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Example 3.5.1. Someone gives us a pencil and two unmarked sticks of lengths \( {52}\mathrm{\\;{cm}} \) and \( {20}\mathrm{\\;{cm}} \) respectively (see Figure 3.5.1). We are told to make measuring sticks by using the pencil to make markings on the sticks. Question: what is the smallest length that we can accurately mea... | Here's one way to look at the situation. Imagine for a moment that we lay the \( {20}\mathrm{\\;{cm}} \) measuring stick next to the \( {52}\mathrm{\\;{cm}} \) stick such that the ends line up. At that point we could make a \( {20}\mathrm{\\;{cm}} \) mark on the \( {52}\mathrm{\\;{cm}} \) stick (see Figure 3.5.2). At t... | Yes |
Let's use algebraic language to express the two-sticks algorithm applied to 52 and 20. Let's start by setting this up as a division problem with a remainder (recall Proposition 3.2.3), since this is effectively what is being done in the stick example above. | \[ 52 = 20 \cdot q_1 + r_1 \] where \( q_1 \) and \( r_1 \) are integers (we put the subscript ’ 1 ’ on the variables \( q_1 \) and \( r_1 \) because we’re going to repeat the process). By division with remainder we find \( q_1 = 2 \) and \( r_1 = 12 \). Now we repeat the process, but this time dividing the remainder 1... | Yes |
Proposition 3.5.5. The Euclidean algorithm applied to two integers will give the gcd of those two integers. | Proof. This proof is broken up into two parts, (A) and (B). Part (A) shows that the algorithm always produces a divisor of the two given integers. Part (B) shows that the produced divisor is indeed the gcd.\n\n(A) Given integers \( a \) and \( b \) and \( a > b \) if we were to plug them into the Euclidean Algorithm we... | Yes |
We'll give another example, giving just the computations and no other words. We find integer solutions to \( {1053x} + {863y} = {245} \) as follows: | \[ {1053} = {863} + {190} \Rightarrow {190} = \left( {1, - 1}\right) \] \[ {863} = 4 \cdot {190} + {103} \Rightarrow {103} = \left( {0,1}\right) - 4 \cdot \left( {1, - 1}\right) = \left( {-4,5}\right) \] \[ {190} = {103} + {87} \Rightarrow {87} = \left( {1, - 1}\right) - \left( {-4,5}\right) = \left( {5, - 6}\right) \]... | Yes |
Proposition 3.5.16. Given the Diophantine equation \( {an} + {bm} = c \), where \( a, b, c \) are integers. Then the equation has integer solutions for \( n \) and \( m \) if and only if \( c \) is a multiple of the gcd of \( a \) and \( b \) . | Proof. Since this is an \ | No |
Example 3.5.24. The Cayley table for \( {\mathbb{Z}}_{3} \smallsetminus \{ 0\} \) is:\n\n<table><thead><tr><th>\( \odot \)</th><th>1</th><th>2</th></tr></thead><tr><td>1</td><td>1</td><td>2</td></tr><tr><td>2</td><td>2</td><td>1</td></tr></table> | Notice that each column has 1, meaning that each element has an inverse. It is also closed, associative and has an identity. Thus \( {\mathbb{Z}}_{3} \smallsetminus \{ 0\} \) is a group under \( \odot \) . | Yes |
The Cayley table for \( {\mathbb{Z}}_{4} \smallsetminus \{ 0\} \) is\n\n<table><thead><tr><th>\( \odot \)</th><th>1</th><th>2</th><th>3</th></tr></thead><tr><td>1</td><td>1</td><td>2</td><td>3</td></tr><tr><td>2</td><td>2</td><td>0</td><td>2</td></tr><tr><td>3</td><td>3</td><td>2</td><td>1</td></tr></table>\n\nNotice t... | The fact that 2 has no inverse is due to 2 being a divisor of 4. This makes all integer multiples of 2 to cycle between the values 0 and 2 (mod 4). | Yes |
Proposition 3.5.28. If \( p \) is a prime number, then all elements in \( {\mathbb{Z}}_{p} \smallsetminus \{ 0\} \) have an inverse under multiplication \( {\;\operatorname{mod}\;p} \) . | Proof. Let \( a, p \) be known integers where \( a < p \) and \( p \) is prime. There exists an inverse to \( a \) under multiplication \( \left( {\;\operatorname{mod}\;p}\right) \) when there is a solution \( k \) to the equation \( {ak} = 1\left( {\;\operatorname{mod}\;p}\right) \) where \( k \) is an integer. By Pro... | Yes |
Let’s find the digit \( {d}_{6} \) for the number \( n = {1928307465} \) (we may note in this case \( {d}_{6} = 8 \) ). | First, we can remove the digits above \( {d}_{6} \) digit taking \( n \) modulo \( {10}^{7} \) :\n\n\[ \n{\;\operatorname{mod}\;\left( {n,{10}^{7}}\right) } = {8307465}.\n\]\n\nOn the other hand, we can obtain all digits below \( {d}_{6} \) by taking \( n \) modulo \( {10}^{6} \) :\n\n\[ \n{\;\operatorname{mod}\;\left(... | Yes |
Find \( {d}_{-3} \) of the decimal number \( x = {0.17428} \) | Since we’re looking for \( {d}_{-3} \), Let’s multiply \( x \) by \( {10}^{3} \). \[ {0.17428} \cdot {10}^{3} = {174.28} \] Then take the floor: \[ \lfloor {174.28}\rfloor = {174} \] Finally, take the modulus base 10 (which is the 1's place of the number, as we've seen before): \[ {\;\operatorname{mod}\;\left( {{174},{... | Yes |
Proposition 4.1.13. Let \( n > 2 \) be an integer such that \( \gcd \left( {n,{10}}\right) = 1 \) . Then there exist a positive integer \( m \) such that \( {\;\operatorname{mod}\;\left( {{10}^{m}, n}\right) } = 1 \) . | Proof. Consider the infinite sequence: \( {\;\operatorname{mod}\;\left( {{10}, n}\right) },{\;\operatorname{mod}\;\left( {{10}^{2}, n}\right) },{\;\operatorname{mod}\;\left( {{10}^{3}, n}\right) },\ldots \) . All of these numbers are between 1 and \( n - 1 \) . Since the sequence is infinite and only can take at most \... | Yes |
Proposition 4.1.15. Let \( n > 1 \) be a positive integer with \( \gcd \left( {{10}, n}\right) = 1 \) , and let \( m \) be the multiplicative order of \( n\left( {\;\operatorname{mod}\;{10}}\right) \) . Then the decimal expansion of \( \frac{1}{n} \) repeats every \( m \) digits. | Proof.\n\nGiven that \( m \) is the multiplicative order of \( n{\;\operatorname{mod}\;{10}} \), from Definition 4.1.14, we get \( {\;\operatorname{mod}\;\left( {\left( {{10}^{m} - 1}\right), n}\right) } = 0 \) . In other words, \( {10}^{m} - 1 \) is divisible by \( n \), so that \( \frac{{10}^{m} - 1}{n} \) is an inte... | Yes |
