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, X} and define µ(0) = 0,µ(E) = p(X - E) = 1/2 and µ(X) = 1. For each natural number n, define fn = n - XE - n - X[X .. E]. The sequence {f n } is uniformly integrable and tight and converges pointwise on X to the function f that takes the constant value no on E and -oo on X - E. The limit function is not integrable ov... |
of f. Define .1 to be the collection of nonnegative measurable functions on X for which fdµ<v(E)for all EEM, J E and then define M = sup fE.F fx .f dµ (34) We show that there is an f E .1 for which fx f dµ = M and (31) holds for any such f. If g and h belong to Y, then so does max(g, h). Indeed, for any measurable set ... |
t f 2 is integrable over X with respect to A. Suppose that the functional 0: L2(X, A) - R is linear, and bounded in the sense that there is some c > 0 such that I+,(f)I2_<c.J f2dAforall fEL2(X,A). X Then there is a function g E L2(X, A) such that t/r(f)=JX Assuming this representation result, verify the following steps... |
at v(Ek) = oo, then, by the monotonicity of v, (46) holds since both sides are infinite. We therefore assume that v(Ek) < oo for all k. By the finite additivity of v, for each natural number n, 0o n v UEk =2v(Ek)+v(U Ek). I k=n+1 k=1 k=1 ao (47) Let E > 0. By the uniform absolute continuity with respect to µ of the seq... |
call 1 the conjugate of oo and oo the conjugate of 1. The proofs of the results in this section are very similar to those of the corresponding results in the case of Lebesgue integration of functions of a real variable. Theorem 1 Let (X, M, µ) be a measure space, 1 < p < oo, and q the conjugate of p. If f belongs to LP... |
metric isomorphism of Lq(X, µ) onto (LP(X, µ))*. Before we prove this theorem, a few words are in order contrasting the proof in the case of a closed, bounded interval with the general proof. In the case of Lebesgue measure m on X = [a, b], a closed, bounded interval of real numbers, the heart of the proof of the Riesz... |
asure space. We wish to define fx f dv for f E LOO (X, µ), now formally viewed as a linear space of equivalence classes of essentially bounded measurable functions with respect to the relation of equality a.e. [µ]. This requires that fx fdv = fx f dv if f = fl a.e. [µ] on X. If there is a set E E M for which µ(E) = 0, ... |
t LP(N, µ) = lP and thereby characterize the dual space of LP for 1 <p<oo. (ii) Discuss the dual of LP(X, µ) for 1 < p < no, where µ is the counting measure on a not necessarily countable set X. 17. Find a measure space (X, M, µ) with the property that all the theorems of this section hold in the case p = 1. 18. Show t... |
we call A X B a measurable rectangle. Lemma 1 Let {Ak X Bk}k 1 be a countable disjoint collection of measurable rectangles whose union also is a measurable rectangle A X B. Then 00 µ(A)Xv(B) _ I µ(Ak)Xv(Bk). k=1 Proof Fix a point x E A. For each y E B, the point (x, y) belongs to exactly one Ak X Bk. Therefore we have... |
ains to prove that klim fX I JY(Pk(x, Y) dv(Y)I du(x) = f V f(x, y) dv(Y)Jdµ(x) oc If we excise from X X Y a set of µ x v-measure zero, then the right-hand side of (7) remains unchanged and, by Lemma 4, so does the left-hand side. Therefore, by possibly excising from X X Y a set of Ax v-measure zero, we may suppose tha... |
ble with respect to the product measure? 10. Let h and g be integrable functions on X and Y, and define f (x, y) = h(x)g(y). Show that f is integrable on X X Y with respect to the product measure, then fXXY f d(AXv) = fXhdµ ffgdv. (Note: We do not need to assume that µ and v are v-finite.) 11. Show that Tonelli's Theor... |
infer that K=K'-OCK'-[K'-E]=K'nECE and therefore K C E. On the other hand, from the inclusion E C K' we infer that and E-'K=E^'[K"-.0}=En0 E n 0 C 0-[K'-El. Therefore, by the excision and monotonicity properties of measure and (19), p (E) - tln(K) = lln(E.'K) <E. Thus (18) is established and the proof is complete. Eac... |
