Math 3450 Homework Solutions

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1 Math 3450 Homework Solutions I have decided to write up all the solutions to prolems NOT assigned from the textook first. There are three more sets to write up and I am doing those now. Once I get the last three done, I will start looking at the ook prolems, and posting solutions to only those that are very difficult. Most if not all of them have solutions anyway. Also, some prolems were done in class, so check your notes too. Please check here again tonight for the remaining 3 sets. Note that we only assigned ook prolems for homework sets 4 and 7, so they are omitted for now. HW 1 1. Prove that n is an even integer if and only if n 2 is an even integer. First, suppose that n is an even integer. Then, n = 2k for some integer k. By squaring oth sides of this equation, we get n 2 = 2(2k 2 ). Since 2k 2 Z, y definition, n 2 is even. Conversely, suppose that n 2 is even. We need to show that n is even. This is too hard, it s etter to prove the contrapositive statement. Suppose that n is odd. Then, n = 2k + 1 for some k Z. But then n 2 = 2(2k 2 + 2k) + 1. Since 2k 2 + 2k Z, y definition, n 2 is odd. 2. Let x, y Z. Prove that if xy is even, then x is even or y is even. We prove the contrapositive statement. Suppose that oth x and y are odd. Then, x = 2m + 1 and y = 2n + 1 for some integers m and n. By multiplying, we get xy = (2m + 1)(2n + 1) = 2(2mn + m + n) + 1. Since 2mn + m + n Z, y definition, xy is odd. 3. Let x, y Z. Prove that if xy is odd, then x and y are odd. We prove the contrapositive statement. Suppose that either x is even or y is even. case 1: x is even By definition, x = 2k for some k Z so that xy = 2ky. Since ky Z, y 1

2 definition, xy is even. case 2: y is even By definition, y = 2k for some k Z so that xy = 2kx. Since kx Z, y definition, xy is even. So in any case, xy is even. 2

3 HW 2 1. Prove that 3 3 is irrational. Suppose y way of contradiction that 3 3 Q. Then, 3 3 = a where a and are integers and 0. If = 1 then 3 3 = a Z which is a contradiction, and we are done. Otherwise, assume that > 1. Fully reduce a so that gcd(a, ) = 1. Here is something you should know. The gcd(a, ) is just the greatest common divisor of a and. If we are assuming that a is reduced, then what we are really saying is that gcd(a, ) = 1. Let n e a positive integer. Assuming that gcd(a, ) = 1, suppose y way of contradiction that gcd(a n, n ) > 1. Then, gcd(a n, n ) must have some prime factor p. Since gcd(a n, n ) divides oth a n and n, p must divides oth a n and n. But then p must divide oth a and (we proved a special case of this in the notes), which is a contradiction ecause gcd(a, ) = 1 y assumption. Hence, gcd(a n, n ) is NOT greater than 1 so that it is less than or equal to 1. But since the gcd(a n, n ) is defined to e the largest POSITIVE integer that divides the numers a n and n without a remainder, it follows that gcd(a n, n ) must equal 1. Now, y cuing oth sides of the equation 3 3 = a a3, we get 3 =. From 3 what we have shown aove, a3 is reduced. But since > 1, a3 3 is not an 3 integer, which contradicts the latter equation. Therefore, 3 3 is irrational. 2. Prove that log 7 is irrational. Suppose y way of contradiction that log 7 Q. Then, log 7 = a where a and are integers and 0. But then 2 a 5 a = 7. And since 5 divides the left hand side, it must divide the right hand side, which is a contradiction. 3. Prove that n r(r!) = (n + 1)! 1. r=1 We will do a y induction. Let n = 1. It is easy to see that the equation is valid for n = 1. Let 3

4 us ASSUME that the equation is valid for n = k for some integer k > 1. Then, we have the equation k r(r!) = (k + 1)! 1. r=1 By adding (k + 1)[(k + 1)!] to oth sides, we get the equation k+1 r(r!) = (k + 1)! 1 + (k + 1)[(k + 1)!]. r=1 By factoring and comining terms, the right hand side ecomes (k+2)! 1. So y induction, n r(r!) = (n + 1)! 1 for all n. r=1 4

5 HW 3 1. For x, y R, define x y to mean that x y Z. Prove that this defines an equivalence relation on R. Let x R. Since x x = 0 Z, x x. This shows that the relation is reflexive. Let x, y R and suppose that x y. Then, x y Z so that y x = (x y) Z. This shows that the relation is symmetric. Finally, let x, y, z R and suppose that x y and y z. Then, x z = (x y) + (y z) Z so that x z. This shows that the relation is transitive. Therefore, defines an equivalence relation. 2. For (a, ), (c, d) R R, define (a, ) (c, d) to mean that 2a = 2c d. Prove that this defines an equivalence relation on R R. Let (a, ) R R. Since 2a = 2a, (a, ) (a, ). This shows that the relation is reflexive. Let (a, ), (c, d) R R and suppose that (a, ) (c, d). Then, 2a = 2c d which means that (c, d) (a, ). This shows that the relation is symmetric. Finally, let (a, ), (c, d), (e, f) R R and suppose that (a, ) (c, d) and (c, d) (e, f). Then, 2a = 2c d. Also, 2c d = 2e f. Then, 2a = 2e f so that (a, ) (e, f). This shows that the relation is transitive. Therefore, defines an equivalence relation. 3. Let R[x] denote the set of all polynomials with real numer coefficients. Let A = {f(x) R[x] : deg(f) is odd} and B = {f(x) R[x] : deg(f) is even}. Prove that R[x] = A B. Let f R[x]. Since the degree of f is a nonnegative integer, it must e either even or odd. So, y definition of the union, f A B. Since f was aritrary, R[x] A B. By how the sets A and B are defined, A B R[x]. Therefore, R[x] = A B. 5