Is 6472 divisible by 11 ? | In the following argument we use the fact that \( {10} \equiv - 1\left( {\;\operatorname{mod}\;{11}}\right) \), which means that we can replace 10\n\nwith -1 whenever we are taking mod's base 10 .\n\n\[ \n{\;\operatorname{mod}\;\left( {{6472},{11}}\right) } = {\;\operatorname{mod}\;\left( {6 \cdot {10}^{3} + 4 \cdot {1... | Yes |
Proposition 4.1.22. A number is divisible by 11 if and only if the alternating sums of the digits is divisible by 11. (Note: alternating sums is where the signs of the number alternate when summing.) | Proof. Given an integer with digits \( {d}_{0}\ldots {d}_{n} \) where the number is writeen as \( {d}_{n}{d}_{n - 1}\ldots {d}_{1}{d}_{0} \) we can write\n\n\[ n = {d}_{m} \cdot {10}^{m} + {d}_{m - 1} \cdot {10}^{m - 1} + \cdots + {d}_{0} \cdot {10}^{0} \]\n\n it follows that:\n\n\[ {\;\operatorname{mod}\;\left( {n,{11... | Yes |
Find 137 in base 6. | \[ {a}_{0} = {137};{d}_{0} = {\;\operatorname{mod}\;\left( {{137},6}\right) } = 5 \] \[ {a}_{1} = \frac{{137} - 5}{6} = {22};{d}_{1} = {\;\operatorname{mod}\;\left( {{22},6}\right) } = 4 \] \[ {a}_{2} = \frac{{22} - 4}{6} = 3;{d}_{2} = {\;\operatorname{mod}\;\left( {3,6}\right) } = 3 \] \[ {a}_{3} = \frac{3 - 3}{6} = 0... | Yes |
Find 121 in base 3. | Once again using the recursive method\n\n\[ \n{a}_{0} = {121};\;{d}_{0} = {\;\operatorname{mod}\;\left( {{121},3}\right) } = 1 \]\n\n\[ \n{a}_{1} = \frac{{121} - 1}{3} = {40};\;{d}_{1} = {\;\operatorname{mod}\;\left( {{40},3}\right) } = 1 \]\n\n\[ \n{a}_{2} = \frac{{40} - 1}{3} = {13};\;{d}_{2} = {\;\operatorname{mod}\... | Yes |
Find the 5th digit of 65432 in base 3. (This is the coefficient of \( {3}^{4} \) in the base 3 representation). | \[ {d}_{4} = \frac{{\;\operatorname{mod}\;\left( {{65432},{3}^{5}}\right) } - {\;\operatorname{mod}\;\left( {{65432},{3}^{4}}\right) }}{{3}^{4}} \] | Yes |
Example 4.2.5. Find 31 in base 2. | \[ {a}_{0} = {31};{d}_{0} = {\;\operatorname{mod}\;\left( {{31},2}\right) } = 1;\;{a}_{1} = \frac{{31} - 1}{2} = {15};{b}_{1} = {\;\operatorname{mod}\;\left( {{15},2}\right) } = 1 \] \[ {a}_{2} = \frac{{15} - 1}{2} = {15};{b}_{2} = {\;\operatorname{mod}\;\left( {{14},2}\right) } = 0;\;{a}_{3} = \frac{{14} - 0}{2} = {15... | Yes |
Example 4.2.8. Convert 121 in base 3 to a number in base 10. | \[ {\left( {121}\right) }_{3} = 1 \cdot {3}^{2} + 2 \cdot {3}^{1} + 1 \cdot {3}^{0} = 1 \cdot 9 + 2 \cdot 3 + 1 \cdot 1 = {\left( {16}\right) }_{10} \] | Yes |
Example 4.2.9. Convert 4752 in base 8 to a number in base 10 | \[ {\left( {4752}\right) }_{8} = 4 \cdot {8}^{3} + 7 \cdot {8}^{2} + 5 \cdot {8}^{1} + 2 \cdot {8}^{0} = 4 \cdot {512} + 7 \cdot {64} + 5 \cdot 8 + 2 \cdot 1 = {\left( {2538}\right) }_{10} \] | Yes |
Example 4.2.12. Convert 1011 in base 2 to a number in base 10. | \[ {\left( {1011}\right) }_{2} = 1 \cdot {2}^{3} + 0 \cdot {2}^{2} + 1 \cdot {2}^{1} + 1 \cdot {2}^{0} = 1 \cdot 8 + 0 \cdot 4 + 1 \cdot 2 + 1 \cdot 1 = {\left( {11}\right) }_{10} \] | Yes |
(a) \( \mathop{\bigcup }\limits_{{i = 1}}^{n}\{ i\} \) | (a) \( \mathop{\bigcup }\limits_{{i = 1}}^{n}\{ i\} = \{ 1\} \bigcup \{ 2\} \bigcup \{ 3\} \bigcup \ldots \bigcup \{ n\} \n\n\( = \{ 1,\ldots, n\} \; \) [list of elements]\n\n\( = \) all integers from 1 to \( n \) . [property] | Yes |
Example 5.1.19. Let \( \mathbb{N} \) be the universal set, and suppose that\n\n\[ A = \{ x \in \mathbb{N} : x\text{ is divisible by }2\} \]\n\n\[ B = \{ x \in \mathbb{N} : x\text{ is divisible by }3\} \]\n\n\[ C = \{ x \in \mathbb{N} : x\text{ is divisible by }6\} \]\n\n\[ D = \{ \text{the odd natural numbers}\} \]\n\n... | (a)\n\n\[ A \cap B = \{ x \in \mathbb{N} : x\text{ is divisible by }2\text{ and }x\text{ is divisible by }3\} \]\n\n\[ = \{ x \in \mathbb{N} : x\text{ is divisible by }6\} \]\n\n\[ = C \] | Yes |
Proposition 5.2.1. Given any sets \( A, B \), It is always true that\n\n\[ A \cap B \subset A \\text{ and } A \subset A \cup B. \]\n | Proof. The style of proof we'll use here is often described as element by element, because the proofs make use of the definitions of \( A \cap B \) and \( A \cup B \) in terms of their elements.\n\nFirst, suppose that \( x \) is an element of \( A \cap B \) . we then have:\n\n\[ x \in A \cap B \] [supposition]\n\n\[ \\... | No |
Proposition 5.2.3. Let \( A, B \), and \( C \) be subsets of a universal set \( U \) . Then\n\n1. \( A \cup {A}^{\prime } = U \) and \( A \cap {A}^{\prime } = \varnothing \) | Proof. We'll prove parts (1), (2), (5), and (7), and leave the rest to you!\n\n(1) From our definitions we have:\n\n\[ A \cup {A}^{\prime } = \left\{ {x : x \in A\text{ or }x \in {A}^{\prime }}\right\} \]\n[def. of \( \cup \) ]\n\n\[ = \{ x : x \in A\text{ or }x \notin A\} \]\n[def. of complement]\n\nBut every \( x \in... | No |
Example 5.2.8. Prove that\n\n\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \varnothing . \] | Proof. To see that this is true, observe that\n\n\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \left( {A \cap {B}^{\prime }}\right) \cap \left( {B \cap {A}^{\prime }}\right) \] [definition of \( \smallsetminus \) ]\n\n\[ = A \cap {A}^{\prime } \cap B \cap {B}^{\prime }\; \] [by Proposi... | Yes |
Proposition 6.1.8. Given any sets \( A \) and \( B \), then:\n\n\[ \left| {A \times B}\right| = \left| A\right| \cdot \left| B\right| \] | Proof. We can prove this formula by some creative arranging. Suppose the sets \( A \) and \( B \) have \( m \) and \( n \) elements, respectively. We may list these elements as follows:\n\n\[ A = \left\{ {{a}_{1},{a}_{2},{a}_{3},\ldots ,{a}_{m}}\right\} \text{ and }B = \left\{ {{b}_{1},{b}_{2},{b}_{3},\ldots ,{b}_{n}}\... | Yes |