ype 2: T Type 3: T (ej) = ej + ej+1, and T(ek) = ek for k 96 j : andT(ek)=ekfork j: ei+1, T (e.i+l) = ej and T(ek) = ek fork # j, j + 1 : That every invertible linear operator may be expressed as the composition of elementary operators is an assertion in terms of linear operator of a property of matrices: every inverti... |
. Proof First let µ be a Borel measure on B(I ). Its cumulative distribution function is certainly increasing and bounded. Let xo belong to [a, b) and (xk) be a decreasing sequence in (xo, b] that converges to xo. Then fl 1(xo, ski = 0 so that, since µ is finite, by the continuity of measure, 0 = µ(0) = lim 1- (x0, xk]... |
es R2k and U R2j, since the latter set is contained k-1 in B2k_2. Consequently, we argue by induction to show that for each k, j=1 k k-1 A* U R2 j = A* (R2k) +A* U R2j j=1 j=1 `k L,= j=1 µ* (R2j) 442 Chapter 20 The Construction of Particular Measures j=1 R2j C B C A, we have Since converges. Similarly, the series 1 µ* ... |
tegration against a signed Borel measure. Furthermore, in each of these representations it is possible to choose the representing measure to belong to a class of Borel measures that we here name Radon, within which the representing measures are unique. The Riesz Representation Theorem provides the opportunity for the a... |
umbers with the topology that has complements of countable sets as a base. Show that X is not locally compact. 10. Provide a proof of Proposition 3 by applying Urysohn's Lemma to the Alexandroff one-point compactification of X. 11. Let f continuously map the locally compact Hausdorff space X onto the topological space ... |
asurable if and only if g(O)>g*(OfE)+g*(O^-E) for each open setOfor which p(O) <oo. (3) Proof Since the union of any collection of open sets is open, (2) follows from the countable monotonicity of g. Let E be a subset of X for which (3) holds. To show that E is g*measurable, let A be a subset of X for which g* (A) < oo... |
. We begin by showing that µ is a premeasure. To establish countable monotonicity, let {Ok}k'=1 be a collection of open subsets of X that covers the open set 0. Let f be a function in Cc (X) with 0 < f < 1 and supp f C 0. Define K = supp f. By the compactness of K there is a finite collection {Ok}k=1 that also covers K... |
the difference, L+ - L_, of two positive linear functionals on C(X). We always have IILII the opposite direction, let cp be any function in C(X) for which 0 < ' and hence IIL+II+IIL- II = L+(1)+L_(1).Toestablish theinequality in 1 1. Then Taking the supremum over all such cp, we have IILII ? L(2(p - 1) = 2L(,p) - L(1).... |
ction g defined on [a, b] by g(a) = 0 and g(x) = µ(a, x] + µ{a} for x E (a, b]. The functon g inherits continuity on the right at each point in (a, b) from the continuity of the measure µ. Thus g belongs to Y. We infer from Proposition 16 that fdµ=f f (x) dg(x) for all f EC[a, b], b f 6 a where µ is the unique Borel me... |
erval that does not contain 0. Finally, since f-1{0} = X- [f-1(-o0, 0) U f_1(0, oo)] we infer that the inverse image under f of any nonempty interval belongs to.F and therefore f is measurable with respect to the Baire a-algebra. Proposition 22 Let X be a compact Hausdorff space. Then every Baire measure on Ba(X) is re... |
led the general linear group of E. We denote its identity element by Id. It also is a topological space with the topology induced by the operator norm. Lemma 1 Let E be a Banach space and the operator C EC(E) have IICII < 1. Then Id-C is invertible and II(Id-C)-111 (1 - IICII)-1 (2) Proof We infer from (1) that for eac... |
functional p: K* - R by p(hi) = sup Ii(ir(g)xo)I for i/i E K*. gEG Since K* is weak-* compact, the Uniform Boundedness Principle tells us that K* is bounded. According to the preceding lemma, p is continuous with respect to the weak-* topology. Therefore, if, for r > 0 and 71 E K*, we define Bo(n,r)={grEK*I p(O -n)<r}a... |
is right-invariant. Proof Theorem 6 tells us that there is a probability functional l E [C(G)]* that is fixed under the adjoint of the regular representation on G on C(G). This means that i/r(1) = 1 and PP(f) = q,(-(g-')f) for all f E C(G) and g E G. (14) On the other hand, according to the Riesz-Markov Theorem, there ... |
