Principles and Practice John Hughes 4.0 out of 5 stars 47 Hardcover $133.97$133.97 FREE Shipping Only 2 left in stock - order soon. 2. Computer Graphics: Principles and Practice HUGHES FOLEY, VAN DAM, FEINER 4.6 out of 5 stars 71 Paperback 34 offers from $3.25 3. Introduction to Computer Graphics James D. Foley 4.3 out of 5 stars 11 Hardcover 38 offers from $1.69 4. Computer Graphics: Principles and Practice (text only) 2nd(Second) edition by J.D.Foley.A.V.Dam.S.K.Feiner J.D.Foley.A.V.Dam.S.K.Feiner 5.0 out of 5 stars 1 Hardcover $69.10$69.10 Get it Sep 9 - 11 $3.99 shipping 5. Computer Graphics: Principles And Practice, 3Rd Edition John F Hughes 4.2 out of 5 stars 78 Paperback $42.20$42.20 Get it Sep 8 - 15 FREE Shipping Only 15 left in stock - order soon. 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Computer Vision: Algorithms and Applications (Texts in Computer Science) Richard Szeliski 4.5 out of 5 stars 80 Hardcover $57.12$57.12 Get it as soon as Thursday, Sep 4 FREE Shipping by Amazon Only 11 left in stock - order soon. Next set of slides Editorial Reviews Amazon.com Review Computer Graphics: Principles and Practice is the most exhaustive overview of computer graphics techniques available. This textbook's 21 chapters cover graphics hardware, user interface software, rendering, and a host of other subjects. Assuming a solid background in computer science or a related field, Computer Graphicsgives example programs in C and provides exercises at the end of each chapter to test your knowledge of the material. The guide has over 100 beautiful, four-color photographs that illustrate important topics and algorithms, such as ray tracing and bump maps, and also inspire you to acquire the skills necessary to produce them. Encyclopedic in its coverage, the book has a good table of contents so that you can immediately turn to information on the z-Buffer algorithm or the chapter on animation. From the Inside Flap "Interactive graphics is a field whose time has come. Until recently it was an esoteric specialty involving expensive display a hardware, substantial computer resources, and idiosyncratic software. In the last few years, however, it has benefited from the steady and sometimes even spectacular reduction in the hardware price/performance ration (E.G., personal computers for home or office with their standard graphics terminals), and from the development of high-level , device -independent graphics packages that help make graphics programming rational and straightforward. Interactive graphics is now finally ready to fulfill its promise to provide us with pictorial communication and thus to become a major facilitator of man/machine interaction." (From preface, Fundamentals of Interactive Computer Graphics, James Foley and Andries van Dam, 1982) This assertion that computer graphics had finally arrived was made before the revolution in computer culture sparked by Apple's Macintosh and the IBM PC and its clones. Now even preschool children are comfortable with interactive-graphics techniques, such as the desktop metaphor for window manipulation and menu and icon selection with a mouse. Graphics-based user interfaces have made productive users of neophytes, and the desk without its graphics computer is increasingly rare. At the same time that interactive graphics has become common in user interfaces and visualization of data and objects, the rendering of 3D objects has become dramatically more realistic, as evidenced by the ubiquitous computer-generated commercials and movie special effects. Techniques that were experimental in the early eighties are now standard practice, and more remarkable "photorealistic" effects are around the corner. The simpler kinds of pseudorealism, which took hours of computer time per image in the early eighties, now are done routinely at animation rates (ten or more frames/second) on personal computers. Thus "real-time" vector displays in 1981 showed moving wire-frame objects made of tens of thousands of vectors without hidden-edge removal; in 1990 real-time raster displays can show not only the same kinds of line drawings but also moving objects composed of as many as one hundred thousand triangles rendered with Gouraud or Phong shading and specular highlights and with full hidden-surface removal. The highest-performance systems provide real-time texture mapping, anitialiasing, atmospheric attenuation for fog and haze, and other advanced effects. Graphics software standards have also advanced significantly since our first edition. The SIGGRAPH Core '79 package, on which the first edition's SGP package was based, has all but disappeared, along with direct-view storage tube and refresh vector displays. The much more powerful PHIGS package, supporting storage and editing of structure hierarchy, has become an official ANSI and ISO standard, and it is widely available for real-time geometric graphics in scientific and engineering applications, along with PHIGS+, which supports lighting, shading, curves, and surfaces. Official graphics standards complement lower-level, more efficient de facto standards, such as Apple's QuickDraw X Window System's Xlip 2D integer raster graphics package, and Silicon Graphics' GL 3D library. Also widely available are implementations of Pixar's RenderMan interface for photorealistic rendering and Post Script interpreters for hardcopy page and screen image description. Better graphics software has been used to make dramatic improvements in the "look and feel" of user interfaces, and we may expect increasing use of 3D effects, both for aesthetic reasons and for providing new metaphors for organizing and presenting, and navigating through information. Perhaps the most important new movement in graphics is the increasing concern for modeling objects, not just for creating their pictures. Furthermore, interest is growing in describing the time-varying geometry and behavior of 3D objects. Thus graphics is increasingly concerned with simulation, animation, and a "back to physics" movement in both modeling and rendering in order to create objects that look and behave as realistically as possible. As the tools and capabilities available become more and more sophisticated and complex, we need to be able to apply them effectively. Rendering is no longer the bottleneck. Therefore researchers are beginning to apply artificial-intelligence techniques to assist in the design of object models, in motion planning, and in the layout of effective 2D and 3D graphical presentations. Today the frontiers of graphics are moving very rapidly, and a text that sets out to be a standard reference work must periodically be updated and expanded. This book is almost a total rewrite of the Fundamentals of Interactive Computer Graphics, and although this second edition contains nearly double the original 623 pages, we remain painfully aware of how much material we have been forced to omit. Major differences from the first edition include the following: The vector-graphics orientation is replaced by a raster orientation. The simple 2D floating-point graphics package (SG) is replaced by two packages--SRGP and SPHIGS--that reflect the two major schools of interactive graphics programming. SRGP combines features of the QuickDraw and Xlib 2D integer raster graphics packages. SPHIGS, based on PHIGS, provides the fundamental features of a 3D floating-point package with hierarchical display lists. We explain how to do applications programming in each of these packages and who how to implement the basic clipping, scan conversion, viewing, and display list traversal algorithms that underlie these systems. User-interface issues are discussed at considerable length, both for 2D desktop metaphors and for 3D interaction devices. Coverage of modeling is expanded to include NURB (nonuniform rational B-spline) curves and surfaces, a chapter on solid modeling, and a chapter on advanced modeling techniques, such as physically based modeling, procedural models, factals, L-grammar systems, and particle systems. Increased coverage of rendering includes a detailed treatment of antialiasing and greatly expanded chapters on visible-surface determination, illumination, and shading, including physically based illumination models, ray tracing, and radiosity. Material is added on advanced raster graphics architectures and algorithms, including clipping and scan-conversion of complex primitives and simple image-processing operations, such as compositing. A brief introduction to animation is added. This text can be used by those without prior background in graphics and only some background in Pascal programming, basic data structures and algorithms, computer architecture, and simple linear algebra. An appendix reviews the necessary mathematical foundations. The book covers enough material for a full-year course, but is partitioned into groups to make selective coverage possible. The reader, therefore, can progress through a carefully designed sequence of units, starting with simple, generally applicable fundamentals and ending with more complex and specialized subjects. Basic Group. Chapter 1 provides a historical perspective and some fundamental issues in hardware, software, and applications. Chapters 2 and 3 describe, respectively, the use and the implementation of SRGP, a simple 2D integer graphics package. Chapter 4 introduces graphics hardware, including some hints about how to use hardware in implementing the operations described in the preceding chapters. The next two chapters, 5 and 6, introduce the ideas of transformations in the plane and 3-space, representations by matrices, the use of homogeneous coordinates unify linear and affine transformations, and the description of 3D views, including the transformations from arbitrary view volumes to canonical view volumes. Finally, Chapter 7 introduces SPHIGS, a 3D floating-point hierarchical graphics package that is a simplified version of the PHIGS standard, and describes its use in some basic modeling operations. Chapter 7 also discusses the advantages and disadvantages of the hierarchy available in PHIGS and the structure of applications that use this graphics package. User Interface Group. Chapters 8-10 describe the current technology of interaction devices and then address the higher-level issues in user-interface design. Various popular user-interface paradigms are described and critiqued. In the final chapter user-interface software, such as window mangers, interaction technique-libraries, and user-interface management systems, is addressed. Model Definition Group. The first two modeling chapters, 11 and 12, describe the current technologies used in geometric modeling: the representation of curves and surfaces by parametric functions, especially cubic splines, and the representation of solids by various techniques, including boundary representations and CSG models. Chapter 13 introduces the human color-vision system, various color-description systems, and conversion from one to another. This chapter also briefly addresses rules for the effective use of color. Image Syntheses Group. Chapter 14, the first in a four-chapter sequence, describes the quest for realism from the earliest vector drawings to state-of-the-art shaded graphics. The artifacts caused by aliasing are of crucial concern in raster graphics, and this chapter discusses their causes and cures in considerable detail by introducing the Fourier transform and convolution. Chapter 15 describes a variety of strategies for visible-surface determination in enough detail to allow the reader to implement some of the most important ones. Illumination and shading algorithms are covered in detail in Chapter 16. The early part of part of this chapter discusses algorithms most commonly found in current hardware, while the remainder treats texture, shadows, transparency, reflections, physically based illumination models, rat tracing, and radiosity methods. The last chapter in this group, Chapter 17, describes both image manipulations, such as scaling, shearing, and rotating pixmaps, and image storage techniques, including various image-compression schemes. Advanced Techniques Group. The last four chapters give an overview of the current state of the art (a moving target, of course). Chapter 18 describes advanced graphics hardware used in high-end commercial and research machines; this chapter was contributed by Steven Molnar and Henry Fuchs, authorities on high-performance graphics architectures. Chapter 19 describes the complex raster algorithms used for such tasks as scan-converting arbitary conics, generating antialiased text, and implementing page-description languages, such as PostScript. The final two chapters survey some of the most important techniques in the fields of high-level modeling and computer animation. The first two groups cover only elementary material and thus can be used for a basic course at the undergraduate level. A follow-on course can then use the more advanced chapters. Alternatively, instructors can assemble customized courses by picking chapters out of the various groups. For example, a course designed to introduce students to primarily 2D graphics would include Chapters 1 and 2, simple scan conversion and clipping from Chapter 3, a technology overview with emphasis on raster architectures and interaction devices from Chapter 4, homogeneous mathematics from Chapter 5, and 3D viewing only from a "how to use it" point of view from Sections 6.1 to 6.3. The User Interface Group, Chapters 8-10, would be followed by selected introductory sections and simple algorithms from the Image Syntheses Group, Chapters 14, 15, and 16. A one-course general overview of graphics would include Chapters 1 and 2, basic algorithms from Chapter 3, raster architectures and interaction devices from Chapter 4, Chapter 5, and most of Chapters 6 and 7 on viewing and SPHIGs. The second half of the course would include sections on modeling from Chapters 11 and 13, on image syntheses from Chapters 14, 15, and 16, and on advanced modeling from Chapter 20 to give breadth of coverage in these slightly more advanced areas. A course emphasizing 3D modeling and rendering would start with Chapter 3 sections on scan converting, clipping of lines and polygons, and introducing antialiasing. The course would then progress to Chapters 5 and 6 on the basic mathematics of transformations and viewing. Chapter 13 on color, and then cover the key Chapters 14, 15 and 16 in the Image Syntheses Group. Coverage would be rounded off by selections in surface and solid modeling. Chapter 20 on advanced modeling, and Chapter 21 on animation from the Advanced Techniques Group. Graphics Packages The SRGP and SPHIGS graphics packages, designed by David Sklar, coauthor of the two chapters on these packages, are available from the publisher for the IBM PC (ISBN 0-201-54700-7), the Macintosh (ISBN 0-201-54701-5), and UNIX workstations running X11, as are many of the algorithms for scan conversion, clipping, and viewing (see page 1175). Acknowledgments This book could not have been produced without the dedicated work and the indulgence of many friends and colleagues. We acknowledge here our debt to those who have contributed significantly to one or more chapters; many others have helped by commenting on individual chapters, and we are grateful to them as well. We regret any inadvertent omissions. Katrina Avery and Lyn Dupré did a superb job of editing. Additional valuable editing on multiple versions of multiple chapters was provided by Debbie van Dam, Melissa Gold, and Clare Campbell. We are especially grateful to our production supervisor Bette Aaronson, our art director, Joe Vetere, and our editor, Keith Wollman, not only for their great help in producing the book, but also for their patience and good humor under admittedly adverse circumstances--if we ever made a promised deadline during these frantic five years, we can't remember it! Computer graphics has become too complex for even a team of four main authors and three guest authors to be expert in all areas. We relied on colleagues and students to amplify our knowledge, catch our mistakes and provide constructive criticism of form and content. We take full responsibility for any remaining sins of omission and commission. Detailed technical readings on one or more chapters were provided by John Airey, Kurt Akeley, Tom Banchoff, Brian Barsky, David Bates, Cliff Beshers, Gary Bishop, Peter J Bono, Marvin Bunker, Bill Buxton, Edward Chang, Norman Chin, Michael F. Cohen, William Cowan, John Dennis, Tom Dewald, Scott Draves, Steve Drucker, Tom Duff, Richard Economy, David Ellsworth, Nick England, Jerry Farrell, Robin Forrest, Alain Fournier, Alan Freiden, Christina Gibbs, Melissa Gold, Mark Green, Cathleen Greenberg, Margaret Hagen, Griff Hamlin, Pat Hanrahan, John Heidema, Rob Jacob, Abid Kamran, Mike Kappel, Henry Kaufman, Karen Kendler, David Kurlander, David Laidlaw, Keith Lantz, Hsien-Che Lee, Aaron Marcus, Nelson Max, Deborah Mayhew, Barbara Meier, Gary Meyer, Jim Michener, Jakob Nielsen, Mark Nodine, Randy Pausch, Ari Requicha, David Rosenthal, David Salesin, Nanan Samet, James Sanford, James Sargent, Robin Schaufler, Robert Scheifler, John Schnizlein, Michael Shantzis, Ben Shneiderman, Ken Shoemake, Judith Schrier, John Sibert, Dave Simons, Jonathan Steinhart, Maureen Stone, Paul Strauss, Seth Tager, Peter Tanner, Bruce Tebbs, Ben Trumbore, Yi Tso, Greg Turk, Jeff Vroom, Colin Ware, Gary Watkins, Chuck Weger, Kevin Weiler, Turner Whitted, George Wolberg, and Larry Wolff. Several colleagues, including Jack Bresenham, Brian Barsky, Jerry Van Aken, Dilip Da Silva (who suggested the uniform midpoint treatment of Chapter 3) and Don Hatfield, not only read chapters closely but also provided detailed suggestions on algorithms. Welcome word-processing relief was provided by Katrina Avery, Barbara Britten, Clare Campbell, Tina Cantor, Joyce Cavatoni, Louisa Hogan, Jenni Rodda, and Debbie van Dam. Drawings for Chapters 1-3 were ably created by Dan Robbins, Scott Snibbe, Tina Cantor, and Clare Campbell. Figure and image sequences created for this book were provided by Beth Cobb, David Kurlander, Allen Paeth, and George Wolberg (with assistance from Peter Karp). Plates II.21-37, showing a progression of rendering techniques, were designed and rendered at Pixar by Thomas Williams and H.B. Siegel, under the direction of M.W. Mantle, using Pixar's PhotoRealistic Renderman software. Thanks to Industrial Light & Magic for the use of their laser scanner to create Plates II.24-37, and to Norman Chin for computing vertex normals for Color Plates II.30-32. L. Lu and Carles Castellsagué wrote programs to make figures. Jeff Vogel implemented the algorithms of Chapter 3, and he and Atul Butte verified the code in Chapters 2 and 7. David Sklar wrote the Mac and X11 implementations of SRGP and SPHIGS with help from Ron Balsys, Scott Boyajian, Atul Butte, Alex Contovounesios, and Scott Draves. Randy Pausch and his students ported the packages to the PC environment. We have installed an automated electronic mail server to allow our readers to obtain machine-readable copies of many of the algorithms, suggest exercises, report errors in the text and in SRGP/SPHIGS, and obtain errata lists for the text and software. Send email to "graphtext @ cs.brown" with a Subject line of "Help" to receive the current list of available services (See page 1175 for information on how to order SRGP and SPHIGS.) Preface to the C Edition This is the C-language version of a book originally written with examples in Pascal. It includes all changes through the ninth printing of the Pascal second edition, as well as minor modifications to several algorithms, and all it s Pascal code has been rewritten in ANSI C. The interfaces to the SRGP and SPHIGS graphics packages are now defined in C, rather than Pascal, and correspond to the new C implementations of these packages. (See page 1175 for information on obtaining the software.) We wish to thank Norman Chin for converting the Pascal code of the second edition to C, proofreading it, and formatting it using the typographic conventions of the original. Thanks to Matt Ayers for careful proofing of Chapters 2, 3, and 7, and for useful suggestions about conversion problems. 0201848406P04062001 From the Back Cover The best-selling book on computer graphics is now available in this C-language version. All code has been converted into C, and changes through the ninth printing of the second edition have been incorporated. The book's many outstanding features continue to ensure its position as the standard computer graphics text and reference. By uniquely combining current concepts and practical applications in computer graphics, four well-known authors provide here the most comprehensive, authoritative, and up-to-date coverage of the field. The important algorithms in 2D and 3D graphics are detailed for easy implementation, including a close look at the more subtle special cases. There is also a thorough presentation of the mathematical principles of geometric transformations and viewing. In this book, the authors explore multiple perspectives on computer graphics: the user's, the application programmer's, the package implementor's, and the hardware designer's. For example, the issues of user-centered design are expertly addressed in three chapters on interaction techniques, dialogue design, and user interface software. Hardware concerns are examined in a chapter, contributed by Steven Molnar and Henry Fuchs, on advanced architectures for real-time, high performance graphics. The comprehensive topic coverage includes: Programming with SRGP, a simple but powerful raster graphics package that combines features of Apple's QuickDraw and the MIT X Window System graphics library. Hierarchical, geometric modeling using SPHIGS, a simplified dialect of the 3D graphics standard PHIGS. Raster graphics hardware and software, including both basic and advanced algorithms for scan converting and clipping lines, polygons, conics, spline curves, and text. Image synthesis, including visible-surface determination, illumination and shading models, image manipulation, and antialiasing. Techniques for photorealistic rendering, including ray tracing and radiosity methods. Surface modeling with parametric polynomials, including NURBS, and solid-modeling representations such as B-reps, CSG, and octrees. Advanced modeling techniques such as fractals, grammar-based models, particle systems. Concepts of computer animation and descriptions of state-of-the-art animation systems. Over 100 full-color plates and over 700 figures illustrate the techniques presented in the book. 0201848406B04062001 About the Author James D. Foley (Ph.D., University of Michigan) is the founding director of the interdisciplinary Graphics, Visualization & Usability Center at Georgia Institute of Technology, and Professor of Computer Science and of Electrical Engineering. Coauthor with Andries van Dam of Fundamentals of Interactive Computer Graphics, Foley is a member of ACM, ACM SIGGRAPH, ACM SIGCHI, the Human Factors Society, IEEE, and the IEEE Computer Society. He recently served as Editor-in-Chief of ACM Transactions on Graphics, and is on the editorial boards of Computers and Graphics, User Modeling and User-Adapted Interaction, and Presence. His research interests include model-based user interface development tools, user interface software, information visualization, multimedia, and human factors of the user interface. Foley is a Fellow of the IEEE, and a member of Phi Beta Kappa, Tau Beta Phi, Eta Kappa Nu, and Sigma Xi. At Georgia Tech, he has received College of Computing graduate student awards as Most Likely to Make Students Want to Grow Up to Be Professors, Most Inspirational Faculty Member, the campus Interdisciplinary Activities Award, and the Sigma Xi Sustained Research Award. In 1997, Foley received the SIGGRAPH Steven A. Coons Award. Andries van Dam (Ph.D., University of Pennsylvania) was the first chairman of the Computer Science Department at Brown University. Currently Thomas J. Watson, Jr. University Professor of Technology and Education and Professor of Computer Science at Brown, he is also Director of the NSF/ARPA Science and Technology Center for Computer Graphics and Scientific Visualization. His research interests include computer graphics, hypermedia systems, and workstations. He is past Chairman of the Computing Research Association, Chief Scientist at Electronic Book Technologies, Chairman of Object Power's Technical Advisory Board, and a member of Microsoft's Technical Advisory Board. A Fellow of both the IEEE Computer Society and of ACM, he is also cofounder of ACM SIGGRAPH. Coauthor of the widely used book Fundamentals of Interactive Computer Graphics with James Foley, and of Object-Oriented Programming in Pascal: A Graphical Approach, with D. Brookshire Conner and David Niguidula, he has, in addition, published over eighty papers. In 1990 van Dam received the NCGA Academic Award, in 1991, the SIGGRAPH Steven A. Coons Award, and in 1993 the ACM Karl V. Karlstrom Outstanding Educator Award. Steven K. Feiner (Ph.D., Brown University) is Associate Professor of Computer Science at Columbia University, where he directs the Computer Graphics and User Interfaces Lab. His current research focuses on 3D user interfaces, virtual worlds, augmented reality, knowledge-based design of graphics and multimedia, animation, visualization, and hypermedia. Dr. Feiner is on the editorial boards of ACM Transactions on Graphics, IEEE Transactions on Visualizations and Computer Graphics, and Electronic Publishing, and is on the executive board of the IEEE Technical Committee on Computer Graphics. He is a member of ACM SIGGRAPH and the IEEE Computer Society. In 1991 he received an ONR Young Investigator Award. Dr. Feiner's work has been published in over fifty papers and presented in numerous talks, tutorials, and panels. John F. Hughes (Ph.D., University of California, Berkeley) is an Assistant Professor of Computer Science at Brown University, where he codirects the computer graphics group with Andries van Dam. His research interests are in applications of mathematics to computer graphics, scientific visualization, mathematical shape description, mathematical fundamentals of computer graphics, and low-dimensional topology and geometry. He is a member of the AMS, IEEE, and ACM SIGGRAPH. His recent papers have appeared in Computer Graphics, and in Visualization Conference Proceedings. He also has a long-standing interest in the use of computer graphics in mathematics education. 0201848406AB04062001 Excerpt. © Reprinted by permission. All rights reserved. Preface "Interactive graphics is a field whose time has come. Until recently it was an esoteric specialty involving expensive display a hardware, substantial computer resources, and idiosyncratic software. In the last few years, however, it has benefited from the steady and sometimes even spectacular reduction in the hardware price/performance ration (E.G., personal computers for home or office with their standard graphics terminals), and from the development of high-level , device -independent graphics packages that help make graphics programming rational and straightforward. Interactive graphics is now finally ready to fulfill its promise to provide us with pictorial communication and thus to become a major facilitator of man/machine interaction." (From preface, Fundamentals of Interactive Computer Graphics, James Foley and Andries van Dam, 1982) This assertion that computer graphics had finally arrived was made before the revolution in computer culture sparked by Apple's Macintosh and the IBM PC and its clones. Now even preschool children are comfortable with interactive-graphics techniques, such as the desktop metaphor for window manipulation and menu and icon selection with a mouse. Graphics-based user interfaces have made productive users of neophytes, and the desk without its graphics computer is increasingly rare. At the same time that interactive graphics has become common in user interfaces and visualization of data and objects, the rendering of 3D objects has become dramatically more realistic, as evidenced by the ubiquitous computer-generated commercials and movie special effects. Techniques that were experimental in the early eighties are now standard practice, and more remarkable "photorealistic" effects are around the corner. The simpler kinds of pseudorealism, which took hours of computer time per image in the early eighties, now are done routinely at animation rates (ten or more frames/second) on personal computers. Thus "real-time" vector displays in 1981 showed moving wire-frame objects made of tens of thousands of vectors without hidden-edge removal; in 1990 real-time raster displays can show not only the same kinds of line drawings but also moving objects composed of as many as one hundred thousand triangles rendered with Gouraud or Phong shading and specular highlights and with full hidden-surface removal. The highest-performance systems provide real-time texture mapping, anitialiasing, atmospheric attenuation for fog and haze, and other advanced effects. Graphics software standards have also advanced significantly since our first edition. The SIGGRAPH Core '79 package, on which the first edition's SGP package was based, has all but disappeared, along with direct-view storage tube and refresh vector displays. The much more powerful PHIGS package, supporting storage and editing of structure hierarchy, has become an official ANSI and ISO standard, and it is widely available for real-time geometric graphics in scientific and engineering applications, along with PHIGS+, which supports lighting, shading, curves, and surfaces. Official graphics standards complement lower-level, more efficient de facto standards, such as Apple's QuickDraw X Window System's Xlip 2D integer raster graphics package, and Silicon Graphics' GL 3D library. Also widely available are implementations of Pixar's RenderMan interface for photorealistic rendering and Post Script interpreters for hardcopy page and screen image description. Better graphics software has been used to make dramatic improvements in the "look and feel" of user interfaces, and we may expect increasing use of 3D effects, both for aesthetic reasons and for providing new metaphors for organizing and presenting, and navigating through information. Perhaps the most important new movement in graphics is the increasing concern for modeling objects, not just for creating their pictures. Furthermore, interest is growing in describing the time-varying geometry and behavior of 3D objects. Thus graphics is increasingly concerned with simulation, animation, and a "back to physics" movement in both modeling and rendering in order to create objects that look and behave as realistically as possible. As the tools and capabilities available become more and more sophisticated and complex, we need to be able to apply them effectively. Rendering is no longer the bottleneck. Therefore researchers are beginning to apply artificial-intelligence techniques to assist in the design of object models, in motion planning, and in the layout of effective 2D and 3D graphical presentations. Today the frontiers of graphics are moving very rapidly, and a text that sets out to be a standard reference work must periodically be updated and expanded. This book is almost a total rewrite of the Fundamentals of Interactive Computer Graphics, and although this second edition contains nearly double the original 623 pages, we remain painfully aware of how much material we have been forced to omit. Major differences from the first edition include the following: The vector-graphics orientation is replaced by a raster orientation. The simple 2D floating-point graphics package (SG) is replaced by two packages--SRGP and SPHIGS--that reflect the two major schools of interactive graphics programming. SRGP combines features of the QuickDraw and Xlib 2D integer raster graphics packages. SPHIGS, based on PHIGS, provides the fundamental features of a 3D floating-point package with hierarchical display lists. We explain how to do applications programming in each of these packages and who how to implement the basic clipping, scan conversion, viewing, and display list traversal algorithms that underlie these systems. User-interface issues are discussed at considerable length, both for 2D desktop metaphors and for 3D interaction devices. Coverage of modeling is expanded to include NURB (nonuniform rational B-spline) curves and surfaces, a chapter on solid modeling, and a chapter on advanced modeling techniques, such as physically based modeling, procedural models, factals, L-grammar systems, and particle systems. Increased coverage of rendering includes a detailed treatment of antialiasing and greatly expanded chapters on visible-surface determination, illumination, and shading, including physically based illumination models, ray tracing, and radiosity. Material is added on advanced raster graphics architectures and algorithms, including clipping and scan-conversion of complex primitives and simple image-processing operations, such as compositing. A brief introduction to animation is added. This text can be used by those without prior background in graphics and only some background in Pascal programming, basic data structures and algorithms, computer architecture, and simple linear algebra. An appendix reviews the necessary mathematical foundations. The book covers enough material for a full-year course, but is partitioned into groups to make selective coverage possible. The reader, therefore, can progress through a carefully designed sequence of units, starting with simple, generally applicable fundamentals and ending with more complex and specialized subjects. Basic Group. Chapter 1 provides a historical perspective and some fundamental issues in hardware, software, and applications. Chapters 2 and 3 describe, respectively, the use and the implementation of SRGP, a simple 2D integer graphics package. Chapter 4 introduces graphics hardware, including some hints about how to use hardware in implementing the operations described in the preceding chapters. The next two chapters, 5 and 6, introduce the ideas of transformations in the plane and 3-space, representations by matrices, the use of homogeneous coordinates unify linear and affine transformations, and the description of 3D views, including the transformations from arbitrary view volumes to canonical view volumes. Finally, Chapter 7 introduces SPHIGS, a 3D floating-point hierarchical graphics package that is a simplified version of the PHIGS standard, and describes its use in some basic modeling operations. Chapter 7 also discusses the advantages and disadvantages of the hierarchy available in PHIGS and the structure of applications that use this graphics package. User Interface Group. Chapters 8-10 describe the current technology of interaction devices and then address the higher-level issues in user-interface design. Various popular user-interface paradigms are described and critiqued. In the final chapter user-interface software, such as window mangers, interaction technique-libraries, and user-interface management systems, is addressed. Model Definition Group. The first two modeling chapters, 11 and 12, describe the current technologies used in geometric modeling: the representation of curves and surfaces by parametric functions, especially cubic splines, and the representation of solids by various techniques, including boundary representations and CSG models. Chapter 13 introduces the human color-vision system, various color-description systems, and conversion from one to another. This chapter also briefly addresses rules for the effective use of color. Image Syntheses Group. Chapter 14, the first in a four-chapter sequence, describes the quest for realism from the earliest vector drawings to state-of-the-art shaded graphics. The artifacts caused by aliasing are of crucial concern in raster graphics, and this chapter discusses their causes and cures in considerable detail by introducing the Fourier transform and convolution. Chapter 15 describes a variety of strategies for visible-surface determination in enough detail to allow the reader to implement some of the most important ones. Illumination and shading algorithms are covered in detail in Chapter 16. The early part of part of this chapter discusses algorithms most commonly found in current hardware, while the remainder treats texture, shadows, transparency, reflections, physically based illumination models, rat tracing, and radiosity