For the function \( f\left( x\right) = {x}^{3} \), the input \( x \) can be any real number. Plugging a value for \( x \) into the formula yields an output value, which is also a real number. For example, using \( x = 2 \) as the input yields the output value \( f\left( 2\right) = {2}^{3} = 8 \) . | In Example 6.2.1, any real number can be used as the input \( x \), so the domain is \( \mathbb{R} \), the set of all real numbers. Similarly, any output is a real number, so the codomain can also be taken as \( \mathbb{R} \) . | Yes |
Example 6.2.3. \( g\left( x\right) = 1/x \) is not a function from \( \mathbb{R} \) to \( \mathbb{R} \) . This is because 0 is an element of \( \mathbb{R} \), but the formula does not define a value for \( g\left( 0\right) \) . Thus, 0 cannot be in the domain of \( g \) . | To correct this problem, one could say that \( g \) is a function from the set \( \{ x \in \mathbb{R} \mid x \neq 0\} \) of nonzero real numbers, to \( \mathbb{R} \) . \( \blacklozenge \) | Yes |
Example 6.2.13. Suppose that the function \( f \) is defined by \( f\left( x\right) = {x}^{2} \), on the domain \( \{ 0,1,2,4\} \) . Then | 1. to represent \( f \) as a set of ordered pairs, each element of the domain must appear exactly once as a first coordinate, with the corresponding output given in the second coordinate. Since there are four elements in the domain, there will be four ordered pairs: \( \{ \left( {0,0}\right) ,\left( {1,1}\right) ,\left... | Yes |
Suppose Inspector Gadget knows two facts:\n\n1. Alice is the thief's wife, and\n\n2. Alice is Bob's wife.\n\nThen the inspector can arrest Bob for theft, because a person cannot (legally) be the wife of more than one husband. | According to U.S. law as of 2017. | No |
Is Temp a one-to-one function? | Not at all: it's very likely that at any given time, at least two points on the equator have exactly the same temperature (to arbitrary precision). Another way to say this is that at any given time, there exists a temperature \( b \) for which we can find two points on earth \( x \) and \( y \) such that \( \operatorna... | Yes |
(a) \( f : \mathbb{R} \rightarrow \mathbb{R} \), defined by \( f\left( x\right) = x + 1 \) . | Let's go back to the definition of one-to-one. Suppose we know that \( f\left( x\right) = f\left( y\right) \), where \( x, y \) are real numbers. Can we conclude that \( x = y \) ? If so, then that means that \( f \) is one-to-one.\n\nSo let’s follow through on this. \( f\left( x\right) = f\left( y\right) \) means that... | Yes |
Let \( f : \mathbb{N} \rightarrow \mathbb{N} \) be defined by \( f\left( n\right) = {\left( n - 2\right) }^{2} + 1 \) . Is \( f \) one-to-one? | First let’s try to prove that \( f \) is one-to-one. Start with arbitrary elements \( m, n \in \mathbb{N} \), and suppose that \( f\left( m\right) = f\left( n\right) \) . By the definition of \( f \), this means that \( {\left( m - 2\right) }^{2} + 1 = {\left( n - 2\right) }^{2} + 1 \), or \( {\left( m - 2\right) }^{2}... | Yes |
Example 6.3.16. We know from calculus that the function \( {e}^{x} : \mathbb{R} \rightarrow \mathbb{R} \) is a strictly increasing function since its derivative is always positive. In mathematical terms, we can say\n\n\[ x > y\\text{ implies }{e}^{x} > {e}^{y}. \]\n\nWe can use this fact and Definition 6.3.15 to prove ... | Take any two real numbers \( {x}_{1} \) and \( {x}_{2} \) where \( {x}_{1} \\neq {x}_{2} \) . If \( {x}_{1} > {x}_{2} \), then by the above equation it follows that \( {e}^{{x}_{1}} > {e}^{{x}_{2}} \) . On the other hand, if \( {x}_{1} < {x}_{2} \) , then by the above equation it follows that \( {e}^{{x}_{1}} < {e}^{{x... | Yes |
Define \( g : \mathbb{R} \rightarrow \mathbb{R} \) by \( g\left( x\right) = {5x} - 2 \) . Determine whether \( g \) is onto. | Given \( y \in \mathbb{R} \), let \( x = \left( {y + 2}\right) /5 \) . Since the reals are closed under addition and non-zero division, it follows that \( x \in \mathbb{R} \) . Then\n\n\[ g\left( x\right) = {5x} - 2 = 5\left( \frac{y + 2}{5}\right) - 2 = \left( {y + 2}\right) - 2 = y.\]\n\nTherefore \( g \) is onto. | Yes |
Define \( h : \left\lbrack {0,2}\right\rbrack \rightarrow \left\lbrack {-7, - 1}\right\rbrack \) by \( h\left( x\right) = - {3x} - 1 \) . Determine whether \( h \) is onto. | Proof. Given \( y \in \mathbb{R} \), let \( x = \frac{y + 1}{-3} \) . By basic algebra, \( - 7 \leq y \leq - 1 \Rightarrow 0 \leq \) \( \frac{y + 1}{-3} \leq 2 \), so \( x \) is in the domain of \( h \) . Also,\n\n\[ h\left( x\right) = - 3\left( \frac{y + 1}{-3}\right) - 1 = y. \]\n\nTherefore \( h \) is onto. | Yes |
Example 6.4.16. Define \( f : \mathbb{C} \rightarrow \mathbb{C} \) by \( f\left( z\right) = {z}^{2} \) . Determine whether \( f \) is onto. | Proof. Given \( z = r\operatorname{cis}\theta \in \mathbb{C} \), let \( w = \sqrt{r}\operatorname{cis}\left( {\theta /2}\right) \) . By the definition of polar form, \( w \in \mathbb{C} \) and we have\n\n\[ f\left( w\right) = {\left( \sqrt{r}\operatorname{cis}\left( \theta /2\right) \right) }^{2} = {\left( \sqrt{r}\rig... | Yes |
Consider a hypothetical country \( \mathrm{Z} \), in which\n\n- every person is married to at least one other person (no singles),\n\n- everyone is married to at most one other person (no polygamists or polyandrists), and\n\n- every marriage is between a man and a woman (no same-sex marriages).\n\nLet \( \mathsf{{Men}}... | - Two different men cannot have the same wife, so we know that wife is one-to-one.\n\n- Every woman is the wife of some man (because everyone is married), so wife is also onto.\n\nSimilarly, the function husband: Women \( \rightarrow \) Men is also a bijection. | Yes |
Example 6.5.6. Define \( f : \left\lbrack {1,3}\right\rbrack \rightarrow \left\lbrack {-2,8}\right\rbrack \) by \( f\left( x\right) = {5x} - 7 \) . Then \( f \) is a bijection. | Proof. It suffices to show that \( f \) is both one-to-one and onto:\n\n- (one-to-one) Given \( {x}_{1},{x}_{2} \in \mathbb{R} \), such that \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right) \), we have\n\n\[ 5{x}_{1} - 7 = 5{x}_{2} - 7 \]\n\nAdding 7 to both sides and dividing by 5 , we have\n\n\[ \frac{\left( {5{x}_... | Yes |