Banach space C(X) of continuous real-valued functions on X with the maximum norm. Since X is a compact metric space, Borsuk's Theorem tells us that C(X) is separable. Let 77 be any Borel probability measure on B(X ). Define the sequence {1lin } of linear functionals on C(X) by /1n(g)=J- go fk di7forallnENandgEC(X). l n... |
mberg, Real and Abstract Analysis, Graduate Texts in Mathematics, Springer, 1975. [Jec06] Thomas Jech, Set Theory, Springer, 2006. [Ke175] John L. Kelley, General Topology, Springer, 1975. [Lax97] Peter D. Lax, Linear Algebra, John Wiley and Sons, 1997. [LaxO2] [Lit4l] , Functional Analysis, Wiley-Interscience, 2002. J... |
292 Identity mapping, 4 Image of a mapping, 5 Increasing function, 50 Increasing sequence, 21 Indefinite integral, 125 Index set, 4 Induced measure, 350 Inductive set, 11 Inequality Cauchy-Schwarz, 142,396 Chebychev's, 80, 367 Holder's, 140,396 Jensen's, 133 Minkowski's, 141,396 triangle, 9, 137, 184 Young's, 140 Infim... |
, 376 Uniform metric, 206 Uniformly bounded sequence, 207 Uniformly Cauchy sequence, 206 Uniformly continuous function, 26,192 Uniformly equicontinuous, 208 Unit function, 137 Unit vector, 255 Upper Darboux sum, 69 Upper derivative, 111 Upper Lebesgue integral, 73 Upper Riemann integral, 69 Urysohn Metrization Theorem,... |
χ is trivial or principal if ρ is the trivial representation. We write χ = 1G. χ is a complete invariant in the sense that it determines ρ up to isomorphism. This is staggering. We reduce the whole matrix into a single number — the trace, and yet we have not lost any information at all! We will prove this later, after ... |
1≤i≤n 1≤j≤n For any linear map ϕ : V → V , we define a new map by averaging by ρ and ρ. ˜ϕ : V → V 1 |G| v → ρ(g−1)ϕρ(g)v We first check ˜ϕ is a G-homomorphism — if h ∈ G, we need to show ρ(h−1) ˜ϕρ(h)(v) = ˜ϕ(v). We have ρ(h−1) ˜ϕρ(h)(v) = = 1 |G| 1 |G| g∈G ρ((gh)−1)ϕρ(gh)v ρ(g−1)ϕρ(g)v g∈G = ˜ϕ(v). (i) Now we first cons... |
4) 6 (1 2) 1 1 −1 −1 1 1 0 0 1 −1 −1 1 1 −1 1 −1 34 7 Permutation representations II Representation Theory For the last representation we can find the dimension by computing 24 − (12 + 12 + 32 + 32) = 22. So it has dimension 2. To obtain the whole of χ5, we can use column orthogonality — for example, we let the entry i... |
can define the tensor product. The definition we will take here is a rather hands-on construction of the space, which involves picking a basis. We will later describe some other ways to define the tensor product. Definition (Tensor product). Let V, W be vector spaces over F. Suppose dim V = m and dim W = n. We fix a basis v... |
s carefully, it is clear that these two actions commute with each other. This rather simple innocent-looking observation is the basis of an important theorem by Schur, and has many many applications. However, that would be for another course. Getting back on track, since the two actions commute, we can decompose V ⊗n a... |
) = |CG(g)| m i=1 ψ(xi) |CH (xi)| . This is all just group theory. Note that some people will think this proof is excessive — everything shown is “obvious”. In some sense it is. Some steps seem very obvious to certain people, but we are spelling out all the details so that everyone is happy. Proof. If m = 0, then {x ∈ ... |
G fixes xH and yH, then we have g ∈ xHx−1 ∩ yHy−1. This implies H ∩ (y−1x)H(y−1x)−1 = 1. Hence xH = yH. Note that J. Thompson (in his 1959 thesis) proved any finite group having a fixed-point free automorphism of prime order is nilpotent. This implies K is nilpotent, which means K is a direct product of its Sylow subgroup... |
cisely, CG = g∈G αgg : αg ∈ C . Borrowing the multiplication of G, this forms a ring, hence an algebra. We now list the conjugacy classes of G as Definition (Class sum). The class sum of a conjugacy class Cj of a group G is {1} = C1, · · · , Ck. Cj = g∈Cj g ∈ CG. 65 13 Integrality in the group algebra II Rep... |