methods. The last chapter in this group, Chapter 17, describes both image manipulations, such as scaling, shearing, and rotating pixmaps, and image storage techniques, including various image-compression schemes. Advanced Techniques Group. The last four chapters give an overview of the current state of the art (a moving target, of course). Chapter 18 describes advanced graphics hardware used in high-end commercial and research machines; this chapter was contributed by Steven Molnar and Henry Fuchs, authorities on high-performance graphics architectures. Chapter 19 describes the complex raster algorithms used for such tasks as scan-converting arbitary conics, generating antialiased text, and implementing page-description languages, such as PostScript. The final two chapters survey some of the most important techniques in the fields of high-level modeling and computer animation. The first two groups cover only elementary material and thus can be used for a basic course at the undergraduate level. A follow-on course can then use the more advanced chapters. Alternatively, instructors can assemble customized courses by picking chapters out of the various groups. For example, a course designed to introduce students to primarily 2D graphics would include Chapters 1 and 2, simple scan conversion and clipping from Chapter 3, a technology overview with emphasis on raster architectures and interaction devices from Chapter 4, homogeneous mathematics from Chapter 5, and 3D viewing only from a "how to use it" point of view from Sections 6.1 to 6.3. The User Interface Group, Chapters 8-10, would be followed by selected introductory sections and simple algorithms from the Image Syntheses Group, Chapters 14, 15, and 16. A one-course general overview of graphics would include Chapters 1 and 2, basic algorithms from Chapter 3, raster architectures and interaction devices from Chapter 4, Chapter 5, and most of Chapters 6 and 7 on viewing and SPHIGs. The second half of the course would include sections on modeling from Chapters 11 and 13, on image syntheses from Chapters 14, 15, and 16, and on advanced modeling from Chapter 20 to give breadth of coverage in these slightly more advanced areas. A course emphasizing 3D modeling and rendering would start with Chapter 3 sections on scan converting, clipping of lines and polygons, and introducing antialiasing. The course would then progress to Chapters 5 and 6 on the basic mathematics of transformations and viewing. Chapter 13 on color, and then cover the key Chapters 14, 15 and 16 in the Image Syntheses Group. Coverage would be rounded off by selections in surface and solid modeling. Chapter 20 on advanced modeling, and Chapter 21 on animation from the Advanced Techniques Group. Graphics Packages The SRGP and SPHIGS graphics packages, designed by David Sklar, coauthor of the two chapters on these packages, are available from the publisher for the IBM PC (ISBN 0-201-54700-7), the Macintosh (ISBN 0-201-54701-5), and UNIX workstations running X11, as are many of the algorithms for scan conversion, clipping, and viewing (see page 1175). Acknowledgments This book could not have been produced without the dedicated work and the indulgence of many friends and colleagues. We acknowledge here our debt to those who have contributed significantly to one or more chapters; many others have helped by commenting on individual chapters, and we are grateful to them as well. We regret any inadvertent omissions. Katrina Avery and Lyn Dupr did a superb job of editing. Additional valuable editing on multiple versions of multiple chapters was provided by Debbie van Dam, Melissa Gold, and Clare Campbell. We are especially grateful to our production supervisor Bette Aaronson, our art director, Joe Vetere, and our editor, Keith Wollman, not only for their great help in producing the book, but also for their patience and good humor under admittedly adverse circumstances--if we ever made a promised deadline during these frantic five years, we can't remember it! Computer graphics has become too complex for even a team of four main authors and three guest authors to be expert in all areas. We relied on colleagues and students to amplify our knowledge, catch our mistakes and provide constructive criticism of form and content. We take full responsibility for any remaining sins of omission and commission. Detailed technical readings on one or more chapters were provided by John Airey, Kurt Akeley, Tom Banchoff, Brian Barsky, David Bates, Cliff Beshers, Gary Bishop, Peter J Bono, Marvin Bunker, Bill Buxton, Edward Chang, Norman Chin, Michael F. Cohen, William Cowan, John Dennis, Tom Dewald, Scott Draves, Steve Drucker, Tom Duff, Richard Economy, David Ellsworth, Nick England, Jerry Farrell, Robin Forrest, Alain Fournier, Alan Freiden, Christina Gibbs, Melissa Gold, Mark Green, Cathleen Greenberg, Margaret Hagen, Griff Hamlin, Pat Hanrahan, John Heidema, Rob Jacob, Abid Kamran, Mike Kappel, Henry Kaufman, Karen Kendler, David Kurlander, David Laidlaw, Keith Lantz, Hsien-Che Lee, Aaron Marcus, Nelson Max, Deborah Mayhew, Barbara Meier, Gary Meyer, Jim Michener, Jakob Nielsen, Mark Nodine, Randy Pausch, Ari Requicha, David Rosenthal, David Salesin, Nanan Samet, James Sanford, James Sargent, Robin Schaufler, Robert Scheifler, John Schnizlein, Michael Shantzis, Ben Shneiderman, Ken Shoemake, Judith Schrier, John Sibert, Dave Simons, Jonathan Steinhart, Maureen Stone, Paul Strauss, Seth Tager, Peter Tanner, Bruce Tebbs, Ben Trumbore, Yi Tso, Greg Turk, Jeff Vroom, Colin Ware, Gary Watkins, Chuck Weger, Kevin Weiler, Turner Whitted, George Wolberg, and Larry Wolff. Several colleagues, including Jack Bresenham, Brian Barsky, Jerry Van Aken, Dilip Da Silva (who suggested the uniform midpoint treatment of Chapter 3) and Don Hatfield, not only read chapters closely but also provided detailed suggestions on algorithms. Welcome word-processing relief was provided by Katrina Avery, Barbara Britten, Clare Campbell, Tina Cantor, Joyce Cavatoni, Louisa Hogan, Jenni Rodda, and Debbie van Dam. Drawings for Chapters 1-3 were ably created by Dan Robbins, Scott Snibbe, Tina Cantor, and Clare Campbell. Figure and image sequences created for this book were provided by Beth Cobb, David Kurlander, Allen Paeth, and George Wolberg (with assistance from Peter Karp). Plates II.21-37, showing a progression of rendering techniques, were designed and rendered at Pixar by Thomas Williams and H.B. Siegel, under the direction of M.W. Mantle, using Pixar's PhotoRealistic Renderman software. Thanks to Industrial Light & Magic for the use of their laser scanner to create Plates II.24-37, and to Norman Chin for computing vertex normals for Color Plates II.30-32. L. Lu and Carles Castellsagu wrote programs to make figures. Jeff Vogel implemented the algorithms of Chapter 3, and he and Atul Butte verified the code in Chapters 2 and 7. David Sklar wrote the Mac and X11 implementations of SRGP and SPHIGS with help from Ron Balsys, Scott Boyajian, Atul Butte, Alex Contovounesios, and Scott Draves. Randy Pausch and his students ported the packages to the PC environment. We have installed an automated electronic mail server to allow our readers to obtain machine-readable copies of many of the algorithms, suggest exercises, report errors in the text and in SRGP/SPHIGS, and obtain errata lists for the text and software. Send email to "graphtext @ cs.brown.edu" with a Subject line of "Help" to receive the current list of available services (See page 1175 for information on how to order SRGP and SPHIGS.) Preface to the C Edition This is the C-language version of a book originally written with examples in Pascal. It includes all changes through the ninth printing of the Pascal second edition, as well as minor modifications to several algorithms, and all it s Pascal code has been rewritten in ANSI C. The interfaces to the SRGP and SPHIGS graphics packages are now defined in C, rather than Pascal, and correspond to the new C implementations of these packages. (See page 1175 for information on obtaining the software.) We wish to thank Norman Chin for converting the Pascal code of the second edition to C, proofreading it, and formatting it using the typographic conventions of the original. Thanks to Matt Ayers for careful proofing of Chapters 2, 3, and 7, and for useful suggestions about conversion problems. Read more Product details ASIN ‏ : ‎ 0201848406 Publisher ‏ : ‎ Addison-Wesley Professional Publication date ‏ : ‎ January 1, 1995 Edition ‏ : ‎ 2nd Language ‏ : ‎ English Print length ‏ : ‎ 1175 pages ISBN-10 ‏ : ‎ 9780201848403 ISBN-13 ‏ : ‎ 978-0201848403 Item Weight ‏ : ‎ 3.7 pounds Dimensions ‏ : ‎ 6.75 x 1.75 x 9.75 inches Best Sellers Rank: #1,582,780 in Books (See Top 100 in Books) 571 in Computer Graphics 867 in Graphics & Multimedia Programming 3,455 in Computer Graphics & Design Customer Reviews: 4.6 4.6 out of 5 stars 71 ratings About the authors Follow authors to get new release updates, plus improved recommendations. 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188392	https://en.wikipedia.org/wiki/Kinetic_theory_of_gases	Jump to content Search Contents 1 History 1.1 Kinetic theory of matter 1.1.1 Antiquity 1.1.2 Modern era 1.1.2.1 "Heat is motion" 1.2 Kinetic theory of gases 2 Assumptions 3 Equilibrium properties 3.1 Pressure and kinetic energy 3.2 Temperature and kinetic energy 3.3 Collisions with container wall 3.4 Speed of molecules 3.5 Mean free path 4 Transport properties 4.1 Viscosity and kinetic momentum 4.2 Thermal conductivity and heat flux 4.3 Diffusion coefficient and diffusion flux 5 Detailed balance 5.1 Fluctuation and dissipation 5.2 Onsager reciprocal relations 6 See also 7 References 7.1 Citations 7.2 Sources cited 8 Further reading 9 External links Kinetic theory of gases العربية Azərbaycanca বাংলা Беларуская Català Чӑвашла Čeština Dansk Deutsch Ελληνικά Español Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Italiano עברית Latviešu Limburgs Magyar Bahasa Melayu Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Português Romnă Русский Scots Simple English Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska தமிழ் ไทย Türkçe Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikiquote Wikidata item Appearance From Wikipedia, the free encyclopedia Understanding of gas properties in terms of molecular motion The kinetic theory of gases is a simple classical model of the thermodynamic behavior of gases. Its introduction allowed many principal concepts of thermodynamics to be established. It treats a gas as composed of numerous particles, too small to be seen with a microscope, in constant, random motion. These particles are now known to be the atoms or molecules of the gas. The kinetic theory of gases uses their collisions with each other and with the walls of their container to explain the relationship between the macroscopic properties of gases, such as volume, pressure, and temperature, as well as transport properties such as viscosity, thermal conductivity and mass diffusivity. The basic version of the model describes an ideal gas. It treats the collisions as perfectly elastic and as the only interaction between the particles, which are additionally assumed to be much smaller than their average distance apart. Due to the time reversibility of microscopic dynamics (microscopic reversibility), the kinetic theory is also connected to the principle of detailed balance, in terms of the fluctuation-dissipation theorem (for Brownian motion) and the Onsager reciprocal relations. The theory was historically significant as the first explicit exercise of the ideas of statistical mechanics. History [edit] See also: Heat § History, Atomism, and History of thermodynamics Kinetic theory of matter [edit] Antiquity [edit] In about 50 BCE, the Roman philosopher Lucretius proposed that apparently static macroscopic bodies were composed on a small scale of rapidly moving atoms all bouncing off each other. This Epicurean atomistic point of view was rarely considered in the subsequent centuries, when Aristotlean ideas were dominant.[citation needed] Modern era [edit] "Heat is motion" [edit] One of the first and boldest statements on the relationship between motion of particles and heat was by the English philosopher Francis Bacon in 1620. "It must not be thought that heat generates motion, or motion heat (though in some respects this be true), but that the very essence of heat ... is motion and nothing else." "not a ... motion of the whole, but of the small particles of the body." In 1623, in The Assayer, Galileo Galilei, in turn, argued that heat, pressure, smell and other phenomena perceived by our senses are apparent properties only, caused by the movement of particles, which is a real phenomenon. In 1665, in Micrographia, the English polymath Robert Hooke repeated Bacon's assertion, and in 1675, his colleague, Anglo-Irish scientist Robert Boyle noted that a hammer's "impulse" is transformed into the motion of a nail's constituent particles, and that this type of motion is what heat consists of. Boyle also believed that all macroscopic properties, including color, taste and elasticity, are caused by and ultimately consist of nothing but the arrangement and motion of indivisible particles of matter. In a lecture of 1681, Hooke asserted a direct relationship between the temperature of an object and the speed of its internal particles. "Heat ... is nothing but the internal Motion of the Particles of [a] Body; and the hotter a Body is, the more violently are the Particles moved." In a manuscript published 1720, the English philosopher John Locke made a very similar statement: "What in our sensation is heat, in the object is nothing but motion." Locke too talked about the motion of the internal particles of the object, which he referred to as its "insensible parts". In his 1744 paper Meditations on the Cause of Heat and Cold, Russian polymath Mikhail Lomonosov made a relatable appeal to everyday experience to gain acceptance of the microscopic and kinetic nature of matter and heat: Movement should not be denied based on the fact it is not seen. Who would deny that the leaves of trees move when rustled by a wind, despite it being unobservable from large distances? Just as in this case motion remains hidden due to perspective, it remains hidden in warm bodies due to the extremely small sizes of the moving particles. In both cases, the viewing angle is so small that neither the object nor their movement can be seen. Lomonosov also insisted that movement of particles is necessary for the processes of dissolution, extraction and diffusion, providing as examples the dissolution and diffusion of salts by the action of water particles on the “molecules of salt”, the dissolution of metals in mercury, and the extraction of plant pigments by alcohol. Also the transfer of heat was explained by the motion of particles. Around 1760, Scottish physicist and chemist Joseph Black wrote: "Many have supposed that heat is a tremulous ... motion of the particles of matter, which ... motion they imagined to be communicated from one body to another." Kinetic theory of gases [edit] In 1738 Daniel Bernoulli published Hydrodynamica, which laid the basis for the kinetic theory of gases. In this work, Bernoulli posited the argument, that gases consist of great numbers of molecules moving in all directions, that their impact on a surface causes the pressure of the gas, and that their average kinetic energy determines the temperature of the gas. The theory was not immediately accepted, in part because conservation of energy had not yet been established, and it was not obvious to physicists how the collisions between molecules could be perfectly elastic.: 36–37 Pioneers of the kinetic theory, whose work was also largely neglected by their contemporaries, were Mikhail Lomonosov (1747), Georges-Louis Le Sage (ca. 1780, published 1818), John Herapath (1816) and John James Waterston (1843), which connected their research with the development of mechanical explanations of gravitation. In 1856 August Krönig created a simple gas-kinetic model, which only considered the translational motion of the particles. In 1857 Rudolf Clausius developed a similar, but more sophisticated version of the theory, which included translational and, contrary to Krönig, also rotational and vibrational molecular motions. In this same work he introduced the concept of mean free path of a particle. In 1859, after reading a paper about the diffusion of molecules by Clausius, Scottish physicist James Clerk Maxwell formulated the Maxwell distribution of molecular velocities, which gave the proportion of molecules having a certain velocity in a specific range. This was the first-ever statistical law in physics. Maxwell also gave the first mechanical argument that molecular collisions entail an equalization of temperatures and hence a tendency towards equilibrium. In his 1873 thirteen page article 'Molecules', Maxwell states: "we are told that an 'atom' is a material point, invested and surrounded by 'potential forces' and that when 'flying molecules' strike against a solid body in constant succession it causes what is called pressure of air and other gases." In 1871, Ludwig Boltzmann generalized Maxwell's achievement and formulated the Maxwell–Boltzmann distribution. The logarithmic connection between entropy and probability was also first stated by Boltzmann. At the beginning of the 20th century, atoms were considered by many physicists to be purely hypothetical constructs, rather than real objects. An important turning point was Albert Einstein's (1905) and Marian Smoluchowski's (1906) papers on Brownian motion, which succeeded in making certain accurate quantitative predictions based on the kinetic theory. Following the development of the Boltzmann equation, a framework for its use in developing transport equations was developed independently by David Enskog and Sydney Chapman in 1917 and 1916. The framework provided a route to prediction of the transport properties of dilute gases, and became known as Chapman–Enskog theory. The framework was gradually expanded throughout the following century, eventually becoming a route to prediction of transport properties in real, dense gases. Assumptions [edit] The application of kinetic theory to ideal gases makes the following assumptions: The gas consists of very small particles. This smallness of their size is such that the sum of the volume of the individual gas molecules is negligible compared to the volume of the container of the gas. This is equivalent to stating that the average distance separating the gas particles is large compared to their size, and that the elapsed time during a collision between particles and the container's wall is negligible when compared to the time between successive collisions. The number of particles is so large that a statistical treatment of the problem is well justified. This assumption is sometimes referred to as the thermodynamic limit. The rapidly moving particles constantly collide among themselves and with the walls of the container, and all these collisions are perfectly elastic. Interactions (i.e. collisions) between particles are strictly binary and uncorrelated, meaning that there are no three-body (or higher) interactions, and the particles have no memory. Except during collisions, the interactions among molecules are negligible. They exert no other forces on one another. Thus, the dynamics of particle motion can be treated classically, and the equations of motion are time-reversible. As a simplifying assumption, the particles are usually assumed to have the same mass as one another; however, the theory can be generalized to a mass distribution, with each mass type contributing to the gas properties independently of one another in agreement with Dalton's law of partial pressures. Many of the model's predictions are the same whether or not collisions between particles are included, so they are often neglected as a simplifying assumption in derivations (see below). More modern developments, such as the revised Enskog theory and the extended Bhatnagar–Gross–Krook model, relax one or more of the above assumptions. These can accurately describe the properties of dense gases, and gases with internal degrees of freedom, because they include the volume of the particles as well as contributions from intermolecular and intramolecular forces as well as quantized molecular rotations, quantum rotational-vibrational symmetry effects, and electronic excitation. While theories relaxing the assumptions that the gas particles occupy negligible volume and that collisions are strictly elastic have been successful, it has been shown that relaxing the requirement of interactions being binary and uncorrelated will eventually lead to divergent results. Equilibrium properties [edit] Pressure and kinetic energy [edit] In the kinetic theory of gases, the pressure is assumed to be equal to the force (per unit area) exerted by the individual gas atoms or molecules hitting and rebounding from the gas container's surface. Consider a gas particle traveling at velocity, , along the -direction in an enclosed volume with characteristic length, , cross-sectional area, , and volume, . The gas particle encounters a boundary after characteristic time The momentum of the gas particle can then be described as We combine the above with Newton's second law, which states that the force experienced by a particle is related to the time rate of change of its momentum, such that Now consider a large number, , of gas particles with random orientation in a three-dimensional volume. Because the orientation is random, the average particle speed, , in every direction is identical Further, assume that the volume is symmetrical about its three dimensions, , such that The total surface area on which the gas particles act is therefore The pressure exerted by the collisions of the gas particles with the surface can then be found by adding the force contribution of every particle and dividing by the interior surface area of the volume, The total translational kinetic energy of the gas is defined as providing the result This is an important, non-trivial result of the kinetic theory because it relates pressure, a macroscopic property, to the translational kinetic energy of the molecules, which is a microscopic property. The mass density of a gas is expressed through the total mass of gas particles and through volume of this gas: . Taking this into account, the pressure is equal to Relativistic expression for this formula is where is speed of light. In the limit of small speeds, the expression becomes . Temperature and kinetic energy [edit] Rewriting the above result for the pressure as , we may combine it with the ideal gas law \| \| \| --- \| \| \| 1 \| where is the Boltzmann constant and is the absolute temperature defined by the ideal gas law, to obtain which leads to a simplified expression of the average translational kinetic energy per molecule, The translational kinetic energy of the system is times that of a molecule, namely . The temperature, is related to the translational kinetic energy by the description above, resulting in \| \| \| --- \| \| \| 2 \| which becomes \| \| \| --- \| \| \| 3 \| Equation (3) is one important result of the kinetic theory: The average molecular kinetic energy is proportional to the ideal gas law's absolute temperature. From equations (1) and (3), we have \| \| \| --- \| \| \| 4 \| Thus, the product of pressure and volume per mole is proportional to the average translational molecular kinetic energy. Equations (1) and (4) are called the "classical results", which could also be derived from statistical mechanics; for more details, see: The equipartition theorem requires that kinetic energy is partitioned equally between all kinetic degrees of freedom, D. A monatomic gas is axially symmetric about each spatial axis, so that D = 3 comprising translational motion along each axis. A diatomic gas is axially symmetric about only one axis, so that D = 5, comprising translational motion along three axes and rotational motion along two axes. A polyatomic gas, like water, is not radially symmetric about any axis, resulting in D = 6, comprising 3 translational and 3 rotational degrees of freedom. Because the equipartition theorem requires that kinetic energy is partitioned equally, the total kinetic energy is Thus, the energy added to the system per gas particle kinetic degree of freedom is Therefore, the kinetic energy per kelvin of one mole of monatomic ideal gas (D = 3) is where is the Avogadro constant, and R is the ideal gas constant. Thus, the ratio of the kinetic energy to the absolute temperature of an ideal monatomic gas can be calculated easily: per mole: 12.47 J/K per molecule: 20.7 yJ/K = 129 μeV/K At standard temperature (273.15 K), the kinetic energy can also be obtained: per mole: 3406 J per molecule: 5.65 zJ = 35.2 meV. At higher temperatures (typically thousands of kelvins), vibrational modes become active to provide additional degrees of freedom, creating a temperature-dependence on D and the total molecular energy. Quantum statistical mechanics is needed to accurately compute these contributions. Collisions with container wall [edit] For an ideal gas in equilibrium, the rate of collisions with the container wall and velocity distribution of particles hitting the container wall can be calculated based on naive kinetic theory, and the results can be used for analyzing effusive flow rates, which is useful in applications such as the gaseous diffusion method for isotope separation. Assume that in the container, the number density (number per unit volume) is and that the particles obey Maxwell's velocity distribution: Then for a small area on the container wall, a particle with speed at angle from the normal of the area , will collide with the area within time interval , if it is within the distance from the area . Therefore, all the particles with speed at angle from the normal that can reach area within time interval are contained in the tilted pipe with a height of and a volume of . The total number of particles that reach area within time interval also depends on the velocity distribution; All in all, it calculates to be: Integrating this over all appropriate velocities within the constraint , , yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time: This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed of the Maxwell's velocity distribution, one has to integrate over , , . The momentum transfer to the container wall from particles hitting the area with speed at angle from the normal, in time interval is: Integrating this over all appropriate velocities within the constraint , , yields the pressure (consistent with Ideal gas law): If this small area is punched to become a small hole, the effusive flow rate will be: Combined with the ideal gas law, this yields The above expression is consistent with Graham's law. To calculate the velocity distribution of particles hitting this small area, we must take into account that all the particles with that hit the area within the time interval are contained in the tilted pipe with a height of and a volume of ; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of : with the constraint , , . The constant can be determined by the normalization condition to be , and overall: Speed of molecules [edit] From the kinetic energy formula it can be shown that where v is in m/s, T is in kelvin, and m is the mass of one molecule of gas in kg. The most probable (or mode) speed is 81.6% of the root-mean-square speed , and the mean (arithmetic mean, or average) speed is 92.1% of the rms speed (isotropic distribution of speeds). See: Average, Root-mean-square speed Arithmetic mean Mean Mode (statistics) Mean free path [edit] Main article: Mean free path In kinetic theory of gases, the mean free path is the average distance traveled by a molecule, or a number of molecules per volume, before they make their first collision. Let be the collision cross section of one molecule colliding with another. As in the previous section, the number density is defined as the number of molecules per (extensive) volume, or . The collision cross section per volume or collision cross section density is , and it is related to the mean free path by Notice that the unit of the collision cross section per volume is reciprocal of length. Transport properties [edit] See also: Transport phenomena The kinetic theory of gases deals not only with gases in thermodynamic equilibrium, but also very importantly with gases not in thermodynamic equilibrium. This means using Kinetic Theory to consider what are known as "transport properties", such as viscosity, thermal conductivity, mass diffusivity and thermal diffusion. In its most basic form, Kinetic gas theory is only applicable to dilute gases. The extension of Kinetic gas theory to dense gas mixtures, Revised Enskog Theory, was developed in 1983-1987 by E. G. D. Cohen, J. M. Kincaid and M. Lòpez de Haro, building on work by H. van Beijeren and M. H. Ernst. Viscosity and kinetic momentum [edit] See also: Viscosity § Momentum transport In books on elementary kinetic theory one can find results for dilute gas modeling that are used in many fields. Derivation of the kinetic model for shear viscosity usually starts by considering a Couette flow where two parallel plates are separated by a gas layer. The upper plate is moving at a constant velocity to the right due to a force F. The lower plate is stationary, and an equal and opposite force must therefore be acting on it to keep it at rest. The molecules in the gas layer have a forward velocity component which increase uniformly with distance above the lower plate. The non-equilibrium flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions. Inside a dilute gas in a Couette flow setup, let be the forward velocity of the gas at a horizontal flat layer (labeled as ); is along the horizontal direction. The number of molecules arriving at the area on one side of the gas layer, with speed at angle from the normal, in time interval is These molecules made their last collision at , where is the mean free path. Each molecule will contribute a forward momentum of where plus sign applies to molecules from above, and minus sign below. Note that the forward velocity gradient can be considered to be constant over a distance of mean free path. Integrating over all appropriate velocities within the constraint , , yields the forward momentum transfer per unit time per unit area (also known as shear stress): The net rate of momentum per unit area that is transported across the imaginary surface is thus Combining the above kinetic equation with Newton's law of viscosity gives the equation for shear viscosity, which is usually denoted when it is a dilute gas: Combining this equation with the equation for mean free path gives Maxwell-Boltzmann distribution gives the average (equilibrium) molecular speed as where is the most probable speed. We note that and insert the velocity in the viscosity equation above. This gives the well known equation (with subsequently estimated below) for shear viscosity for dilute gases: and is the molar mass. The equation above presupposes that the gas density is low (i.e. the pressure is low). This implies that the transport of momentum through the gas due to the translational motion of molecules is much larger than the transport due to momentum being transferred between molecules during collisions. The transfer of momentum between molecules is explicitly accounted for in Revised Enskog theory, which relaxes the requirement of a gas being dilute. The viscosity equation further presupposes that there is only one type of gas molecules, and that the gas molecules are perfect elastic and hard core particles of spherical shape. This assumption of elastic, hard core spherical molecules, like billiard balls, implies that the collision cross section of one molecule can be estimated by The radius is called collision cross section radius or kinetic radius, and the diameter is called collision cross section diameter or kinetic diameter of a molecule in a monomolecular gas. There are no simple general relation between the collision cross section and the hard core size of the (fairly spherical) molecule. The relation depends on shape of the potential energy of the molecule. For a real spherical molecule (i.e. a noble gas atom or a reasonably spherical molecule) the interaction potential is more like the Lennard-Jones potential or Morse potential which have a negative part that attracts the other molecule from distances longer than the hard core radius. The radius for zero Lennard-Jones potential may then be used as a rough estimate for the kinetic radius. However, using this estimate will typically lead to an erroneous temperature dependency of the viscosity. For such interaction potentials, significantly more accurate results are obtained by numerical evaluation of the required collision integrals. The expression for viscosity obtained from Revised Enskog Theory reduces to the above expression in the limit of infinite dilution, and can be written as where is a term that tends to zero in the limit of infinite dilution that accounts for excluded volume, and is a term accounting for the transfer of momentum over a non-zero distance between particles during a collision. Thermal conductivity and heat flux [edit] See also: Thermal conductivity Following a similar logic as above, one can derive the kinetic model for thermal conductivity of a dilute gas: Consider two parallel plates separated by a gas layer. Both plates have uniform temperatures, and are so massive compared to the gas layer that they can be treated as thermal reservoirs. The upper plate has a higher temperature than the lower plate. The molecules in the gas layer have a molecular kinetic energy which increases uniformly with distance above the lower plate. The non-equilibrium energy flow is superimposed on a Maxwell-Boltzmann equilibrium distribution of molecular motions. Let be the molecular kinetic energy of the gas at an imaginary horizontal surface inside the gas layer. The number of molecules arriving at an area on one side of the gas layer, with speed at angle from the normal, in time interval is These molecules made their last collision at a distance above and below the gas layer, and each will contribute a molecular kinetic energy of where is the specific heat capacity. Again, plus sign applies to molecules from above, and minus sign below. Note that the temperature gradient can be considered to be constant over a distance of mean free path. Integrating over all appropriate velocities within the constraint , , yields the energy transfer per unit time per unit area (also known as heat flux): Note that the energy transfer from above is in the direction, and therefore the overall minus sign in the equation. The net heat flux across the imaginary surface is thus Combining the above kinetic equation with Fourier's law gives the equation for thermal conductivity, which is usually denoted when it is a dilute gas: Similarly to viscosity, Revised Enskog theory yields an expression for thermal conductivity that reduces to the above expression in the limit of infinite dilution, and which can be written as where is a term that tends to unity in the limit of infinite dilution, accounting for excluded volume, and is a term accounting for the transfer of energy across a non-zero distance between particles during a collision. Diffusion coefficient and diffusion flux [edit] See also: Fick's laws of diffusion Following a similar logic as above, one can derive the kinetic model for mass diffusivity of a dilute gas: Consider a steady diffusion between two regions of the same gas with perfectly flat and parallel boundaries separated by a layer of the same gas. Both regions have uniform number densities, but the upper region has a higher number density than the lower region. In the steady state, the number density at any point is constant (that is, independent of time). However, the number density in the layer increases uniformly with distance above the lower plate. The non-equilibrium molecular flow is superimposed on a Maxwell–Boltzmann equilibrium distribution of molecular motions. Let be the number density of the gas at an imaginary horizontal surface inside the layer. The number of molecules arriving at an area on one side of the gas layer, with speed at angle from the normal, in time interval is These molecules made their last collision at a distance above and below the gas layer, where the local number density is Again, plus sign applies to molecules from above, and minus sign below. Note that the number density gradient can be considered to be constant over a distance of mean free path. Integrating over all appropriate velocities within the constraint , , yields the molecular transfer per unit time per unit area (also known as diffusion flux): Note that the molecular transfer from above is in the direction, and therefore the overall minus sign in the equation. The net diffusion flux across the imaginary surface is thus Combining the above kinetic equation with Fick's first law of diffusion gives the equation for mass diffusivity, which is usually denoted when it is a dilute gas: The corresponding expression obtained from Revised Enskog Theory may be written as where is a factor that tends to unity in the limit of infinite dilution, which accounts for excluded volume and the variation chemical potentials with density. Detailed balance [edit] Fluctuation and dissipation [edit] Main article: Fluctuation-dissipation theorem The kinetic theory of gases entails that due to the microscopic reversibility of the gas particles' detailed dynamics, the system must obey the principle of detailed balance. Specifically, the fluctuation-dissipation theorem applies to the Brownian motion (or diffusion) and the drag force, which leads to the Einstein–Smoluchowski equation: where D is the mass diffusivity; μ is the "mobility", or the ratio of the particle's terminal drift velocity to an applied force, μ = vd/F; kB is the Boltzmann constant; T is the absolute temperature. Note that the mobility μ = vd/F can be calculated based on the viscosity of the gas; Therefore, the Einstein–Smoluchowski equation also provides a relation between the mass diffusivity and the viscosity of the gas. Onsager reciprocal relations [edit] Main article: Onsager reciprocal relations The mathematical similarities between the expressions for shear viscocity, thermal conductivity and diffusion coefficient of the ideal (dilute) gas is not a coincidence; It is a direct result of the Onsager reciprocal relations (i.e. the detailed balance of the reversible dynamics of the particles), when applied to the convection (matter flow due to temperature gradient, and heat flow due to pressure gradient) and advection (matter flow due to the velocity of particles, and momentum transfer due to pressure gradient) of the ideal (dilute) gas. See also [edit] \| Statistical mechanics \| \| Thermodynamics Kinetic theory \| \| Particle statistics Spin–statistics theorem Indistinguishable particles Maxwell–Boltzmann Bose–Einstein Fermi–Dirac Parastatistics Anyonic statistics Braid statistics \| \| Thermodynamic ensembles NVE Microcanonical NVT Canonical µVT Grand canonical NPH Isoenthalpic–isobaric NPT Isothermal–isobaric \| \| Debye Einstein Ising Potts \| \| Internal energy Enthalpy Helmholtz free energy Gibbs free energy Grand potential / Landau free energy \| \| Maxwell Boltzmann Helmholtz Bose Gibbs Einstein Dirac Ehrenfest von Neumann Tolman Debye Fermi Synge Ising Landau \| \| v t e \| Bogoliubov-Born-Green-Kirkwood-Yvon hierarchy of equations Boltzmann equation Chapman–Enskog theory Collision theory Critical temperature Gas laws Heat Interatomic potential Magnetohydrodynamics Maxwell–Boltzmann distribution Mixmaster universe Thermodynamics Vicsek model Vlasov equation References [edit] Citations [edit] ^ Maxwell, J. C. (1867). "On the Dynamical Theory of Gases". Philosophical Transactions of the Royal Society of London. 