Example 6.6.4. Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) and \( g : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = {3x} \) and \( g\left( x\right) = {x}^{2} \) . Then \( g \circ f \) and \( f \circ g \) are functions from \( \mathbb{R} \) to \( \mathbb{R} \) . For all \( x \in \mathbb{R} \) , we... | \[ g \circ f\left( x\right) = g\left( {f\left( x\right) }\right) = g\left( {3x}\right) = {\left( 3x\right) }^{2} = 9{x}^{2} \] and \[ f \circ g\left( x\right) = f\left( {g\left( x\right) }\right) = f\left( {x}^{2}\right) = 3\left( {x}^{2}\right) = 3{x}^{2}. \] Notice that (in this example) \( f \circ g \neq g \circ f \... | Yes |
Example 6.6.7. Figure 6.6.1 provides an arrow diagram to illustrate the composition \( g \circ f \) . | - Starting from any point of \( A \), follow the arrow (for the function \( f \) that starts there to arrive at some point of \( B \) .\n- Then follow the arrow (for the function \( g \) ) that starts there to arrive at a point of \( C \) .\nFor example, the \( f \)-arrow from \( a \) leads to \( m \) and the \( g \)-a... | Yes |
Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \), where \( A \subset C \). Show that if\n\n\[ g \circ f\left( a\right) = a\text{, for every}a \in A\text{,} \]\n\nthen \( f \) is one-to-one. | Proof. Given that \( g \circ f\left( a\right) = a \), for every \( a \in A \), by the definition of composition, this means that, for any \( {a}_{1},{a}_{2} \in A \) we have\n\n\[ g\left( {f\left( {a}_{1}\right) }\right) = {a}_{1}\text{ and }g\left( {f\left( {a}_{2}\right) }\right) = {a}_{2}. \]\n\nNow suppose \( f\lef... | Yes |
Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) . Show that if \( f \) and \( g \) are onto, then \( g \circ f \) is onto. | Proof. Let \( c \) be an arbitrary element of \( C \) . Since \( g \) is onto, there exists a \( b \) in \( B \) such that \( g\left( b\right) = c \) . Since \( f \) is onto, there exists a \( a \) in \( A \) such that \( f\left( a\right) = b \) . It follows that \( g \circ f\left( a\right) = g\left( {f\left( a\right) ... | Yes |
Example 6.6.17. Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) . Show that if \( g \circ f \) is one-to-one, and the range of \( f \) is \( B \), then \( g \) is one-to-one. | Proof. Suppose \( {b}_{1} \) and \( {b}_{2} \) are distinct elements of \( B \) . Since the range of \( f \) is \( B \), it follows that there exist \( {a}_{1} \neq {a}_{2} \) such that \( f\left( {a}_{1}\right) = {b}_{1} \) and \( f\left( {a}_{2}\right) = {b}_{2} \) . Since \( g \circ f \) is one-to-one, it follows th... | Yes |
Example 6.6.20. Suppose \( f : A \rightarrow B \) and \( g : B \rightarrow C \) and \( g \) is not onto. Then \( g \circ f \) is not onto. | Proof. We will prove by contradiction. Suppose on the contrary that \( g \circ f \) is onto. Then given any \( c \in C \), there exists a \( a \in A \) such that \( g \circ f\left( a\right) = c \) . By the definition of composition, it follows that \( g\left( b\right) = c \) where \( {bf}\left( a\right) \) . Since \( c... | Yes |
In Example 6.5.6, we showed that \( f\left( x\right) = {5x} - 7 \) is a bijection. A quick look at the proof reveals that the formula\n\n\[ x = \frac{y + 7}{5} \]\n\nplays a key role. This formula is obtained by replacing \( f\left( x\right) \) in \( f\left( x\right) = {5x} - 7 \) with \( y \), and solving for \( x \) ... | In order to see \( x = \frac{y + 7}{5} \) as an \ | No |
Example 6.7.2. Let \( f : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) be defined by: \( f\left( x\right) = {x}^{2} \) . We may define \( g : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) by: \( g\left( y\right) = \sqrt{y} \) . Note that in this case the domains and ranges are restricted to positive real... | \[ y = {x}^{2} \Leftrightarrow x = \sqrt{y} \] In view of the definitions of \( f \) and \( g \), we may see that this is the same formula as in the previous example: \( y = f\left( x\right) \Leftrightarrow x = g\left( y\right) \). | Yes |
Proposition 6.7.4. Suppose that \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) are functions such that\n\n\[ \forall x \in X,\forall y \in Y,\left( {y = f\left( x\right) \Leftrightarrow x = g\left( y\right) }\right) . \]\n\nThen the following statements are also true:\n\n(a) \( g\left( {f\left( x\right) }\righ... | Proof. The proof of (a) runs as follows. Suppose that \( y = f\left( x\right) \Leftrightarrow x = g\left( y\right) \) for all \( x, y \) in the respective domains of \( f \) and \( g \) . Then for any \( x \in X \), we may define \( z \) as \( z = f\left( x\right) \) . By the \( \Leftrightarrow \) statement it follows ... | Yes |
Example 6.7.8. The husband of the wife of any married man is the man himself - in other words,\n\n\[ \n\\text{husband}\\left( {\\operatorname{wife}\\left( y\\right) }\\right) = y\\text{.}\n\]\n\nAlso, the wife of the husband of any married woman is the woman herself, so that\n\n\[ \n\\text{wife(husband(x))} = x\\text{.... | It follows that the wife function is an inverse of the husband function. In fact, it's pretty clear that husband is the only inverse of wife. | No |
Proposition 6.7.11. Suppose \( f : X \rightarrow Y \) . Then \( f \) has an inverse \( g : Y \rightarrow X \) if and only if \( f \) is a bijection. | Proof. (forward direction) Assume there is a function \( g : Y \rightarrow X \) that is an inverse of \( f \) . Then by the definition of an inverse function,\n\n(a) \( f\left( {g\left( y\right) }\right) = y \) for all \( y \in Y \), and\n\n(b) \( g\left( {f\left( x\right) }\right) = x \) for all \( x \in X \) .\n\nSup... | Yes |