. Example. The group S1 = {z ∈ C : |z| = 1} under multiplication is a compact group. This is known as the circle group, for mysterious reasons. Thus the torus S1 × S1 × · · · × S1 is also a compact group. Example. The orthogonal group O(n) ≤ GLn(R) is compact, and so is SO(n) = {A ∈ O(n) : det A = 1}. These are compact... |
ble representations of G. Then χV , χW = 1 V ∼= W 0 V ∼= W . 76 15 Representations of compact groups II Representation Theory Now do irreducible characters form a basis for C(G)? Example. We take the only (infinite) compact group we know about — G = S1. We have found that the one-dimensional representations are ρn : z →... |
1 √ 2 1 −1 1 1 xn−iyi = 1 √ 2 (x + y)n−i(−x + y)i ∈ W. It is clear that the coefficient of xn is non-zero. So we can use the claim to deduce xn ∈ W . Finally, for general a, b = 0, we apply a −¯b ¯a b ρn xn = (ax + by)n ∈ W, and the coefficient of everything is non-zero. So basis vectors are in W . So W = Vn. This proof is... |
diffeomorphism in B(0, ε). By looking at the proof of the local minimizing of length, and using the same notation, we know that we have equality iff τ = 1 and ρ(t)2|(d expp)ψ(t)ψ(t)u(t)|2 = 0 for all t. Since d expp is regular, this requires u(t) = 0 for all t (since ρ(t) = 0 when t = 0, or else we can remove the loop to... |
. Note that the length of a curve is independent of parametrization. Thus, if we are interested in critical points, then the critical points cannot possibly be isolated, as we can just re-parametrize to get a nearby path with the same length. On the other hand, the energy E does depend on parametrization. This does hav... |
because Jacobi fields do not. Proposition. (i) If γ(t) = expp(ta), and q = expp(βa) is conjugate to p, then q is a singular value of exp. (ii) Let J be as in the definition. Then J must be pointwise normal to ˙γ. Proof. (i) We wlog [α, β] = [0, 1]. So J(0) = 0 = J(1). We a = ˙γ(0) and w = J (0). Note that a, w are both n... |
get an extension to within V , and then we can just lift directly, and extend it to ε1. So ε1 ∈ I. So we are done. Corollary. Let f : M → N be a local isometry onto N , and M be complete. Then f is a covering map. Note that since N is (assumed to be) connected, we know f is necessarily surjective. To see this, note th... |
d right hand side live in different parts of the Hodge decomposition. So they must be individually zero. Alternatively, we can compute dβ2 g = dβ, α1 − α2g = β, δα1 − δα2g = 0 since harmonic forms are co-closed. To prove existence, let α ∈ Ωp(M ) be such that dα = 0. We write α = α1 + dα2 + δα3 ∈ Hp ⊕ dΩp−1(M ) ⊕ δΩp+1(... |
t ∇ωg = 0. So it follows that ∇Y (i(X) ωg) = i(∇Y X) ωg. Therefore we obtain d(i(X)ωg) = = = n k=1 n k=1 n k=1 dxk ∧ ∇k(i(X)ωg) dxk ∧ i(∇kX)ωg dxk ∧ i(∇kX)(|g|dx1 ∧ · · · ∧ dxn) = dxk(∇kX) ωg = (divX) ωg. Note that this requires us to think carefully how wedge products work (i(X)(α∧β) is not just α(X)β, or else α ∧ β w... |
k=1 − ek(dα(ek, X)) + dα(ek, ∇ek X) − ek(∇ek α, X − ∇X α, ek) + ∇ek α, ∇ek X − ∇∇ek X α, ek − ∇ek ∇ek α, X − ∇ek α, ∇ek X + ∇ek ∇X α, ek) n = − ∇ek ∇ek α, X + + ∇ek α, ∇ek X − ∇∇ek X α, ek n n ∇ek ∇X α, ek − ∇∇ek X α, ek. k=1 k=1 k=1 What does this get us? The first term on the right is exactly the ∇∗∇ term we wanted. ... |
f M is compact, then we have a decomposition H k dR(M ) = H k i,dR(M ), where H k i,dR(M ) = {[α] : α ∈ Ωk i (M ), ∆α = 0}. The dimensions of these groups are known as the refined Betti numbers. We have only proved this for k = 1, but the same proof technique can be used to do it for arbitrary k. Our treatment is rather... |
a geodesic, 32 sectional curvature, 9 symmetric connection, 5 Synge’s theorem, 34 torsion-free connection, 5 vector field left invariant, 4 volume form, 44 weak solution, 50 Whitney embedding theorem, 4 72 |
ided by -7. Show that if a and D are positive integers, then there are integers q,r and ] . Show that if a and b are positive real numbers, then labl 2 Laltbl . What is the corresponding inequality when both a and b are negative? When one is negative and the other positive.2 Divisibilitv 23 17. What is the value of [a ... |