157: 49–88. doi:10.1098/rstl.1867.0004. S2CID 96568430. ^ Bacon, F. (1902) . Dewey, J. (ed.). Novum Organum: Or True Suggestions for the Interpretation of Nature. P. F. Collier & son. p. 153. ^ Bacon, F. (1902) . Dewey, J. (ed.). Novum Organum: Or True Suggestions for the Interpretation of Nature. P. F. Collier & son. p. 156. ^ Galilei 1957, p. 273-4. ^ Adriaans, Pieter (2024), "Information", in Zalta, Edward N.; Nodelman, Uri (eds.), The Stanford Encyclopedia of Philosophy (Summer 2024 ed.), Metaphysics Research Lab, Stanford University, p. 3.4 Physics ^ Hooke, Robert (1665). Micrographia: Or Some Physiological Descriptions of Minute Bodies Made by Magnifying Glasses with Observations and Inquiries Thereupon. Printed by Jo. Martyn, and Ja. Allestry, Printers to the Royal Society. p. 12. (Facsimile, with pagination){{cite book}}: CS1 maint: postscript (link) ^ Hooke, Robert (1665). Micrographia: Or Some Physiological Descriptions of Minute Bodies Made by Magnifying Glasses with Observations and Inquiries Thereupon. Printed by Jo. Martyn, and Ja. Allestry, Printers to the Royal Society. p. 12. (Machine-readable, no pagination){{cite book}}: CS1 maint: postscript (link) ^ Boyle, Robert (1675). Experiments, notes, &c., about the mechanical origine or production of divers particular qualities: Among which is inserted a discourse of the imperfection of the chymist's doctrine of qualities; together with some reflections upon the hypothesis of alcali and acidum. Printed by E. Flesher, for R. Davis. pp. 61–62. ^ Chalmers, Alan (2019), "Atomism from the 17th to the 20th Century", in Zalta, Edward N. (ed.), The Stanford Encyclopedia of Philosophy (Spring 2019 ed.), Metaphysics Research Lab, Stanford University, p. 2.1 Atomism and the Mechanical Philosophy ^ Hooke, Robert (1705) . The posthumous works of Robert Hooke ... containing his Cutlerian lectures, and other discourses, read at the meetings of the illustrious Royal Society ... Illustrated with sculptures. To these discourses is prefixt the author's life, giving an account of his studies and employments, with an enumeration of the many experiments, instruments, contrivances and inventions, by him made and produc'd as Curator of Experiments to the Royal Society. Publish'd by Richard Waller. Printed by Sam. Smith and Benj. Walford, (Printers to the Royal Society). p. 116. ^ Locke, John (1720). A collection of several pieces of Mr. John Locke, never before printed, or not extant in his works. Printed by J. Bettenham for R. Francklin. p. 224 – via Internet Archive. ^ Locke, John (1720). A collection of several pieces of Mr. John Locke, never before printed, or not extant in his works. Printed by J. Bettenham for R. Francklin. p. 224 – via Google Play Books. ^ Lomonosov, Mikhail Vasil'evich (1970) . "Meditations on the Cause of Heat and Cold". In Leicester, Henry M. (ed.). Mikhail Vasil'evich Lomonosov on the Corpuscular Theory. Harvard University Press. p. 100. ^ Lomonosov, Mikhail Vasil'evich (1970) . "Meditations on the Cause of Heat and Cold". In Leicester, Henry M. (ed.). Mikhail Vasil'evich Lomonosov on the Corpuscular Theory. Harvard University Press. pp. 102–3. ^ Black, Joseph (1807). Robinson, John (ed.). Lectures on the Elements of Chemistry: Delivered in the University of Edinburgh. Mathew Carey. p. 80. ^ L.I Ponomarev; I.V Kurchatov (1 January 1993). The Quantum Dice. CRC Press. ISBN 978-0-7503-0251-7. ^ Lomonosov 1758 ^ Le Sage 1780/1818 ^ Herapath 1816, 1821 ^ Waterston 1843 ^ Krönig 1856 ^ Clausius 1857 ^ See: Maxwell, J.C. (1860) "Illustrations of the dynamical theory of gases. Part I. On the motions and collisions of perfectly elastic spheres," Philosophical Magazine, 4th series, 19 : 19–32. Maxwell, J.C. (1860) "Illustrations of the dynamical theory of gases. Part II. On the process of diffusion of two or more kinds of moving particles among one another," Philosophical Magazine, 4th series, 20 : 21–37. ^ Mahon, Basil (2003). The Man Who Changed Everything – the Life of James Clerk Maxwell. Hoboken, NJ: Wiley. ISBN 0-470-86171-1. OCLC 52358254. ^ Gyenis, Balazs (2017). "Maxwell and the normal distribution: A colored story of probability, independence, and tendency towards equilibrium". Studies in History and Philosophy of Modern Physics. 57: 53–65. arXiv:1702.01411. Bibcode:2017SHPMP..57...53G. doi:10.1016/j.shpsb.2017.01.001. S2CID 38272381. ^ Maxwell 1873 ^ Einstein 1905 ^ Smoluchowski 1906 ^ Chang, Raymond; Thoman, John W. Jr. (2014). Physical Chemistry for the Chemical Sciences. New York, NY: University Science Books. p. 37. ^ van Enk, Steven J.; Nienhuis, Gerard (1991-12-01). "Inelastic collisions and gas-kinetic effects of light". Physical Review A. 44 (11): 7615–7625. Bibcode:1991PhRvA..44.7615V. doi:10.1103/PhysRevA.44.7615. PMID 9905900. ^ McQuarrie, Donald A. (1976). Statistical Mechanics. New York, NY: University Science Press. ^ Cohen, E. G. D. (1993-03-15). "Fifty years of kinetic theory". Physica A: Statistical Mechanics and Its Applications. 194 (1): 229–257. Bibcode:1993PhyA..194..229C. doi:10.1016/0378-4371(93)90357-A. ISSN 0378-4371. ^ Fedosin, Sergey G. (2021). "The potentials of the acceleration field and pressure field in rotating relativistic uniform system". Continuum Mechanics and Thermodynamics. 33 (3): 817–834. arXiv:2410.17289. Bibcode:2021CMT....33..817F. doi:10.1007/s00161-020-00960-7. S2CID 230076346. ^ The average kinetic energy of a fluid is proportional to the root mean-square velocity, which always exceeds the mean velocity - Kinetic Molecular Theory[usurped] ^ Configuration integral (statistical mechanics) Archived 2012-04-28 at the Wayback Machine ^ Chang, Raymond; Thoman, John W. Jr. (2014). Physical Chemistry for the Chemical Sciences. New York: University Science Books. pp. 56–61. ^ "5.62 Physical Chemistry II" (PDF). MIT OpenCourseWare. ^ Lòpez de Haro, M.; Cohen, E. G. D.; Kincaid, J. M. (1983). "The Enskog theory for multicomponent mixtures. I. Linear transport theory". The Journal of Chemical Physics. 78 (5): 2746–2759. Bibcode:1983JChPh..78.2746L. doi:10.1063/1.444985. ^ Kincaid, J. M.; Lòpez de Haro, M.; Cohen, E. G. D. (1983). "The Enskog theory for multicomponent mixtures. II. Mutual diffusion". The Journal of Chemical Physics. 79 (9): 4509–4521. doi:10.1063/1.446388. ^ Lòpez de Haro, M.; Cohen, E. G. D. (1984). "The Enskog theory for multicomponent mixtures. III. Transport properties of dense binary mixtures with one tracer component". The Journal of Chemical Physics. 80 (1): 408–415. Bibcode:1984JChPh..80..408L. doi:10.1063/1.446463. ^ Kincaid, J. M.; Cohen, E. G. D.; Lòpez de Haro, M. (1987). "The Enskog theory for multicomponent mixtures. IV. Thermal diffusion". The Journal of Chemical Physics. 86 (2): 963–975. Bibcode:1987JChPh..86..963K. doi:10.1063/1.452243. ^ van Beijeren, H.; Ernst, M. H. (1973). "The non-linear Enskog-Boltzmann equation". Physics Letters A. 43 (4): 367–368. Bibcode:1973PhLA...43..367V. doi:10.1016/0375-9601(73)90346-0. hdl:1874/36979. ^ a b c Sears, F.W.; Salinger, G.L. (1975). "10". Thermodynamics, Kinetic Theory, and Statistical Thermodynamics (3 ed.). Reading, Massachusetts, USA: Addison-Wesley Publishing Company, Inc. pp. 286–291. ISBN 978-0201068948. ^ Hildebrand, J.H. (1976). "Viscosity of dilute gases and vapors". Proc Natl Acad Sci U S A. 76 (12): 4302–4303. Bibcode:1976PNAS...73.4302H. doi:10.1073/pnas.73.12.4302. PMC 431439. PMID 16592372. ^ Dill, Ken A.; Bromberg, Sarina (2003). Molecular Driving Forces: Statistical Thermodynamics in Chemistry and Biology. Garland Science. p. 327. ISBN 9780815320517. Sources cited [edit] Clausius, R. (1857), "Ueber die Art der Bewegung, welche wir Wärme nennen", Annalen der Physik, 176 (3): 353–379, Bibcode:1857AnP...176..353C, doi:10.1002/andp.18571760302 de Groot, S. R., W. A. van Leeuwen and Ch. G. van Weert (1980), Relativistic Kinetic Theory, North-Holland, Amsterdam. Galilei, Galileo (1957) . "The Assayer". In Drake, Stillman (ed.). Discoveries and Opinions of Galileo (PDF). Doubleday. Einstein, A. (1905), "Über die von der molekularkinetischen Theorie der Wärme geforderte Bewegung von in ruhenden Flüssigkeiten suspendierten Teilchen" (PDF), Annalen der Physik, 17 (8): 549–560, Bibcode:1905AnP...322..549E, doi:10.1002/andp.19053220806 Grad, Harold (1949), "On the Kinetic Theory of Rarefied Gases.", Communications on Pure and Applied Mathematics, 2 (4): 331–407, doi:10.1002/cpa.3160020403 Herapath, J. (1816), "On the physical properties of gases", Annals of Philosophy, Robert Baldwin: 56–60 Herapath, J. (1821), "On the Causes, Laws and Phenomena of Heat, Gases, Gravitation", Annals of Philosophy, 9, Baldwin, Cradock, and Joy: 273–293 Krönig, A. (1856), "Grundzüge einer Theorie der Gase", Annalen der Physik, 99 (10): 315–322, Bibcode:1856AnP...175..315K, doi:10.1002/andp.18561751008 Le Sage, G.-L. (1818), "Physique Mécanique des Georges-Louis Le Sage", in Prévost, Pierre (ed.), Deux Traites de Physique Mécanique, Geneva & Paris: J.J. Paschoud, pp. 1–186 Liboff, R. L. (1990), Kinetic Theory, Prentice-Hall, Englewood Cliffs, N. J. Lomonosov, M. (1970) , "On the Relation of the Amount of Material and Weight", in Henry M. Leicester (ed.), Mikhail Vasil'evich Lomonosov on the Corpuscular Theory, Cambridge: Harvard University Press, pp. 224–233 Mahon, Basil (2003), The Man Who Changed Everything – the Life of James Clerk Maxwell, Hoboken, New Jersey: Wiley, ISBN 0-470-86171-1 Maxwell, James Clerk (1873), "Molecules", Nature, 8 (204): 437–441, Bibcode:1873Natur...8..437., doi:10.1038/008437a0 Smoluchowski, M. (1906), "Zur kinetischen Theorie der Brownschen Molekularbewegung und der Suspensionen", Annalen der Physik, 21 (14): 756–780, Bibcode:1906AnP...326..756V, doi:10.1002/andp.19063261405 Waterston, John James (1843), Thoughts on the Mental Functions (reprinted in his Papers, 3, 167, 183.) Williams, M. M. R. (1971). Mathematical Methods in Particle Transport Theory. Butterworths, London. ISBN 9780408700696.{{cite book}}: CS1 maint: location missing publisher (link) Further reading [edit] Sydney Chapman and Thomas George Cowling (1939/1970), The Mathematical Theory of Non-uniform Gases: An Account of the Kinetic Theory of Viscosity, Thermal Conduction and Diffusion in Gases, (first edition 1939, second edition 1952), third edition 1970 prepared in co-operation with D. Burnett, Cambridge University Press, London Joseph Oakland Hirschfelder, Charles Francis Curtiss, and Robert Byron Bird (1964), Molecular Theory of Gases and Liquids, revised edition (Wiley-Interscience), ISBN 978-0471400653 Richard Lawrence Liboff (2003), Kinetic Theory: Classical, Quantum, and Relativistic Descriptions, third edition (Springer), ISBN 978-0-387-21775-8 Behnam Rahimi and Henning Struchtrup Archived 2021-07-25 at the Wayback Machine (2016), "Macroscopic and kinetic modelling of rarefied polyatomic gases", Journal of Fluid Mechanics, 806, 437–505, DOI 10.1017/jfm.2016.604 External links [edit] Wikiquote has quotations related to Kinetic theory of gases. PHYSICAL CHEMISTRY – Gases[usurped] Early Theories of Gases Thermodynamics Archived 2017-02-28 at the Wayback Machine - a chapter from an online textbook Temperature and Pressure of an Ideal Gas: The Equation of State on Project PHYSNET. Introduction to the kinetic molecular theory of gases, from The Upper Canada District School Board Java animation illustrating the kinetic theory from University of Arkansas Flowchart linking together kinetic theory concepts, from HyperPhysics Interactive Java Applets allowing high school students to experiment and discover how various factors affect rates of chemical reactions. A demonstration apparatus for the thermal agitation in gases. \| Authority control databases \| \| International \| \| \| National \| United States France BnF data Japan Czech Republic Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Gases Thermodynamics Classical mechanics Hidden categories: CS1 maint: postscript Webarchive template wayback links Articles with short description Short description is different from Wikidata All articles with unsourced statements Articles with unsourced statements from December 2024 CS1 maint: location missing publisher Kinetic theory of gases Add topic
188393	https://math.stackexchange.com/questions/412585/inequality-for-sums-of-reciprocals	Skip to main content Inequality for sums of reciprocals Ask Question Asked Modified 12 years, 2 months ago Viewed 1k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. Prove or disprove :: If a0,…,an≥m≥1 then ∑ni=01ai≤n+m/∏ni=0ai. [[ This is a generalization I formulated from this question :: Inequality Of Four Variables. ]] Thank-you, Moses inequality summation integral-inequality Share CC BY-SA 3.0 Follow this question to receive notifications edited Apr 13, 2017 at 12:20 CommunityBot 1 asked Jun 6, 2013 at 2:14 Musa Al-hassyMusa Al-hassy 2,22222 gold badges1313 silver badges1515 bronze badges Add a comment \| 3 Answers 3 Reset to default This answer is useful 1 Save this answer. Show activity on this post. I have a proof by induction on n. Note that verifying the base case n=0 is trivial from m≥1. Let us assume that the inequality is true for n−1, Thus ∑i=0n−11ai≤n−1+m∏n−1i=0ai Let us add 1an on both sides and denote λk=∏ki=0ai. Note that λn−1an=λn. We get ∑i=0n−11ai+1an≤n−1+mλn−1+1an So to prove our statement it suffices to show n−1+mλn−1+1an≤n+mλn By taking common denominators and using λn−1an=λn, λn−1+anm−anλn−1<m Rearranging it, we have to prove: (λn−1−m)(1−an)≤0 It is given that an≥1 and therefore (1−an)≤0. It is also given that m≥1 and ai≥m, therefore we have λn−1=∏n−1i=0ai≥mn≥m. This implies that (λn−1−m)≥0. Since (1−an)≤0 and (λn−1−m)≥0, we can multiply the two inequalities to obtain (λn−1−m)(1−an)≤0. Therefore by induction, we have proved our inequality for all natural numbers. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jun 6, 2013 at 16:05 answered Jun 6, 2013 at 5:01 IsomorphismIsomorphism 5,98022 gold badges2626 silver badges4848 bronze badges 2 (+1) I could easily follow your reasoning but could you please clearly spell-out the paragraph starting ``From the data...''. Thank-you. – Musa Al-hassy Commented Jun 6, 2013 at 14:02 Hope it is clear now! – Isomorphism Commented Jun 6, 2013 at 16:06 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. If n=0 I think the case is trivial, thus assume that n>0. If m≥2 then ∑i=0n1ai≤∑i=0n1m=n+1m≤n. Thus we are reduced to proving that if a0,…,an≥1 then ∑i=0n1ai≤n+1∏ni=0ai. Well, if ax,ay≥2 for some x and y, then ∑i=0n1ai≤∑i≠x,y1ai+12+12≤(n−1)+1=n≤n+1∏ni=0ai. Thus we may assume that ax≥2 for some x, but ai=1 for all i≠x. In this case ∑i=0n1ai=n+1ax=n+1∏ni=0ai. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jun 6, 2013 at 2:40 Bobby OceanBobby Ocean 3,28711 gold badge1515 silver badges1313 bronze badges 3 Oh, sorry, I was under the impression we were dealing with integers. – Bobby Ocean Commented Jun 6, 2013 at 2:46 (+1) Just confirming: you took m to be an integer? Also your casing seems unclear to me: what if no ax,ay≤2?. Why are we justified in ``we may assume ax≥2...''? Thanks! – Musa Al-hassy Commented Jun 6, 2013 at 14:08 I actually took, m to be an integer as well as ai to be integers. Sorry, I left out the case of when all ai=1. – Bobby Ocean Commented Jun 6, 2013 at 18:56 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Take f(a0,…,an)=(∑i=0n1ai)−ma0a1⋯an=1aj(1−ma0a1⋯aj^⋯an)nonnegative+c where c is constant with respect to aj. f is maximized when aj is minimal, i.e. aj=m. Hence the overall maximum of f is at f(m,m,…,m)=n+1m−mmn+1=n+1m−m−n. If m=1, this is equal to n. We have ∂∂m(n+1m−m−n)=−n+1m2+nmn+1=nmn+1(1−n+1nmn−1)negative for m≥1 Hence f(m,m,…,m) achieves its global maximum at m=1. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jun 6, 2013 at 2:55 answered Jun 6, 2013 at 2:41 vadim123vadim123 84k99 gold badges120120 silver badges224224 bronze badges 2 (+1) You're reasoning is clear but, I personally see that, using calculus in this context is a bit more than I'd like. That is to say, I would prefer a discrete-math approach. Also, I'm not sure how this proves the inequality. Anyhow, thank-you. – Musa Al-hassy Commented Jun 6, 2013 at 14:05 @Moses, calculus is only needed to prove that f(m,…,m)≤n for m∈(1,2), because outside this range f≤n. The reason it is a proof is that f(1,…,1)=n and f is decreasing for m≥1. – vadim123 Commented Jun 6, 2013 at 14:10 Add a comment \| You must log in to answer this question. 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188394	https://www.khanacademy.org/math/geometry-home/geometry-lines/points-lines-planes/v/specifying-planes-in-three-dimensions	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
188395	https://www.mooc-list.com/course/precalculus-relations-and-functions-coursera	Published Time: Wed, 23 Jul 2025 18:33:36 GMT Precalculus: Relations and Functions (Coursera) \| MOOC List Skip to main content User account menu Log in Advanced Course Search (Multiple Criteria) Sort options Sort by Course Title Contains Categories Subjects/Skills Initiative/Provider University/Entity Start Date Length Estimated Effort Course Level Language Subtitles Country Exam Certificate Breadcrumb Home Course Precalculus: Relations and Functions (Coursera) Precalculus: Relations and Functions (Coursera) Start Date Jul 28th 2025 facebook twitter envelope print Course Auditing Coursera Johns Hopkins University Categories Sci: Mathematics Effort Beginner 4 Weeks 1-4 Hours/Week Certification Yes, Paid Exam and/or Final Project Paid Certificate 41.00 EUR/month Languages English French Portuguese Russian English Spanish Some familiarity with algebraic expressions and function basics at the high school level. 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188396	https://thenextrecession.wordpress.com/2015/09/11/corbynomics-extreme-or-moderate/	Corbynomics – extreme or moderate? – Michael Roberts Blog Skip to content Michael Roberts Blog blogging from a marxist economist [x] Menu +×expanded collapsed Home Blog Books About Contact Facebook YouTube Corbynomics – extreme or moderate? Posted bymichael robertsSeptember 11, 2015 September 13, 2015Posted incapitalism, economics This post may not interest many of my blog readers as it relates to politics in the UK, not a very important capitalist state these days. In the UK, after the election defeat of May 2015 for the opposition Labour Party, an election for a new leader has been under way. The previous leader, Ed Miliband, resigned within minutes of losing. The ‘neo-liberal’ right-wing dominates the parliamentary group of the Labour Party. But they made a big mistake. They allowed onto the ballot paper a left-wing MP, Jeremy Corbyn, as a ‘gesture’. He was expected to get a derisory vote. But the right-wing made another mistake: they decided to adopt an American-style ‘primary’ that allowed non-Labour party members to vote if they paid a small fee. Within weeks thousands signed up and most were backing Corbyn. As I write, with just a day to go before the result is announced, all the polls show that Corbyn is heading for a surprise win. This has produced a barrage of media and right-wing attacks. In particular, the capitalist media and the other Labour candidates have attacked Corbyn’s economic policies. During the campaign 40 economists (mainly Keynesians and heterodox leftist economists) signed a letter to the British Guardian newspaper stating that ‘Corbynomics’ wasnot extreme. “The accusation is widely made that Jeremy Corbyn and his supporters have moved to the extreme left on economic policy. But this is not supported by the candidate’s statements or policies. His opposition to austerity is actually mainstream economics, even backed by the conservative IMF.” And the signatories are right: the Corbyn programme is not extreme – except for the supporters of Capital. The Corbyn campaign for Labour leader is one that should be supported by all those wanting a better life for the majority and an end to the ‘austerity’ still being imposedby the recently victorious Conservative with ever-increasing gusto. The recent Conservative Welfare Bill to cut public services to the most vulnerable and weak in British society is particularly vicious and should have been opposed by the Labour party. But it was not by any of the four candidates for the leadership, except Corbyn, a longstanding fighter for labour causes and for socialism. Corbyn’s economic policy did not only stimulate support from leftist and heterodox economists, it also provoked opposition from right-wing mainstream economists. Professors Tony Yates of Birmingham University and Paul Levine of Surrey University rustled up a letter from 55 economists ‘from across the political spectrum’ who agreed that Corbyn’s economic plans are “likely to be highly damaging” and represent thinking that is far from mainstream economics.“It is hard to think where mainstream economics and Corbynism sit together at all,” said Yates. Most of these signatories were on the pro-austerity wing of the ‘mainstream’ like Patrick Minford of Cardiff Business School, or hedge fund advisor George Magnus, the former chief economist of UBS and Kitty Ussher, a former Treasury minister. Their criticisms were as limited in their explanation as the supporters of the letter backing Corbynomics were. Apparently, Corbyn’s popular plan to renationalise the railways would a ‘waste of public money’ and ‘make things worse’. And the idea that the Bank of England could print money to fund public infrastructure at all times, whether the economy was in recession or not, was “highly damaging” and “unnecessary” since public investment could be financed conventionally, the letter said. David Smith, the right-wing economics pundit for the Murdoch-owned Times newspaper praised the 55 mainstream economists for exposing Corbyn’s ‘incoherent’ economic policies. Renationalising the energy sector would be “a hugely expensive folly”,Smith said “The privatized utilities have not been perfect but they have been infinitely better than what went before.” Really? Smith quietly forgets the disastrous franchising and fragmenting of the rail network under successive governments, creating high prices (highest in Europe), big profits, low investment and still large taxpayer subsidies to private monopolies. The privatised utilities (gas, electricity, water, telecoms) have also delivered high prices and huge profits; but low investment and massive manager bonuses. Martin Wolf, leading Keynesian guru in the FT, reckons that “Mr Corbyn’s economic ideasare also muddled.”… Some proposals — notably higher public investment at a time of low interest rates — make sense. Some, such as letting the Bank of England inject the money it creates directly into the economy, make sense in quite restricted circumstances. Some — such as nostalgia for nationalisation and the idea that £120bn a year in lost tax revenues can be readily found — make no sense at all. Presumably, the signatories of the ‘mainstream’ letter consider their approach to be ‘rational’ and based on ‘good economics’ But as Steve Keen, the head of Kingston University economics and a prominent heterodox economist, points out, these neoclassical orthodox economists work from an economic model of equilibrium, free markets and marginalism that has no basis in reality, in the same way that the Ptolemaic system of the universe seemed perfectly constructed but “the model appeared to fit the data (except for comets, which it dismissed as “atmospheric phenomena”), but it was completely wrong about the structure of the Universe.” But I digress. In my view, the problem with Corbynomics is that opposing austerity is not enough. The question is whether Corbyn’s main economic policies, as he has been advised to adopt, would work to transform the British economy, if he ever gets the chance to implement them, and deliver an irreversible change for the better in the lives and conditions of the majority of British people. And there I have my doubts. The main planks of the Corbynomics are: ending the tax gap; providing cheap money for investment through what is called a ‘People’s QE’; an investment bank for funding infrastructure projects; renationalising the railway network, returning the postal service to majority public ownership and keeping the majority stake in the main British bank, RBS. With these economic measures, Corbyn would scrap tuition fees and restore maintenance grants for students to be funded by higher taxes on richer earners and a higher corporation tax rate. He would also create a “national education service”, offering free universal childcare — which theIPPR think-tank estimateswould cost about £6.7bn — paid for again by some of the money from raising corporation tax. And Corbyn proposes building 240,000 new homes a year and changing the “right to buy” so it applies to the private sector but not council houses or housing association properties. The new homes could be built through higher borrowing or funded by imposing higher taxes on unused land with planning permission and unoccupied properties. Corbyn also plans to end ‘the market’ in the health service, outsourcing of public services and costly funding through private finance initiatives. All this is good news for the interests of the majority and Labour. But let’s consider the efficacy of these policy measures in achieving these aims. First, the tax gap. The failure of the UK inland revenue services to collect tax from those companies that should pay it; the huge evasion and avoidance of tax by companies using corporate accountants, often previously employed as tax inspectors, is a major scandal.Richard Murphy, who has been a tireless campaigner for ‘tax justice’and is now Corbyn’s main economic adviser, has calculated that up to £120bn a year in tax revenues has gone missing because of tax avoidance, evasion and non-collection. If a Labour government had that sort of annual extra revenue in its hands, it could transform public finances and services. I have reported on this potential source of government revenue before. But could it be collected while corporations remain in private hands? Even Richard Murphy reckons it would be difficult to extract and thinks £20bn would be the most likely pick-up. Keynesian economist, Jonathan Portes, director of the National Institute of Economic and Social Research, said:“Any government expecting to fund a significant proportion of its extra spending plans on the basis of closing the tax gap would find itself with a large hole.” But that does not mean a Labour government should not try with new tax laws and anti-evasion measures. But the problem remains that as long as corporations are private entities beholden to their shareholders, both domestic and foreign, and are not publicly owned, they will seek to maximise their profits. Avoiding and evading tax is a big part of doing that. Indeed, evidence shows that if it were not for governments continually lowering corporate taxes (not raising them as Corbyn plans) and turning a blind eye to abuses, then corporate profitability would be seriously impaired and would thus reduce even the level of investment that is currently taking place. The same concern applies to the idea of closing down the £93bn a year of corporate tax reliefs and subsidies that Kevin Farnsworth at the University of York has identified– way more than the social welfare bill that the Conservative government aims to cut. Farnsworth has shown that corporations get £44bn in tax breaks for buying equipment etc and £16bn of working tax credits. This corporate welfare shows that British capitalism has to be subsidised to boost its profitability and encourage it to invest productively. But if these reliefs were to be cut, would that not just reduce profitability and lower investment further? Would there be sufficient public investment to replace lower private sector investment? Then there is the idea of a People’s ‘Quantitative Easing’ (QE) programme. Instead of the Bank of England buying government and corporate bonds and printing money to do so (the current QE), the Corbyn proposal is that the BoE would buy bonds and other assets directly from local councils, regional agencies etc so that they can invest the money in projects for more housing, schools and services.“People’s QE is fundamentally different. [It] does have the Bank of England print new money, which is identical to the process that is used by ordinary banks when they lend to business, but it gives that money to people like housing authorities, to local councils, to a green investment bank to build houses, to schools, to build hospitals.” Richard Murphy. This has been attacked by the right-wing of the Labour party and by mainstream economists as likely to fuel inflation. This is nonsense in an economy which has only just got back to the level of 2007 and where investment to GDP is near a 50-year low. There is plenty of room to boost productive investment and GDP. Indeed, inflation is currently zero. The only ‘inflation’ around is in stock and bond prices, fuelled by the BoE’s QE funding of the banks.“Any system of people’s QE would be turned off if we got to a situation of high wages and full employment, but we are so far from that at the moment that we have to tackle the low-wage economy and the lack of productivity in the UK by creating new investment, which is the foundation for new prosperity.” (Murphy). The other argument against People’s QE has come from mainstream economists, including Keynesians, who argue that it would mean ending the ‘independence’ of the Bank of England. Apparently, this independence from government control introduced by then Labour PM Gordon Brown, is such a good thing that it needs to be preserved at all costs. This again is nonsense. First, the ‘independent’ BoE does not have a good record in helping the economy and/or avoiding financial crashes. The BoE failed to spot the global financial crash and the ensuing Great Recession. It panicked when it came along and did nothing to sort out the banks out. Its independence was a fake: it really means the BoE is at the beck and call of the major banks and financial institutions in the City of London rather than to the government, parliament and the electorate. We now know that the BoE failed to impose recapitalisation and restructuring of the foreign-owned HSBC and Barclays Bank during the financial crash. As a result, the British taxpayer will never see back all the money invested in the banking system.The BoE decides its interest rate policy and its financial supervision in the interests of the City of London not the wider economy. It only has an inflation target (which it seldom meets) and no growth or employment target that would be in the interests of the people. What is worrying about People’s QE is none of this. It is whether it can do the trick. Can it help deliver more growth, employment and incomes any more than the traditional mainstream funding governments issue bonds for investment? That depends on whether the Corbyn-proposed National Investment Bank (NIB) can change things. People’s QE would be used to buy NIB bonds to fund investment. A state investment bank is certainly not extreme, as Keynesian biographer Robert Skidelsky has pointed out, it “is neither extreme nor new. There is a European Investment Bank, a Nordic Investment Bank and many others, all capitalised by states or groups of states for the purpose of financing mandated projects by borrowing in the capital markets”. And as Corbyn himself puts it: “If I was putting forward these ideas in Germany I’d be called depressingly moderate, depressingly old fashioned as they have a national investment bank already and they invest in public services.” I would add the NIB looks very much like Brazil’s BNDES, which has been mightily successful in driving investment projects at lower borrowing costs during the Great Recession so that Brazil got some investment. Indeed, during the Great Recession, those countries that suffered least were precisely those countries that were bolstered by state-owned investment banks that supported infrastructure projects to keep jobs and create investment. Brazil’s INDES investment bank was very successful in doing that, despite the cries of foul by the privately-owned and foreign banks operating in Brazil. It is no accident, for example, that Brazil had a very mild recession because the government there plunged huge resources through its state-owned development bank for infrastructure spending. But the experience of the Brazilian investment bank also shows the problems. The BNDES has taken most of the investment business away from Brazil’s private banks, both domestic and foreign. So they have concentrated on mortgage loans and speculation in commodities and financial assets, spawning a property and credit bubble. Their huge asset power has not been used for developing the economy because there is no profit in it for them. This is the risk in the UK too. The NIB will do projects for productive investment while the ‘big five’ multinational banks will sit on the sidelines. They are already failing to lend to small businesses and for investment (only 3% of all assets are in manufacturing sectors).So I’m afraid that Skidelsky’s claim that “a state-led investment programme offers a way to rebalance the British economy away from private speculative activity to long-term investment in sustainable growth” won’t happen if the only instrument is the NIB. Surely, just using the NIB and perhaps the state-owned RBS alone cannot turn the credit institutions into vehicles for funding faster investment and employment? There is crying need to take over the big five UK banks and use their financial resources in a national plan for investment and growth. This is actually Trade Union Congress official policy (although that is ignored by the trade union leaders). The case for public ownership is overwhelming. Moreover, how can Corbyn’s plan to end the grotesque salaries and bonuses paid to top executives and bankers be implemented without proper public control and ownership of the banks? Then, as Marxist economist, Chris Dillon recently put it: “who will do the actual investing? There’s a case for the state to invest in infrastructure. But we also need corporate investment, to raise private sector productivity.” I don’t think we need corporate investment, if that means relying on private companies for profit to deliver the investment that the NIB offers to fund. Take a major project just announced for London. The London ‘super sewer is set to begin next year. Bazalgette Tunnel Limited, a special purpose company has been formed to lead the project. Balfour Beatty, the UK construction company, has been awarded a £416m contract to build part of London’s new “super sewer.” This is a private company with shareholders and private investors. Surely this is a project for a publicly owned and controlled entity, not for profit? It brings us back to the need for any effective (if extreme) economic policy to include as one of its main planks public ownership under democratic control of strategic industries, or what used to be called in Old Labour parlance, the ‘commanding heights’ of the economy. Corbynomics includes the vital measure of renationalising the railways after the disgraceful and incompetent breaking up of the state rail system into private monopolies with franchises, and now asking the highest rail prices in Europe and yet still subsidised by the taxpayer. It has taken the British railways back to the 1830s. The most pathetic argument against renationalisation has come from Kate Hoey, the maverick ex-Labour MP. You see it can’t be done because it is against EU directives!