6 HW 5 1. Show that the usual multiplication [x] [y] = [x y] in Z n does not depend on the choice of representative. Let [a], [], [c], [d] Z n, and suppose that [a] = [c] and [] = [d]. Then y definition, n divides oth a c and d. This means that there exist integers k 1 and k 2 such that nk 1 = a c and nk 2 = d. Now, a cd = a c + c cd = (a c) + c( d) = n(k 1 + ck 2 ). Since k 1 + ck 2 Z, y definition, n divides a cd so that [a] = [cd]. 2. Let f : A B e a function. Prove that f(a) = B iff f is surjective. By definition, we know that f(a) B. Suppose that f is surjective. Then for every element B, there exists an element a A such that f(a) =, which means that f(a). This means that B f(a), which proves that f(a) = B. Conversely, suppose that f(a) = B so that B f(a). If B then it lives in f(a) which means that there exists an element a A such that f(a) =. Hence, f is surjective. 6

7 HW 6 1. Let g : A B and f : B C e two functions. Prove that If f and g are ijections, then so is f g. Suppose that (f g)(a) = (f g)(). Then y definition, f(g(a)) = f(g()). Since f is injective, we get that g(a) = g(). And ecause g is injective, a =. This proves that f g is injective. Now let c C. Since f is surjective, there exists an element B such that f() = c. But since g surjective, there exists an element a A such that g(a) =. So, we found an element a A such that (f g)(a) = f(g(a)) = f() = c. This proves that f g is surjective. Therefore, f g is a ijection. 2. Let g : A B and f : B C e two functions. Prove that if f g is surjective, then so is f. Let c C. Since f g is surjective, there exists an element a A such that (f g)(a) = f(g(a)) = c. So, we have shown that for every c C, there exists an element B, namely = g(a), such that f() = c. This proves that f is surjective. Note that g is NOT necessarily surjective. Give a counterexample and think simple! The point here is to try and come up with a slick example. In class we proved the following theorem: Let f : X Y e a function. And let A X and B Y. Then, f(a B) f(a) f(b). 3. Prove that if f is injective, the inclusion in the theorem aove ecomes an equality. You should also give a counterexample showing that the reverse inclusion does NOT hold in general. Again, rememer to think simple. Let y f(a) f(b). By definition, there exist elements a A and B such that y = f(a) = f(). Since f is assumed to e injective, a = so that a A B. Since y = f(a), y definition, y f(a B). This proves that f(a) f(b) f(a B). 7

8 HW 8 1. Prove that a finite set is countale. Let X = {a 1, a 2,..., a n } e a finite set. To show that this set is countale, I need to cook up a function f : X N, and then prove it is injective. The function f(a i ) = i works just fine. Make sure you understand how this function is defined, and why it is injective. Recall the following proposition from the notes: Let X and Y e nonempty sets. Then, Card(X) Card(Y ) iff Card(Y ) Card(X). 2. Read over the of the proposition. Show that the function g : X f 1 ({x}) at the end of the is injective. x X Suppose that g(a) = g(). Since g(a) x X f 1 ({x}), it is in f 1 ({x 1 }) for SOME x 1 X. Similarly, g() f 1 ({x 2 }) for some x 2 X. But then the element g(a), eing equal to g(), lives in f 1 ({x 1 }) f 1 ({x 2 }). And since the collection {f 1 ({x}) : x X} are disjoint, it must e true that f 1 ({x 1 }) = f 1 ({x 2 }) so that x 1 = x 2. This proves that g is injective. Make sure that you read the entire in the notes. It s always good to see how the axiom of choice is utilized. Recall the following corollary from the notes: Z and Q are countale. 3. Assuming that the set S = { n : n N} is countale, we were ale to prove that Z is countale. Show that S is countale. Define the function f : S N y f(s) = s. Suppose that f(s 1 ) = f(s 2 ). Then, s 1 = s 2 so that y multiplying y -1 on oth sides, s 1 = s 2. This proves that f is injective. Therefore y definition, Card(S) Card(N) which means that S is countale. Recall the following fact from the notes: The interval (0, 1) is uncountale. 4. The fact aove was partially proven in class. Finish the y showing that the function f at the end is not surjective. 8

9 The function f : N (0, 1) is defined y f(n) = 0.a n1 a n2 a n3.... Assuming that f is a ijection, in particular a surjection, there exists a numer n N such that f(n) = 0.a n1 a n2 a n3... a nn... = n... But then this means that a nn = n which, y how we constructed the element, cannot e true. This gives the contradiction required at the end of the. 5. For any set X, prove that Card(X) Card(P(X)) Define the function f : X P(X) y f(x) = {x}. Supposes that f(x) = f(y). Then, {x} = {y} which means that x = y. This proves that the function f is injective. Therefore y definition, Card(X) Card(P(X)). 9