asured using weights of 1,2,22,...,2ft-1, when all the weights are placed in one pan. Show that every integer can be uniquely represented in the form where €i balanced ternary expansion. : -1,0, or I for ,/:0,1 ,2, ..., k. This expansion is called a 9. Use problem 8 to show that any weight not exceeding $k -t) /Z may b... |
, but also it is less than brn-t, since 2 qiri g rn-l-l. Therefore, we know that j - 0 This tells us that (-": Tt , -tn.' ' We can obtain Qn-r by successively subtracting br"-l result is obtained, and then qn-1is one less than the number of subtractions. from a until a negative To find the other digits of q,, we define... |
base 28 digit 8, where I is a positive integer. Show that the base 28 expansion of the integer (ana,-1...afl0)z,a starts with the digits of the base 28 expansion of the integer (anana...aflo)zn l(anan-1...ap0)zn * ll and ends with the digits Bl2 and 0 when B is even, and the digits G-l)12 and.B when I is odd. 1.4 Comp... |
tance, it is unknown whether there are infinitlly many primes of the form n2 + | where n is a positive integer. Questions such as this may be easy to state, but are sometimes extremely difficult to resolve. We conclude this section by discussing perhaps the most notorious conjecture about primes. Goldbach's Conjecture.... |
ivisor of Example. To 105, 140, and 350, we use Lemma 2.1 to see that (105, 140. 350 . the three integers Definition. We say that the integers a1.e2,..., e1 are mutually relatively prime if (a1, e2,..., an) : l. These integers 4re called pairwise relatively prime if for each pair of integers 4; and a; from the set, (ai... |
owing result is a consequence of Lam6's theorem. Corollary 2.1. The number of bit operations needed to find the greatest common divisor of two positive integers a and, yy ir;;i.:f$;:ri?', Proof. We know from Lam6's theorem that O Qogra) divisions, each taking O(log2a)2) bit operations, are needed to find fu, b). Hence,... |
integers are given by . Note that it is convenient to combine all the factors of a particular prime into a power of this prime, such as in the previous example. There, for the factorization of 240, all the fdctors of 2 were combined to form 24. Factorizations of integers in which the factors of primes are combined to f... |
form 4n * l, there exist integers r and s such that a : 4r * 1 and D : 4s * 1. Hence , which is again of the form 4n * 1. tr We now prove the desired result. Proof. Let us assume that there are only a finite number of primes of the . Then, there is at least one prime in the factorization of Q of the form 4n * 3. Other... |
Numbers 81 where G+b)12 and b-b)/2 are both integers since a and b are both odd. Conversely, if n is the difference of two squares, say n: s2 - /2, then we can factor n by noting that n : (s-l)(s+t). tr To carry out the method of Fermat factorization, we look for solutions of the equation ,, : *2 - yz by searching for ... |
n is an integer. We see that thi s pai r (x ,y) i s a $liln y:Y0 - solution, since V rfi"v g rof14 . We now show that every solution of the equation ax * by : c must be of the form described in the theorern. Suppose that x and y are integers with ax I bY : c. Since a x s * b y o : , , by subtraction we find that which ... |
h cb-b): have G /il G-b) follows that m/d I Q-b). Hence, a :- b (mod m/il. c( a-b). H ence, km. By dividing both sides by d, we : k fu /d). Since (m /d ,c /d) : 1, from Proposition 2.1 it a Example. Since 50 = 20 (mod 15) and 5 0/10 : 20/10 (mod l5/il, or 5 = 2 (m od 3). (10,5) : 5, we see that The following corollary,... |
viding the pile they find into five equal parts leaving one coconut for the monkey and hiding his portion. In the morning, the men 102 Congruences gather and split the remaining pile of coconuts into five parts and one is left over for the monkey. What is the minimum number of coconuts the men could have collected for ... |
solution to the system of : fttlll2. . . tytk_rntk+l . mr. congruences. To do this, let Mk : M/mt we know that from problem 8 of Section 2.1, since (mi, mp) : I whenever i I k. Hence, from Theorem 3.'7, we can find an inverse ./r of M1 modulo mp, so that Mt lr, = I (mod mt). We now form the sum (Mr, mt * * arMry, integ... |