“It would be hyperbole to say that all efforts to renationalise the railways would be blocked by the EU, but it would be equally naïve to dismiss the problem”. But EU directives are not law but guidelines and can be interpreted or applied as member states wish. Corbyn apparently also is considering restoring the state majority stake in the Royal Mail postal service, after its recent privatisation at a ludicrously low price by the current government. This is good but that still leaves swathes of key British economic operations in the hands of profit-seeking companies. What about the rest of transport: deregulated buses in the big cities; and all the British companies that used to be part of the public sector? What about British Petroleum, British Airways; British Telecom; British Gas; British Aerospace; the electricity and water boards; Transco; Rolls Royce, British Steel, let alone British Coal? And there are the major strategic sectors that should be part of what is called in Labour parlance “the commanding heights of the economy”: the major pharma and auto companies, now mostly in foreign hands where any profits for end up overseas. Public ownership of the key national and regional airports would ensure a proper service that did not cause environmental pain from noise and pollution . “I think the third runway is a problem for noise pollution and so on across west London . . . I also think there is an under-usage of the other airports around London.” (Corbyn). Of course, such a programme would be extreme, not just for the capitalist media, mainstream economists and the Labour leadership, but also in the minds of many of the signatories to the leftist letter. But without these measures, in my view, there will not be a “rebalancing of the British economy away from private speculative activity to long-term investment in sustainable growth” (Skidelsky). The key is investment. As Marxist economist, Mick Burke has pointed in a series of excellent posts,the capitalist sector has failed to deliver decent incomes and sustained growth because it had failed to invest. It seems that the private sector cannot deliver decent incomes and employment for all.Burke points outthat “per capita GDP is still below where it was before the crisis began in 2008. This remains the weakest recovery on record and the year-on-year growth rate has slowed from 3% to 2.6%. This follows a period from the end of 2012 onwards when no new austerity measures were imposed. Renewed austerity on the same scale as in 2010 to 2012 means there is likely to be a similar slowdown.” He goes on: “The level of investment in the British economy was £295 billion in 2014, exactly the same as the pre-crisis level of 2007. But the economy is actually larger 4.2% larger (keeping pace with population growth, but no more than that). Therefore investment is declining as a proportion of GDP. Consumption, not investment is leading very weak growth and this is not sustainable.” The profit rate has only barely returned to its pre-crisis level and is well below profitability prior to this century. The same is true for business investment. Both of these are a recipe for continued slow growth. A Corbyn-led Labour government (in 2020?) would be a step forward for the majority. And it would be necessary because, by then, capitalism, and British capital, would probably have entered yet another major slump and crisis. But that prospect alone shows that Corbynomics is just not extreme enough in reversing the failure of British capitalist production. For 30 years, British Labour leaders have been firmly in the camp of Capital on one bank of the river in the class struggle. Now Corbynomics looks to leave the camp of capital and cross over to the camp of Labour. Thus it is too extreme for the interests of Capital but too moderate for the interests of Labour. Being caught in the middle of a fast stream (capitalist crisis) can be dangerous. PS: see the rabid attack on Corbyn in the FT by Philip Stephens. Share this: Share Click to share on Facebook (Opens in new window)Facebook Click to share on X (Opens in new window)X Click to share on Reddit (Opens in new window)Reddit Like Loading... Related Posted bymichael robertsSeptember 11, 2015 September 13, 2015Posted incapitalism, economics Post navigation Previous Post Previous post: From value and profit to crisis and back Next Post Next post: China: a weird beast 20 thoughts on “Corbynomics – extreme or moderate?” Janet Bsays: September 11, 2015 at 8:33 am Thank you so much Michael for this blog. As a UK pensioner you put everything in perspective for me. I and no doubt many others have been waiting decades for ‘Corbynomics’ . Janet Beale Reply 2. larsjdksays: September 11, 2015 at 8:34 am I do not know, if you are familiar with William (Bill) Mitchell’s work and blog. But I think he must be one of the most important economists for a progressive politics today – MMT Modern Monetary Theory where he even progresses from Keynes’ thinking. He is an Australian but he has also researched and written an important new book on European political economy: Eurozone Dystopia (2015). And on his interesting blog he has also commented on Corbyn’s economic thinking. Here is one of them: “Corbyn should stop saying he will eliminate the deficit” Posted on Thursday, August 13, 2015 by bill Link: Reply 3. Chris Jonessays: September 11, 2015 at 9:42 am Typos in this sentence make it difficult to comprehdn ‘They are already failing to lend to sm“a state-led investment programme offers a way to rebalance the British economy away from private speculative activity to long-term investment in sustainable growth” all businesses and for investment (only 3% of all assets are in manufacturing sectors).’ Reply 1. michael robertssays: September 11, 2015 at 10:02 am Thanks Chris – will fix. Reply SimonHsays: September 11, 2015 at 10:49 am I think criticism by mainstream economists should be worn like a badge of honour. It’s like a climate scientist being attacked by some hack from a Koch funded think tank. Now if the hack was saying nice things then the scientist should start getting worried! Reply Michael S.says: September 11, 2015 at 12:09 pm I’ve been reading this blog for a while now, and though Michael Roberts shows an astonishingly profound understanding of economic developments which most of the times is consistent with Marx’ theories, there are aspects that strike me as odd. In proposing a more radical alternative to Corbyn’s contradictory (and, in my opinion, impossible) left-keynesian programme, M.R. ends up demanding a broad renationalisation. But there is no discussion of something that I always considered to be crucial whenever Marx/Engels talk about revolutionary change: the difference between a “socialisation” (“Vergesellschaftung”) in a literal sense (that is: a self-governed and democratic disposal of the means of production) and the taking over of capitalist functions by the state (“Verstaatlichung”). Marx/Engels were somewhat naïve about that when they wrote the “Manifesto”, but they became increasingly aware of this difference and developed a fragmentary critique of the modern state and state-capitalist ideas later on, which is for example very obvious in Engels’ “Socialism: Utopian and Scientific”: “And the modern State, again, is only the organization that bourgeois society takes on in order to support the external conditions of the capitalist mode of production against the encroachments as well of the workers as of individual capitalists. The modern state, no matter what its form, is essentially a capitalist machine — the state of the capitalists, the ideal personification of the total national capital. The more it proceeds to the taking over of productive forces, the more does it actually become the national capitalist, the more citizens does it exploit. The workers remain wage-workers — proletarians. The capitalist relation is not done away with. It is, rather, brought to a head. But, brought to a head, it topples over. State-ownership of the productive forces is not the solution of the conflict, but concealed within it are the technical conditions that form the elements of that solution.“ ( (One could also discuss whether Engels claim of an indirect relation between state-ownership and a revolutionary solution of the contradictions of capitalism is not far too optimistic given the terrible failures of Bolshevism…) One thing should be clear: as communists, we should not reject the accusations of mainstream economists that our “economic measures” will be destructive by proving that a socialist economy will be more productive in capitalist terms, that it will create employment, increase growth etc etc – instead, we should ensure that this “destruction” opens up a space for new, self-organised ways of cooperation: and that is (in my opinion) something no state-run policy will ever achieve on its own. I hope to hear other views on the points I raised, maybe even by M. Roberts himself… Greetings from Germany Michael Reply 6. Γιώργοςsays: September 11, 2015 at 10:44 pm Another excellent article ! Reply 7. Edgarsays: September 12, 2015 at 2:31 pm “(and, in my opinion, impossible) left-keynesian programme” Saying it doesn’t make it so. If right wing neo liberalism is possible then as sure as night becomes day left Keynesian policies are possible. The impossible/possible line of reasoning is a direct contradiction of the Marxist method, so I am surprised you brought them up. The victory of Corbyn is a great development. Reply 1. sartesiansays: September 13, 2015 at 2:19 pm As was the victory of Syriza……oh wait, never mnd. Reply 1. Edgarsays: September 14, 2015 at 7:46 am Syriza always needed allies to be in a position to push forward with their agenda. You are too harsh. Patience comrade! 2. sartesiansays: September 14, 2015 at 3:33 pm “Harsh”? I believe that is precisely the “Marxist method”– that old line about “merciless criticism of everything in its existence”? Too harsh? No such thing. “Patience” for what? So the left can perform another of the rituals in its repetition compulsion? What was it Hooper says to Mayor Vaughn in Jaws? “I’m not going to waste my time arguing with a man who’s lining up to be a hot lunch.” Yeah, that’s it. 3. Edgarsays: September 14, 2015 at 5:21 pm “merciless criticism of everything in its existence” The definition of criticism isn’t saying bad things, it is an objective study of the thing in consideration. Giving both the good and bad aspects and contextualising it in the given conditions and the historical period. The word merciless means means you do not let any bias impinge your view, as Marx rightly said merciless criticism includes that of oneself. 4. sartesiansays: September 15, 2015 at 3:01 am Really? Can’t wait to read your merciless criticism of Corbyn 2. Image 17: Edgar's avatarEdgarsays: September 15, 2015 at 5:18 pm I will not hold my breath for your merciless criticism of Syriza. Reply 1. Image 18: sartesian's avatarsartesiansays: September 15, 2015 at 9:10 pm You don’t have to. It’s already been written. 2. Image 19: Edgar's avatarEdgarsays: September 16, 2015 at 5:24 pm I have and it didn’t pass the test of merciless criticism. More like merciless moaning. Now let me breath out. 3. sartesiansays: September 17, 2015 at 3:26 am Oh Edgar, you’re such a card. Breathe out? Sure thing. Bleed out too. Fine by me. Nemosays: September 13, 2015 at 12:42 am No, Edgar, you are wrong. The marxist method does speak of contradiction, and the necessity of contradiction and for it to be solved. Keynesianism is, as said by keynes itself, an attempt to soften contradiction; to save capital from it’s own self-destructive tendencies. To direct it in a way that it’s compatible with an integrated society. But capital is all-overreaching, it doesn’t care for a balanced society. It is predatory and it’s apetite is endless (thus it is a “vampire” that feeds off living labor until there can be nothing) – Capital rules society, and unless this is overturned, it can and will destroy any and all “left keynesianism”. In fact, look at the EU: it has already done it. That is what auterity is. Trying again left keynesianism in lieu of living in the consequences of it’s defeat is madness (“doing the same thing and expecting different results”). The key battle is losing the fear of radical economics… and of revolution. Reply 9. Edgarsays: September 13, 2015 at 7:13 am Nemo – you are wrong. Being possible and impossible is not a contradiction, they are just dogmatic absolutes. Left Keynesianism is perfectly possible, as history has shown numerous times. Whether it is desirable is a different question. “Trying again left keynesianism in lieu of living in the consequences of it’s defeat is madness” What if the conditions have changed? You can’t make such absolute proclamations, well you can, but it won’t wash with me. Reply 10. lenny webersays: December 2, 2015 at 1:54 pm To hold that capitalism can be operated in the interests of the working class, as leftists like Corbyn do, is as rational as holding that abattoirs can be operated in the interests of the livestock they process. Society is composed of two contending classes, a non producing master/controlling class that forms a miniscule minority, and a majority producing/working servant class. The contradictions that arise out of this artificial division of society (war, terrorism, dislocation, dispossession, exploitation etc) can only be resolved when the natural and industrial resources of our planet are held as the common heritage of all humans and these resources are democratically administered by and in the interests of the whole community. Echoing Tom Paine “These are the times that try men’s souls.” The challenges humanity faces can’t be met by ignoring the big picture and the the root cause it portrays and makes clear to those inclined to stand back and survey. We all share the same ancestors, we are one family, therefore to be true to our common identity we urgently need to acknowledge the ethic that is the foundation of family, “from each according to ability, to each according to need”. Now more than ever reformism, tinkering with the symptoms of our social malaise needs to be abandoned if favour of a radical revolutionary change in the basis of our society, anything other is to continue the denial of our common humanity. Reply I have restored comments but very long ones (as per subjective opinion) will be rejected Cancel reply Δ Follow Michael Robert's Blog Enter your email address to subscribe to this blog and receive notifications of new posts by email. Email Address: Sign me up! Search the blog Search for: Michael Roberts Blog,Blog at WordPress.com. Comment Reblog SubscribeSubscribed Michael Roberts Blog Join 8,415 other subscribers Sign me up Already have a WordPress.com account? Log in now. Michael Roberts Blog SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... 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188397	https://math.answers.com/math-and-arithmetic/What_is_the_relation_between_a_polygon%27s_number_of_sides_and_sum_of_its_exterior_angle	What is the relation between a polygon's number of sides and sum of its exterior angle? - Answers Create 0 Log in Subjects>Math>Math & Arithmetic What is the relation between a polygon's number of sides and sum of its exterior angle? Anonymous ∙ 11 y ago Updated: 8/8/2025 The sum of the exterior angles of any polygon, regardless of the number of sides, is always 360 degrees. This holds true for polygons with three sides (triangles) all the way to polygons with many sides. Each exterior angle can be calculated by subtracting the interior angle from 180 degrees, but the total remains constant at 360 degrees for all polygons. AnswerBot ∙ 1 mo ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Math & Arithmetic ### What polygons has the larger exterior angle? A polygon with any number of sides can have an exterior angle that is larger than the corresponding interior angle. However, a triangle is the only regular polygon where the exterior angle is larger. ### How do you find the number of side in a polygon if all you know is the sum of exterior angles? No way. All polygons have the same sum of exterior angles.You need the sum of interior angles. ### What is the formulas to finding interior and exterior angles of regular polygons? Interior Angles: n-2 (n is number of sides) ____ 180 Exterior angles are always 360 degrees. ### Why does all regular polygons of the same number of sides are similar to each other? Because their interior and exterior angles are the same measurements. ### How is the number of sides related to the sum of the exterior angles in a polygon? The sum of the exterior angles of any polygon, regardless of the number of sides, is always 360 degrees. This holds true whether the polygon is regular or irregular. Therefore, the number of sides does not affect the total sum of the exterior angles; it remains constant at 360 degrees for all polygons. Related Questions Trending Questions How do you write to netanyahu?What is 5 times -2?What number is 60 percent of 980?How many hours is 9.8days?What is 398.52 minus 10 percent?What percent is 1342 of 4026?If a corporation sells certain capital equipment for more than their initial purchase price the difference between the sales price and the purchase price is called a?What is 50p as a percentage of 10?When 5 triangles are combined how many corners?Is 2184 divisible by 3?What does one trillion dollar look like in numbers?What is the greatest integer less than -8?What Natwest branch has sort code 16-60-51?Do sundials have to be round?What is a perfect square between the numbers 70-85?Who has the most number of inventions to their credit?What is 60 percent off of 119?Why does the area of 70 times 30 equal 2100 but the area of 50 times 50 equal 2500 square feet?How many cells are there after 1 hour?Is bromfed dm a narcotic? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
188398	https://www.cazoommaths.com/us/math-worksheet/percents-finding-the-original-worksheet/	Percents: Finding the Original Worksheet \| PDF Printable Number & Operations Worksheet \| Cazoom Math BY TOPICBY CCSS BY GRADE Kindergarten1st Grade2nd Grade3rd Grade4th Grade5th Grade6th Grade7th Grade8th GradeHigh SchoolHigh School GeometryAlgebra 1Algebra 2Integrated Math 1Integrated Math 2Integrated Math 3 PRICINGSEARCHHELP US UK US LOGINSIGN UP LOGINSIGN UP Menu BY TOPICBY CCSS BY GRADE PRICINGSEARCHHELP US UK US LOGINSIGN UP Back to: Percentages Percents: Finding the Original WORKSHEET Suitable for Grades:7th Grade CCSS:7.RP.A.3 CCSS Description:Use proportional relationships to solve multistep ratio and percent problems, including simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, and percent error. Percents: Finding the Original WORKSHEET DESCRIPTION This worksheet is targeting the topic of Reverse Percentages. It provides some excellent practice questions covering the many different ways this topic could be phrased in an examination setting. Section A requires students to complete a table stating what 100% would be based on the value of a given percentage. It doesn’t lead students through a fixed method, allowing them to get to their solutions using whichever method they prefer or have been taught. Section B then asks them to find the original amounts after some percentage increases and percentage decreases. The table provided limits the variety of percentage changes to give students time to practice the skill and build confidence. Finally, Section C has a large selection of word problems mixing up increases and decreases. These questions are written in the style of exam questions to give excellent preparation for state exams. DOWNLOAD WORKSHEET DOWNLOAD SOLUTION DOWNLOAD WORKSHEET DOWNLOAD SOLUTION RELATED TO Percents: Finding the Original WORKSHEET ### Percent Change Grades: 7th Grade ### Percent Increase and Decrease Grades: 7th Grade ### Converting Percents to Decimals Grades: 7th Grade ### Percent Word Problems Grades: 7th Grade ABOUT About Us Visual Maths Resources Ltd is a company registered in England and Wales with a company no. 10607102 with a registered office at 71-75 Shelton Street, WC2H 9JQ. Cazoom Maths is the trading name of Visual Maths Resources Ltd. CONNECT support@cazoommaths.com +44 1829 750002 Contact Us Privacy Policy GET STARTED Free Worksheets Worksheet of the Week Plans & Prices Sitemap FAQs MEMBERS AREA Members Area School Access © All rights reserved \| Visual Maths Resources Ltd Need Help?
188399	https://pubchem.ncbi.nlm.nih.gov/compound/Zinc	Zinc \| Zn \| CID 23994 - PubChem An official website of the United States government Here is how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. NIH National Library of Medicine NCBI PubChem About Docs Submit Contact Search PubChem compound Summary Zinc PubChem CID 23994 Structure Primary Hazards Laboratory Chemical Safety Summary (LCSS) Datasheet Molecular Formula Zn Synonyms ZINC 7440-66-6 Zinc dust Zinc powder Zinc, elemental View More... Molecular Weight 65.4 g/mol Computed by PubChem 2.2 (PubChem release 2025.04.14) Element Name Zinc Dates Create: 2004-09-16 Modify: 2025-09-27 Description Zinc is one of the most common elements in the earth's crust. It is found in air, soil, and water, and is present in all foods. Pure zinc is a bluish-white shiny metal. Zinc has many commercial uses as coatings to prevent rust, in dry cell batteries, and mixed with other metals to make alloys like brass, and bronze. A zinc and copper alloy is used to make pennies in the United States. Zinc combines with other elements to form zinc compounds. Common zinc compounds found at hazardous waste sites include zinc chloride, zinc oxide, zinc sulfate, and zinc sulfide. Zinc compounds are widely used in industry to make paint, rubber, dyes, wood preservatives, and ointments. Agency for Toxic Substances and Disease Registry (ATSDR) Zinc ashes appears as a grayish colored powder. May produce toxic zinc oxide fumes when heated to very high temperatures or when burned. Insoluble in water. Used in paints, bleaches and to make other chemicals. CAMEO Chemicals Zinc dust appears as a grayish powder. Insoluble in water. May produce toxic zinc oxide fumes when heated to very high temperatures or when burned. Used in paints, bleaches and to make other chemicals. CAMEO Chemicals View More... See also: Copper (related); Silver (related); Nickel (related) ... View More ... 1 Structures 1.1 2D Structure Structure Search Get Image Download Coordinates Chemical Structure Depiction Full screen Zoom in Zoom out PubChem 1.2 Crystal Structures COD records with this CID as component 1504514 1512230 1512968 1515217 1515219 1515492 1520497 1522875 1534571 1536876 1536877 1536878 1536880 1539843 1540028 1542312 1543516 1543517 1543518 1545987 1546182 1548308 1548759 1548761 1548762 1550070 1550170 1551518 1552204 1552207 1552217 1554685 1555240 1556026 1559308 1559309 2001817 2015245 2015599 2015657 2016069 2103022 2104608 2106141 2200577 2202428 2203969 2207304 2207819 2208334 2210417 2216714 2216776 2217683 2229724 2238371 2238395 2241720 2243182 4029735 4032623 4032624 4032625 4036417 4038858 4064336 4064929 4064930 4064931 4068910 4069330 4069592 4073425 4074228 4077871 4081144 4082207 4083487 4086169 4101591 4105280 4105281 4105285 4105289 4107525 4107532 4107658 4108985 4108986 4110841 4112741 4113442 4114200 4115069 4117019 4120877 4121121 4121162 4121245 4123391 4124134 4124135 4124137 4124138 4130618 4131537 4131539 4132331 4132474 4132475 4134299 4134453 4301621 4304579 4309869 4318150 4318812 4318991 4320376 4320597 4321715 4322640 4323514 4323733 4323738 4323746 4324502 4325560 4325562 4330551 4333719 4337015 4338823 4338888 4338976 4339261 4340468 4341122 4345930 4348662 4501279 4503879 4504094 4504095 4508364 4517514 7001717 7010543 7012618 7014853 7017415 7017417 7017419 7019116 7019263 7021427 7022103 7022252 7022253 7023363 7023553 7023779 7029075 7029076 7029492 7029984 7030101 7030190 7035600 7039732 7040079 7040080 7040081 7040691 7048364 7049052 7049058 7054011 7055346 7057244 7057254 7100694 7102257 7106172 7109174 7109341 7109409 7109411 7112917 7112918 7116269 7116825 7116826 7117411 7117801 7118306 7120173 7124165 7211795 7215411 (the list is too long and has been truncated) Crystallography Open Database (COD) 2 Names and Identifiers 2.1 Computed Descriptors 2.1.1 IUPAC Name zinc Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14) PubChem 2.1.2 InChI InChI=1S/Zn Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.3 InChIKey HCHKCACWOHOZIP-UHFFFAOYSA-N Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.4 SMILES [Zn] Computed by OEChem 2.3.0 (PubChem release 2025.04.14) PubChem 2.2 Molecular Formula Zn Computed by PubChem 2.2 (PubChem release 2025.04.14) Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; PubChem 2.3 Other Identifiers 2.3.1 CAS 7440-66-6 Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; DrugBank; EPA Chemical Data Reporting (CDR); EPA Chemicals under the TSCA; EPA DSSTox; EPA Integrated Risk Information System (IRIS); European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA); NJDOH RTK Hazardous Substance List; Risk Assessment Information System (RAIS) 15176-26-8 ChemIDplus; EPA DSSTox View More... 2.3.2 Deprecated CAS 118102-83-3, 12793-53-2, 1865801-44-0, 195161-85-4, 199281-21-5, 2443494-44-6, 298688-49-0, 91741-63-8 ChemIDplus; EPA Chemicals under the TSCA 195161-85-4, 199281-21-5, 298688-49-0 EPA DSSTox 2.3.3 European Community (EC) Number 231-175-3 European Chemicals Agency (ECHA) 2.3.4 UNII J41CSQ7QDS FDA Global Substance Registration System (GSRS) 2.3.5 UN Number 1435 (ZINC ASHES) CAMEO Chemicals; Emergency Response Guidebook (ERG) 1436 (ZINC DUST) CAMEO Chemicals; Emergency Response Guidebook (ERG) 1436 (Zinc powder) Emergency Response Guidebook (ERG) 1436 ILO-WHO International Chemical Safety Cards (ICSCs) 2.3.6 AIDS Number 114928 NIAID ChemDB 2.3.7 ChEBI ID CHEBI:27363 ChEBI 2.3.8 ChEMBL ID CHEMBL1201279 ChEMBL 2.3.9 DrugBank ID DB01593 DrugBank 2.3.10 DSSTox Substance ID DTXSID7035012 EPA DSSTox DTXSID101316732 EPA DSSTox DTXSID201316735 EPA DSSTox 2.3.11 ICSC Number 1205 ILO-WHO International Chemical Safety Cards (ICSCs) 2.3.12 NCI Thesaurus Code C948 NCI Thesaurus (NCIt) C68295 NCI Thesaurus (NCIt) 2.3.13 Nikkaji Number J1.458.431E Japan Chemical Substance Dictionary (Nikkaji) J3.735D Japan Chemical Substance Dictionary (Nikkaji) 2.3.14 PharmGKB ID PA451956 PharmGKB 2.3.15 RXCUI 11416 NLM RxNorm Terminology 2.3.16 Wikidata Q758 Wikidata 2.3.17 Wikipedia Zinc Wikipedia Merrillite Wikipedia 2.4 Synonyms 2.4.1 MeSH Entry Terms Zinc Medical Subject Headings (MeSH) 2.4.2 Depositor-Supplied Synonyms ZINC 7440-66-6 Zinc dust Zinc powder Zinc, elemental Zincum metallicum Rheinzink Jasad Granular zinc Blue powder Emanay zinc dust Asarco L 15 ZN Lead refinery vacuum zinc CHEBI:30185 J41CSQ7QDS CHEBI:27363 Zinc Met Zyncum Metallicum ZINCUM MET RefChem:6357 Zincum Metallicum 12X NEURALGIA HEADACHE Zincum metallicum 200C ZINCUM METALLICUM 30c ZINCUM METALLICUM 200CK Bestmade Natural Products Zyncum Met 231-175-3 Dietary Zinc Zinc(2+) ion cinc Zinc, ion (Zn1+) MFCD00011291 Zinc(1+) Zinc(1-) zincide zincum zinc atom Zink Zinc, ion (Zn 1-) zinc(0) 30Zn Zinc foil Zinc shot Zinc rod CHEMBL1201279 Zn Standard Solution Merrillite Zinc, SAJ first grade 14018-82-7 Zinc powder; Cinc; Merrillite Zinc, p.a., 99.8% Zinc (fume or dust) Zinc rod, 1.27cm (0.5in) dia Zinc wire, 1.0mm (0.04in) dia Zinc (metallic) Zinc wire, 0.25mm (0.01in) dia Zinc, suitable for arsenic determination CCRIS 1582 LS 2 LS 6 Zinc (dust or fume) HSDB 1344 Zinc Standard: Zn @ 10 microg/mL in 2% HNO3 ZINC, DUST Zinc Standard: Zn @ 1000 microg/mL in 5% HNO3 EINECS 231-175-3 UNII-J41CSQ7QDS UN1435 UN1436 Zinc Standard: Zn @ 10000 microg/mL in 5% HNO3 Zinc AA Standard: Zn @ 1000 microg/mL in 5% HNO3 Zinc Standard: Zn @ 1000 microg/g in Hydrocarbon Oil zinc anion Zinc granules Zinc element Element Zinc Element:Zinc Sulfur-Free Zinc Concentrate: Zn @ approx. 18 wt% in Hydrocarbons Sulfur-Free Zinc Standard: Zn @ 1000 microg/g in Hydrocarbon Oil Sulfur-Free Zinc Standard: Zn @ 5000 microg/g in Hydrocarbon Oil Zinc ashes Zinc mossy Zinc, lump, 10 mm max. lump size, weight 100 g, purity 99.99% Zinc, lump, 10 mm max. lump size, weight 200 g, purity 99.99% Zinc wire Atomic Zinc Element-Zn Element:Zn Zinc, pellets DTXSID7035012 Zinc (element) zinc(I) cation Zinc, rods Zinc, Mossy Zinc (atomic) Zn (element) zincide(-I) Zinc single crystal disc, 10mm (0.39in) dia, 2-3mm (0.08-0.1in) thick, (0001) orientation, +/-0.5 degrees zincide(1-) Zinc, purum, powder zinc(1+) ion Zinc, dust 44 zinc ion (1+) Zinc, ion (Zn1-) Zinc Powder 325 mesh SCHEMBL50 ZINC [VANDF] ZINC [EMA EPAR] Zinc powder or zinc dust Zinc, JIS special grade Zinc pieces 2-14 mesh Zinc powder 6-9 microns ZINC [WHO-DD] Epitope ID:114075 ZINC [MI] ZINC66 EC 231-175-3 Zinc powder, -100 mesh Zinc, mossy, >=99% Zinc dust, <10 micrometers Zinc powder (-325mesh) Zinc, 99.5% Zn(+) Zn(-) Zn(0) 7440-66-6(powder) Zinc metal, NIST SRM 683 Zinc nanopowder (40~60 nm) ZINC, ELEMENTAL [HSDB] Zinc, p.a., 99.9% SCHEMBL29735010 SCHEMBL29754791 SCHEMBL29825032 SCHEMBL30134638 Zinc (Zn), 99.5% Nano Zinc rod, 7mm (0.3in) dia Zinc, LR, dust, >=90% ZINCUM METALLICUM [HPUS] CHEBI:37254 CHEBI:37255 Zinc, shot, 99.9999% Zinc nanopowder (80-100 nm) Zinc powder or zinc dust [UN1436] [Dangerous when wet] Zinc, LR, granular, >=99% Zinc powder, -140+325 mesh Zinc rod, 8mm (0.31in) dia DTXSID101316732 DTXSID201316735 High-purity zinc, NIST SRM 682 Zinc rod, 10mm (0.39in) dia Zinc, AR, dust, >=99.5% Rieke Zinc 2.5g in 25ml of THF BDBM50259154 Zinc wire, 0.5mm (0.02in) dia Zinc wire, 2.0mm (0.08in) dia Zinc, dust, <10 mum, >=98% AKOS015907403 Zinc ingot, 345mm (13.6in) long Zinc, SAJ first grade, >=85.0% DB01593 Fine Zinc powder, 3.4 - 3.9 ?m FZ07896 Zinc, SAJ special grade, >=90.0% Zinc rod, 3.18mm (0.125in) dia Zinc wire, 3.18mm (0.125in) dia Zinc shot, 1-5mm (0.04-0.2in) Zinc, granular, -10-+50 mesh, ACS BP-11393 Q758 Zinc mossy, 2.5cm (0.98in) & down Zinc Powder 325 mesh High Grade Material Zinc, mossy, 99.99% trace metals basis Zinc - Zn @ 10 microg/mL in 2% HNO3 Zinc, granular, 20-30 mesh, ACS reagent NS00099584 Z0015 Zinc, foil, thickness 0.1 mm, 99.999% Zinc, foil, thickness 0.5 mm, 99.999% Zinc, granular, -30 mesh+/-100, 99% Zinc ashes [UN1435] [Dangerous when wet] Zinc, wire, diam. 0.25 mm, 99.999% Zinc - Zn @ 1000 microg/mL in 5% HNO3 Zinc - Zn @ 1000 microg/g in Hydrocarbon Oil Zinc - Zn @ 5000 microg/g in Hydrocarbon Oil Zinc, ACS reagent, -30-+100 mesh, granular F044742 Zinc foil, 1.0mm (0.04in) thick, Puratronic? Zinc wire, 0.25mm (0.01in) dia, Puratronic ? Zinc, shot, 5 mm, 99.999% trace metals basis Zinc shot, 1-6mm (0.04-0.24in), Puratronic? Zinc, foil, 4mm disks, thickness 0.01mm, 99.9% Zinc, foil, 4mm disks, thickness 0.02mm, 99.9% Zinc, foil, 6mm disks, thickness 0.01mm, 99.9% Zinc, foil, 6mm disks, thickness 0.02mm, 99.9% Zinc, foil, 8mm disks, thickness 0.01mm, 99.9% Zinc, foil, 8mm disks, thickness 0.02mm, 99.9% Zinc, shot, <12 mm, 99.99% trace metals basis Zinc, wire reel, 1m, diameter 1.0mm, 99.99+% Zinc, wire reel, 1m, diameter 1.0mm, 99.999% Zinc, wire reel, 1m, diameter 2.0mm, 99.99+% Zinc, wire reel, 2m, diameter 1.0mm, 99.99+% Zinc, wire reel, 2m, diameter 1.0mm, 99.999% Zinc, wire reel, 2m, diameter 2.0mm, 99.99+% Zinc, wire reel, 5m, diameter 1.0mm, 99.99+% Zinc, wire reel, 5m, diameter 2.0mm, 99.99+% Q27117082 Q27117083 Zinc freezing point standard 419.527 C, NIST SRM Zinc Powder, 99.9+% (metal basis, O<10%) Nano Zinc, foil, 10mm disks, thickness 0.01mm, 99.9% Zinc, foil, 10mm disks, thickness 0.02mm, 99.9% Zinc, foil, 15mm disks, thickness 0.01mm, 99.9% Zinc, foil, 15mm disks, thickness 0.02mm, 99.9% Zinc, foil, 4mm disks, thickness 0.005mm, 99.9% Zinc, foil, 4mm disks, thickness 0.015mm, 99.9% Zinc, foil, 6mm disks, thickness 0.005mm, 99.9% Zinc, foil, 6mm disks, thickness 0.015mm, 99.9% Zinc, foil, 8mm disks, thickness 0.005mm, 99.9% Zinc, foil, 8mm disks, thickness 0.015mm, 99.9% Zinc, pieces, 2-14 mesh, 99.9% trace metals basis Zinc, wire reel, 0.1m, diameter 1.0mm, 99.999% Zinc, wire reel, 0.2m, diameter 1.0mm, 99.999% Zinc, wire reel, 0.5m, diameter 1.0mm, 99.99+% Zinc, wire reel, 0.5m, diameter 1.0mm, 99.999% Zinc, wire reel, 0.5m, diameter 2.0mm, 99.99+% Zinc, wire reel, 10m, diameter 1.0mm, 99.99+% Zinc, wire reel, 10m, diameter 2.0mm, 99.99+% Zinc, wire reel, 1m, diameter 6.35mm, 99.99+% Zinc shot, 10mm (0.4in) dia x 2mm (0.08in) thick Zinc Standard: Zn @ 5000 microg/g in Hydrocarbon Oil Zinc, foil, 10mm disks, thickness 0.0025mm, 99.9% Zinc, foil, 10mm disks, thickness 0.005mm, 99.9% Zinc, foil, 10mm disks, thickness 0.015mm, 99.9% Zinc, foil, 15mm disks, thickness 0.0025mm, 99.9% Zinc, foil, 15mm disks, thickness 0.005mm, 99.9% Zinc, foil, 15mm disks, thickness 0.015mm, 99.9% Zinc, foil, 4mm disks, thickness 0.0025mm, 99.9% Zinc, foil, 6mm disks, thickness 0.0025mm, 99.9% Zinc, foil, 8mm disks, thickness 0.0025mm, 99.9% Zinc, granular, 20-30 mesh, ACS reagent, >=99.8% Zinc, powder, < 150 um, 99.995% trace metals grade Zinc, powder, <150 mum, 99.995% trace metals basis Zinc, wire reel, 0.1m, diameter 6.35mm, 99.99+% Zinc, wire reel, 0.2m, diameter 6.35mm, 99.99+% Zinc, wire reel, 0.5m, diameter 6.35mm, 99.99+% Zinc, wire reel, 1m, diameter 0.05mm, as drawn, 99% Zinc, wire reel, 2m, diameter 0.05mm, as drawn, 99% Zinc, wire reel, 5m, diameter 0.05mm, as drawn, 99% Zinc foil, 0.25mm (0.01in) thick, 30cm (12in) wide Zinc foil, 0.62mm (0.024in) thick, 30cm (12in) wide Zinc, ACS reagent, -30-+100 mesh, >=99.8%, granular Zinc, foil, thickness 0.25 mm, 99.9% trace metals basis Zinc, foil, thickness 1.0 mm, 99.99% trace metals basis Zinc, granular, -10-+50 mesh, >=99.8%, ACS reagent Zinc, rod, 19 mm diameter, length 100 mm, purity 99.9% Zinc, rod, 19 mm diameter, length 200 mm, purity 99.9% Zinc, rod, 19 mm diameter, length 900 mm, purity 99.9% Zinc, rod, length 500 mm, 19 mm diameter, purity 99.9% Zinc, sticks, diam. 7-10 mm, 99.97% trace metals basis Zinc, wire reel, 100m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 100m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 10m, diameter 0.025mm, as drawn, 99% Zinc, wire reel, 