MATH10040: Numbers and Functions Homework 1: Solutions

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1 MATH10040: Numbers and Functions Homework 1: Solutions 1. Prove that a Z and if 3 divides into a then 3 divides a. Solution: The statement to be proved is equivalent to the statement: For any a N, if 3 does not divide a then 3 does not divide a. Now, if 3 does not divide a, there are two possibilities: (a) a leaves remainder 1 on division by 3; i.e. a = 3k + 1 for some k N. (b) a = 3k + We consider both possibilities separately: (a) If a = 3k + 1, then a = (3k + 1) = 9k + k + 1 = 3 (3k + k) + 1 = 3m + 1 leaves remainder 1 also. (b) If a = (3k + ), then a = 9k + 1k + 4 = 3 (3k + 4k + 1) + 1 = 3t + 1 leaves remainder 1. This proves the result.. Prove that 3 is an irrational number. Solution: Suppose, FTSOC, that 3 were rational. Then (as in the case of ) we can write 3 = m n where m, n are natural numbers, not both even. Now it follows that n 3 = m 3 and hence that m 3 is even. Since the cube of an odd number is odd, m is necessarily even. But then m = t for some t N, and thus n 3 = 8t 3 = n 3 = 4t 3, so that n 3 is even, and hence n also is even. This contradicts the fact that not both n and m are even. This contradiction completes the proof that 3 is rational. 3. Prove that is irrational. Solution: Suppose, FTSOC, that were rational. So we can write = m n where m, n are natural numbers, not both even. Now it follows that n = m and hence that m is even. Since the square of an odd number is odd, m is necessarily even.

2 But then m = t for some t N, and thus n = 4t = 3n = t, so that 3n is even, and hence n is even ( since if a is odd then 3a is also). Thus n itself is also even. This contradicts the fact that not both n and m are even. This contradiction completes the proof that is rational. 4. (a) Show that the sum or difference of two rational numbers is rational. (b) Suppose that a 0 is rational and b is irrational. Show that a + b and ab are irrational. Solution: (a) Let a, b Q. Then a = m n and b = r s for some m, n, r, s Z. Thus a ± b = m n ± r s = ms ± nr ns is also a ratio of two integers (since ns, ms ± nr Z) and hence is rational. (b) Suppose, FTSOC, that c = a + b were rational. Then, since c is rational and a is rational, it follows from (a) that c a = b is rational, a contradiction. Prove that if r, s are rational, then so are rs and r/s (if s 0). Now suppose, FTSOC, that d = ab were rational. Then since d is rational and a 0 is rational, we deduce that d/a = b is rational, a contradiction. 5. Prove that + 3 is irrational. Solution: Suppose, FTSOC, that a = + 3 were rational. Since the square of a rational number is rational it follows that a = 5+ is rational. Therefore b = a 5 = is rational, and thus b/ = is rational, a contradiction.. Give an example of two distinct irrational numbers a and b which have the property that a + b is rational. Prove your assertions. Solution: Let a = and b =. (a b since a > 1 and b < 1). Show that b is irrational. But a + b = is rational. (Exercise: Use this idea to create 10 more examples.)

3 7. Show that there exist positive irrational numbers a and b with the property that a b is rational. [Hint: Consider and ( ).] Solution: Either is rational or it isn t. (I don t know which.) If it is rational, we can take a = b =. If it is not rational, take a = and b =. Then is rational. a b = ( ) = ( ) = ( ) = Either way, we have shown that there must be a, b R irrational with a b rational. 8. (a) Let n Z be of the form 4k + 3 for some k Z; i.e. n leaves remainder 3 on division by 4. Suppose that n = ab with a, b Z. Prove that exactly one of a and b is of the form 4t + 3. (b) Show that if a is a positive integer of the form 4k + 3, then a has a prime factor of the form 4k + 3. (c) Prove that there are infinitely many primes of the form 4k + 3. Solution: (a) a and b must both be odd, since otherwise ab = n would be even. Thus a and b are each either of the form 4n + 1 or 4n + 3. We must rule out the possibility that they are both of the same type. i. Suppose that a = 4n + 1 and b = 4m + 1. Then n = ab = 1nm + 4n + 4m + 1 = 4(4nm + n + m) + 1, a contradiction. ii. Suppose that a = 4n + 3 and b = 4m + 3. Then n = ab = 1nm + 1n + 1m + 9 = 4(4nm + 3n + 3m + ) + 1, a contradiction. Thus one of them is of the form 4n + 1 and the other of the form 4m + 3. (b) We will prove this statement by (complete) induction on k 1: If k = 1, then a = 3 and is prime. So the statement is true in this case. Suppose that we have proved it for all numbers k and consider a = 4(k + 1) + 3. If a is prime we re done. Otherwise a = bc where b, c < a and one of them, b say, is of the form 4n + 3 for some n k. By our inductive hypothesis, b has a prime factor p of the form 4m + 3. But p is also a factor of a. So we re done.