Example 7.2.1. One of the first and most famous private key cryptosystems was the shift code used by Julius Caesar. We first represent the alphabet numerically by letting \( \mathrm{A} = 0,\mathrm{\;B} = 1,\ldots ,\mathrm{Y} = {24},\mathrm{Z} = {25} \) . This means for example that the word BAY would be represented num... | Suppose we receive the encoded message DOJHEUD. To decode this message, we first represent it numerically:\n\n\[ 3,{14},9,7,4,{20},3\text{.} \]\n\nNext we apply the decryption function to get\n\n\[ 0,{11},6,4,1,{17},0, \]\n\nwhich is the numerical representation of ALGEBRA. Notice here that there is nothing special abo... | Yes |
Example 7.2.5. Suppose we receive a message that we know was encrypted by using a shift transformation on single letters of the 26-letter alphabet. To find out exactly what the shift transformation was, we must compute \( b \) in the equation \( f\left( n\right) = n + b{\;\operatorname{mod}\;{26}} \) . We can do this u... | \[ f\left( n\right) = n \oplus {14} \]\n\nThe corresponding decoding function is\n\n\[ {f}^{-1}\left( m\right) = m \oplus {12}. \]\n\nIt is now easy to determine whether or not our guess is correct. | Yes |
Example 7.2.12. Let's consider the affine cryptosystem encoding function \( f\left( n\right) = \left( {a \odot n}\right) \oplus b \), where \( \odot \) and \( \oplus \) are multiplication and addition mod 26 respectively. For this cryptosystem to work we must choose an \( a \in {\mathbb{Z}}_{26} \) that is invertible. ... | The decryption function will be\n\n\[ \n{f}^{-1}\left( n\right) = \left( {{21} \odot n}\right) \ominus \left( {{21} \odot 3}\right) = \left( {{21} \odot n}\right) \oplus {15}. \n\] | Yes |
Example 7.3.1. Before exploring the theory behind the RSA cryptosystem or attempting to use large integers, we will use some small integers just to see that the system does indeed work. Suppose that we wish to send some message, which when digitized is 395 . Let \( p = {23} \) and \( q = {29} \) . Then\n\n\[ \n n = {pq... | First of all, we know that \( {DE} \equiv 1{\;\operatorname{mod}\;m} \) ; so there exists a \( k \) such that\n\n\[ \n{DE} = {km} + 1\text{.} \]\n\nThis means that\n\n\[ \n{y}^{D} = {\left( {x}^{E}\right) }^{D} = {x}^{DE} = {x}^{{km} + 1} = {\left( {x}^{m}\right) }^{k}x. \]\n\nAt this point we need Euler's theorem from... | No |
Consider:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{{10}}\left( {i + 2}\right) \] | In this case, the \( \sum \) symbol lets us know that this is a sum. The \( i = 1 \) serves two functions. It tells us that the index variable is \( i \), and that \( i \) has a starting value of 1 . The 10 is the final value, and the \( \left( {i + 2}\right) \) to the right of the \( \sum \) is the formula. The \( i \... | Yes |
Consider the product of the sums \( \mathop{\sum }\limits_{{i = 0}}^{2}{3}^{i} \) and \( \mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j} \) . | By the distributive law and additive commutivity we have:\n\n\[ \left( {\mathop{\sum }\limits_{{i = 0}}^{2}{3}^{i}}\right) \left( {\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j}}\right) = \mathop{\sum }\limits_{{i = 0}}^{2}\left( {{3}^{i}\left( {\mathop{\sum }\limits_{{j = 0}}^{2}{3}^{-j}}\right) }\right) \]\n\n\[ = \math... | Yes |
Let \( \mathbf{v} \) be the \( {10} \times 1 \) matrix given by\n\n\[ \n{v}_{j,1} = j, j = 1\ldots {10}.\n\]\n\n(Note that \( \mathbf{v} \) is essentially a column vector.) Let us compute \( C\mathbf{v} \), where the matrix \( C \) defined by\n\n\[ \n{c}_{i, j} \mathrel{\text{:=}} \frac{1}{2}\left( {-{\delta }_{i - 1, ... | The summation notation expression for the product is:\n\n\[ \n{\left\lbrack C\mathbf{v}\right\rbrack }_{ij} = \mathop{\sum }\limits_{{k = 1}}^{{10}}{c}_{i, k}{v}_{k, j}\n\]\n\nThe first thing to notice is that the second index \( j \) must be 1 since \( \mathbf{v} \) is a \( {10} \times 1 \) matrix. We may also substit... | Yes |
Example 8.5.10. This time we'll compute the entries of the matrix product \( {FV} \), where the entries \( {f}_{ij} \) of \( F \) and \( {v}_{ij} \) of \( V \) are given by:\n\n\[ \n{f}_{ij} \mathrel{\text{:=}} {\delta }_{i + 1, j} - {\delta }_{i, j};\;{v}_{i, j} \mathrel{\text{:=}} {2}^{i + j},\;1 \leq i, j \leq {20}.... | We may begin once again with the matrix product formula:\n\n\[ \n\left\lbrack {F{V}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{{20}}{f}_{ik}{v}_{kj}\;}\right. \text{[Matrix mulitplication formula ]} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{{20}}\left( {{\delta }_{i + 1, k} - {\delta }_{i, k}}\right) {2}^{k + j}\;\t... | Yes |
Proposition 8.5.15. If \( A \) and \( B \) are matrices such that the matrix product is defined, then\n\n\[{\left( AB\right) }^{\mathrm{T}} = {B}^{\mathrm{T}}{A}^{\mathrm{T}}\] | Proof. We’ll prove this by expressing the \( \left( {i, j}\right) \) entry of the left-hand side in summation notation, doing some algebraic hocus-pocus, and showing that it agrees with the \( \left( {i, j}\right) \) entry of the right side. First we make things clear by specifying that \( A \) has \( n \) columns and ... | Yes |
Proposition 8.6.16. Given any \( 3 \times 3 \) matrix \( A \), then\n\n\[ \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{ij}{a}_{k\ell } = \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{ji}{a}_{k\ell } \]\n\n(Observe the minute difference between the two sides: there’s an \( {a}_{ij} \... | Proof. Let us consider the case \( i = 1 \) :\n\n\[ \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{1j}{a}_{k\ell } = \mathop{\sum }\limits_{{j, k,\ell }}{\epsilon }_{{jk}\ell }{a}_{j1}{a}_{k\ell } \]\n\nand we’ll leave the cases \( i = 2,3 \) as exercises.\n\nOn both right and left sides there are term... | No |