10m, diameter 0.05mm, as drawn, 99% Zinc, wire reel, 10m, diameter 0.125mm, as drawn, 99% Zinc, wire reel, 10m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 10m, diameter 3.0mm, extruded, 99.9% Zinc, wire reel, 1m, diameter 0.025mm, as drawn, 99% Zinc, wire reel, 1m, diameter 0.125mm, as drawn, 99% Zinc, wire reel, 200m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 20m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 20m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 20m, diameter 3.0mm, extruded, 99.9% Zinc, wire reel, 25m, diameter 0.05mm, as drawn, 99% Zinc, wire reel, 25m, diameter 0.125mm, as drawn, 99% Zinc, wire reel, 2m, diameter 0.025mm, as drawn, 99% Zinc, wire reel, 2m, diameter 0.125mm, as drawn, 99% Zinc, wire reel, 2m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 2m, diameter 3.0mm, extruded, 99.9% Zinc, wire reel, 500m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 50m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 50m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 50m, diameter 3.0mm, extruded, 99.9% Zinc, wire reel, 5m, diameter 0.025mm, as drawn, 99% Zinc, wire reel, 5m, diameter 0.125mm, as drawn, 99% Zinc, wire reel, 5m, diameter 1.0mm, extruded, 99.9% Zinc, wire reel, 5m, diameter 2.0mm, extruded, 99.9% Zinc, wire reel, 5m, diameter 3.0mm, extruded, 99.9% Zinc, wire, diam. 1.0 mm, 99.999% trace metals basis Zinc foil, 0.5mm (0.02in) thick, 99.994% (metals basis) Zinc foil, 1.6mm (0.063in) thick, 15x15cm (5.9x5.9in) Zinc foil, 2mm (0.08in) thick, 99.999% (metals basis) Zinc Hollow Cathode Lamp: 2.0" Diameter, 4-pin, Cableless Zinc Hollow Cathode Lamp: 2.0" Diameter, 9-pin, Non-Coded Zinc wire, 0.5mm (0.02in) dia, 99.994% (metals basis) Zinc wire, 1.0mm (0.04in) dia, 99.9985% (metals basis) Zinc wire, 2.0mm (0.08in) dia, 99.999% (metals basis) Zinc, foil, 100x100mm, thickness 3mm, as rolled, 99.99+% Zinc, foil, 100x100mm, thickness 6mm, as rolled, 99.99+% Zinc, foil, 1m coil, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 200x200mm, thickness 0.5mm, as rolled, 99.7% Zinc, foil, 200x200mm, thickness 1.0mm, as rolled, 99.7% Zinc, foil, 25x25mm, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 25x25mm, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 25x25mm, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 3mm, as rolled, 99.99+% Zinc, foil, 25x25mm, thickness 6mm, as rolled, 99.99+% Zinc, foil, 2m coil, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 4mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 3mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 6mm, as rolled, 99.99+% Zinc, foil, 5m coil, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 6mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, thickness 0.25 mm, 99.999% trace metals basis Zinc, rod, 10.0 mm diameter, length 100 mm, purity 99.9% Zinc, rod, 10.0 mm diameter, length 1000 mm, purity 99.9% Zinc, rod, 10.0 mm diameter, length 200 mm, purity 99.9% Zinc, rod, 10.0 mm diameter, length 500 mm, purity 99.9% Zinc, rod, 12.0 mm diameter, length 100 mm, purity 99.9% Zinc, rod, 12.0 mm diameter, length 1000 mm, purity 99.9% Zinc, rod, 12.0 mm diameter, length 200 mm, purity 99.9% Zinc, rod, 12.0 mm diameter, length 500 mm, purity 99.9% Zinc, rod, 7.62 mm diameter, length 100 mm, purity 99.9% Zinc, rod, 7.62 mm diameter, length 1000 mm, purity 99.9% Zinc, rod, 7.62 mm diameter, length 200 mm, purity 99.9% Zinc, rod, 7.62 mm diameter, length 500 mm, purity 99.9% Zinc, wire reel, 0.5m, diameter 0.5mm, as drawn, 99.99+% Zinc, wire reel, 10m, diameter 0.25mm, as drawn, 99.99+% Zinc, wire reel, 10m, diameter 0.5mm, as drawn, 99.99+% Zinc, wire reel, 1m, diameter 0.5mm, as drawn, 99.99+% Zinc, wire reel, 25m, diameter 0.25mm, as drawn, 99.99+% Zinc, wire reel, 2m, diameter 0.5mm, as drawn, 99.99+% Zinc, wire reel, 50m, diameter 0.25mm, as drawn, 99.99+% Zinc, wire reel, 5m, diameter 0.25mm, as drawn, 99.99+% Zinc, wire reel, 5m, diameter 0.5mm, as drawn, 99.99+% Zinc (impurities), manufactured by MBH Analytical Ltd (41X Z2) Zinc foil, 0.01+/-0.0025mm (0.0004+/-0.0001in) thick Zinc foil, 0.1mm (0.004in) thick, 99.994% (metals basis) Zinc foil, 0.25mm (0.01in) thick, 99.994% (metals basis) Zinc, foil, 0.2m coil, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 0.3m coil, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 0.3m coil, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 0.5m coil, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 0.5m coil, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 100x100mm, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 100x100mm, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 100x100mm, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 100x100mm, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 100x100mm, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 10mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 10mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 10mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 10mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 10mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 10mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 150x150mm, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 150x150mm, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 150x150mm, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 15mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 15mm disks, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 15mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 15mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 15mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 15mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 1m coil, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 1m coil, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 1m coil, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 25mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 25mm disks, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 25mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 25mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 25mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 25mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 25x25mm, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 25x25mm, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 25x25mm, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 2m coil, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 2m coil, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 300x300mm, thickness 1.5mm, as rolled, 99.95+% Zinc, foil, 300x300mm, thickness 2.0mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 4mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 4mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 4mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 4mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 4mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 500x1000mm, thickness 1.0mm, as rolled, 99.7% Zinc, foil, 50mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 50mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.1mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 50mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.5mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 0.5mm, as rolled, 99.99+% Zinc, foil, 50mm disks, thickness 1.0mm, as rolled, 99.95+% Zinc, foil, 50mm disks, thickness 1.0mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 50x50mm, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 50x50mm, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 5m coil, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 6mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 6mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 6mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 6mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 6mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.125mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.125mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 0.15mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.175mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.20mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.25mm, as rolled, 99.95+% Zinc, foil, 8mm disks, thickness 0.25mm, as rolled, 99.99+% Zinc, foil, 8mm disks, thickness 0.35mm, as rolled, 99.95+% Zinc, foil, not light tested, 25x25mm, thickness 0.01mm, 99.9% Zinc, foil, not light tested, 25x25mm, thickness 0.02mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.01mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.02mm, 99.9% Zinc, foil, thickness 0.5 mm, size 100 x 100 mm, purity 99.7% Zinc, foil, thickness 0.5 mm, size 565 x 600 mm, purity 99.7% Zinc, lump, 10 mm max. lump size, weight 1000 g, purity 99.99% Zinc, lump, 10 mm max. lump size, weight 2000 g, purity 99.99% Zinc, lump, 10 mm max. lump size, weight 50 g, purity 99.99% Zinc, lump, 10 mm max. lump size, weight 500 g, purity 99.99% Zinc, lump, 2 mm max. lump size, weight 100 g, purity 99.98% Zinc, lump, 2 mm max. lump size, weight 1000 g, purity 99.98% Zinc, lump, 2 mm max. lump size, weight 200 g, purity 99.98% Zinc, lump, 2 mm max. lump size, weight 2000 g, purity 99.98% Zinc, lump, 2 mm max. lump size, weight 500 g, purity 99.98% Zinc, rod, 2.0 mm diameter, length 100 mm, high purity 99.99+% Zinc, rod, 2.0 mm diameter, length 200 mm, high purity 99.99+% Zinc, rod, 4.8 mm diameter, length 100 mm, high purity 99.99+% Zinc, rod, 4.8 mm diameter, length 200 mm, high purity 99.99+% Zinc, rod, 4.8 mm diameter, length 500 mm, high purity 99.99+% Zinc, rod, 7.0 mm diameter, length 500 mm, high purity 99.99+% Zinc, rod, 8 mm diameter, length 100 mm, high purity 99.999% Zinc, rod, 8 mm diameter, length 245 mm, high purity 99.999% Zinc, rod, 8 mm diameter, length 50 mm, high purity 99.999% Zinc, rod, length 100 mm, 7.0 mm diameter, high purity 99.99+% Zinc, rod, length 500 mm, 2.0 mm diameter, high purity 99.99+% Zinc, shots, 99.99999% (7N), diameter 1-4 mm, vacuum bags Zinc, wire reel, 100m, diameter 0.25mm, as drawn, 99.99+% Sulfur-Free Magnesium Concentrate: Mg @ approx. 2 wt% in Hydrocarbons Zinc preparation, 5 g/dL Zn+2 in THF, highly reactive Rieke(R)metal Zinc sputtering target, 50.8mm (2.0in) dia x 3.18mm (0.125in) thick Zinc sputtering target, 50.8mm (2.0in) dia x 6.35mm (0.250in) thick Zinc sputtering target, 76.2mm (3.0in) dia x 3.18mm (0.125in) thick Zinc, foil, not light tested, 100x100mm, thickness 0.0025mm, 99.9% Zinc, foil, not light tested, 100x100mm, thickness 0.005mm, 99.9% Zinc, foil, not light tested, 100x100mm, thickness 0.015mm, 99.9% Zinc, foil, not light tested, 100x100mm, thickness 0.01mm, 99.9% Zinc, foil, not light tested, 100x100mm, thickness 0.02mm, 99.9% Zinc, foil, not light tested, 25x25mm, thickness 0.001mm, 99.9% Zinc, foil, not light tested, 25x25mm, thickness 0.0025mm, 99.9% Zinc, foil, not light tested, 25x25mm, thickness 0.005mm, 99.9% Zinc, foil, not light tested, 25x25mm, thickness 0.015mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.001mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.0025mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.005mm, 99.9% Zinc, foil, not light tested, 50x50mm, thickness 0.015mm, 99.9% Zinc, foil, thickness 0.25 mm, size 150 x 150 mm, high purity 99.99+% Zinc, foil, thickness 0.5 mm, size 600 x 1130 mm, purity 99.7% Zinc, lump, 6 mm max. lump size, weight 10 g, high purity 99.9998% Zinc, lump, 6 mm max. lump size, weight 100 g, high purity 99.9998% Zinc, lump, 6 mm max. lump size, weight 20 g, high purity 99.9999% Zinc, lump, 6 mm max. lump size, weight 30 g, high purity 99.9998% Zinc, lump, 6 mm max. lump size, weight 50 g, high purity 99.9998% Zinc, nanopowder, 40-60 nm avg. part. size, >=99% trace metals basis Zinc, powder, max. particle size 150 micron, weight 100 g, purity 99.9% Zinc, powder, max. particle size 150 micron, weight 200 g, purity 99.9% Zinc, powder, max. particle size 150 micron, weight 500 g, purity 99.9% Zinc, powder, mean particle size 7.5 micron, weight 100 g, purity 98.8% Zinc, powder, mean particle size 7.5 micron, weight 200 g, purity 98.8% Zinc, rod, 10.0 mm diameter, length 1000 mm, high purity 99.99+% Zinc, rod, 10.0 mm diameter, length 200 mm, high purity 99.99+% Zinc, rod, 10.0 mm diameter, length 500 mm, high purity 99.99+% Zinc, rod, 19.0 mm diameter, length 200 mm, high purity 99.99+% Zinc, rod, 19.0 mm diameter, length 500 mm, high purity 99.99+% Zinc, rod, 2.0 mm diameter, length 1000 mm, high purity 99.99+% Zinc, rod, 4.8 mm diameter, length 1000 mm, high purity 99.99+% Zinc, rod, length 100 mm, 10.0 mm diameter, high purity 99.99+% Zinc, rod, length 100 mm, 19.0 mm diameter, high purity 99.99+% Zinc single crystal, 15mm (0.59in) dia, 50mm (2.0in) long, random orientation Zinc slug, 6.35mm (0.25in) dia x 6.35mm (0.25in) length, 99.993% (metals basis) Zinc standard for AAS, analytical standard, ready-to-use, traceable to BAM, in nitric acid Zinc, foam, thickness 10 mm, size 100 x 100 mm, bulk density 0.95 g/cm3 Zinc, foam, thickness 10 mm, size 150 x 150 mm, bulk density 0.28 g/cm3 Zinc, foam, thickness 20 mm, size 150 x 150 mm, bulk density 0.28 g/cm3 Zinc, foam, thickness 20 mm, size 300 x 300 mm, bulk density 0.28 g/cm3 Zinc, foil, 100x100mm, thickness 6mm, as rolled, polished on both sides, 99.99+% Zinc, foil, light tested, 100x100mm, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, light tested, 100x100mm, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, light tested, 100x100mm, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, light tested, 100x100mm, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, light tested, 150x150mm, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, light tested, 150x150mm, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, light tested, 150x150mm, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, light tested, 25x25mm, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, light tested, 25x25mm, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, light tested, 25x25mm, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, light tested, 25x25mm, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, light tested, 25x25mm, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, light tested, 50x50mm, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, light tested, 50x50mm, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, light tested, 50x50mm, thickness 0.05mm, as rolled, 99.95+% Zinc, foil, light tested, 50x50mm, thickness 0.05mm, as rolled, 99.99+% Zinc, foil, light tested, 50x50mm, thickness 0.075mm, as rolled, 99.95+% Zinc, foil, not light tested, 100x100mm, thickness 0.025mm, as rolled, 99.99+% Zinc, foil, not light tested, 150x150mm, thickness 0.025mm, as rolled, 99.95+% Zinc, foil, not light tested, 150x150mm, thickness 0.025mm, as rolled, 99.99+% Zinc, ingot, 99.9999% (6N), diameter 15 mm, length 80 mm, approx. 101 g Zinc, ingot, 99.99999% (7N), diameter 18 mm, length 50-55 mm, min. 90g Zinc, ingot, tapered, 99.99999% (7N), diameter 20-22 mm, length 55 mm, approx. 132 g Zinc, powder, 150 max. part. size (micron), weight 1000 g, purity 99.9% Zinc, powder, 150 max. part. size (micron), weight 50 g, high purity 99.999% Zinc, powder, 5 max. part. size (micron), weight 100 g, min. particle size 1 micron Zinc, powder, 5 max. part. size (micron), weight 500 g, min. particle size 1 micron Zinc, powder, 7.5 mean particle size (micron), weight 500 g, purity 98.8% Zinc, powder, max. particle size 150 micron, weight 100 g, high purity 99.999% Zinc, powder, max. particle size 150 micron, weight 20 g, high purity 99.999% Zinc, powder, max. particle size 5 micron, weight 200 g, min. particle size 1 micron Zinc, powder, mean particle size 7.5 micron, weight 1000 g, purity 98.8% Zinc, powder, mean particle size 7.5 micron, weight 2000 g, purity 98.8% Zinc, puriss. p.a., ACS reagent, reag. ISO, reag. Ph. Eur., >=99.9%, granular Zinc single crystal, 15mm (0.59in) dia, 50mm (2.0in) long, (0001) orientation, +/-2 degrees Zinc sputtering target, 76.2mm (3.0in) dia x 6.35mm (0.250in) thick, 99.99% (metals basis) Zinc, sputtering target, diam. x thickness 3.00 in. x 0.125 in., 99.995% trace metals basis PubChem 3 Chemical and Physical Properties 3.1 Computed Properties Property Name Property Value Reference Property Name Molecular Weight Property Value 65.4 g/mol Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Hydrogen Bond Donor Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Hydrogen Bond Acceptor Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Rotatable Bond Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Exact Mass Property Value 63.929142 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Monoisotopic Mass Property Value 63.929142 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Topological Polar Surface Area Property Value 0 Å² Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Heavy Atom Count Property Value 1 Reference Computed by PubChem Property Name Formal Charge Property Value 0 Reference Computed by PubChem Property Name Complexity Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Isotope Atom Count Property Value 0 Reference Computed by PubChem Property Name Defined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Defined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Covalently-Bonded Unit Count Property Value 1 Reference Computed by PubChem Property Name Compound Is Canonicalized Property Value Yes Reference Computed by PubChem (release 2025.04.14) PubChem 3.2 Experimental Properties 3.2.1 Physical Description Zinc ashes appears as a grayish colored powder. May produce toxic zinc oxide fumes when heated to very high temperatures or when burned. Insoluble in water. Used in paints, bleaches and to make other chemicals. CAMEO Chemicals Zinc dust appears as a grayish powder. Insoluble in water. May produce toxic zinc oxide fumes when heated to very high temperatures or when burned. Used in paints, bleaches and to make other chemicals. CAMEO Chemicals Dry Powder, Water or Solvent Wet Solid; Other Solid; Dry Powder; Liquid; Pellets or Large Crystals, Other Solid; Pellets or Large Crystals; Dry Powder, Pellets or Large Crystals, Water or Solvent Wet Solid, Liquid EPA Chemical Data Reporting (CDR) Gray powder; [CAMEO] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases GREY-TO-BLUE POWDER. ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.2 Color / Form Bluish-white, lustrous metal; distorted hexagonal close-packed structure; when heated to 100-150 °C becomes malleable, at 210 °C becomes brittle and pulverizable O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1809 Hazardous Substances Data Bank (HSDB) Brittle at ordinary temperatures Lide, D.R. CRC Handbook of Chemistry and Physics 86TH Edition 2005-2006. CRC Press, Taylor & Francis, Boca Raton, FL 2005, p. 4-42 Hazardous Substances Data Bank (HSDB) 3.2.3 Boiling Point 907 MSDS DrugBank 907 °C Lide, D.R. CRC Handbook of Chemistry and Physics 86TH Edition 2005-2006. CRC Press, Taylor & Francis, Boca Raton, FL 2005, p. 4-95 Hazardous Substances Data Bank (HSDB) 907Â °C ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.4 Melting Point 419 MSDS DrugBank 787 °F Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 419.53 °C Lide, D.R. CRC Handbook of Chemistry and Physics 86TH Edition 2005-2006. CRC Press, Taylor & Francis, Boca Raton, FL 2005, p. 4-95 Hazardous Substances Data Bank (HSDB) 419Â °C ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.5 Solubility Soluble in acids and alkalies; insoluble in water Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 14th Edition. John Wiley & Sons, Inc. New York, NY 2001., p. 1190 Hazardous Substances Data Bank (HSDB) Solubility in water: reaction ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.6 Density 7.133 g/cu cm at 25 °C; 6.830 g/cu cm at 419.5 °C (solid); 6.620 g/cu cm at 419.5 °C (liquid); 6.250 g/cu cm at 800 °C Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V25: 793 (1998) Hazardous Substances Data Bank (HSDB) 7.1 g/cmÂ³ ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.7 Vapor Pressure 1.47X10-6 Pa (1.10X10-8 mm Hg) at 400 K (127 °C); 0.653 Pa (4.9X10-3 mm Hg) at 600 K (327 °C) Lide, D.R. CRC Handbook of Chemistry and Physics 86TH Edition 2005-2006. CRC Press, Taylor & Francis, Boca Raton, FL 2005, p. 4-130 Hazardous Substances Data Bank (HSDB) 3.2.8 Stability / Shelf Life Stable in dry air; becomes covered with white coating of basic carbonate on exposure to moist air. The Merck Index. 10th ed. Rahway, New Jersey: Merck Co., Inc., 1983., p. 1455 Hazardous Substances Data Bank (HSDB) Zinc is attacked by carbon dioxide&sulfur dioxide resulting chiefly in a coating of hydrated basic carbonate of variable composition; hydrogen peroxide may be formed in the process Clayton, G. D. and F. E. Clayton (eds.). Patty's Industrial Hygiene and Toxicology: Volume 2A, 2B, 2C: Toxicology. 3rd ed. New York: John Wiley Sons, 1981-1982., p. 2037 Hazardous Substances Data Bank (HSDB) 3.2.9 Autoignition Temperature 460Â °C ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.10 Decomposition When heated to decomposition it emits toxic fumes of /zinc oxide/. Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 3717 Hazardous Substances Data Bank (HSDB) 3.2.11 Heat of Vaporization 114.8 kJ/mol at 907 °C Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V25: 793 (1998) Hazardous Substances Data Bank (HSDB) 3.2.12 Other Experimental Properties Natural zinc is composed of five stable isotopes: zinc-64 (48.6%); zinc-66 (27.9%); zinc-67 (4.1%); zine-68 (18.8%); zinc-70 (0.6%); 18 artificial radioactive isotopes are known, most with very short (millisecond, seconds) or short (minutes) half-lives; zinc-65: half-life = 243.8 days Ohnesorge FK, Wilhelm M; pp. 1309-42 in Metals and Their Compounds in the Environment. Merian E, ed. Weinheim, Germany: VCH (1991) Hazardous Substances Data Bank (HSDB) Valence: 2; stable in dry air; becomes covered with a white coating of basic carbonate on exposure to moist air; Mohs' hardness: 2.5; burns in air with a bluish-green flame; loses electrons in aqueous systems to form Zn2+; Reduction potential (aq) Zn/Zn2+: 0.763 V O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1809 Hazardous Substances Data Bank (HSDB) Slowly attacked by H2SO4 or HCl; oxidizing agents or metal ions (e.g., Cu2+, Ni2+, Co2+) accelerate the process; reacts slowly with ammoniawater and acetic acid; rapidly with HNO3; reacts with alkali hydroxides to form "zincates", (ZnO2)2-, which are actually hydroxo complexes, such as (Zn(OH)3)-, (Zn(OH)4)2-, ((Zn(OH)4(H2O)2))2- O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1809 Hazardous Substances Data Bank (HSDB) Naturally occurring zinc contains 5 stable isotopes; 25 other unstable isotopes and isomers are recognized; fair conductor of electricity; burns in air at high red heat with evolution of white clouds of the oxide Lide, D.R. CRC Handbook of Chemistry and Physics 86TH Edition 2005-2006. CRC Press, Taylor & Francis, Boca Raton, FL 2005, p. 4-42 Hazardous Substances Data Bank (HSDB) For more Other Experimental Properties (Complete) data for ZINC, ELEMENTAL (9 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 3.3 SpringerMaterials Properties thermal expansion coefficient electron density of states dielectric constant electromagnetic induction optical coefficient crystal structure Debye frequency isoelectronic shift deformation parameter dielectricity diffusion enthalpy radiation electrode potential isothermal section Q value Gibbs energy Lorenz number impurity solubility phase equilibrium lattice strain positional coordinate alpha-particle emission cross section hole conductivity absorption coefficient spectral factor work function temperature-composition section electron conductivity orbital angular momentum space group phase diagram unit cell parameter ion scattering spectroscopy melting temperature electron diffraction heat capacity neutron binding energy dissociation energy adsorption third-order stiffness half-width sound velocity capacitance reflectance photoemission spectroscopy excitation energy transition enthalpy displacement parameter LEED atomic radius surface anisotropy sound propagation isotopic spin spin neutron resonance energy concentration thermal diffusivity high frequency properties sound absorption phase transition branching ratio lineshape Young's modulus vibrational properties chemical diffusion elastic stiffness self-diffusion diffusion of impurities cyclic voltammetry film energy band nuclear spin boiling point band structure formation entropy condensation electronic surface structure exchange current density X-ray diffraction interplanar spacing surface structure plasticity natural abundance adsorption energy coordination distance atomic environment atomic mass quadrupole coupling electron-atom scattering phase stability cell voltage transition entropy particle size liquidus surface formation enthalpy pressure coefficient atomic defect properties thermopower reflection high-energy electron diffraction nucleon separation energy corrosion density surface state volume coefficient resonance cross-section mechanical strain superconductivity stress polarizability crystal structure type melting transition electron scattering elastic scattering ionization coefficient stress-strain curve surface determination diamagnetic susceptibility Fermi energy formation energy resonance parameter energy level nuclear quadrupole resonance spectroscopy formula unit unit cell axes electron state compressibility structure type nuclear properties refractive index rate of conversion transition pressure elasticity radiative width entropy defect concentration activity surface energy parity isentropic compressibility shear modulus shock waves electrochemical conversion electrical resistivity strain dependence composition thermal conductivity unit cell phonon dispersion gamma-ray transition Drude parameter viscosity diamagnetic anisotropy diffusive flux partial wave amplitude scanning tunneling microscopy SpringerMaterials 3.4 Chemical Classes Metals -> Elements, Metallic Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Inorganics NIAID ChemDB Flammable agents - 3rd degree Reactive agents - 1st degree NJDOH RTK Hazardous Substance List Trace element USGS Health-Based Screening Levels for Evaluating Water-Quality Data 3.4.1 Drugs 3.4.1.1 Human Drugs Breast Feeding; Lactation; Milk, Human; Minerals; Elements Drugs and Lactation Database (LactMed) 3.4.1.2 Animal Drugs Pharmaceuticals -> UK Veterinary Medicines Directorate List S104 \| UKVETMED \| UK Veterinary Medicines Directorate's List \| DOI:10.5281/zenodo.7802119 NORMAN Suspect List Exchange 3.4.2 Cosmetics Cosmetics ingredient -> Other (Specify) California Safe Cosmetics Program (CSCP) Product Database 3.4.3 Food Contact Substances FCS -> FDA Inventory of Food Contact Substances Listed in 21 CFR FDA Packaging & Food Contact Substances (FCS) 3.4.4 Pesticides Pesticide -> EPA IRIS EPA Integrated Risk Information System (IRIS) 4 Spectral Information 4.1 Raman Spectra 1 of 4 items View All Type Raman Spectrum Mineral Merrillite RRUFF ID R150063 Sample Unoriented Raman Data Processed Wave Length 532 nm Thumbnail RRUFF Project 2 of 4 items View All Type Raman Spectrum Mineral Merrillite RRUFF ID R150063 Sample Unoriented Raman Data Processed Wave Length 780 nm Thumbnail RRUFF Project 5 Related Records 5.1 Related Compounds with Annotation Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here. PubChem 5.2 Related Compounds Same Connectivity Count 32 Mixtures, Components, and Neutralized Forms Count 65125 Similar Compounds (2D) View in PubChem Search Similar Conformers (3D) View in PubChem Search PubChem 5.3 Related Element Element Name Zinc Element Symbol Zn Atomic Number 30 PubChem Elements 5.4 Substances 5.4.1 PubChem Reference Collection SID 481107697 PubChem 5.4.2 Related Substances All Count 137481 Same Count 968 Mixture Count 136513 PubChem 5.4.3 Substances by Category PubChem 5.5 Other Relationships Copper (related) Silver (related) Nickel (related) Mercury (related) Cadmium (related) Iron (related) Cobalt (related) Manganese (related) Molybdenum (related) Chromium (related) Gold (related) Titanium (related) Thallium (related) Bismuth (related) Indium (related) Strontium (related) Hafnium (related) Yttrium (related) Rubidium (related) Gallium (related) Germanium (related) PubChem 5.6 Entrez Crosslinks PubMed Count 718 Protein Structures Count 1 Taxonomy Count 40 OMIM Count 58 Gene Count 3021 PubChem 5.7 NCBI LinkOut NCBI 6 Chemical Vendors PubChem 7 Drug and Medication Information 7.1 Drug Indication Zinc can be used for the treatment and prevention of zinc deficiency/its consequences, including stunted growth and acute diarrhea in children, and slowed wound healing. It is also utilized for boosting the immune system, treating the common cold and recurrent ear infections, as well as preventing lower respiratory tract infections. DrugBank Zinc is a trace mineral, second only to iron in its concentration in the body. Adult humans contain 2 to 3 grams of zinc, but it is difficult to measure an individual's zinc status, particularly during acute illness. Zinc is necessary for the immune system to function correctly. Zinc is involved in cell division, cell growth, wound healing, breakdown of carbohydrates, enhancing action insulin, and it is necessary for the sense of smell and taste. During pregnancy, infancy, and childhood, zinc is a requirement for proper growth and development. StatPearls 7.2 LiverTox Summary Zinc is an essential mineral and heavy metal that is included in most over-the-counter multivitamin and mineral supplements, and is used therapeutically in higher doses because of its ability to block copper absorption as maintenance therapy of Wilson disease. Zinc has not been associated with worsening of serum enzyme elevations during therapy or with clinically apparent liver injury. LiverTox 7.3 Drug Classes Breast Feeding; Lactation; Milk, Human; Minerals; Elements Drugs and Lactation Database (LactMed) Trace Elements and Metals; Chelating Agents LiverTox 7.4 FDA National Drug Code Directory National Drug Code (NDC) Directory 7.5 Drug Labels Drug and label DailyMed View More... 7.6 Clinical Trials 7.6.1 ClinicalTrials.gov ClinicalTrials.gov 7.6.2 EU Clinical Trials Register EU Clinical Trials Register 7.7 Drug Warnings Seven hospital patients undergoing intravenous feeding with fluid containing elemental zinc at a concentration of 227 ug/100 mL were inadvertently given fluid containing 10 times that amount (2270 ug/100 mL) for 26 to 60 days. Mortality was high (5/7). While the clinical manifestation of the zinc overdose was hyperamylasaemia (unaccompanied by clinical signs of pancreatitis), the authors concluded that all deaths had resulted from septic complications already present before the appearance of this symptom. WHO; Environ Health Criteria 221: Zinc p.139 (2001) Hazardous Substances Data Bank (HSDB) 8 Food Additives and Ingredients 8.1 Food Additive Classes JECFA Functional Classes Food Contaminant -> METALS; Joint FAO/WHO Expert Committee on Food Additives (JECFA) 8.2 FDA Food Contact Substances (FCS) Substance ZINC Document Number (21 eCFR) 172.898 175.390 FDA Packaging & Food Contact Substances (FCS) 8.3 Associated Foods FooDB 8.4 Evaluations of the Joint FAO / WHO Expert Committee on Food Additives - JECFA Chemical Name ZINC Evaluation Year 1982 Comments Zinc is an essential element in the nutrition of man and animals. Studies with experimental animals have shown that high levels of dietary zinc can cause anaemia as well as decreased levels of copper and iron absorption, and reduction in the activities of several important enzymes in various tissues. These effects also occur at lower levels of dietary zinc when the diet is deficient in copper. Zinc was not teratogenic, had no effect on the reproductive performance of test animals, & was not mutagenic in a number of bacterial and mammalian systems. The required daily intake for adult humans is about 15 mg/day. The average dietary zinc intake in adults is 14-20 mg/day and is thus sufficient for nutritional needs. There is a wide margin between nutritionally required amounts of zinc and toxic levels. Clinical studies in which up to 600 mg of zinc sulfate (equivalent to 200 mg elemental zinc) has been administered daily in divided doses for a period of several months is the basis for the PMTDI for Zn of 0.3-1.0 mg/kg bw/d. Report TRS 683-JECFA 26/32 Tox Monograph FAS 17-JECFA 26/320 Joint FAO/WHO Expert Committee on Food Additives (JECFA) 9 Agrochemical Information 9.1 Agrochemical Category Fungicide EPA Pesticide Ecotoxicity Database 10 Minerals 1 of 8 items Name MERRILLITE Formula Ca9Na(Mg,Fe)(PO4)7 System Rhomboedral (trigonal) Athena Minerals 2 of 8 items Name ZINC Formula Zn System Hexagonal Athena Minerals 3 of 8 items Name Merrillite IMA Number / Year 1976 s.p. IMA Symbol Mer Status Rd - Redefined Formula Ca9NaMg(PO4)7 Reference(s) American Mineralogist 2 (1917), 119 American Mineralogist 107 (2022), 1652 Commission on New Minerals, Nomenclature and Classification (CNMNC) 4 of 8 items Name Zinc IMA Number / Year ? IMA Symbol Zn Status G - Grandfathered Formula Zn Reference(s) original paper? Zapiski Vsesoyuznogo Mineralogicheskogo Obshchestva 110 (1981), 186 Commission on New Minerals, Nomenclature and Classification (CNMNC) 5 of 8 items Name zinc Link PDF Handbook of Mineralogy 6 of 8 items Name Merrillite IMA Number IMA1976 s.p. Formula Ca 9 NaMg(PO 4)7 IMA Symbol Mer Crystal Structure Data Link American Mineralogist Crystal Structure Database RRUFF Project 7 of 8 items Name Zinc Formula Zn IMA Symbol Zn PDF Link zinc.pdf Crystal Structure Data Link American Mineralogist Crystal Structure Database RRUFF Project 8 of 8 items Mineral Description Zinc is the 23rd most abundant element in the earth's crust. Sphalerite, zinc sulfide, is and has been the principal ore mineral in the world. Zinc is necessary to modern living, and, in tonnage produced, stands fourth among all metals in world production - being exceeded only by iron, aluminum, and copper. Zinc uses range from metal products to rubber and medicines. About three-fourths of zinc used is consumed as metal, mainly as a coating to protect iron and steel from corrosion (galvanized metal), as alloying metal to make bronze and brass, as zinc-based die casting alloy, and as rolled zinc. The remaining one-fourth is consumed as zinc compounds mainly by the rubber, chemical, paint, and agricultural industries. Zinc is also a necessary element for proper growth and development of humans, animals, and plants; it is the second most common trace metal, after iron, naturally found in the human body. USGS Mineral Commodity Summaries (PDF links) 2010 \| 2011 \|2012 \|2013 \|2014 \|2015 \|2016 \|2017 \|2018 \| 2019 \| 2020 \| 2021 \| 2022 \| 2023 \| 2024 \| 2025 USGS Mineral Yearbook (PDF links) 2010 \| 2011 \| 2012 \| 2013 \| 2014 \| 2015 \| 2016 \| 2017 \| 2018 \| 2019 \| 2020 \| 2021 USGS National Minerals Information Center 11 Pharmacology and Biochemistry 11.1 Pharmacodynamics Zinc is involved in various aspects of cellular metabolism. It has been estimated that approximately 10% of human proteins may bind zinc, in addition to hundreds of proteins that transport and traffic zinc. It is required for the catalytic activity of more than 200 enzymes, and it plays a role in immune function wound healing, protein synthesis, DNA synthesis, and cell division. Zinc is an essential element for a proper sense of taste and smell and supports normal growth and development during pregnancy, childhood, and adolescence. It is thought to have antioxidant properties, which may be protective against accelerated aging and helps to speed up the healing process after an injury; however, studies differ as to its effectiveness. Zinc ions are effective antimicrobial agents even if administered in low concentrations. Studies on oral zinc for specific conditions shows the following evidence in various conditions: Colds: Evidence suggests that if zinc lozenges or syrup are taken within 24 hours after cold symptoms start, the supplement may shorten the length of colds. The use intranasal zinc has been associated with the loss of the sense of smell, in some cases long-term or permanently. Wound healing: Patients with skin ulcers and decreased levels of zinc may benefit from oral zinc supplements. Diahrrea: Oral zinc supplements can reduce the symptoms of diarrhea in children with low levels of zinc, especially in cases of malnutrition. DrugBank 11.2 MeSH Pharmacological Classification Trace