m2,..., ffi,l. (Hint: Use problem 7 and mathematical induction.) for all pairs of integers (i,7 10. Using problem 9, solve the following systems of congruences ) c) x = 2 (mod 14 : l0 (mod 30) x = 2 x = 8 x = l0 ( mod 25) d) .r = 2 (mod 6mod l0 . What is the smallest number of eggs in a basket if one egg is left over ... |
a formula for an inverse of an nxn matrix where n is a positive integer, we need a result from linear algebra. This result may be found in Anton [60; page 791. It involves the notion of the adjoint of a matrix, which is defined as follows. Definition. The adjoint of an nxn malrix A is the n\n matrix with (i,;)th entry ... |
e by d if a n d only if (-I)kap * -ar * a6 is divisi ble by d. Proof. Since d I ft + 1), we have g: (mod d). H ence, bi = (-l )/ (mod d), and consequently, n : (a1 , ...a ps)b (-t )k a1, + * ao (mod d). Hence, d I n if and only if d | ((-l )o oo + * a s ) . n -l - o1 -a1 : (7F28A6)16 (in hex notation). Then, since zl t... |
to September l: from September 1, to October I : from October l, to November l: from November 1, to December 1: from December l, to January l: from January 1, to February 1: 3 daYs 2 daYs 3 daYs 2 daYs 3 daYs 3 daYs 2 daYs 3 days 2 days 3 daYs 3 daYs. We need a formula that gives us the same increments. Notice that we ... |
les. When this occurs, we say the there is a collision. We need a method to resolve collisions, so that files are assigned to different memory locations. There are two kinds of collision resolution policies. In the first kind, when a collision occurs. extra memory locations are linked together to the first memory locat... |
ntegers. Hence, is congru ent modu lo p to th e produc t of the fi rst p-l are a set of p- ) . Therefore, using Corollary 3.1, we can cel Q-l )! to obtai n aP-t = I (mod p). tr We illustrate the ideas of the proof with an example. Ex ample. Let p:7 and a:3. Then, l'3 = 3(mod 7), 2' 3 = 6 (mod 7), 3 .3 = 2 (mod 7), 4'3 ... |
w th at b'-t : 6\Q'-r)tt'-t t-oO qrl. There fore, by C oroll ary 3.2, w e see that bn-t : I (mod n), and we conclude that n is a Carmichael number. D Exa mple. Theorem 5.5 shows th at 6601 :7 '23'41 b ecause J , 23 , and 4I Ql - t) | oooo, and 4o: (+t - t) | oooo. are all pri me, 6 : i s a Carmichael number : The conv... |
then n is a pseudoprime to less than or equal 6 Ah) different bases a with I ( a ( n. (Hint: Show that the sets c t, o2,..., a, and ba1,ba2,..., ba, have no common elements, where ot, o2, ..., ar are the bases less than n to which n is a pseudoprime.) 12. Show that 25 is a strong pseudoprime to the base 7. 13. Show th... |
strate the idea behind the proof with the following example to 36 in a rectangular chart, as shown in Figure 6.1. 1 6 8 Multiplicative Functions OOe@@2,@@33 l 0 t 4 1 8 22 34 ,O@,5@@27@@ t2 l 6 20 24 28 32 3 6 Figure 6.1. Neither the second nor fourth row contains integers relatively prime to 36, since each element in ... |
et Ffu) d l n 1 7 6 Multiplicative Functions r(60) : r(4)F(15). Each of the divisors of 60 may be written as the product of a divisor of 4 and a divisor of 15 i n the fol lo wi ng w ay: l :1.1 . 20 :4'5, 30 : 2'15, 60 : 4-15 (in each product, the first factor is the divisor of 4 , and the second is the divisor of I 5).... |
Mersenne num ber M5:25 - I - 3l ' Then r,: prime, i s The Lucas-Lehmer test can be performed quite rapidly as the following corollary states. Corollary 6.1. Let p be prime and let Mp : 2p - | denote the pth Mersenne number. It is possible to determine whether Mo is prime using OQ3) bit operations. Proof. To determine ... |
quivalent of the corresponding ciphertext letter. Then . The correspondence between plaintext and ciphertext is given in Table 7.2. 1 9 0 Cryptology plaintext ciphertext 20 V 21 w 22 X 23 Y 24 Z 25 20 U 2 l V 22 w 23 X 24 Y 25 z 0 A I B 2 c Table 7.2. The Correspondence of Letters for the Caesar Cipher. To encipher a m... |