Example 8.7.1. Evaluate \( \mathop{\sum }\limits_{{k = 1}}^{n}{2}^{k - 1}k \) . | In order to use the summation by parts formula, we need to define \( {a}_{k} \) and \( {B}_{k} \) so that the summand \( {2}^{k - 1}k \) is the product of \( {a}_{k} \) and \( {B}_{k} \) . Just as in integration by parts, we want to make our choice based on what makes the calculations easiest. Note that \( {B}_{k} \) i... | Yes |
Example 9.3.3. Express the following polynomials in summation notation:\n\n(a) \( {p}_{1}\left( x\right) = 1 + x + {x}^{2} + {x}^{3} \) | (a) \( {p}_{1}\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{3}{x}^{j} \) | Yes |
Proposition 9.5.3. Given that \( R \) satisfies conditions (I)-(V) listed above, then addition in \( R\left\lbrack x\right\rbrack \) is both commutative:\n\n\[ p\left( x\right) + q\left( x\right) = q\left( x\right) + p\left( x\right) \]\n\nand associative:\n\n\[ \left( {p\left( x\right) + q\left( x\right) }\right) + r\... | Proof. First, we show commutativity: Given two polynomials \( p\left( x\right) \) and \( q\left( x\right) \) where\n\n\[ p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i};\;q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{b}_{i}{x}^{i}, \]\n\nthen\n\n\[ p\left( x\right) + q\left( x\right) = \ma... | Yes |
Proposition 9.5.4. Given that \( R \) satisfies properties (I)-(V), then the additive identity of \( R\left\lbrack x\right\rbrack \) is \( 0{x}^{0} \), where 0 denotes the additive identity of \( R \) . | Proof. The proof has two parts: (i) \( p\left( x\right) + 0{x}^{0} = p\left( x\right) \) and (ii) \( 0{x}^{0} + p\left( x\right) = \) \( p\left( x\right) ,\forall p\left( x\right) \in R\left\lbrack x\right\rbrack \) . We’ll prove (i) and leave (ii) as an exercise.\n\n(i) Given an arbitrary polynomial \( p\left( x\right... | No |
Proposition 9.5.7. Let \( p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) be a polynomial in \( R\left\lbrack x\right\rbrack \), where \( R \) satisfies properties (I)-(V). Then the additive inverse of \( p\left( x\right) \) is \( q\left( x\right) = \) \( \mathop{\sum }\limits_{{i = 0}}^{n}\left... | Exercise 9.5.8. Prove Proposition 9.5.7 by showing that \( p\left( x\right) + q\left( x\right) \) and \( q\left( x\right) + p\left( x\right) \) both sum to the additive identity of \( R\left\lbrack x\right\rbrack \) . | No |
Proposition 9.5.12. Multiplication in \( R\left\lbrack x\right\rbrack \) is associative: | Proof. We’ve seen that the product of two polynomials \( p\left( x\right) \) and \( q\left( x\right) \) may be written in summation notation as:\n\n\[ p\left( x\right) q\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{m}\mathop{\sum }\limits_{{j = 0}}^{n}{a}_{i}{b}_{j}{x}^{i + j} \]\n\nNow we multiply a third polynom... | No |
Proposition 9.5.14. Polynomials in \( R\left\lbrack x\right\rbrack \) have both right distributivity across addition:\n\n\[ \left( {q\left( x\right) + r\left( x\right) }\right) p\left( x\right) = q\left( x\right) p\left( x\right) + r\left( x\right) p\left( x\right) ,\]\n\nand left distributivity across addition:\n\n\[ ... | Proof. To show right distributivity, we have:\n\n\[ \left( {q\left( x\right) + r\left( x\right) }\right) p\left( x\right) = \left( {\mathop{\sum }\limits_{{j = 0}}^{n}{b}_{j}{x}^{j} + \mathop{\sum }\limits_{{j = 0}}^{\ell }{c}_{j}{x}^{j}}\right) \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}^{i} \]\n\n\[ = \left( {\math... | No |
Dividing polynomials in \( \mathbb{R}\left\lbrack x\right\rbrack \) is very similar to long division of real numbers. For example, suppose that we divide \( {x}^{3} - {x}^{2} + {2x} - 3 \) by \( x - 2 \) . | In the example, we need to take the leading power term of \( x \) in the divisor and multiply by something that will make it equal to the the leading power term in the dividend. In this case it’s \( {x}^{2} \) . This gives \( {x}^{2} \cdot \left( {x - 2}\right) = {x}^{3} - 2{x}^{2} \) Subtract from the dividend to yiel... | Yes |
Example 9.6.3. Divide \( \left( {2{x}^{3} + 3{x}^{2} + x + 4}\right) \) by \( \left( {x + 2}\right) \) where both polynomials are in \( {\mathbb{Z}}_{5}\left\lbrack x\right\rbrack \) . |  | No |
Proposition 9.6.7. (Division algorithm for polynomials) Suppose that the set \( F \) has addition and multiplication operations that satisfy (I)-(VII). Let \( f\left( x\right) \) and \( g\left( x\right) \) be nonzero polynomials in \( F\left\lbrack x\right\rbrack \), where the degree of \( g\left( x\right) \) is greate... | Proof. We will first prove the existence of \( q\left( x\right) \) and \( r\left( x\right) \) . We define a set \( S \) as follows: \[ S = \{ f\left( x\right) - g\left( x\right) h\left( x\right) ,\text{ for all }h\left( x\right) \in F\left\lbrack x\right\rbrack \} . \] This set is nonempty since \( f\left( x\right) \in... | Yes |
Suppose that we would like to find the gcd of \( a\left( x\right) = {x}^{4} - 5{x}^{3} + 5{x}^{2} + {5x} - 6 \) and \( b\left( x\right) = {x}^{4} + 5{x}^{3} + 5{x}^{2} - {5x} - 6 \) . | We first divide \( a\left( x\right) \) by \( b\left( x\right) \) to determine the remainder, \( {r}_{1} \).\n\n\n\nSo \( {r}_{1} = - {10}{x}^{3} + {10x} \). We then divide \( b\left( x\right) \) by \( {r}_{1} \) to d... | Yes |
Suppose that we would like to find the gcd of \( a\left( x\right) = {x}^{4} + 2{x}^{3} + 5{x}^{2} + {5x} + 1 \) and \( b\left( x\right) = {x}^{4} + 5{x}^{3} + 5{x}^{2} + {2x} + 1 \) in \( {\mathbb{Z}}_{7}\left\lbrack x\right\rbrack \) . | We first divide \( a\left( x\right) \) by \( b\left( x\right) \) to determine the remainder, \( {r}_{1} \) . \n\nSo \( {r}_{1} = 4{x}^{3} + {3x} \) . We then divide \( b\left( x\right) \) by \( {r}_{1} \) to determin... | Yes |