Elements A group of chemical elements that are needed in minute quantities for the proper growth, development, and physiology of an organism. (From McGraw-Hill Dictionary of Scientific and Technical Terms, 4th ed) Medical Subject Headings (MeSH) 11.3 Absorption, Distribution and Excretion Absorption Zinc is absorbed in the small intestine by a carrier-mediated mechanism. Under regular physiologic conditions, transport processes of uptake do not saturate. The exact amount of zinc absorbed is difficult to determine because zinc is secreted into the gut. Zinc administered in aqueous solutions to fasting subjects is absorbed quite efficiently (at a rate of 60-70%), however, absorption from solid diets is less efficient and varies greatly, dependent on zinc content and diet composition. Generally, 33% is considered to be the average zinc absorption in humans. More recent studies have determined different absorption rates for various populations based on their type of diet and phytate to zinc molar ratio. Zinc absorption is concentration dependent and increases linearly with dietary zinc up to a maximum rate. Additionally zinc status may influence zinc absorption. Zinc-deprived humans absorb this element with increased efficiency, whereas humans on a high-zinc diet show a reduced efficiency of absorption. DrugBank Route of Elimination The excretion of zinc through gastrointestinal tract accounts for approximately one-half of all zinc eliminated from the body. Considerable amounts of zinc are secreted through both biliary and intestinal secretions, however most is reabsorbed. This is an important process in the regulation of zinc balance. Other routes of zinc excretion include both urine and surface losses (sloughed skin, hair, sweat). Zinc has been shown to induce intestinal metallothionein, which combines zinc and copper in the intestine and prevents their serosal surface transfer. Intestinal cells are sloughed with approximately a 6-day turnover, and the metallothionein-bound copper and zinc are lost in the stool and are thus not absorbed. Measurements in humans of endogenous intestinal zinc have primarily been made as fecal excretion; this suggests that the amounts excreted are responsive to zinc intake, absorbed zinc and physiologic need. In one study, elimination kinetics in rats showed that a small amount of ZnO nanoparticles was excreted via the urine, however, most of the nanoparticles were excreted via the feces. DrugBank Volume of Distribution A pharmacokinetic study was done in rats to determine the distribution and other metabolic indexes of zinc in two particle sizes. It was found that zinc particles were mainly distributed to organs including the liver, lung, and kidney within 72 hours without any significant difference being found according to particle size or rat gender. DrugBank Clearance In one study of healthy patients, the clearance of zinc was found to be 0.63 ± 0.39 μg/min. DrugBank In mice injected SC with finely dispersed zinc (Zn) powder (particle size 0.05-0.1 mu) increased amounts of Zn were found in the liver. Fatullina LD et al; Izve Akade Nauk SSSR, Ser Biol (1): 130-3 (1984) Hazardous Substances Data Bank (HSDB) 11.4 Metabolism / Metabolites Zinc is released from food as free ions during its digestion. These freed ions may then combine with endogenously secreted ligands before their transport into the enterocytes in the duodenum and jejunum.. Selected transport proteins may facilitate the passage of zinc across the cell membrane into the hepatic circulation. With high intake, zinc may also be absorbed through a passive paracellular route. The portal system carries absorbed zinc directly into the hepatic circulation, and then it is released into systemic circulation for delivery to various tissues. Although, serum zinc represents only 0.1% of the whole body zinc, the circulating zinc turns over rapidly to meet tissue needs. DrugBank 11.5 Biological Half-Life The half-life of zinc in humans is approximately 280 days. DrugBank 11.6 Mechanism of Action Zinc has three primary biological roles: catalytic, structural, and regulatory. The catalytic and structural role of zinc is well established, and there are various noteworthy reviews on these functions. For example, zinc is a structural constituent in numerous proteins, inclusive of growth factors, cytokines, receptors, enzymes, and transcription factors for different cellular signaling pathways. It is implicated in numerous cellular processes as a cofactor for approximately 3000 human proteins including enzymes, nuclear factors, and hormones. Zinc promotes resistance to epithelial apoptosis through cell protection (cytoprotection) against reactive oxygen species and bacterial toxins, likely through the antioxidant activity of the cysteine-rich metallothioneins. In HL-60 cells (promyelocytic leukemia cell line), zinc enhances the up-regulation of A20 mRNA, which, via TRAF pathway, decreases NF-kappaB activation, leading to decreased gene expression and generation of tumor necrosis factor-alpha (TNF-alpha), IL-1beta, and IL-8. There are several mechanisms of action of zinc on acute diarrhea. Various mechanisms are specific to the gastrointestinal system: zinc restores mucosal barrier integrity and enterocyte brush-border enzyme activity, it promotes the production of antibodies and circulating lymphocytes against intestinal pathogens, and has a direct effect on ion channels, acting as a potassium channel blocker of adenosine 3-5-cyclic monophosphate-mediated chlorine secretion. Cochrane researchers examined the evidence available up to 30 September 2016. Zinc deficiency in humans decreases the activity of serum thymulin (a hormone of the thymus), which is necessary for the maturation of T-helper cells. T-helper 1 (Th(1)) cytokines are decreased but T-helper 2 (Th(2)) cytokines are not affected by zinc deficiency in humans. The change of Th(1) to Th(2) function leads to cell-mediated immune dysfunction. Because IL-2 production (Th(1) cytokine) is decreased, this causes decreased activity of natural-killer-cell (NK cell) and T cytolytic cells, normally involved in killing viruses, bacteria, and malignant cells. In humans, zinc deficiency may lead to the generation of new CD4+ T cells, produced in the thymus. In cell culture studies (HUT-78, a Th(0) human malignant lymphoblastoid cell line), as a result of zinc deficiency, nuclear factor-kappaB (NF-kappaB) activation, phosphorylation of IkappaB, and binding of NF-kappaB to DNA are decreased and this results in decreased Th(1) cytokine production. In another study, zinc supplementation in human subjects suppressed the gene expression and production of pro-inflammatory cytokines and decreased oxidative stress markers. In HL-60 cells (a human pro-myelocytic leukemia cell line), zinc deficiency increased the levels of TNF-alpha, IL-1beta, and IL-8 cytokines and mRNA. In such cells, zinc was found to induce A20, a zinc finger protein that inhibited NF-kappaB activation by the tumor necrosis factor receptor-associated factor pathway. This process decreased gene expression of pro-inflammatory cytokines and oxidative stress markers. The exact mechanism of zinc in acne treatment is poorly understood. However, zinc is considered to act directly on microbial inflammatory equilibrium and facilitate antibiotic absorption when used in combination with other agents. Topical zinc alone as well as in combination with other agents may be efficacious because of its anti-inflammatory activity and ability to reduce P. acnes bacteria by the inhibition of P. acnes lipases and free fatty acid levels. DrugBank 11.7 Biochemical Reactions PubChem 12 Use and Manufacturing 12.1 Uses EPA CPDat Chemical and Product Categories The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125 EPA Chemical and Products Database (CPDat) Sources/Uses Used in galvanizing and alloying; Also used in dry cell batteries and in organic chemistry, bronze deoxidizing, gold extracting, fat purifying, and bone bleaching; An essential dietary mineral; [Merck Index] Zinc dust is used as a pigment in rust-resistant coatings. Also used as a reducing agent, an additive to plastics and lubricants, a purification agent in electrolytic zinc plants, and a component of primary batteries and abrasives; it is used in spray metallizing and mechanical plating. [CHEMINFO] Spalerite is the most important Zn ore; Workers are exposed to zinc in mining, smelting, and welding; [Nordberg, p. 1370, 1374] Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013. Nordberg - Nordberg GF, Fowler BA, Nordberg M (eds). Handbook on the Toxicology of Metals, 4th Ed. Boston: Elsevier, 2015., p. 1370, 1374 Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Industrial Processes with risk of exposure Metal Preparation and Pouring [Category: Foundry] Welding [Category: Weld] Electroplating [Category: Plate] Metal Thermal Spraying [Category: Plate] Battery Manufacturing [Category: Industry] Painting (Pigments, Binders, and Biocides) [Category: Paint] Gas Welding and Cutting [Category: Weld] Mining [Category: Industry] Welding Over Coatings [Category: Weld] Metal Extraction and Refining [Category: Industry] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases For zinc, elemental (USEPA/OPP Pesticide Code: 129015) ACTIVE products with label matches. /SRP: Registered for use in the U.S. but approved pesticide uses may change periodically and so federal, state and local authorities must be consulted for currently approved uses./ National Pesticide Information Retrieval System's USEPA/OPP Chemical Ingredients Database on Zinc, Elemental (7440-66-6). Available from, as of April 21, 2006: Hazardous Substances Data Bank (HSDB) Galvanizing sheet iron; as ingredient of alloys such as bronze, brass, Babbitt metal, German silver, and special alloys for die-casting. O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1809 Hazardous Substances Data Bank (HSDB) Protective coating for metals to prevent corrosion; for electrical apparatus, especially dry cell batteries, household utensils, casting, printing plates, building materials, railroad car lining, automotive equipment; reducing agent in organic chemistry; deoxidizing bronze; extracting gold by cyanide process, purifying fats for soaps; bleaching bone glue; manufacture sodium hydrosulfite; insulin zinc salts; reagent in analytical chemistry O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1810 Hazardous Substances Data Bank (HSDB) As negative electrode in alkaline cell electrode Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V25: 795 Hazardous Substances Data Bank (HSDB) For more Uses (Complete) data for ZINC, ELEMENTAL (15 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) Substance listed with specific concentration in tattoo ink and/or permanent make up according to EU Commission Regulation 2020/2081. The concentration limit (by weight) is 0.2%. S86 \| TATTOOINK \| Regulated Tattoo Ink Ingredients as per EU regulation 2020/2081 \| A list of regulated ingredients for tattoo ink and permanent make up ( Appendix 13 added to Commission Regulation (EU) 2020/2081, 14 December 2020 amending Annex XVII of REACH ( Dataset DOI:10.5281/zenodo.5710243 NORMAN Suspect List Exchange 12.1.1 Use Classification Chemical Classes -> Inorganic substances Agency for Toxic Substances and Disease Registry (ATSDR) Food Contaminant -> METALS; -> JECFA Functional Classes Joint FAO/WHO Expert Committee on Food Additives (JECFA) 12.1.2 Industry Uses Alloys Intermediates Bleaching agents Intermediate Corrosion inhibitors and anti-scaling agents Not Known or Reasonably Ascertainable Solvents (which become part of product formulation or mixture) Pigment Other (specify) Surface modifier Paint additives and coating additives not described by other categories Agricultural chemicals (non-pesticidal) Lubricating agent Plating agents and surface treating agents Other Pigments Corrosion inhibitor Preservative Processing aids not otherwise specified EPA Chemical Data Reporting (CDR) 12.1.3 Consumer Uses Corrosion inhibitor Plating agent Other Pigments Cleaning agent Agricultural chemicals (non-pesticidal) Paint additives and coating additives not described by other categories Plating agents and surface treating agents Adhesion/cohesion promoter Other (specify) Intermediate Corrosion inhibitors and anti-scaling agents Sealant (barrier) Not Known or Reasonably Ascertainable Alloys Bleaching agents EPA Chemical Data Reporting (CDR) 12.1.4 Household Products California Safe Cosmetics Program (CSCP) Cosmetics product ingredient: Zinc Product count: 13 California Safe Cosmetics Program (CSCP) Product Database Household & Commercial/Institutional Products Information on 33 consumer products that contain Zinc dust in the following categories is provided: • Auto Products • Hobby/Craft • Home Maintenance • Inside the Home • Landscaping/Yard • Personal Care Consumer Product Information Database (CPID) 12.2 Methods of Manufacturing Zinc ore is mined using both underground mining and open pit mining. The mined zinc ores are too low in zinc content for direct reduction to refined metal; thus, they are first concentrated. Production of concentrates requires crushing and grinding followed by gravity or magnetic methods of separation or flotation. These processes may be combined, depending on the complexity of the ore. A caustic-leach process is used to decrease the extent of metal loss during the concentration process. In this process, the metal is leached by caustic soda, the resulting electrolyte is purified with zinc dust and lime, and the zinc is electrodeposited. The crude zinc may be dissolved in sulfuric acid and purified by electrodeposition. DHHS/ATSDR; Toxicological Profile for Zinc p.129 PB2009-100008 (August 2005) Hazardous Substances Data Bank (HSDB) Concentration of zinc ore, roasting the concentrate, followed by thermal smelting (reduction with carbon), or followed by leaching of the roasted concentrate to extract soluble zinc, purification, and electrolytic refining. SRI Hazardous Substances Data Bank (HSDB) The hydrometallurgical or electrolytical process where the zinc oxide is leached from the roasted or calcined material with sulfuric acid to form zinc sulfate solution which is electrolyzed in cells to deposit zinc on cathodes. Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 14th Edition. John Wiley & Sons, Inc. New York, NY 2001., p. 1191 Hazardous Substances Data Bank (HSDB) Two processes are used to produce metallic zinc from the ore concentrates that are not subjected to caustic soda leaching. In one process, the ore concentrate containing zinc sulfide is roasted in the presence of air to produce zinc oxide, which is combined with coke or coal and retorted to approximately 1,100 °C to produce metallic zinc. In the other process, the roasted zinc oxide is leached with sulfuric acid, and the solution is electrolyzed to produce zinc of >99.9% purity. DHHS/ATSDR; Toxicological Profile for Zinc p129 PB2006-100008 (August 2005) Hazardous Substances Data Bank (HSDB) 12.3 Impurities The effect of small amt of common impurities is to incr corrosion resistance to solutions, but not in the atmosphere ... Brittleness /of ordinary zinc/ is thought to be assoc with impurities such as tin. Clayton, G.D., F.E. Clayton (eds.) Patty's Industrial Hygiene and Toxicology. Volumes 2A, 2B, 2C, 2D, 2E, 2F: Toxicology. 4th ed. New York, NY: John Wiley & Sons Inc., 1993-1994., p. 2333 Hazardous Substances Data Bank (HSDB) Lead contaminates special high grade zinc at 0.003%; high grade zinc at 0.07%; intermediate grade at 0.2%; brass special at 0.6%; prime western at 1.6% CONSIDINE. CHEMICAL AND PROCESS TECHNOL ENCYC 1974 p.1179 Hazardous Substances Data Bank (HSDB) Iron contaminates special high grade zinc at 0.003%; high grade at 0.02%; intermediate at 0.03%; brass special at 0.03%; prime western at 0.08% CONSIDINE. CHEMICAL AND PROCESS TECHNOL ENCYC 1974 p.1179 Hazardous Substances Data Bank (HSDB) Cadmium contaminates special high grade zinc at 0.003%; high grade at 0.03%; intermediate grade at 0.4%; brass special at 0.5% CONSIDINE. CHEMICAL AND PROCESS TECHNOL ENCYC 1974 p.1179 Hazardous Substances Data Bank (HSDB) For more Impurities (Complete) data for ZINC, ELEMENTAL (6 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.4 Formulations / Preparations Zinc is available in many commercial forms, including ingots, lumps, sheets, wire, shot, strips, sticks, granules, granulated zinc (obtained when molten metal is poured into cold water), and powder. DHHS/ATSDR; Toxicological Profile for Zinc p.130 PB2006-100008 (August 2005) Hazardous Substances Data Bank (HSDB) Forms available: Slab, rolled (strip, sheet, rod, tubing), wire, mossy zinc, zinc dust powder (99% pure); single crystals; zinc anodes Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 14th Edition. John Wiley & Sons, Inc. New York, NY 2001., p. 1191 Hazardous Substances Data Bank (HSDB) Commercial forms: ingots, lumps, sheets, wire, shot, strips, sticks, granules, mossy, powder (dust). O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. 13th Edition, Whitehouse Station, NJ: Merck and Co., Inc., 2001., p. 1809 Hazardous Substances Data Bank (HSDB) GRADES: Special high-grade (99.990%); high-grade (99.95%); intermediate (99.5%); brass special (99%); prime western (98%) Lewis, R.J. Sr.; Hawley's Condensed Chemical Dictionary 14th Edition. John Wiley & Sons, Inc. New York, NY 2001., p. 1191 Hazardous Substances Data Bank (HSDB) For more Formulations/Preparations (Complete) data for ZINC, ELEMENTAL (7 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.5 Consumption Patterns Zinc-base alloy, 38%; galvanizing, 38%; brass products, 15%; rolled zinc, 3%; zinc oxide, 3%; other, 3% SRI Hazardous Substances Data Bank (HSDB) Refined zinc (expressed in metric tons): 1991, 790,000; 1992, 814,000; 1993, 828,000; 1994, 843,000; 1995, 838,000. United States Department of the Interior/U.S. Geological Survey. Minerals Yearbook. Metals and Minerals Volume 1. Washington, D.C. 1995., p. 926 Hazardous Substances Data Bank (HSDB) Slab zinc = 1,205,000 tons (1996) Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V25 814 Hazardous Substances Data Bank (HSDB) ... In 2002, the reported consumption of zinc by industry was 265,000 metric tons (53.4% of total consumption) for galvanizing; 103,000 metric tons (20.8% of total consumption) for zinc-based alloys; and 86,800 metric tons (17.5% of total consumption) for brass and bronze. DHHS/ATSDR; Toxicological Profile for Zinc p.135 PB2006-100008 (August 2005) Hazardous Substances Data Bank (HSDB) For more Consumption Patterns (Complete) data for ZINC, ELEMENTAL (14 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.6 U.S. Production Aggregated Product Volume 2019: 1,000,000,000 - <5,000,000,000 lb 2018: 1,000,000,000 - <5,000,000,000 lb 2017: 1,000,000,000 - <5,000,000,000 lb 2016: 1,000,000,000 - <5,000,000,000 lb EPA Chemical Data Reporting (CDR) (1972) 6.4x10+11 g (slab, not secondary zinc) SRI Hazardous Substances Data Bank (HSDB) (1975) 4.5x10+11 g (slab, not secondary zinc) SRI Hazardous Substances Data Bank (HSDB) (1986) >1 million-10 million pounds US EPA; Non-confidential Production Volume Information Submitted by Companies for Chemicals Under the 1986-2002 Inventory Update Rule (IUR). Zinc (7440-66-6). Available from, as of May 26, 2006: Hazardous Substances Data Bank (HSDB) (1990) >10 thousand-500 thousand pounds US EPA; Non-confidential Production Volume Information Submitted by Companies for Chemicals Under the 1986-2002 Inventory Update Rule (IUR). Zinc (7440-66-6). Available from, as of May 26, 2006: Hazardous Substances Data Bank (HSDB) For more U.S. Production (Complete) data for ZINC, ELEMENTAL (23 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.7 U.S. Imports (1972) 4.75x10+11 g (slab zinc) SRI Hazardous Substances Data Bank (HSDB) (1975) 3.45x10+11 g (slab zinc) SRI Hazardous Substances Data Bank (HSDB) Slab zinc = 827,000; zinc dust = 2,000 tons (1996) Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V25 814 Hazardous Substances Data Bank (HSDB) (Data expressed as metric tons) ... Slab zinc, 1991, 549,000; 1992, 644,000; 1993, 724,000; 1994, 793,000; 1995, 856,000. Rolled zinc, 1991, 537; 1992, 171; 1993, 135; 1994, 475; 1995, 332. United States Department of the Interior/U.S. Geological Survey. Minerals Yearbook. Metals and Minerals Volume 1. Washington, D.C. 1995., p. 926 Hazardous Substances Data Bank (HSDB) For more U.S. Imports (Complete) data for ZINC, ELEMENTAL (10 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.8 U.S. Exports (1972) 3.93x10+9 g (slab zinc) SRI Hazardous Substances Data Bank (HSDB) (1975) 6.26x10+9 g (slab zinc) SRI Hazardous Substances Data Bank (HSDB) (Data expressed as metric tons) ... Slab zinc, 1991, 1,250; 1992, 565; 1993, 1,410; 1994, 6,310; 1995, 3,080. Rolled zinc, 1991, 10,400; 1992, 5,430; 1993, 6,600; 1994, 6,680; 1995, 5,180. United States Department of the Interior/U.S. Geological Survey. Minerals Yearbook. Metals and Minerals Volume 1. Washington, D.C. 1995., p. 926 Hazardous Substances Data Bank (HSDB) (2002) An estimated 822,000 metric tons of ores and concentrates, 1,160 metric tons of slab zinc, and 7,200 metric tons of rolled zinc were exported from the United States. DHHS/ATSDR; Toxicological Profile for Zinc p.130 TP- PB2006-100008 (2005). Hazardous Substances Data Bank (HSDB) For more U.S. Exports (Complete) data for ZINC, ELEMENTAL (9 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 12.9 General Manufacturing Information Industry Processing Sectors Not Known or Reasonably Ascertainable Soap, Cleaning Compound, and Toilet Preparation Manufacturing Services Construction Primary Metal Manufacturing All Other Chemical Product and Preparation Manufacturing Rubber Product Manufacturing Fabricated Metal Product Manufacturing Plastics Product Manufacturing Computer and Electronic Product Manufacturing Wood Product Manufacturing Paper Manufacturing Other (requires additional information) Transportation Equipment Manufacturing All Other Basic Inorganic Chemical Manufacturing Pesticide, Fertilizer, and Other Agricultural Chemical Manufacturing Electrical Equipment, Appliance, and Component Manufacturing Machinery Manufacturing Paint and Coating Manufacturing Textiles, apparel, and leather manufacturing Wholesale and Retail Trade Miscellaneous Manufacturing Petroleum Lubricating Oil and Grease Manufacturing EPA Chemical Data Reporting (CDR) EPA TSCA Commercial Activity Status Zinc: ACTIVE EPA Chemicals under the TSCA The effect of small amt of common impurities is to incr corrosion resistance to solutions, but not in the atmosphere. Ordinary zinc is too brittle to roll at ordinary temperatures, but becomes ductile at elevated temperatures; brittleness is thought to be assoc with impurities such as tin. Clayton, G.D., F.E. Clayton (eds.) Patty's Industrial Hygiene and Toxicology. Volumes 2A, 2B, 2C, 2D, 2E, 2F: Toxicology. 4th ed. New York, NY: John Wiley & Sons Inc., 1993-1994., p. 2333 Hazardous Substances Data Bank (HSDB) The principal zinc ore is in the form of sulfides, such as sphalerite and wurtzite (cubic and hexagonal ZnS) and willemite (Zn2SiO4). To obtain metallic zinc, the zinc ores which are relatively low in zinc content are concentrated. Zinc smelting is gradually being replaced by the electrolytic processes. During smelting there are often large emissions of zinc, and other heavy metals contained in the zinc ore such as lead and cadmium, into the air. Bingham, E.; Cohrssen, B.; Powell, C.H.; Patty's Toxicology Volumes 1-9 5th ed. John Wiley & Sons. New York, N.Y. (2001)., p. 2:255 Hazardous Substances Data Bank (HSDB) 12.10 Sampling Procedures Analyte: Zinc (metals and oxides); Matrix: air; Sampler: filter (0.8 um cellulose ester membrane); Flow rate: 1 L/min; Vol: min: 10 L, max: 400 L; Stability: At least 1 yr at 25 °C /Welding and brazing fumes/ U.S. Department of Health and Human Services, Public Health Service. Centers for Disease Control, National Institute for Occupational Safety and Health. NIOSH Manual of Analytical Methods, 3rd ed. Volumes 1 and 2 with 1985 supplement, and revisions. Washington, DC: U.S. Government Printing Office, February 1984., p. V2 7200-1 Hazardous Substances Data Bank (HSDB) 13 Identification 13.1 Analytic Laboratory Methods Atomic absorption spectrometry (AAS) is most often used for the determination of zinc in the workplace atmosphere. In the method described by NIOSH Method 7030 air samples are collected on 0.8-mm cellulose ester membrane filters at an airflow rate of 1 to 3 L/min. The filters are mineralized with concn HNO3 and the final solutions in 1% HNO3 are analyzed using flame AAS at 213.9 nm. The estimated limit of quantitation (LOD) is 3 ug per sample. Bingham, E.; Cohrssen, B.; Powell, C.H.; Patty's Toxicology Volumes 1-9 5th ed. John Wiley & Sons. New York, N.Y. (2001)., p. 2:255 Hazardous Substances Data Bank (HSDB) Atmospheric aerosols are collected on cellulose filters and after digestion can be determined by means of anodic stripping voltametry. Sample detection limit was 13.7 ug/L. Water and wastewater samples are digested with HNO3, mineralized in muffle furnace, or extracted with ammonium tetramethylene dithiocarbamate (APDC)/methyl isobutyl ketone (MIBK). The final solution can be analyzed by means of flame atomic absorption spectrometry (FAAS) (sample detection limit 5 ug/L), AAS, furnace technique (sample detection limit 0.05 ug/L), or flow-injection analysis (FIA) (sample detection limit 3 ug/L). After APDC-MIBK extraction of seawater sample and determination by means of FAAC /(F-actin affinity chromatography)/, the sample detection limit amounted to 0.05 ug/L. Zinc concentration in soil samples was determined after extraction with diethylenetriaminepentaacetate (DTPA) and NH4HCO3-DTPA by means of inductively coupled argon plasma spectroscopy (ICP). Bingham, E.; Cohrssen, B.; Powell, C.H.; Patty's Toxicology Volumes 1-9 5th ed. John Wiley & Sons. New York, N.Y. (2001)., p. 2:255 Hazardous Substances Data Bank (HSDB) Method: EPA-NERL 200.8; Procedure: inductively coupled plasma/mass spectrometry; Analyte: zinc; Matrix: ground waters, surface waters, and drinking water, and total recoverable elements in these waters as well as wastewaters, sludges, and soils samples; Detection Limit: 1.8 ug/L. National Environmental Methods Index; Analytical, Test and Sampling Methods. Zinc (7440-66-6). Available from, as of July 17, 2006: Hazardous Substances Data Bank (HSDB) Method: EPA-NERL 200.7; Procedure: inductively coupled plasma/atomic emission spectrometry; Analyte: zinc; Matrix: water, wastewater, and solid wastes; Detection Limit: 2 ug/L. National Environmental Methods Index; Analytical, Test and Sampling Methods. Zinc (7440-66-6). Available from, as of July 17, 2006: Hazardous Substances Data Bank (HSDB) For more Analytic Laboratory Methods (Complete) data for ZINC, ELEMENTAL (25 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 13.2 Clinical Laboratory Methods Atomic absorption spectrometry (AAS) is a common and simple laboratory technique capable of routine zinc analysis of biological samples including bone, liver, hair, blood, and urine. Graphite furnace AAS (GF-AAS) is more sensitive than flame AAS and has been used to determine very low levels of zinc (detection limit, 0.052 umol/L) in human milk ... /and/ semen. DHHS/ATSDR; Toxicological Profile for Zinc p.192 PB2006-100008 (2005) Hazardous Substances Data Bank (HSDB) Method: USGS-NWQL B-9001-95; Procedure: inductively coupled plasma/mass spectrometry; Analyte: zinc; Matrix: aquatic biological tissue and aquatic plant material only and is not applicable to bivalve shells, bones, and sediment material contained in bivalves; Detection Limit: 1.2 ug/g. National Environmental Methods Index; Analytical, Test and Sampling Methods. Zinc (7440-66-6). Available from, as of July 17, 2006: Hazardous Substances Data Bank (HSDB) Method: USGS-NWQL B-9001-95; Procedure: inductively coupled plasma/atomic emission spectrometry; Analyte: zinc; Matrix: aquatic biological tissue and aquatic plant material only and is not applicable to bivalve shells, bones, and sediment material contained in bivalves; Detection Limit: not provided. National Environmental Methods Index; Analytical, Test and Sampling Methods. Zinc (7440-66-6). Available from, as of July 17, 2006: Hazardous Substances Data Bank (HSDB) Method: NOAA-NST 151.1; Procedure: flame atomic absorption; Analyte: zinc; Matrix: marine animal tissue; Detection Limit: 1.2 ug/g. National Environmental Methods Index; Analytical, Test and Sampling Methods. Zinc (7440-66-6). Available from, as of July 17, 2006: Hazardous Substances Data Bank (HSDB) For more Clinical Laboratory Methods (Complete) data for ZINC, ELEMENTAL (9 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 13.3 NIOSH Analytical Methods ELEMENTS by ICP (Nitric/Perchloric Acid Ashing) 7300 NIOSH Manual of Analytical Methods ELEMENTS by ICP 7301 NIOSH Manual of Analytical Methods ELEMENTS by ICP (Hot Block/HCl/HNO3 Digestion) 7303 NIOSH Manual of Analytical Methods ELEMENTS in blood or tissue 8005 NIOSH Manual of Analytical Methods ELEMENTS ON WIPES, 9102 NIOSH Manual of Analytical Methods ZINC and compounds, as Zn 7030 NIOSH Manual of Analytical Methods METALS in Urine 8310 NIOSH Manual of Analytical Methods 14 Safety and Hazards 14.1 Hazards Identification 14.1.1 GHS Classification 1 of 6 items View All Pictogram(s) Signal Danger GHS Hazard Statements H250: Catches fire spontaneously if exposed to air [Danger Pyrophoric liquids] H260: In contact with water releases flammable gases which may ignite spontaneously [Danger Substances and mixtures which in contact with water, emit flammable gases] H400: Very toxic to aquatic life [Warning Hazardous to the aquatic environment, acute hazard] H410: Very toxic to aquatic life with long lasting effects [Warning Hazardous to the aquatic environment, long-term hazard] Precautionary Statement Codes P210, P222, P223, P231, P231+P232, P233, P273, P280, P302+P335+P334, P370+P378, P391, P402+P404, and P501 (The corresponding statement to each P-code can be found at the GHS Classification page.) Regulation (EC) No 1272/2008 of the European Parliament and of the Council 14.1.2 Hazard Classes and Categories Pyr. Sol. 1 (26.7%) Water-react. 1 (26.7%) Aquatic Acute 1 (84.9%) Aquatic Chronic 1 (79%) European Chemicals Agency (ECHA) View More... 14.1.3 DOT Hazard Classification 1 of 2 items View All Substance (Descriptions/Shipping Name) Zinc ashes DOT ID (UN/NA Number) UN1435 Hazard Class/Label Code(s) Div 4.3 Dangerous when wet material (49 eCFR § 173.124) Packing Group PG III: the degree of danger presented by the material is minor For more information about the packing group assignment, please visit 49 eCFR § 173 Placard/Label(s) US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172 14.1.4 Health Hazards Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: Inhalation or contact with vapors, substance or decomposition products may cause severe injury or death. May produce corrosive solutions on contact with water. Fire will produce irritating, corrosive and/or toxic gases. Runoff from fire control or dilution water may cause environmental contamination. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals ERG 2024, Guide 138 (Zinc dust) · Inhalation or contact with vapors, substance or decomposition products may cause severe injury or death. · May produce corrosive solutions on contact with water. · Fire will produce irritating, corrosive and/or toxic gases. · Runoff from fire control or dilution water may cause environmental contamination. Emergency Response Guidebook (ERG) 14.1.5 Fire Hazards Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: Produce flammable gases on contact with water. May ignite on contact with water or moist air. Some react vigorously or explosively on contact with water. May be ignited by heat, sparks or flames. May re-ignite after fire is extinguished. Some are transported in highly flammable liquids. Runoff may create fire or explosion hazard. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals ERG 2024, Guide 138 (Zinc dust) · Produce flammable gases on contact with water. · May ignite on contact with water or moist air. · Some react vigorously or explosively on contact with water. · May be ignited by heat, sparks or flames. · May re-ignite after fire is extinguished. · Some are transported in highly flammable liquids. · Runoff may create fire or explosion hazard. Emergency Response Guidebook (ERG) Highly flammable. May ignite spontaneously on contact with air. Many reactions may cause fire or explosion. Finely dispersed particles form explosive mixtures in air. Risk of fire and explosion on contact with water or incompatible substances. See Chemical Dangers. ILO-WHO International Chemical Safety Cards (ICSCs) 14.1.6 Hazards Summary Zinc is one of the most common elements in the earth's crust. It is found in air, soil, and water, and is present in all foods. Pure zinc is a bluish-white shiny metal. Zinc has many commercial uses as coatings to prevent rust, in dry cell batteries, and mixed with other metals to make alloys like brass, and bronze. A zinc and copper alloy is used to make pennies in the United States. Zinc combines with other elements to form zinc compounds. Common zinc compounds found at hazardous waste sites include zinc chloride, zinc oxide, zinc sulfate, and zinc sulfide. Zinc compounds are widely used in industry to make paint, rubber, dyes, wood preservatives, and ointments. Agency for Toxic Substances and Disease Registry (ATSDR) Zinc is essential to plant and animal life; It is in more than 300 enzymes involved in all aspects of metabolism; Zn is also needed for the proper structural function of proteins; Zn deficiency causes failed growth and development; Zn is necessary for healthy immunity and wound healing; Zn is depleted in some soils; Plant ligands, e.g., phytate and lignin, can interfere with Zn absorption; 20% of the world's people are at risk for Zn deficiency; Zn supplements are the most common source of excess Zn; Excess intake may precipitate copper deficiency; Excess Zn in drinks, up to 2500 mg/L, caused vomiting, diarrhea, and abdominal pain; Zinc oxide fume causes metal fume fever, and zinc chloride fume causes acute pneumonitis; [Nordberg, p. 1369-82] Produces flammable hydrogen when in contact with moisture; [CAMEO] Zinc powder or dust are usually coated with zinc carbonate to reduce the risk of spontaneous combustion. Zinc dust is not known to be harmful after inhalation or contact with the skin or eyes. [CHEMINFO] Zinc is a hepatotoxic agent. [Zimmerman, p. 4] See Zinc chloride fume. See Zinc oxide. Nordberg - Nordberg GF, Fowler BA, Nordberg M (eds). Handbook on the Toxicology of Metals, 4th Ed. Boston: Elsevier, 2015., p. 1369-82 Zimmerman - Zimmerman HJ. Hepatotoxicity. Philadelphia: Lippincott Williams & Wilkins, 1999., p. 4 Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 14.1.7 Fire Potential Zinc powder or dust in contact with water or damp air evolves hydrogen. The heat of reaction is sufficient that the hydrogen may ignite. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-207 Hazardous Substances Data Bank (HSDB) ... In compact form does not burn readily until it is heated above 500 °C. International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 2340 Hazardous Substances Data Bank (HSDB) 14.1.8 Skin, Eye, and Respiratory Irritations A human skin irritant. Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 3717 Hazardous Substances Data Bank (HSDB) 14.2 Safety and Hazard Properties 14.2.1 Physical Dangers Ignites in air when finely divided. If dry, it can be charged electrostatically by swirling, pneumatic transport, pouring, etc. ILO-WHO International Chemical Safety Cards (ICSCs) 14.2.2 Explosive Limits and Potential A cloud of zinc dust generated by sieving the hot dried material exploded violently, apparently after initiation by a spark from the percussive sieve-shaking mechanism ... The possibility of explosions of zinc dust suspended in air is presented as a serious hazard ... Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1469 Hazardous Substances Data Bank (HSDB) 14.3 First Aid Measures Inhalation First Aid Fresh air, rest. Seek medical attention if you feel unwell. ILO-WHO International Chemical Safety Cards (ICSCs) Skin First Aid First rinse with plenty of water for at least 15 minutes, then remove contaminated clothes and rinse again. ILO-WHO International Chemical Safety Cards (ICSCs) Eye First Aid Rinse with plenty of water (remove contact lenses if easily possible). ILO-WHO International Chemical Safety Cards (ICSCs) Ingestion First Aid Rinse mouth. Refer for medical attention . ILO-WHO International Chemical Safety Cards (ICSCs) 14.3.1 First Aid Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: Refer to the "General First Aid" section. Specific First Aid: In case of contact with substance, wipe from skin immediately; flush skin or eyes with running water for at least 20 minutes. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals ERG 2024, Guide 138 (Zinc dust) General First Aid: · Call 911 or emergency medical service. · Ensure that medical personnel are aware of the material(s) involved, take precautions to protect themselves and avoid contamination. · Move victim to fresh air if it can be done safely. · Administer oxygen if breathing is difficult. · If victim is not breathing: -- DO NOT perform mouth-to-mouth resuscitation; the victim may have ingestedor inhaled the substance. -- If equipped and pulse detected, wash face and mouth, then give artificial respiration using a proper respiratory medical device (bag-valve mask, pocket mask equipped with a one-way valve or other device). -- If no pulse detected or no respiratory medical device available, provide continuouscompressions. Conduct a pulse check every two minutes or monitor for any signs of spontaneous respirations. · Remove and isolate contaminated clothing and shoes. · For minor skin contact, avoid spreading material on unaffected skin. · In case of contact with substance, remove immediately by flushing skin or eyes with running water for at least 20 minutes. · For severe burns, immediate medical attention is required. · Effects of exposure (inhalation, ingestion, or skin contact) to substance may be delayed. · Keep victim calm and warm. · Keep victim under observation. · For further assistance, contact your local Poison Control Center. · Note: Basic Life Support (BLS) and Advanced Life Support (ALS) should be done by trained professionals. Specific First Aid: · In case of contact with substance, wipe from skin immediately; flush skin or eyes with running water for at least 20 minutes. In Canada, an Emergency Response Assistance Plan (ERAP) may be required for this product. Please consult the shipping paper and/or the "ERAP" section. Emergency Response Guidebook (ERG) 14.4 Fire Fighting Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: DO NOT USE WATER OR FOAM. SMALL FIRE: Dry chemical, soda ash, lime or sand. LARGE FIRE: DRY sand, dry chemical, soda ash or lime or withdraw from area and let fire burn. If it can be done safely, move undamaged containers away from the area around the fire. FIRE INVOLVING METALS OR POWDERS (ALUMINUM, LITHIUM, MAGNESIUM, ETC.): Use dry chemical, DRY sand, sodium chloride powder, graphite powder or class D extinguishers; in addition, for Lithium you may use Lith-XÂ® powder or copper powder. Also, see ERG Guide 170. FIRE INVOLVING TANKS, RAIL TANK CARS OR HIGHWAY TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Do not get water inside containers. Cool containers with flooding quantities of water until well after fire is out. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals Use special powder, dry sand. NO water. NO foam, carbon dioxide. NO other agents. In case of fire: keep drums, etc., cool by spraying with water. NO direct contact of the substance with water. ILO-WHO International Chemical Safety Cards (ICSCs) 14.4.1 Fire Fighting Procedures To fight fire, use special mixtures of dry chemical. Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 3717 Hazardous Substances Data Bank (HSDB) 14.4.2 Firefighting Hazards Burning zinc reacts chemically with halon and CO2 gas extinguishers. Sullivan, J.B., Krieger G.R. (eds). Clinical Environmental Health and Toxic Exposures. Second edition. Lippincott Williams and Wilkins, Philadelphia, Pennsylvania 1999., p. 989 Hazardous Substances Data Bank (HSDB) 14.5 Accidental Release Measures Public Safety: ERG 2024, Guide 138 (Zinc dust) · CALL 911. Then call emergency response telephone number on shipping paper. If shipping paper not available or no answer, refer to appropriate telephone number listed on the inside back cover. · Keep unauthorized personnel away. · Stay upwind, uphill and/or upstream. · Ventilate closed spaces before entering, but only if properly trained and equipped. Emergency Response Guidebook (ERG) Spill or Leak: ERG 2024, Guide 138 (Zinc dust) · ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. · Do not touch or walk through spilled material. · Stop leak if you can do it without risk. · Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. · DO NOT GET WATER on spilled substance or inside containers. Small Spill · Cover with DRY earth, DRY sand or other non-combustible material followed with plastic sheet to minimize spreading or contact with rain. · Dike for later disposal; do not apply water unless directed to do so. Powder Spill · Cover powder spill with plastic sheet or tarp to minimize spreading and keep powder dry. · DO NOT CLEAN-UP OR DISPOSE OF, EXCEPT UNDER SUPERVISION OF A SPECIALIST. Emergency Response Guidebook (ERG) 14.5.1 Isolation and Evacuation Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area in all directions for at least 50 meters (150 feet) for liquids and at least 25 meters (75 feet) for solids. LARGE SPILL: Consider initial downwind evacuation for at least 300 meters (1000 feet). FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals Evacuation: ERG 2024, Guide 138 (Zinc dust) Immediate precautionary measure · Isolate spill or leak area in all directions for at least 50 meters (150 feet) for liquids and at least 25 meters (75 feet) for solids. Large Spill · Consider initial downwind evacuation for at least 300 meters (1000 feet). Fire · If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions. Emergency Response Guidebook (ERG) 14.5.2 Spillage Disposal Remove all ignition sources. Consult an expert! Personal protection: particulate filter respirator adapted to the airborne concentration of the substance. Do NOT let this chemical enter the environment. Do NOT wash away into sewer. Sweep spilled substance into covered dry containers. Then store and dispose of according to local regulations. ILO-WHO International Chemical Safety Cards (ICSCs) 14.5.3 Cleanup Methods Extinguish or remove all ignition sources. Do NOT wash away into sewer. Sweep spilled substance into containers, then remove to safe place. Personal protection: self-contained breathing apparatus. /Zinc powder/ IPCS, CEC; International Chemical Safety Card on Zinc. (October 1994). Available from, as of May 17, 2006: Hazardous Substances Data Bank (HSDB) 14.5.4 Disposal Methods SRP: The most favorable course of action is to use an alternative chemical product with less inherent propensity for occupational exposure or environmental contamination. Recycle any unused portion of the material for its approved use or return it to the manufacturer or supplier. Ultimate disposal of the chemical must consider: the material's impact on air quality; potential migration in soil or water; effects on animal, aquatic, and plant life; and conformance with environmental and public health regulations. Hazardous Substances Data Bank (HSDB) Chemical Treatability of Zinc; Concentration Process: Activated carbon; Chemical Classification: Metals; Scale of Study: Full scale continuous flow; Type of Wastewater Used: (not stated); Results of Study: 81% reduction; 124 ppb effluent concentration; Carbon used as advanced treatment of biologically and chemically treated wastewater. Plant capacity 0.66 cu m/sec. Data presented for two time periods. USEPA; Management of Hazardous Waste Leachate, EPA Contract No. 68-03-2766 p.E-166 (1982) Hazardous Substances Data Bank (HSDB) Chemical Treatability of Zinc; Concentration Process: Miscellaneous sorbents; Chemical Classification: Metals; Scale of Study: Literature review; Type of Wastewater Used: Unknown; Results of Study: Final concentration reduced to 0.1 ppb; SiO2 + CaO slags used. USEPA; Management of Hazardous Waste Leachate, EPA Contract No. 68-03-2766 p.E-202 (1982) Hazardous Substances Data Bank (HSDB) The proprietary sulfex process (Permutit Co) has been applied to zinc wastes. The process involves addition of ferrous sulfide, which gradually releases sulfide to precipitate the zinc ... . Patterson JW; Industrial Wastewater Treatment Technolgy 2nd Edition p.444 (1985) Hazardous Substances Data Bank (HSDB) For more Disposal Methods (Complete) data for ZINC, ELEMENTAL (6 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 14.5.5 Preventive Measures SRP: The scientific literature for the use of contact lenses in industry is conflicting. The benefit or detrimental effects of wearing contact lenses depend not only upon the substance, but also on factors including the form of the substance, characteristics and duration of the exposure, the uses of other eye protection equipment, and the hygiene of the lenses. However, there may be individual substances whose irritating or corrosive properties are such that the wearing of contact lenses would be harmful to the eye. In those specific cases, contact lenses should not be worn. In any event, the usual eye protection equipment should be worn even when contact lenses are in place. Hazardous Substances Data Bank (HSDB) Prevention ... /of metal fume fever/ is a matter of keeping exposure of workers below level of concn currently accepted as satisfactory for working with the metal in industry, preferably by employment of proper local exhaust ventilation to collect fumes at their source. Acceptable respirators are avail commercially but should be used only under suitable conditions. /Zinc/ International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 1340 Hazardous Substances Data Bank (HSDB) In all cases where zinc is heated to the point where fume is produced, it is most important to ensure that adequate ventilation is provided. Individual protection is best ensured by education of the worker concerning metal-fume fever & the provision of local exhaust ventilation, or, in some situations by wearing of supplied-air hood or mask. International Labour Office. Encyclopaedia of Occupational Health and Safety. 4th edition, Volumes 1-4 1998. Geneva, Switzerland: International Labour Office, 1998., p. 63.45 Hazardous Substances Data Bank (HSDB) 14.6 Handling and Storage 14.6.1 Nonfire Spill Response Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. Do not touch or walk through spilled material. Stop leak if you can do it without risk. Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. DO NOT GET WATER on spilled substance or inside containers. SMALL SPILL: Cover with DRY earth, DRY sand or other non-combustible material followed with plastic sheet to minimize spreading or contact with rain. Dike for later disposal; do not apply water unless directed to do so. POWDER SPILL: Cover powder spill with plastic sheet or tarp to minimize spreading and keep powder dry. DO NOT CLEAN-UP OR DISPOSE OF, EXCEPT UNDER SUPERVISION OF A SPECIALIST. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals 14.6.2 Safe Storage Fireproof. Well closed. Separated from incompatible materials and : see Chemical Dangers. Store in an area without drain or sewer access. Provision to contain effluent from fire extinguishing. ILO-WHO International Chemical Safety Cards (ICSCs) 14.6.3 Storage Conditions Fireproof. Separated from acids, bases oxidants. Dry. /Zinc powder/ IPCS, CEC; International Chemical Safety Card on Zinc. (October 1994). Available from, as of May 17, 2006: Hazardous Substances Data Bank (HSDB) Finely divided ... zinc compounds, can be fire and explosion hazard if stored in damp places, sources of spontaneous combustion. /Zinc cmpd/ International Labour Office. Encyclopaedia of Occupational Health and Safety. 4th edition, Volumes 1-4 1998. Geneva, Switzerland: International Labour Office, 1998., p. 63.45 Hazardous Substances Data Bank (HSDB) Store the metal dry, and keep residues thoroughly wet until disposal. Hydrogen is evolved, especially under acid and alkaline conditions. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1472 Hazardous Substances Data Bank (HSDB) 14.7 Exposure Control and Personal Protection Protective Clothing: ERG 2024, Guide 138 (Zinc dust) · Wear positive pressure self-contained breathing apparatus (SCBA). · Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE. · Structural firefighters' protective clothing provides thermal protection but only limited chemical protection. Emergency Response Guidebook (ERG) Maximum Allowable Concentration (MAK) 0.1 [mg/m3] (respirable fraction), 2 mg/m3 (inhalable fraction) for Zn inorg. cmpnds[German Research Foundation (DFG)] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 14.7.1 Occupational Exposure Limits (OEL) MAK (Maximale Arbeitsplatz Konzentration) (as Zn, respirable fraction): 0.1 mg/m ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.2 Emergency Response Planning Guidelines Emergency Response: ERG 2024, Guide 138 (Zinc dust) · DO NOT USE WATER OR FOAM. Small Fire · Dry chemical, soda ash, lime or sand. Large Fire · DRY sand, dry chemical, soda ash or lime or withdraw from area and let fire burn. · If it can be done safely, move undamaged containers away from the area around the fire. Fire Involving Metals or Powders (Aluminum, Lithium, Magnesium, etc.) · Use dry chemical, DRY sand, sodium chloride powder, graphite powder or class D extinguishers; in addition, for Lithium you may use Lith-X® powder or copper powder. Also, see GUIDE 170. Fire Involving Tanks, Rail Tank Cars or Highway Tanks · Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. · Do not get water inside containers. · Cool containers with flooding quantities of water until well after fire is out. · Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. · ALWAYS stay away from tanks in direct contact with flames. Emergency Response Guidebook (ERG) 14.7.3 Inhalation Risk A harmful concentration of airborne particles can be reached quickly when dispersed, especially if powdered or as fumes. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.4 Effects of Short Term Exposure May cause mechanical irritation to the eyes and respiratory tract. Inhalation of the respirable fraction may cause metal fume fever. This may result in influenza-like symptoms. The effects may be delayed up to 48 hours. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.5 Effects of Long Term Exposure Repeated or prolonged contact with skin may cause dermatitis. Repeated or prolonged inhalation may cause effects on the lungs. This may result in reduced lung function . ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.6 Personal Protective Equipment (PPE) Excerpt from ERG Guide 138 [Substances - Water-Reactive (Emitting Flammable Gases)]: Wear positive pressure self-contained breathing apparatus (SCBA). Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE. Structural firefighters' protective clothing provides thermal protection but only limited chemical protection. (ERG, 2024) 2024 Emergency Response Guidebook, 2024 Emergency Response Guidebook, CAMEO Chemicals 14.7.7 Fire Prevention NO open flames, NO sparks and NO smoking. NO contact with oxidizing agents, acids, bases, water or incompatible substances. Closed system, ventilation, explosion-proof electrical equipment and lighting. Prevent build-up of electrostatic charges (e.g., by grounding). Prevent deposition of dust. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.8 Exposure Prevention PREVENT DISPERSION OF DUST! ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.9 Inhalation Prevention Use local exhaust. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.10 Skin Prevention Protective gloves. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.11 Eye Prevention Wear safety spectacles. ILO-WHO International Chemical Safety Cards (ICSCs) 14.7.12 Ingestion Prevention Do not eat, drink, or smoke during work. Wash hands before eating. ILO-WHO International Chemical Safety Cards (ICSCs) 14.8 Stability and Reactivity 14.8.1 Air and Water Reactions If it becomes damp it may spontaneously heat and ignite. Insoluble in water. CAMEO Chemicals Can evolve gaseous hydrogen in contact with water or damp air. The heat of the reaction may be sufficient to ignite the hydrogen produced [Haz. Chem. Data 1966. p. 171]. Flammable. May form an explosive mixture with air [Hawley]. CAMEO Chemicals 14.8.2 Reactive Group Metals, Elemental and Powder, Active CAMEO Chemicals 14.8.3 Reactivity Alerts Strong Reducing Agent Water-Reactive Pyrophoric CAMEO Chemicals Strong Reducing Agent Known Catalytic Activity Water-Reactive Pyrophoric CAMEO Chemicals 14.8.4 Reactivity Profile In the presence of moisture, it may ignite spontaneously on contact with air; reacts violently with oxidants and powdered sulfur, causing fire and explosion hazards. [Handling Chemicals Safely 1980. p. 966]. CAMEO Chemicals ZINC DUST is a reducing agent. Reacts violently with oxidants causing fire and explosion hazards [Handling Chemicals Safely 1980. p. 966]. In the presence of carbon, the combination of chlorine trifluoride with zinc results in a violent reaction [Mellor 2, Supp. 1: 1956]. Sodium peroxide oxidizes zinc with incandescence [Mellor 2:490-93 1946-47]. Zinc powder or dust in contact with acids forms hydrogen. The heat generated by the reaction is sufficient to ignite the hydrogen evolved [Lab. Govt. Chemist 1965]. A mixture of powdered zinc and an oxidizing agent such as potassium chlorate or powdered sulfur can be exploded by percussion. Zinc burns in moist chlorine. A mixture of zinc and carbon disulfide reacts with incandescence. Zinc powder reacts explosively when heated with manganese chloride. The reaction between zinc and selenium or tellurium is accompanied by incandescence [Mellor 4:476-480 1946-47]. When zinc and ammonium nitrate are mixed and wetted with a minimum of water, a violent reaction occurs with evolution of steam and zinc oxide. When hydrazine mononitrate is heated in contact with zinc a flaming decomposition occurs at temperatures a little above its melting point. Hydroxylamine is reduced when heated with zinc dust, unpredictably it may either ignite and burn or explode [Mellor 8 1946-47]. CAMEO Chemicals 14.8.5 Hazardous Reactivities and Incompatibilities Incompatible with NH4NO3, barium oxide, Ba(NO3)2, Cadmium, CS2, chlorates, Cl2, CrO3, (ethyl acetoacetate + tribromoneopentyl alcohol), F2, hydrazine mononitrate, hydroxylamine, Pb(N3)2, (Mg + Ba(NO3)2 + BaO2), MnCl2, HNO3, performic acid, KClO3, KNO3, K2O2, Selenium, NaClO3, Na2O2, Sulfur, Te, water, (NH4)2S, As2O3, CS2, CaCl2, NaOH, chlorinated rubber, catalytic metals, halocarbons, o-nitroanisole, nitrobenzene, nonmetals, oxidants, paint primer base, pentacarbonyliron, transition metal halides, seleninyl bromide. Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 3717 Hazardous Substances Data Bank (HSDB) Zinc powder or dust in contact with acids evolves hydrogen. The heat of reaction is sufficient that the hydrogen may ignite. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-206 Hazardous Substances Data Bank (HSDB) Alloys of zinc with iridium, platinum, or rhodium, after extraction with acid, leave residues which explode on warming in air, owing to the presence of occluded hydrogen (or oxygen) in the catalytic metal powders so produced. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1470 Hazardous Substances Data Bank (HSDB) Ball-milling aluminium-zinc (not stated if alloy or mixture) with inadequate inerting arrangements led to fires during operation or discharge of the mill. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 32 Hazardous Substances Data Bank (HSDB) For more Hazardous Reactivities and Incompatibilities (Complete) data for ZINC, ELEMENTAL (32 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 14.9 Transport Information 14.9.1 DOT Emergency Guidelines /GUIDE 138: SUBSTANCES - WATER-REACTIVE (EMITTING FLAMMABLE GASES)/ Fire or Explosion: Produce flammable gases on contact with water. May ignite on contact with water or moist air. Some react vigorously or explosively on contact with water. May be ignited by heat, sparks or flames. May re-ignite after fire is extinguished. Some are transported in highly flammable liquids. Runoff may create fire or explosion hazard. /Zinc dust; Zinc powder/ U.S. Department of Transportation. 2004 Emergency Response Guidebook. A Guide book for First Responders During the Initial Phase of a Dangerous Goods/Hazardous Materials Incident. Washington, D.C. 2004 Hazardous Substances Data Bank (HSDB) /GUIDE 138: SUBSTANCES - WATER-REACTIVE (EMITTING FLAMMABLE GASES)/ Health: Inhalation or contact with vapors, substance, or decomposition products may cause severe injury or death. May produce corrosive solutions on contact with water. Fire will produce irritating, corrosive and/or toxic gases. Runoff from fire control may cause pollution. /Zinc dust; Zinc powder/ U.S. Department of Transportation. 2004 Emergency Response Guidebook. A Guide book for First Responders During the Initial Phase of a Dangerous Goods/Hazardous Materials Incident. Washington, D.C. 2004 Hazardous Substances Data Bank (HSDB) /GUIDE 138: SUBSTANCES - WATER-REACTIVE (EMITTING FLAMMABLE GASES)/ Public Safety: CALL Emergency Response Telephone Number ... . As an immediate precautionary measure, isolate spill or leak area in all directions for at least 50 meters (150 feet) for liquids and at least 25 meter (75 feet) for solids. Keep unauthorized personnel away. Stay upwind. Keep out of low areas. Ventilate the area before entry. /Zinc dust; Zinc powder/ U.S. Department of Transportation. 2004 Emergency Response Guidebook. A Guide book for First Responders During the Initial Phase of a Dangerous Goods/Hazardous Materials Incident. Washington, D.C. 2004 Hazardous Substances Data Bank (HSDB) /GUIDE 138: SUBSTANCES - WATER-REACTIVE (EMITTING FLAMMABLE GASES)/ Protective Clothing: Wear positive pressure self-contained breathing apparatus (SCBA). Wear chemical protective clothing that is specifically recommended by the manufacturer. It may provide little or no thermal protection. Structural firefighters' protective clothing will only provide limited protection in fire situations ONLY; it is not effective in spill situations where direct contact with the substance is possible. /Zinc dust; Zinc powder/ U.S. Department of Transportation. 2004 Emergency Response Guidebook. A Guide book for First Responders During the Initial Phase of a Dangerous Goods/Hazardous Materials Incident. Washington, D.C. 2004 Hazardous Substances Data Bank (HSDB) For more DOT Emergency Guidelines (Complete) data for ZINC, ELEMENTAL (8 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 14.9.2 Shipping Name / Number DOT/UN/NA/IMO UN 1436; Zinc, powder or dust Hazardous Substances Data Bank (HSDB) IMO 4.3; Zinc, powder or dust; Zinc ashes Hazardous Substances Data Bank (HSDB) 14.9.3 Shipment Methods and Regulations No person may /transport,/ offer or accept a hazardous material for transportation in commerce unless that person is registered in conformance ... and the hazardous material is properly classed, described, packaged, marked, labeled, and in condition for shipment as required or authorized by ... /the hazardous materials regulations (49 CFR 171-177)./ 49 CFR 171.2; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of February 15, 2006: Hazardous Substances Data Bank (HSDB) The International Air Transport Association (IATA) Dangerous Goods Regulations are published by the IATA Dangerous Goods Board pursuant to IATA Resolutions 618 and 619 and constitute a manual of industry carrier regulations to be followed by all IATA Member airlines when transporting hazardous materials. International Air Transport Association. Dangerous Goods Regulations. 47th Edition. Montreal, Quebec Canada. 2006., p. 272-3 Hazardous Substances Data Bank (HSDB) The International Maritime Dangerous Goods Code lays down basic principles for transporting hazardous chemicals. Detailed recommendations for individual substances and a number of recommendations for good practice are included in the classes dealing with such substances. A general index of technical names has also been compiled. This index should always be consulted when attempting to locate the appropriate procedures to be used when shipping any substance or article. International Maritime Organization. International Maritime Dangerous Goods Code. London, UK. 2004., p. 65 Hazardous Substances Data Bank (HSDB) 14.9.4 DOT Label Dangerous When Wet CAMEO Chemicals Dangerous When Wet Spontaneously Combustible CAMEO Chemicals 14.9.5 Packaging and Labelling Airtight. Marine pollutant. ILO-WHO International Chemical Safety Cards (ICSCs) 14.9.6 EC Classification H250; H260; H400 / H400; H410 ILO-WHO International Chemical Safety Cards (ICSCs) 14.9.7 UN Classification UN Hazard Class: 4.3; UN Subsidiary Risks: 4.2 ILO-WHO International Chemical Safety Cards (ICSCs) 14.10 Regulatory Information The Australian Inventory of Industrial Chemicals Chemical: Zinc Australian Industrial Chemicals Introduction Scheme (AICIS) California Safe Cosmetics Program (CSCP) Reportable Ingredient Hazard Traits - Bioaccumulation; Environmental Persistence; Hazard Trait Under Review Authoritative List - CWA 303(c); CWA 303(d) Report - if used as a fragrance or flavor ingredient California Safe Cosmetics Program (CSCP) Product Database REACH Registered Substance Status: Active Update: 27-04-2023 Status: Active Update: 31-10-2012 European Chemicals Agency (ECHA) New Zealand EPA Inventory of Chemical Status Zinc: Does not have an individual approval but may be used under an appropriate group standard New Zealand Environmental Protection Authority (EPA) New Jersey Worker and Community Right to Know Act The New Jersey Worker and Community Right to Know Act requires public and private employers to provide information about hazardous substances at their workplaces. (N.J.S.A. 34:5A-1 et. seq.) NJDOH RTK Hazardous Substance List 14.10.1 Atmospheric Standards ... Substances for which a Federal Register notice has been published that included consideration of the serious health effects, including cancer, from ambient air exposure to the substance. Zinc (52 FR 32597, August 1987) is included on this list. 40 CFR 61.01(b); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of June 21, 2006: Hazardous Substances Data Bank (HSDB) 14.10.2 Federal Drinking Water Guidelines EPA 5000 ug/l /Secondary MCL/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) EPA 2000 ug/l /Lifetime health advisory/ USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) 14.10.3 State Drinking Water Guidelines (AZ) ARIZONA 5000 ug/l USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) (MN) MINNESOTA 2000 ug/l USEPA/Office of Water; Federal-State Toxicology and Risk Analysis Committee (FSTRAC). Summary of State and Federal Drinking Water Standards and Guidelines (11/93) To Present Hazardous Substances Data Bank (HSDB) 14.10.4 Clean Water Act Requirements Toxic pollutant designated pursuant to section 307(a)(1) of the Federal Water Pollution Control Act and is subject to effluent limitations. /Zinc and compounds/ 40 CFR 401.15; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of June 21, 2006: Hazardous Substances Data Bank (HSDB) The secondary maximum contaminant level for public water systems: zinc 5 mg/L. 40 CFR 143.3; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of June 21, 2006: Hazardous Substances Data Bank (HSDB) 14.10.5 CERCLA Reportable Quantities Persons in charge of vessels or facilities are required to notify the National Response Center (NRC) immediately, when there is a release of this designated hazardous substance, in an amount equal to or greater than its reportable quantity of 1,000 lb or 454 kg. The toll free number of the NRC is (800) 424-8802. The rule for determining when notification is required is stated in 40 CFR 302.4 (section IV. D.3.b). No reporting of release of this hazardous substance is required if the diameter of the pieces of the solid metal released is larger than 100 micrometers (0.004 inches). 40 CFR 302.4; U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of June 21, 2006: Hazardous Substances Data Bank (HSDB) 14.10.6 FIFRA Requirements As the federal pesticide law FIFRA directs, EPA is conducting a comprehensive review of older pesticides to consider their health and environmental effects and make decisions about their future use. Under this pesticide reregistration program, EPA examines health and safety data for pesticide active ingredients initially registered before November 1, 1984, and determines whether they are eligible for reregistration. In addition, all pesticides must meet the new safety standard of the Food Quality Protection Act of 1996. Pesticides for which EPA had not issued Registration Standards prior to the effective date of FIFRA, as amended in 1988, were divided into three lists based upon their potential for human exposure and other factors, with List B containing pesticides of greater concern and List D pesticides of less concern. Zinc, elemental is found on List D. Case No: 4099; Pesticide type: fungicide, herbicide, antimicrobial; Case Status: RED Approved 09/92; OPP has made a decision that some/all uses of the pesticide are eligible for reregistration, as reflected in a Reregistration Eligibility Decision (RED) document.; Active ingredient (AI): Zinc; Data Call-in (DCI) Date(s): 04/01/93; AI Status: OPP has completed a Reregistration Eligibility Decision (RED) document for the case/AI. United States Environmental Protection Agency/ Prevention, Pesticides and Toxic Substances; Status of Pesticides in Registration, Reregistration, and Special Review. (1998) EPA 738-R-98-002, p. 346 Hazardous Substances Data Bank (HSDB) 14.11 Other Safety Information Chemical Assessment IMAP assessments - Zinc: Human health tier I assessment Australian Industrial Chemicals Introduction Scheme (AICIS) 14.11.1 Toxic Combustion Products When zinc or one of its alloys is burned, melted, or heated to temperatures above 930 °F, zinc metal oxide fume of particle diameter 1 u & below is formed. Clayton, G. D. and F. E. Clayton (eds.). Patty's Industrial Hygiene and Toxicology: Volume 2A, 2B, 2C: Toxicology. 3rd ed. New York: John Wiley Sons, 1981-1982., p. 2037 Hazardous Substances Data Bank (HSDB) 14.11.2 Other Hazardous Reactions The residues from zinc dust acetic acid reduction operations may ignites after a long delay if discarded into waste bins with paper. Small amounts appear to ignite more rapidly than larger portions. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1469 Hazardous Substances Data Bank (HSDB) Arsenic ... /and zinc/ react with incandescence on heating. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1471 Hazardous Substances Data Bank (HSDB) /Zinc or its alloys/ are unsuitable for service with bromomethane because of the formation of pyrophoric Grignard-type cmpd ... Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 155 Hazardous Substances Data Bank (HSDB) The reaction between zinc and selenium or tellurium is accompanied by incandescence (cadmium less so). Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-207 Hazardous Substances Data Bank (HSDB) A mixture of zinc and carbon disulfide reacts with incandescence. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-207 Hazardous Substances Data Bank (HSDB) The rapid autocatalytic dissolution of ... zinc in 9:1 methanol-carbon tetrachloride mixtures is sufficiently vigorous to be rated as potentially hazardous. Dissolution of zinc powder is subject to an induction period of 2 hr, which is eliminated by traces of copper(II) chloride, mercury(II) chloride or chromium(III) bromide. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 174 Hazardous Substances Data Bank (HSDB) A paste of zinc powder and carbon tetrachloride (with kieselguhr as thickener) will readily burn after ignition by a high-temperature primer. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1470 Hazardous Substances Data Bank (HSDB) Zinc burns in moist chlorine. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-207 Hazardous Substances Data Bank (HSDB) Causes of spontaneous combustion and other hazards of silver-zinc batteries were investigated. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 3 Hazardous Substances Data Bank (HSDB) Zinc burns in moist fluorine. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-89 Hazardous Substances Data Bank (HSDB) Presence of zinc ... causes flaming decomp /of hydrazinium nitrate/ above the melting point (70 °C). Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1259 Hazardous Substances Data Bank (HSDB) Lead azide, when in contact with ... zinc, or alloys containing ... zinc, forms, over a period of time, the extremely sensitive ... zinc azides which on slight disturbance can set off the main body of lead azide. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-65 Hazardous Substances Data Bank (HSDB) Addition of concn nitric acid to molten zinc (419 °C) causes it to incandesce ... Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1168 Hazardous Substances Data Bank (HSDB) Zinc residues from reduction of nitrobenzene to N-phenylhydroxylamine are often pyrophoric and must be kept wet during disposal. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1471 Hazardous Substances Data Bank (HSDB) When nitryl fluoride is passed at /mild warming/ temp over ... /zinc/, glowing or white incandescence occurs. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1094 Hazardous Substances Data Bank (HSDB) Oxidation of ... zinc /by potassium dioxide/ proceeds with incandescence. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1306 Hazardous Substances Data Bank (HSDB) Potassium peroxide oxidizes many metals with incandescence; eg ... zinc ... Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-156 Hazardous Substances Data Bank (HSDB) Sodium peroxide oxidizes zinc with incandescence. Fire Protection Guide to Hazardous Materials. 13 ed. Quincy, MA: National Fire Protection Association, 2002., p. 491-207 Hazardous Substances Data Bank (HSDB) Reduction of /titanium(IV)/ oxide by ... zinc is accompanied by more or less incandescence. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1417 Hazardous Substances Data Bank (HSDB) In contact with atmospheric oxygen and limited amounts of water, zinc dust will generate heat, and may become incandescent. Presence of acetic acid and copper shortens the induction period. Store the metal dry, and keep residues thoroughly wet until disposal. Hydrogen is evolved, especially under acid and alkaline conditions. Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1472 Hazardous Substances Data Bank (HSDB) 15 Toxicity 15.1 Toxicological Information CDC-ATSDR Toxicological Profile Agency for Toxic Substances and Disease Registry (ATSDR) 15.1.1 Toxicity Summary Although zinc is considered relatively nontoxic, an extremely high intake of zinc can manifest with symptoms including nausea, vomiting, epigastric pain, lethargy, and fatigue. There is no listed treatment for zinc overdose other than to cease using the supplement. StatPearls 15.1.2 EPA IRIS Information Substance Zinc and Compounds Critical Effect Systems Hematologic Immune Reference Dose (RfD), chronic 3 x 10 ^-1 mg/kg-day EPA Integrated Risk Information System (IRIS) 15.1.3 RAIS Toxicity Values Oral Chronic Reference Dose (RfDoc) (mg/kg-day) 0.3 Oral Chronic Reference Dose Reference IRIS Current Oral Subchronic Chronic Reference Dose (RfDos) (mg/kg-day) 0.3 Oral Subchronic Chronic Reference Dose Reference ATSDR Final Short-term Oral Reference Dose (RfDot) (mg/kg-day) 0.3 Short-term Oral Reference Dose Reference ATSDR Final Risk Assessment Information System (RAIS) 15.1.4 USGS Health-Based Screening Levels for Evaluating Water-Quality Chemical Zinc USGS Parameter Code 01090 Chemical Classes Trace element Noncancer HBSL (Health-Based Screening Level)[μg/L] 2000 Benchmark Remarks Due to lack of human health toxicity and low exposure from pesticidal uses, risks associated from dietary uses of zinc are not of concern; registered uses of zinc salts are not expected to significantly increase the natural concentrations of zinc in surface water or finished water used for drinking water Reference Smith, C.D. and Nowell, L.H., 2024. Health-Based Screening Levels for evaluating water-quality data (3rd ed.). DOI:10.5066/F71C1TWP USGS Health-Based Screening Levels for Evaluating Water-Quality Data 15.1.5 Hepatotoxicity In case series and small trials of zinc therapy in patients with Wilson disease, adverse events have included gastrointestinal upset, but serum enzyme elevations and clinically apparent liver injury were not reported. Patients with Wilson disease frequently have liver injury and, in most case series, zinc therapy has been associated with slow improvement in the hepatic manifestations. Occasional patients with Wilson disease who are switched from penicillamine to zinc therapy develop mild elevations in serum aminotransferase levels, but these are more likely due to worsening control of copper balance and the underlying Wilson disease rather than direct hepatotoxicity of zinc. Zinc overdose can be associated with liver injury, jaundice and even hepatic failure, usually arising after several days and resembling injury from copper or iron overdose. The injury is clearly direct toxicity. Likelihood score: E (unlikely cause of clinically apparent liver injury in normal therapeutic doses). Drug Class: Trace Elements and Metals; Chelating Agents, Wilson Disease Agents Other Drugs in the Subclass, Wilson Disease: Dimercaprol, Penicillamine, Trientine LiverTox 15.1.6 Evidence for Carcinogenicity Cancer Classification: Group D Not Classifiable as to Human Carcinogenicity USEPA Office of Pesticide Programs, Health Effects Division, Science Information Management Branch: "Chemicals Evaluated for Carcinogenic Potential" (April 2006) Hazardous Substances Data Bank (HSDB) CLASSIFICATION: D; not classifiable as to human carcinogenicity. BASIS FOR CLASSIFICATION: No human data and no animal data. HUMAN CARCINOGENICITY DATA: None. ANIMAL CARCINOGENICITY DATA: None. U.S. Environmental Protection Agency's Integrated Risk Information System (IRIS). Summary on Zinc and compounds (7440-66-6). Available from, as of March 15, 2000: Hazardous Substances Data Bank (HSDB) 15.1.7 Carcinogen Classification Substance Dietary Zinc NTP Technical Report TR-592: Toxicology and Carcinogenesis Study of Dietary Zinc (CASRN 5263-02-5) in Sprague Dawley Rats (Hsd:Sprague Dawley SD) (Feed Studies) (2019 ) Peer Review Date 07/13/17 Conclusion for Male Rat Equivocal Evidence Conclusion for Female Rat No Evidence Summary Under the conditions of this 2-year dietary study, there was equivocal evidence of carcinogenic activity (see a summary of the Peer Review Panel comments in Appendix I) of diets deficient in zinc in male Hsd:Sprague Dawley SD rats based on higher incidences of adenoma of the pancreas and increased incidences of animals with multiple pancreatic adenomas. There was no evidence of carcinogenic activity of diets deficient in zinc (3.5 or 7 ppm) in female Hsd:Sprague Dawley SD rats. There was no evidence of carcinogenic activity of diets containing excess zinc (250 or 500 ppm) in male or female Hsd:Sprague Dawley SD rats. NTP Technical Reports 15.1.8 Effects During Pregnancy and Lactation ◉ Summary of Use during Lactation Zinc is a normal component in human milk. Typical daily doses of 15 mg or less of oral zinc from prenatal vitamins or other multimineral supplements do not alter milk zinc levels in lactating women. Mothers may therefore take zinc supplementation during lactation to achieve the recommended daily intake of 12 to 13 mg. Daily oral doses between 15 and 25 mg have negligible effects on milk zinc levels. Treatment of patients with Wilson’s disease with zinc acetate up to 100 mg daily may increase zinc levels in milk, but not above the normal range. Sublingual zinc lozenges and nasal sprays used to prevent or treat adult viral upper respiratory tract infections have not been studied during lactation. Maternal use of these remedies several times daily for short time periods, as they are typically intended to be used, would not be expected to cause harm to the breastfed infant. Zinc deficiency in exclusively breastfed infants, whether due to inadequate maternal zinc status or to infant premature birth or other causes, should be treated with direct zinc supplementation of the infant. ◉ Effects in Breastfed Infants Zinc deficiency in exclusively breastfed infants can occur. Clinical features include facial and groin rash, diarrhea, hair loss, disinterest in feeding, and failure to thrive. One known cause is below-normal milk zinc levels due to maternal genetic mutations affecting mammary zinc transport proteins. Zinc deficiency may also occur in infants born very preterm who are not supplemented with special human milk fortifiers designed for premature babies. With both causes, direct administration of zinc drops to the infant quickly corrects the deficiency and alleviate the infant’s symptoms. Acrodermatitis enteropathica is a congenital zinc deficiency disorder caused by genetic mutations affecting the infant’s intestinal zinc transporter proteins. Breastmilk is protective against this disorder, and symptoms typically develop after weaning from breastmilk feeding.[19,21] Resuming breastmilk feeding, if possible, and direct infant supplementation with zinc drops are the recommended treatments. ◉ Effects on Lactation and Breastmilk Certain variations in genes encoding mammary epithelial cell zinc transporter proteins are associated with reduced milk volume and altered milk content beyond zinc. This is not thought to be triggered by maternal zinc intake or zinc status and is not correctable with maternal zinc supplementation. Drugs and Lactation Database (LactMed) ◈ What is zinc? Zinc is an essential nutrient. This means the body cannot make zinc, so people need to get zinc from other sources. Zinc is naturally found in some foods such as meats and seafood or fish, and it is available as a dietary supplement. Zinc has also been added to some cereals and can be found in some prenatal vitamins. Zinc has also been found in homeopathic products marketed for colds. Denture adhesives may also contain zinc.Because zinc is an essential nutrient, people who are pregnant will need to continue to get zinc from food, drinks, and supplements. Talk with your healthcare providers about all supplements/vitamins that you take. Have the bottles or photos of the labels with you so that all ingredients and their amounts can be reviewed. Products that contain herbals are typically not recommended during pregnancy. For more information on herbal products please see our fact sheet at: ◈ What are the Dietary Reference Intakes of zinc for people who are pregnant? Dietary Reference Intakes (DRI) help people know how much of each vitamin or mineral they should aim to get each day. DRIs include the Recommended Daily Allowance (RDA) and a Tolerable Upper Intake Level (UL). The Recommended Dietary Allowance (RDA) is the amount people should aim to get each day from food, drinks, and supplements. The Tolerable Upper Intake Level (UL) is the dose at which people can start to have side effects. RDAs and ULs are there to help guide us in getting enough of a good thing but also to keep us from getting too much of a good thing.Recommended Daily Allowance (RDA)Upper Limit (UL) 14 to 18 years old and pregnant12 mg per day34 mg per day19 years old or older and pregnant11 mg per day40 mg per dayMost people can get enough zinc from a balanced diet. It is unlikely that you will get too much zinc if your only source of zinc is in your food and drink. There are resources available online that list amounts of zinc typically found in foods. Labels on supplements will list the amount of zinc in the product.People who do not eat meat or seafood, have had bariatric surgery (such as gastric bypass), have medical conditions that might affect how their body absorbs nutrients (such as cancer, eating disorders, kidney disease, malabsorption, or substance misuse), or have exposure to cigarette smoke, should talk with their healthcare providers about their specific nutritional needs. ◈ I take zinc. Can it make it harder for me to get pregnant? If a person is getting recommended amounts of zinc (not too much and not too little) it would be unlikely to make it harder to get pregnant. It might be harder for a person to get pregnant if their zinc levels are too low. ◈ Does taking zinc increase the chance of miscarriage? Miscarriage is common and can occur in any pregnancy for many different reasons. Studies have not been done to see if zinc intake below the RDA (too little) or at doses higher than the UL (too much) can increase the chance of miscarriage. ◈ Does taking zinc increase the chance of birth defects? Every pregnancy starts out with a 3-5% chance of having a birth defect. This is called the background risk. Zinc intake between the RDA and the UL is not expected to increase the chance for birth defects. Some studies have suggested that having blood zinc levels that are too high or too low might be linked to a higher chance of spina bifida (opening in the spine) in the fetus. ◈ Could taking zinc increase the chance of other pregnancy-related problems? Some studies found that low blood levels of zinc in the person who is pregnant might increase the chance of pregnancy-related problems, such as: low birth weight (weighing less than 5 pounds 8 ounces [2500 grams] at birth), high blood pressure (pregnancy-induced hypertension), preeclampsia (a serious pregnancy-related condition that can cause symptoms such as high blood pressure or fluid retention), and stillbirth.There are some studies looking at zinc supplementation (between 20mg to 30mg) in pregnancy among people known to have low blood zinc levels. These studies were looking to see if zinc supplementation might reduce the chance of some pregnancy-related problems for people with low levels of zinc. These studies did not consistently find either benefit or harm to the baby from this supplementation.Pregnancy-related problems from high zinc levels have not been well studied. ◈ Does taking zinc in pregnancy affect future behavior or learning for the child? Studies have not been done to see if zinc intake below the RDA (too little) or at doses higher than the UL (too much) can cause behavior or learning issues. ◈ Breastfeeding while taking zinc: Zinc is a normal part of breastmilk. The RDA for breastfeeding is different from pregnancy. See the chart below for the Recommended Daily Allowance (RDA) and a Tolerable Upper Intake Level (UL) of zinc while breastfeeding.Recommended Daily Allowance (RDA)Upper Limit (UL)14 to 18 years old and breastfeeding13 mg per day34 mg per day19 years old or older and breastfeeding12 mg per day40 mg per dayPeople who do not eat meat or seafood, have had bariatric surgery (such as gastric bypass), have medical conditions that might affect how their body absorbs nutrients (such as cancer, eating disorders, kidney disease, malabsorption, or substance misuse), or have exposure to cigarette smoke, should talk with their healthcare providers about their specific nutritional needs. ◈ If a male takes zinc, could it affect fertility or increase the chance of birth defects? A few studies have found that low zinc levels are linked to infertility (ability to get a partner pregnant). Zinc supplementation might help to improve fertility for some males but not for others. It is not clear how too much zinc would affect fertility. In general, exposures that fathers or sperm donors have are unlikely to increase the risks to a partner’s pregnancy. For more information, please see the MotherToBaby fact sheet Paternal Exposures at Mother To Baby Fact Sheets 15.1.9 Exposure Routes The substance can be absorbed into the body by inhalation. ILO-WHO International Chemical Safety Cards (ICSCs) 15.1.10 Signs and Symptoms Inhalation Exposure Metallic taste. Sore throat. Cough. Weakness. Fever. See Effects of short-term exposure. ILO-WHO International Chemical Safety Cards (ICSCs) Skin Exposure No acute symptoms expected. ILO-WHO International Chemical Safety Cards (ICSCs) Eye Exposure Redness. ILO-WHO International Chemical Safety Cards (ICSCs) Ingestion Exposure Abdominal pain. Nausea. Vomiting. ILO-WHO International Chemical Safety Cards (ICSCs) 15.1.11 Target Organs Gastrointestinal (Digestive), Hematological (Blood Forming), Respiratory (From the Nose to the Lungs) Agency for Toxic Substances and Disease Registry (ATSDR) Hematologic Immune EPA Integrated Risk Information System (IRIS) 15.1.12 Adverse Effects Occupational hepatotoxin - Secondary hepatotoxins: the potential for toxic effect in the occupational setting is based on cases of poisoning by human ingestion or animal experimentation. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Zinc taken in large amounts may cause diarrhea, abdominal cramps, and vomiting within 3 to 10 hours of swallowing the supplement. The symptoms usually alleviate within a short period. An excess intake of zinc can result in copper or anemia, iron deficiency, or copper deficiency. Nasal sprays and gels containing zinc may have side effects such as loss of sense of smell. StatPearls 15.1.13 Acute Effects ChemIDplus 15.1.14 Antidote and Emergency Treatment Basic treatment: Establish a patent airway. Suction if necessary. Watch for signs of respiratory insufficiency and assist ventilations if necessary. Administer oxygen by nonrebreather mask at 10 to 15 L/min. Monitor for pulmonary edema and treat if necessary ... . Anticipate seizures and treat if necessary ... . Monitor for shock and treat if necessary ... . For eye contamination, flush eyes immediately with water. Irrigate each eye continuously with normal saline during transport ... . Do not use emetics. For ingestion, rinse mouth and administer 5 ml/kg up to 200 ml of water for dilution if the patient can swallow, has a strong gag reflex, and does not drool. Administer activated charcoal ... . /Zinc and related compounds/ Bronstein, A.C., P.L. Currance; Emergency Care for Hazardous Materials Exposure. 2nd ed. St. Louis, MO. Mosby Lifeline. 1994., p. 381 Bronstein, A.C., P.L. Currance; Emergency Care for Hazardous Materials Exposure. 2nd ed. St. Louis, MO. Mosby Lifeline. 1994., p. 381 Hazardous Substances Data Bank (HSDB) Advanced treatment: Consider orotracheal or nasotracheal intubation for airway control in the patient who is unconscious or has severe pulmonary edema. Positive pressure ventilation techniques with a bag valve mask device may be beneficial. Monitor cardiac rhythm and treat arrhythmias if necessary ... . Start an IV with D5W /SRP: "To keep open", minimal flow rate/. Use lactated Ringer's if signs of hypovolemia are present. Watch for signs of fluid overload. Consider drug therapy for pulmonary edema ... . For hypotension with signs of hypovolemia, administer fluids cautiously. Consider vasopressors for hypotension with a normal fluid volume. Watch for signs of fluid overload ... . Treat seizures with diazepam ... . Use proparacaine, hydrochloride to assist eye irrigation ... . /Zinc and related compounds/ Bronstein, A.C., P.L. Currance; Emergency Care for Hazardous Materials Exposure. 2nd ed. St. Louis, MO. Mosby Lifeline. 1994., p. 382 Hazardous Substances Data Bank (HSDB) Advanced treatment: Consider orotracheal or nasotracheal intubation for airway control in the patient who is unconscious or has severe pulmonary edema. Positive-pressure ventilation techniques with a bag-valve-mask device may be beneficial. Monitor cardiac rhythm and treat arrhythmias if necessary ... . Start an IV with D5W TKO /SRP: "To keep open", minimal flow rate/. Use lactated Ringer's if signs of hypovolemia are present. Watch for signs of fluid overload. Consider drug therapy for pulmonary edema ... . For hypotension with signs of hypovolemia, administer fluids cautiously. Consider vasopressors for hypotension with a normal fluid volume. Watch for signs of fluid overload ... . Treat seizures with diazepam (Valium) ... . Use proparacaine hydrochloride to assist eye irrigation ... . /Zinc and related compounds/ Bronstein, A.C., P.L. Currance; Emergency Care for Hazardous Materials Exposure. 2nd ed. St. Louis, MO. Mosby Lifeline. 1994., p. 382 Hazardous Substances Data Bank (HSDB) For more Antidote and Emergency Treatment (Complete) data for ZINC, ELEMENTAL (7 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 15.1.15 Medical Surveillance The assessment of zinc exposure can be accomplished through measurement of zinc. The presence of excess zinc ... can indicate high exposure to zinc; however, no information was found in the literature regarding the accuracy of these levels in predicting possible health effects. Blood Reference Ranges: Normal - average level 1200 ug/dl; Exposed - not established; Toxic - not established. Serum or Plasma Reference Ranges: Normal - average levels 100 ug/dl; Exposed - not established; Toxic - not established. Urine Reference Ranges: Normal - average levels 0.5 mg/g creatinine or 150 to 1200 ug/24 hours; Exposed - urinary concentrations of 600 to 700 ug/l were found in workers exposed to zinc oxide at levels of 3 to 5 mg/cu m; Toxic - greater than 1200 ug/24 hours has been indicated as a toxic urinary level of zinc; however no information was located relating to the severity of symptoms seen with high urinary zinc levels. /Zinc/ Ryan, R.P., C.E. Terry (eds.). Toxicology Desk Reference 4th ed. Volumes 1-3. Taylor & Francis, Washington, D.C. 1997., p. 2317 Hazardous Substances Data Bank (HSDB) Respiratory Symptom Questionnaires: Questionnaires have been published by the American Thoracic Society and the British Medical Research Council. These questionnaires have been found to be useful in identification of people with chronic bronchitis, however certain pulmonary function tests such as FEV1 have been found to be better predictors of chronic airflow obstruction. /Zinc/ Ryan, R.P., C.E. Terry (eds.). Toxicology Desk Reference 4th ed. Volumes 1-3. Taylor & Francis, Washington, D.C. 1997., p. 2318 Hazardous Substances Data Bank (HSDB) Chest Radiography: This test is widely used for assessing pulmonary disease. Chest radiographs have been found to be useful for detection of early lung cancer in asymptomatic people, especially for detection of peripheral tumors such as adenocarcinomas. However, even though OSHA mandates this test for exposure to some toxicants such as asbestos, there are conflicting views on its efficacy in detection of pulmonary disease. /Zinc/ Ryan, R.P., C.E. Terry (eds.). Toxicology Desk Reference 4th ed. Volumes 1-3. Taylor & Francis, Washington, D.C. 1997., p. 2318 Hazardous Substances Data Bank (HSDB) Pulmonary Function Tests: The tests that have been found to be practical for population monitoring include: Spirometry and expiratory flow-volume curves; Determination of lung volumes; Diffusing capacity for carbon monoxide; Single-breath nitrogen washout; Inhalation challenge tests; Serial measurements of peak expiratory flow; Exercise testing. /Zinc/ Ryan, R.P., C.E. Terry (eds.). Toxicology Desk Reference 4th ed. Volumes 1-3. Taylor & Francis, Washington, D.C. 1997., p. 2319 Hazardous Substances Data Bank (HSDB) For more Medical Surveillance (Complete) data for ZINC, ELEMENTAL (8 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 15.1.16 Human Toxicity Excerpts /HUMAN EXPOSURE STUDIES/ ... 250 women (before 20 weeks of gestation) were given 20 mg elemental zinc daily until the end of pregnancy. The control group received placebo. Various adverse outcomes were tested, including maternal bleeding, hypertension, complications of delivery, gestational age, Apgar scores, and neonatal abnormalities. The main outcome under study was the birth weight. There were no differences between mothers receiving zinc and controls. /Zinc/ Bingham, E.; Cohrssen, B.; Powell, C.H.; Patty's Toxicology Volumes 1-9 5th ed. John Wiley & Sons. New York, N.Y. (2001)., p. 2:268 Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ ... Oral ingestion of 12 g of elemental zinc (800 times the RDA) resulted only in pronounced lethargy. American Medical Association, Department of Drugs. Drug Evaluations. 6th ed. Chicago, Ill: American Medical Association, 1986., p. 859 Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ ... Metal fume fever results from inhalation of fumes of zinc oxide produced when zinc is heated to high temperatures, such as during welding, metal cutting, or smelting zinc alloys. Victims complain of nausea and vomiting, chills and fever, muscular aches and pains, and weakness. Gossel, T.A., J.D. Bricker. Principles of Clinical Toxicology. 3rd ed. New York, NY: Raven Press, Ltd., 1994., p. 202 Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ There is a reciprocal relationship between plasma levels of zinc and copper, to the degree that ... large doses of elemental zinc result in negative copper balance in patients with Wilson's disease. Klaassen, C.D. (ed). Casarett and Doull's Toxicology. The Basic Science of Poisons. 6th ed. New York, NY: McGraw-Hill, 2001., p. 847 Hazardous Substances Data Bank (HSDB) For more Human Toxicity Excerpts (Complete) data for ZINC, ELEMENTAL (12 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 15.1.17 Non-Human Toxicity Excerpts /ALTERNATIVE IN VITRO TESTS/ Hemolytic reactions, adsorption tests, and microscopic evidence provided information about the interactions between either zinc, zinc oxide, or zinc sulfide dust particles and human red blood cells. In vitro, zinc dust extensively hemolyzed red blood cells and absorbed the liberated hemoglobin. Metallic zinc had the greatest hemolytic effect and the largest hemoglobin binding capacity; it was followed by zinc oxide and zinc sulfide. DELBECK G, DELBECK M; RES EXP MED 160 (4): 255-60 (1973) Hazardous Substances Data Bank (HSDB) /OTHER TOXICITY INFORMATION/ ... Intratracheal instillation or intrapleural injection of zinc powder ... resulted in an increased incidence of testicular seminomas, as well as localized lung reticulosarcomas. Chang, L.W. (ed.). Toxicology of Metals. Boca Raton, FL: Lewis Publishers, 1996, p. 272 Hazardous Substances Data Bank (HSDB) /OTHER TOXICITY INFORMATION/ Poisoning ... /has been observed/ ... in cattle and horses ... from food stuffs containing particles of metal, and in pigs and hens from use of zinc plated funnels ... It is evident that young animals are much more susceptible to poisoning by zinc than mature animals. Clarke, M. L., D. G. Harvey and D. J. Humphreys. Veterinary Toxicology. 2nd ed. London: Bailliere Tindall, 1981., p. 76 Hazardous Substances Data Bank (HSDB) /OTHER TOXICITY INFORMATION/ /Researchers/ reported that 20 of 25 ferrets (Mustela putorius furo) being used in an experiment died of renal failure after eating raw meat that was accidentally contaminated with zinc from the wire cages. Zinc poisoning was diagnosed after autopsy and laboratory investigation. WHO; Environ Health Criteria 221: Zinc p.179 (2001) Hazardous Substances Data Bank (HSDB) 15.1.18 Non-Human Toxicity Values LD50 Rats oral 630 mg/kg Ullmann's Encyclopedia of Industrial Chemistry. 6th ed.Vol 1: Federal Republic of Germany: Wiley-VCH Verlag GmbH & Co. 2003 to Present, p. V. 26 640 (2003) Hazardous Substances Data Bank (HSDB) 15.1.19 Protein Binding Approximately 60-70% of the zinc in circulation is bound to albumin. Any condition that alters serum albumin concentration may have a secondary effect on serum zinc levels. DrugBank 15.2 Ecological Information 15.2.1 EPA Ecotoxicity Pesticide Ecotoxicity Data from EPA EPA Pesticide Ecotoxicity Database 15.2.2 Ecotoxicity Excerpts /BIRDS and MAMMALS/ /Researchers/ dosed mallard ducks (A. platyrhynchos) orally with eight No. 6 zinc shot and observed them for 30 days: three of the 15 birds died within the observation period. The average weight loss among surviving birds was 22%, significantly more than in control birds. Only three of the mallards dosed retained the zinc shot until the end of the study. Signs of intoxication in order of increasing severity were stumbling, an inability to run, complete loss of muscular control of the legs, loss of swimming ability and spasmodic wing movements. WHO; Environ Health Criteria 221: Zinc p.179 (2001) Hazardous Substances Data Bank (HSDB) /ACCIDENTAL POISONINGS/ /Researchers/ reported that 20 of 25 ferrets (Mustela putorius furo) being used in an experiment died of renal failure after eating raw meat that was accidentally contaminated with zinc from the wire cages. Zinc poisoning was diagnosed after autopsy and laboratory investigation. WHO; Environ Health Criteria 221: Zinc p.179 (2001) Hazardous Substances Data Bank (HSDB) /ACCIDENTAL POISONINGS/ Poisoning ... /has been observed/ in ferrets and mink from chewing corroded cages ... It is evident that young animals are much more susceptible to poisoning by zinc than mature animals. Clarke, M. L., D. G. Harvey and D. J. Humphreys. Veterinary Toxicology. 2nd ed. London: Bailliere Tindall, 1981., p. 76 Hazardous Substances Data Bank (HSDB) /ACCIDENTAL POISONINGS/ ... Horses, sheep, and cattle have been intoxicated by grazing on forage in the vicinity of zinc smelters. Horses are the most zinc-sensitive farm animals and react with lameness, osteochondrosis (possibly caused by an abnormal collagen metabolism due to an inhibition of lysyl oxidase followed by a zinc provoked copper deficiency), and lymphoid hyperplasia of the spleen and lymph nodes. European Chemicals Bureau; IUCLID Dataset, Zinc (7440-66-6) (2000 CD-ROM edition). Available from, as of July 11, 2006: Hazardous Substances Data Bank (HSDB) 15.2.3 US EPA Regional Screening Levels for Chemical Contaminants Resident Soil (mg/kg) 2.30e+04 Industrial Soil (mg/kg) 3.50e+05 Tapwater (ug/L) 6.00e+03 MCL (ug/L) 1.00e+04 Risk-based SSL (mg/kg) 3.70e+02 Chronic Oral Reference Dose (mg/kg-day) 3.00e-01 Volatile Volatile Mutagen Mutagen Fraction of Contaminant Absorbed in Gastrointestinal Tract 1 EPA Regional Screening Levels for Chemical Contaminants at Superfund Sites 15.2.4 US EPA Regional Removal Management Levels for Chemical Contaminants Resident Soil (mg/kg) 7.00e+04 Industrial Soil (mg/kg) 1.10e+06 Tapwater (ug/L) 1.80e+04 MCL (ug/L) 1.00e+04 Chronic Oral Reference Dose (mg/kg-day) 3.00e-01 Volatile Volatile Mutagen Mutagen Fraction of Contaminant Absorbed in Gastrointestinal Tract 1 EPA Regional Screening Levels for Chemical Contaminants at Superfund Sites 15.2.5 ICSC Environmental Data The substance is very toxic to aquatic organisms. The substance may cause long-term effects in the aquatic environment. ILO-WHO International Chemical Safety Cards (ICSCs) 15.2.6 Natural Pollution Sources Zinc is not found free in nature, but rather occurs in the +2 oxidation state often as zinc sulfide or zinc oxide. (SRC) Hazardous Substances Data Bank (HSDB) 15.2.7 Atmospheric Concentrations SOURCE DOMINATED: Geometric mean zinc concns in air for various jobs, melting, die-casting, and fettling, at Swedish aluminum die-casting foundries were 0.0023, 0.0014, and 0.024 mg/cu m, respectively(1). A geometric zinc concn in air at Swedish aluminum remelting plants was 0.0043 mg/cu m(1). A geometric zinc concn in air at aluminum remelting plants for two job titles, melting and scrap sorting were 0.0067 and 0.00095 mg/cu m, respectively(1). PMID:11202030 (1) Westberg HB et al; Appl Occup Environ Hyg 16: 66-77 (2001) Hazardous Substances Data Bank (HSDB) 15.2.8 Probable Routes of Human Exposure NIOSH (NOES Survey 1981-1983) has statistically estimated that 412,702 workers (48,684 of these are female) are potentially exposed to zinc in the US(1). (1) NIOSH; International Safety Cards. Zinc. CAS No. 7440-66-6. Available at as of June 21, 2006. Hazardous Substances Data Bank (HSDB) 16 Associated Disorders and Diseases Comparative Toxicogenomics Database (CTD) Associated Occupational Diseases with Exposure to the Compound Metal fume fever [Category: Acute Poisoning] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 17 Literature 17.1 Consolidated References PubChem 17.2 NLM Curated PubMed Citations Medical Subject Headings (MeSH) 17.3 Thieme References Thieme Chemistry Thieme Chemistry 17.4 Wiley References Wiley 17.5 Chemical Co-Occurrences in Literature PubChem 17.6 Chemical-Gene Co-Occurrences in Literature PubChem 17.7 Chemical-Disease Co-Occurrences in Literature PubChem 17.8 Chemical-Organism Co-Occurrences in Literature PubChem 18 Patents 18.1 Depositor-Supplied Patent Identifiers PubChem Link to all deposited patent identifiers PubChem 18.2 Chemical Co-Occurrences in Patents PubChem 18.3 Chemical-Disease Co-Occurrences in Patents PubChem 18.4 Chemical-Gene Co-Occurrences in Patents PubChem 18.5 Chemical-Organism Co-Occurrences in Patents PubChem 19 Interactions and Pathways 19.1 Protein Bound 3D Structures View 1 protein in NCBI Structure PubChem 19.2 Chemical-Target Interactions Comparative Toxicogenomics Database (CTD) DrugBank 19.3 Drug-Drug Interactions DrugBank 19.4 Drug-Food Interactions Avoid milk and dairy products. Separate the use of zinc from these products by at least 2 hours before administration. Separate for 2 hours after administration if these products also contain phosphorus. Do not take with bran and high fiber foods. For optimal absorption, take zinc at least 2 hours before or after eating high-fiber foods. Take on an empty stomach. Take at least 1 hour before and 2 hours after eating for optimal absorption. Zinc can be taken with food to reduce gastrointestinal upset. DrugBank 19.5 Pathways PubChem 20 Biological Test Results 20.1 BioAssay Results PubChem 21 Taxonomy E. coli Metabolome Database (ECMDB); LOTUS - the natural products occurrence database 22 Classification 22.1 MeSH Tree Medical Subject Headings (MeSH) 22.2 NCI Thesaurus Tree NCI Thesaurus (NCIt) 22.3 ChEBI Ontology ChEBI 22.4 ChemIDplus ChemIDplus 22.5 CAMEO Chemicals CAMEO Chemicals 22.6 ChEMBL Target Tree ChEMBL 22.7 Household Products Database Tree Consumer Product Information Database (CPID) 22.8 UN GHS Classification GHS Classification (UNECE) 22.9 EPA CPDat Classification EPA Chemical and Products Database (CPDat) 22.10 NORMAN Suspect List Exchange Classification NORMAN Suspect List Exchange 22.11 EPA DSSTox Classification EPA DSSTox 22.12 Consumer Product Information Database Classification Consumer Product Information Database (CPID) 22.13 EPA TSCA and CDR Classification EPA Chemicals under the TSCA 22.14 LOTUS Tree LOTUS - the natural products occurrence database 22.15 FDA Drug Type and Pharmacologic Classification National Drug Code (NDC) Directory 22.16 EPA Substance Registry Services Tree EPA Substance Registry Services 22.17 MolGenie Organic Chemistry Ontology MolGenie 22.18 Chemicals in PubChem from Regulatory Sources PubChem 22.19 ATCvet Classification WHO ATCvet - Classification of Veterinary Medicines 23 Information Sources Filter by Source Agency for Toxic Substances and Disease Registry (ATSDR)LICENSE The information provided using CDC Web site is only intended to be general summary information to the public. It is not intended to take the place of either the written law or regulations. Zinc EPA Integrated Risk Information System (IRIS)LICENSE Zinc and Compounds Joint FAO/WHO Expert Committee on Food Additives (JECFA)LICENSE Permission from WHO is not required for the use of WHO materials issued under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Intergovernmental Organization (CC BY-NC-SA 3.0 IGO) licence. ZINC Haz-Map, Information on Hazardous Chemicals and Occupational DiseasesLICENSE Copyright (c) 2022 Haz-Map(R). All rights reserved. Unless otherwise indicated, all materials from Haz-Map are copyrighted by Haz-Map(R). No part of these materials, either text or image may be used for any purpose other than for personal use. Therefore, reproduction, modification, storage in a retrieval system or retransmission, in any form or by any means, electronic, mechanical or otherwise, for reasons other than personal use, is strictly prohibited without prior written permission. 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These documents may be freely distributed and used for non-commercial, scientific and educational purposes. Zinc EPA Chemicals under the TSCALICENSE Zinc EPA TSCA Classification EPA DSSToxLICENSE Zinc Zinc, ion (Zn1-) Zinc, ion (Zn1+) CompTox Chemicals Dashboard Chemical Lists European Chemicals Agency (ECHA)LICENSE Use of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for non-commercial purposes provided that ECHA is acknowledged as the source: "Source: European Chemicals Agency, Such acknowledgement must be included in each copy of the material. ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. Zinc Zinc (EC: 231-175-3) FDA Global Substance Registration System (GSRS)LICENSE Unless otherwise noted, the contents of the FDA website (www.fda.gov), both text and graphics, are not copyrighted. They are in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from FDA. Credit to the U.S. Food and Drug Administration as the source is appreciated but not required. ZINC Hazardous Substances Data Bank (HSDB)LICENSE ZINC, ELEMENTAL ILO-WHO International Chemical Safety Cards (ICSCs)LICENSE Creative Commons CC BY 4.0 ZINC POWDER (pyrophoric) New Zealand Environmental Protection Authority (EPA)LICENSE This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International licence. Zinc NJDOH RTK Hazardous Substance ListLICENSE zinc Risk Assessment Information System (RAIS)LICENSE This work has been sponsored by the U.S. Department of Energy (DOE), Office of Environmental Management, Oak Ridge Operations (ORO) Office through a joint collaboration between United Cleanup Oak Ridge LLC (UCOR), Oak Ridge National Laboratory (ORNL), and The University of Tennessee, Ecology and Evolutionary Biology, The Institute for Environmental Modeling (TIEM). All rights reserved. Zinc and Compounds California Safe Cosmetics Program (CSCP) Product DatabaseLICENSE Zinc Zinc Consumer Product Information Database (CPID)LICENSE Copyright (c) 2024 DeLima Associates. All rights reserved. Unless otherwise indicated, all materials from CPID are copyrighted by DeLima Associates. No part of these materials, either text or image may be used for any purpose other than for personal use. Therefore, reproduction, modification, storage in a retrieval system or retransmission, in any form or by any means, electronic, mechanical or otherwise, for reasons other than personal use, is strictly prohibited without prior written permission. Zinc dust Household Products Classification Consumer Products Category Classification Emergency Response Guidebook (ERG)LICENSE Zinc dust ChEMBLLICENSE Access to the web interface of ChEMBL is made under the EBI's Terms of Use ( The ChEMBL data is made available on a Creative Commons Attribution-Share Alike 3.0 Unported License ( ChEMBL Protein Target Tree ClinicalTrials.govLICENSE The ClinicalTrials.gov data carry an international copyright outside the United States and its Territories or Possessions. Some ClinicalTrials.gov data may be subject to the copyright of third parties; you should consult these entities for any additional terms of use. 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This work is available under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported license (CC BY-NC-ND 3.0). zinc EPA Chemical and Products Database (CPDat)LICENSE EPA CPDat Classification NORMAN Suspect List ExchangeLICENSE Data: CC-BY 4.0; Code (hosted by ECI, LCSB): Artistic-2.0 NORMAN Suspect List Exchange Classification EPA Pesticide Ecotoxicity DatabaseLICENSE EPA Regional Screening Levels for Chemical Contaminants at Superfund SitesLICENSE Zinc and Compounds Zinc and Compounds EU Clinical Trials Register Regulation (EC) No 1272/2008 of the European Parliament and of the CouncilLICENSE The copyright for the editorial content of this source, the summaries of EU legislation and the consolidated texts, which is owned by the EU, is licensed under the Creative Commons Attribution 4.0 International licence. zinc powder - zinc dust (pyrophoric) Hazardous Chemical Information System (HCIS), Safe Work Australiazinc powder - zinc dust (pyrophoric) zinc powder - zinc dust (stabilised) NITE-CMCzinc powder - zinc dust (stabilized) - FY2008 (New/original classication) FDA Packaging & Food Contact Substances (FCS)LICENSE ZINC FooDBLICENSE FooDB is offered to the public as a freely available resource. 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ZINC NIOSH Manual of Analytical MethodsLICENSE The information provided using CDC Web site is only intended to be general summary information to the public. It is not intended to take the place of either the written law or regulations. 7440-66-6 7440-66-6 7440-66-6 7440-66-6 7440-66-6 7440-66-6 7440-66-6 NLM RxNorm TerminologyLICENSE The RxNorm Terminology is created by the National Library of Medicine (NLM) and is in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from NLM. Credit to the U.S. National Library of Medicine as the source is appreciated but not required. 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