the caesar cipher. Decipher messages that have been enciphered using the transformation C = P+k (mod 26), where ft is a given integer. Decipher messages that have been enciphered using the transformation c = aP+6 (mod 26), where a and b are integers with (a,26) : r. Cryptanalyze, using frequency counts, ciphertext tha... |
a digraphic Hill cipher, by analyzing the frequency of digraphs in the ciphertext. 7.3 Exponentiation Ciphers In this section, we discuss a cipher, based on modular exponentiation, that was invented in 1978 by Pohlig and Hellman [9t1. We will see that ciphers produced by this system are resistant to cryptanalysis. : Le... |
rify that the other player was 7.3 Exponentiation CiPhers 2 1 1 actually dealt the cards claimed. A description of a possible weakness in this scheme, and how it may be overcome, may be found in problem 38 of Section 9.1. 7.3 Problems l . 2 . Using the prime p - GOOD MORNING using modular exponentiation' l0l and enciph... |
uggested that work in general are equivalent to factoring n, and as we have remarked, factoring large integers Seems to be an intractable problem, requiring tremendous amounts of computer time. : b-l )Q-l ) (see Mill.r A few extra precautions should be taken in choosing the primes p and q to be used in the RSA cipher s... |
co 41, a 2,..., an, i.e. to fi nd th e values of xt, x2, ... , xn w ith ,S : atx l * a2x 2 following algorithm. First, we find x, by noting that . Then, we find xn-r, xn-2,..., x1, in succession, using the equations 7.5 Knapsack Ciphers 221 x j - n t-i+l n if s- .s- ;-;+ . To see that this, algorirhm works, first note ... |
7 6. Solve multiplicative knapsack problems involving sequences of mutually relatively prime integers (see Problem 10). 7.6 Some Applications to Computer Science In this section we describe two applications of cryptography to computer science. The Chinese remainder theorem is used in both applications. The first applic... |
ord175. Since 58 = 16 (mod 17), 516 = I (m od l7), ( m o d l 7 ) , we conclude that ord175 - 16. 234 Primitive Roots The following theorem will be useful in our subsequent discussions. Theorem 8.2. rf a and n are relatively prime integers with n ) ai = aj , (mod n) where r and 7 are nonnegative integers, if and i = j ... |
root of g(x) modulo p' This shows that the polynomial g(x), which is of degree n - | and has a leading incongruent roots modulo p' This coefficient not divisible by P, has n contradicts the induction hypothesis. Hence, f G) must have no more than n incongruent roots modulo p. The induction argument is complete' tr tha... |
e, ordot r : pk- t b-D Consequently, r is also a prirnitive root modulo pk. : oeo). All that remains is to prove (8.3) using mathematical induction. The case follows from (8.2). Let us assume the assertion is true for the positive of k: 7nt-t(t_t) # l (modpk ). since G,p) : theorem, we know that l, we know that (r,pk-t... |
ositive integer. (i) (ii) (iii) -- Proof of G). From Euler's theorem, we know that ,6(m): Since r congruent to 1 modulo rn. Hence, ind,l : 6(m) = O (mod Qfu)) . is a primitive root modulo m, no smaller positive power of r is I (mod z). To prove this congruence, note that from the definition of ,ind'Qil : ab (mod ,,, ) ... |
a) Find the index systems of 7 and 9 modulo 16. b) Develop rules for the index systems modulo 2& of products and powers analogous to the rules for indices. c) Use the index system modulo 32 to find all solutions of j xs = I I (mo d 32) and 3' = 17 (mod 32). 12. Let n : 2"p\'pj ' ' ' ph be the prime-power factorization... |
1 : 20 is a universal exponent of From Euler's theorem, we know that d(n) have already demonstrated, the intege r (J - IAQ\),,0|'il,...,ybh)l universal exponent of n: p'ip'; smallest positive universal exponent of n. p';. We are interested in finding the is a universal exponent. As we is also a Definition. The least un... |
e rePeating. the The most commonly used method for generating pseudo-random numbers is called the linear congruential method which works as follows. A set of integers t/t, e, c, and xs is chosen so that m ) 0, 2 < a 4' m, 0 < c 4 m' The sequence of pseudo-random numbers is defined and 0 ( xo ( z. 276 recursively by Pri... |