Proposition 9.6.12. Let \( F \) satisfy properties (I)-(VII), \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), and \( a \in F \) . When \( f\left( x\right) \) is divided by \( x - a \), the remainder is \( f\left( a\right) \) . | Proof. According to Proposition 9.6.7, if we divide \( f\left( x\right) \) by \( x - a \), it will produce two unique polynomials \( q\left( x\right) \) and \( r\left( x\right) \) such that \( f\left( x\right) = \left( {x - a}\right) q\left( x\right) + \) \( r\left( x\right) \) . Since the degree of \( x - a \) is 1, t... | Yes |
Proposition 9.6.14. Let \( F \) satisfy properties (I)-(VII), \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), and \( a \in F \) . Then \( x - a \) divides \( f\left( x\right) \) if and only if \( f\left( a\right) = 0 \) . | Proof. From Proposition 9.6.17 \( f\left( x\right) = \left( {x - a}\right) \cdot q\left( x\right) + f\left( a\right) \) . Therefore \( f\left( a\right) = 0 \) if and only if \( f\left( x\right) = \left( {x - a}\right) \cdot q\left( x\right) \), which is true if and only if \( x - a \) divides \( f\left( x\right) \) . | Yes |
Proposition 9.6.15. Suppose that \( F \) satisfies properties (I)-(VII). Given \( a, b \in F \) and \( {ab} = 0 \), then either \( a = 0 \) or \( b = 0 \) . | Exercise 9.6.16. Prove Proposition 9.6.15. (Hint: you may follow the steps given in Exercise 2.2.12 | No |
Proposition 9.6.17. Suppose \( F \) satisfies Properties (I)-(VII), and suppose \( p\left( x\right), q\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( p\left( x\right) q\left( x\right) = 0 \) iff either \( p\left( x\right) = 0 \) or \( q\left( x\right) = 0 \) . | Proof. Since this is a \ | No |
Proposition 9.6.19. Suppose \( F \) satisfies properties (I)-(VII), and let \( c \) be any element \( F \) . Then \( c \) has at most \( n{n}^{\text{th }} \) roots. | Proof. Given \( C \in F \), then the polynomial \( {x}^{n} - c \) is an element of \( F\left\lbrack x\right\rbrack \) . By Proposition 9.6.18, the equation \( {x}^{n} - c = 0 \) has at most \( n \) solutions. This is exactly the same thing as saying that \( c \) has at most \( n{n}^{\text{th }} \) roots. | Yes |
Find the roots of \( p\left( x\right) = 2{x}^{2} + {2x} + 5 \) . | Since this is a quadratic polynomial we can use the famous quadratic formula:\n\n\[ x = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \]\n\nIn \( p\left( x\right), a = 2, b = 2 \), and \( c = 5 \) . We substitute those values into the formula and obtain the following:\n\n\[ x = \frac{-2 \pm \sqrt{{2}^{2} - 4 \cdot 2 \cdot 5... | Yes |
Find the roots of \( f\left( x\right) = 3{x}^{3} + {10}{x}^{2} + {11x} + 6 \) . | Since this is a cubic polynomial, we can't use the quadratic formula, at least not to begin with. The coefficients are integers, so we may use Proposition 9.6.22, which says that any rational roots of \( p\left( x\right) \) have numerators that are factors of \( {a}_{0} \) and denominators that are factors of \( {a}_{n... | Yes |
Example 9.6.29. We begin with an example of a linear binomial in \( \mathbb{C}\left\lbrack x\right\rbrack \) . Let \( p\left( x\right) = \left( {3 + {2i}}\right) x + \left( {2 - i}\right) \) . Find the root of \( p\left( x\right) \) (since \( p\left( x\right) \) is linear, it will only have one root). | First we set \( p\left( x\right) \) equal to zero and then proceed to find the root as follows. Begining with \( \left( {3 + {2i}}\right) x + \left( {2 - i}\right) = 0 \), we may rearrange to obtain\n\n\[ x = \frac{-2 + i}{3 + {2i}} \]\n\nand multiplying numerator and denominator by \( 3 - {2i} \) and simplifying gives... | Yes |
Let \( p\left( x\right) = \left( {1 + i}\right) {x}^{2} + \left( {2 - i}\right) x + \left( {3 + {3i}}\right) \). Find the roots of \( p\left( x\right) \). | Since this is a quadratic polynomial we can use the quadratic formula and obtain the following:\n\n\[ x = \frac{-\left( {2 - i}\right) \pm \sqrt{{\left( 2 - i\right) }^{2} - 4\left( {1 + i}\right) \left( {3 + {3i}}\right) }}{2\left( {1 + i}\right) } = \frac{\left( {-2 + i}\right) \pm \sqrt{3 - {4i} - {24i}}}{\left( 2 +... | Yes |
Proposition 9.6.32. Any polynomial \( p\\left( x\\right) \) of degree \( n \) in \( \\mathbb{C}\\left\\lbrack x\\right\\rbrack \) can be completely factored as a constant times a product of \( n \) linear terms, as follows:\n\n\[ p\\left( x\\right) = b\\left( {x - {a}_{1}}\\right) \\left( {x - {a}_{2}}\\right) \\ldots ... | Proof. Let \( p\\left( x\\right) \) be an arbitrary polynomial of degree \( n \) in \( \\mathbb{C}\\left\\lbrack x\\right\\rbrack \) . By Proposition 9.6.28, \( p\\left( x\\right) \) has at least one complex root \( a \) . So \( \\left( {x - a}\\right) \) is a factor of \( p\\left( x\\right) \) and we can write \( p\\l... | Yes |
Example 9.6.35. Let \( f\left( x\right) = {x}^{4} - 4{x}^{3} + {10}{x}^{2} - {24x} + {24} \) be a polynomial in \( \mathbb{C}\left\lbrack x\right\rbrack \) . Notice that the coefficients of \( f\left( x\right) \) are integers, so \( f\left( x\right) \) is also in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Therefore ... | After testing all possibilities we find the following rational root: \( f\left( 2\right) = {2}^{4} - 4{\left( 2\right) }^{3} + {10}{\left( 2\right) }^{2} - {24}\left( 2\right) + {24} = 0 \) . Therefore, \( x = 2 \) is a root of \( f\left( x\right) \) and \( \left( {x - 2}\right) \) is a factor of \( f\left( x\right) \)... | Yes |
Example 10.1.5. Figure 10.1.2 shows a \( {60}^{ \circ } \) counterclockwise rotation of a regular hexagon where the vertices of the hexagon are labeled \( A, B, C, D, E, F \) . (Notice how the letters run counterclockwise around the hexagon. We will consistently follow this pattern. The reason is that in mathematical c... | The rotation moves \( A \) to \( B, B \) to \( C \), and so on. Now of course there are other points on our figure, namely all the points on the line segments between the vertices. But notice that if we account for where the vertices are moved to, then the movement of the line segments is automatically accounted for. I... | No |