4 (c) We carefully adapt Euclid s proof of the infinity of primes: Suppose, FTSOC, that there are only finitely many primes, 3 = q 1, 7 = q,..., q t, of type 4n + 3. Then consider a = 4q 1 q q t 1 = 4(q 1 q q t 1) + 3. a is of the type 4n + 3. By the previous result it has a prime factor, p say, of the form 4n + 3. But none of the q i divide a. (For a = q i s 1 for some s N.) Thus p is not one of the q i. This contradicts the supposition that we have already listed all primes of this form. 9. Prove by induction on n that for all natural numbers n n = n(n + 1)(n + 1). Solution: For n = 1 the statement reads and so is true. 1 = 1 ( + 1) = = 1 Suppose we have proved the statement for some n. Then n + (n + 1) n(n + 1)(n + 1) = + (n + 1) (by our inductive hyp.) [ ] [ ] n(n + 1) n + 7n + = (n + 1) + n + 1 = (n + 1) (n + 3)(n + ) = (n + 1) which is precisely the statement for n + 1. So we re done, by induction. 10. Suppose that x is number other than 1. Prove by induction on n that for all natural numbers n x + x + + x n = xn+1 1. Solution: For n = 1, the statement reads which is clearly true. 1 =

5 Suppose that we have proved the statement for some n. Then 1 + x + + x n + x n+1 = xn+1 1 = xn x n+1 () which is the statement for n Prove by induction on n that ( for all integers n. ) ( Solution: When n = the statement is + x n+1 (by our inductive hyp.) = xn x n+ x n+1 = xn+ 1 ) (1 1n ) = n + 1 n which is visibly true = 3 4 Suppose the statement is true for n. Then ( 1 1 ) (1 1n ) ( ) (n + 1) = n + 1 n which is the statement for n + 1. ( 1 = n + 1 n (n + 1) 1 = (n + 1) 1 (n + 1) n(n + 1) = n + n n(n + 1) 1. Prove by induction that n > n for all n 4. = n + (n + 1) ) 1 (n + 1) Solution: For n = 4 we have 4 = 1 = 4, so the statement is true. Suppose the statement is true for some n 4. Then by our inductive hypothesis. But n+1 = n > n n (n+1) = n n 1 = (n 1) 3 > 0 = n > (n+1) and thus n+1 > (n + 1) as required. (by our inductive hyp.)

6 13. Let p 1 =, p = 3, p 3 = 5, p 4 = 7,... be the sequence of primes. What is p 13? Prove that p n n + 15 for all n 13. Solution: We prove the statement by induction on n 13. p 13 = 41 = So the statement is true for 13. Suppose now that the statement is true for some n 13. Then p n+1 p n + (why?) and p n n + 15 by our inductive hypothesis. Thus p n+1 (n + 15) + = (n + 1) + 15 as required. 14. An infinite sequence of integers is defined by a 1 = 7 and a n+1 = 5a n 8 for all n 1. Find and prove a formula for a n. Solution: We prove that a n = 5 n + by induction on n. The case n = 1 is clear. Suppose the result is proved for n. Then a n+1 = 5a n 8 = 5 (5 n + ) 8 = 5 n = 5 n+1 + as required.