that 282 Primitive Roots o)'tu)/2 # -r (mod ru). it that az, = I (mod z). Since o" = + I (mod rn ), follows From :- +1 (mod ln), but o),@)/2 * o>'(-)/z # -t (mod z) o rd. a : )r(m), a has +l - exponent e, and a, = _l (mod z and , by hypothesis. Therefore, we can conclude that if then m od rn), si nce ord^ o:\(m) , We ... |
tion, say x : ro. ob-r)/2 - : t Then, the congruence x2 : a (mod p) r l l* | l p ) Using Fermat's little theorem, we see that ) . Hence, if know that - ob -t)/2(m od p). Now consider the case where : - t l* I Then, the congruence x.2 = a (mod p) has no solutions. o-i?{.orem 3.7, for each integer i such that I S t < p-1... |
n-x, he has no way to factor n in a reasonable length of time. Consequently, Bob wins the coin flip if he can factor n, whereas Alice wins if Bob cannot factor n. From previous comments, we know that there is an equal chance for Bob to receive a solution of x2 = a (mod n) that helps him rapidly factor n, or a solution ... |
reciprocity, we know Theorem 9.2, we see that t+ I l . / ) = q = :- 3 (mod 4) , from the law of r ) I 12 l. From Dart (i) of L 7 ) Again' using the iaw of quadratic Sincerp that lil : l+l 306 Quadratic Residues reciprocity, since 5 = l(mod 4) and 7 = j(mod 4), part ., (i) of Theorem 2.2 and Theorem f-T l+l - l?l : -' H... |
d #l :[+*l :[+l [+l [+] :[+l [+l l*l : [+]'[+l'[+] : '- D2 t2(-'l): -r When r is prime, the Jacobi symbol is the same as the Legendre symbol' However, when n is composite, the value of the Jacobi symbol lq I Oott nor tell us whether the congruence x2 = a (mod n) has solutions.., *. do know l r ) ' that if the congruenc... |
e integer is called an Euler pseudoprime to the base b. An Euler pseudoprime to the base b masquerades as a prime by satisfying the congruence given in the definition. is a composite integer that Since '1105 = I (mod 8), we see that ) . Hence, I | (-oa l 105). Because I r 05 is composite, it is an Euler : t. - l+] l l ... |
w that if n is an Euler pseudoprime pseudoprime to the base n-b. to the base b, then n is also an Euler Show that if n= 5 (mod 8) and n is an is a strong pseudoprime to the base 2. Euler pseudoprime to the base 2, then r 6. Show that if n = 5 (mod 12) and n is an Euler pseudoprime to the base 3, then n is a strong pseu... |
ions. Moreover, the theorem gives the lengths of the pre-period and periods of base b expansions of rational numbers. Theorem 10.6. Let b be a positive integer. Then a periodic base b expansion represents a rational number. Conversely, the base b expansion of a rational number either terminates or is periodic. Furthero... |
ematical induction. For n : 1 we have [ which is rational. Now assume. that for the positive integer k the simple continued fraction [ag;at,e2,...,ekl is rational whlnevst as,or,...,ok are integers with a r,...,ak positive. Let as,at,...,ek+t be integers with er,...,ek+t positive. Note that [ By the induction hypothesi... |
ment of the theorem is called the value of the infinite simple continued fraction [as;a t,o 2,...1 . To prove Theorem 10.13, we will show that the infinite sequence of evennumbered convergents is increasing and has an upper bound and that the infinite sequence of odd-numbered convergents is decreasing and has a lower b... |
ngle inequality th at to obtain the second inequality Consequently, t/2sqp I t/2s2 Zsqp ) 2s2, which implies that q1, ) s, contradicting the assumption. tr 10.3 Problems L Find the simple continued fractions of the following real numbers a) b) ,rf2 ^f3 c) d) -,/i r+.6 . 2' Find the first five partial quotients of the s... |
ect square, and -,tr\. Now let p : alcl, e We now present an algorithm for finding the sample continued fractions of quadratic irrationals. Theorem 10.19. Let a be a quadratic irrational, so are integers Ps,Qs, and d such that that by Lemma 10.5 ther e @o+,/7) /Qo , where Q0*0,d > 0, d is not a perfect square, and eel ... |
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