Proposition 10.1.7. If \( S \) is the set of points that represent a figure, all symmetries of the figure are bijections from \( S \rightarrow S \) . | Proof. Since the result of any symmetry acting on \( S \) must be all of \( S \), then every point of \( S \) must be in the range of \( S \). Thus any symmetry is onto.\n\nFurthermore, the symmetry must map two different points to two different points, since the distance between points must be left unchanged by the sy... | Yes |
Suppose we wanted to find \( {r}_{180} \circ {s}_{v} \) using the tableau forms for \( {r}_{180} \) and \( {s}_{v} \) above. That is\n\n\[ \n{r}_{180} \circ {s}_{v} = \left( \begin{array}{llll} A & B & C & D \\ C & D & A & B \end{array}\right) \circ \left( \begin{array}{llll} A & B & C & D \\ D & C & B & A \end{array}\... | To see how this works, let’s \ | No |
Proposition 10.2.5. Suppose \( f \) and \( g \) are both symmetries of a figure. Then \( f \circ g \) is itself a symmetry of the same figure. | Proof. Recall that composition works from right to left. Since \( g \) is a symmetry, \( g \) takes the points of the figure and rearranges them so that the angles and distances of points in the figure are preserved. The symmetry \( f \) then takes the points of this preserved figure and moves them in such a way that t... | Yes |
Proposition 10.3.5. The set of symmetries \( S \) of any figure under composition is associative. | Proof. By definition, we know any symmetry of a figure is a function. From the Functions chapter, we know | No |
Proposition 10.3.6. The set of symmetries \( S \) of any figure has an identity. | Proof. By the definition of a symmetry, the \ | No |
Proposition 10.3.7. All elements of the set \( S \) of symmetries of any figure have inverses. | Proof. Given a symmetry \( s \in S \), by definition \( s \) is a bijection. In the Functions chapter, we showed that every bijection has an inverse \( {s}^{-1} \) . It remains to show that \( {s}^{-1} \) is itself a symmetry. This means that we have to show:\n\n(i) \( {s}^{-1} \) leaves distances unchanged between poi... | No |
Proposition 10.3.9. The set \( S \) of symmetries of any figure forms a group. | Exercise 10.3.10. Prove Proposition 10.3.9 (make use of the propositions that we've proved previously.) | No |
Proposition 10.4.1. The dihedral group, \( {D}_{n} \), is a group of order \( {2n} \) . | Let us try to characterize these \( {2n} \) elements of the dihedral group \( {D}_{n} \) .\n\n\n\nFigure 10.4.1. Rotations and reflections of a regular \( n \) -gon\n\nFirst, we know that the elements of the dihedral... | No |
Example 10.4.16. In \( {D}_{5} \), to compute \( \left( {s \circ {r}^{3}}\right) \circ \left( {s \circ {r}^{4}}\right) \) we have (using Proposition 10.4.15 and associativity): | \[ \left( {s \circ {r}^{3}}\right) \circ \left( {s \circ {r}^{4}}\right) = s \circ \left( {{r}^{3} \circ s}\right) \circ {r}^{4}\text{by associativity} \] \[ = s \circ \left( {s \circ {r}^{2}}\right) \circ {r}^{4}\text{by Prop. 10.4.15(c)} \] \[ = \left( {s \circ s}\right) \circ r \circ {r}^{5}\text{by associativity} \... | Yes |
Example 11.1.2. Let us recall for a moment the equilateral triangle \( \bigtriangleup {ABC} \) from the Symmetries chapter. Let \( T \) be the set of vertices of \( \bigtriangleup {ABC} \) ; i.e. \( T = \{ A, B, C\} \) . We may list the permutations of \( T \) as follows. For input \( A \) , we have 3 possible outputs;... | \[ \left( \begin{array}{lll} A & B & C \\ A & B & C \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ C & A & B \end{array}\right) \;\left( \begin{array}{lll} A & B & C \\ B & C & A \end{array}\right) \] \[ \left( \begin{array}{lll} A & B & C \\ A & C & B \end{array}\right) \;\left( \begin{array}{lll} A & B &... | Yes |
Proposition 11.2.1. Given any set \( X,{S}_{X} \) is a group under function composition. | Proof.\n\n- First then, if \( f, g \in {S}_{X} \), then \( f \circ g \) would be, by definition of composition, a function from \( X \rightarrow X \) . Further, since it is a composition of two bijections, \( f \circ g \) would be a bijection (proved in Functions chapter). Therefore by definition \( f \circ g \) is per... | Yes |
Example 11.3.1. Suppose \( \rho \in {S}_{6} \) and \( \rho = \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 5 & 6 & 1 \end{array}\right) \) . Then | \[ \rho \left( 1\right) = 2,\rho \left( 2\right) = 3,\rho \left( 3\right) = 4,\rho \left( 4\right) = 5,\rho \left( 5\right) = 6\text{, and}\rho \left( 6\right) = 1\text{.} \] A shorter way to represent this is \[ 1 \rightarrow 2,2 \rightarrow 3,3 \rightarrow 4,4 \rightarrow 5,5 \rightarrow 6\text{and}6 \rightarrow 1\te... | Yes |
Example 11.3.9. Suppose \( \tau \in {S}_{6} \) and \( \tau = \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 4 & 2 & 3 & 5 & 6 \end{array}\right) \) . Then | - \( 1 \rightarrow 1 \), which means that 1 \ | No |
Suppose we want to form the product (that is, composition) \( {\sigma \tau } \), where \( \sigma ,\tau \in {S}_{6} \) and \( \sigma = \left( {1532}\right) ,\tau = \left( {126}\right) \) . | Figure 11.3.2 provides a visual representation of how the product \( {\sigma \tau } \) acts on 1. Remember that we operate from right to left, so the figure shows '1' coming in from the right. The action of \( \tau \) takes 1 to 2 . (For convenience we have \ | No |
Consider the product \( {\sigma \tau } \) where \( \sigma = \left( {AEDBF}\right) \) and \( \tau = \left( {ABDFE}\right) \) . | - \( \tau \) takes \( A \rightarrow B \) and \( \sigma \) takes \( B \rightarrow \mathrm{F} \) ; hence \( {\sigma \tau } \) takes \( A \rightarrow F \) .\n\n- \( \tau \) takes \( F \rightarrow E \), and \( \sigma \) takes \( E \rightarrow D \) ; hence \( {\sigma \tau } \) takes \( F \rightarrow D \) .\n\n- \( \tau \) t... | Yes |
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