Math1a Set 1 Solutions

Math1a Set 1 Solutions October 15, 2018 Problem 1. (a) For all x, y, z Z we have (i) x x since x x = 0 is a multiple of 7. (ii) If x y then there is a k Z such that x y = 7k. So, y x = (x y) = 7k is also a multiple of 7, i.e. y x. (iii) If x y and y z then there exist k, l Z such that x y = 7k, y z = 7l and consequently x z = (x y)+(y z) = 7k +7l = 7(k +l) is also a multiple of 7. So, x z. Therefore, satisfies all axioms of an equivalence relation. (b) We will show that S = {0, 1, 2, 3, 4, 5, 6} is a set of representatives for Y = X/. To prove that S is a set of representatives, one has to show 1. Elements of S are pairwise inequivalent; 2. Every integer is equivalent to one element of S. Now we show these two for S 1. The difference between two different elements of S is never a multiple of 7. This means that elements of S are pairwise inequivalent. 2. Given any integer n, we may write (by division by 7) n = 7a + r, with a, r Z, such that the remainder r satisfies: 0 r 6, i.e. r S. Since n r is a multiple of 7, we know that n r. So, any integer n is equivalent to one element of S. We showed that S is a set of representatives for Y = X/. Hence the cardinality of Y is equal to the cardinality of S, which is 7. Problem 2. The method of contrapositive is to prove P = Q by showing Q = P. Let a, b Z. We are to show that if 1

(i) ab even either a or b is even (ii) ab odd both a and b are odd (i): To use the method of contrapositive for this problem, we let P be the statement that ab is even, and let Q be the statement that either a or b is even. Note that: 1. P is the statement that ab is not even, or equivalently ab is odd. 2. Q is the statement that neither a nor b is even, or equivalently a and b are both even. By the method of contrapositive, to prove this problem we have to show that Q = P, or equivalently if both a and b are odd, then ab is odd. We show this now. If both a and b are odd, then there are k, l Z such that a = 2k + 1, b = 2l + 1. Then ab = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1 is also odd. (ii): To use the method of contrapositive for this problem, we let P be the statement that ab is odd, and let Q be the statement that a and b are both odd. Note that: 1. P is the statement that ab is not odd, or equivalently ab is even. 2. Q is the statement that either a or b is not odd, or equivalently either a or b is even. By the method of contrapositive, to prove this problem we have to show that Q = P, or equivalently if either a or b is even, then ab is even. We show this now. If either a or b is even, we may assume a is even (the case of b being even can be done identically swapping the names of a and b). So, there is a k Z such that a = 2k. This means ab = 2kb, i.e. ab is even. Problem 3. Set up for the proof by contradiction: to prove the statement of the problem by contradiction, we keep the conditions (i.e. x and y are positive integers), but we assume the opposite of what we are trying to prove (ie. that there exist such x and y satisfying the equation x 2 y 2 = 2). 2

Factorization: We have x 2 y 2 = 2, so we can factorize both the left and right hand sides. The left side factorizes as (x y)(x + y) and since both x and y are positive integers, it follows x y is an integer (could be negative!) and x+y is a positive integer. Thus, we need to factorize 2 into two integers at least one of which is positive. This can only be done as 1 2 or 2 1. Reaching a contradiction: The two factorisations of 2 lead to the two systems of equations: x y = 2, x + y = 1, or x y = 1, x + y = 2. Both of them yield 2x = 3, so there is no solution where x is an integer, so we contradict our condition on x. Therefore, we conclude by saying our assumption must have been wrong and therefore, there do not exist positive integers x and y such that x 2 y 2 = 2. Replacing 2 by 2n: We proceed exactly as before. Let x and y be positive integers and assume there exists a solution to x 2 y 2 = 2n for some n odd. Factorizing as before gives (x y)(x+y) = 2n. Again, x y and x+y are integers. As the left hand side is the product of two integers, we should factorize the 2n into two integers, say p q. (Be careful here, there may be more factorisations than just 2 n or n 2.) Since 2n is even, at least one of p or q is even. If both p and q are even then 2n = p q is divisible by 4 which is not the case since n is odd. So, exactly one of p and q is even and the other is odd. Recall that the sum of an even integer and an odd integer is odd. But then 2x = (x y) + (x + y) = p + q is odd. So, x is again not an integer contradicting the condition on x. So, we conclude that our assumption was wrong and indeed there do not exist positive integers x and y such that x 2 y 2 = 2n where n is odd. Problem 4. [Note: a key fact that is used in the problem is the following. m is a positive integer such that m 2 n = m n for all positive integer n, iff m is square-free. In particular, this applies when m is a prime number. It is important in your solutions to make note that you understand that this is true for primes as it is not true for all numbers, for example m = 4.] To prove the statement of the problem by contradiction, we assume the opposite of what we are trying to prove (ie. that 14 is rational). We now use this fact to reach a contradiction. Since we have assumed 14 is rational, we can write it as the ratio of two integers, ie. 14 = a where a, b Z and b 0. If a is negative, then b must be too, so we can multiply a and b by 1 to make it a ratio of positive integers. In the same vein, if a and b have a positive integer divisor d in common, we can divide the numerator and denominator by d and still have 14 as the ratio of positive integers. Thus, b 3

the natural conclusion is to divide through by the greatest common divisor of a and b leaving 14 as the ratio of two positive integers with no positive integer factor in common other than 1. Hence, 14 = A B where A, B N and gcd(a, B) = 1 (We have ignored the possibility A = 0 since 14 0). Rearranging and squaring yields 14B 2 = A 2. The left hand side is divisible by 2, so the right hand side must be as well. Thus, 2 divides A 2 and hence 2 divides A as 2 is prime. We may write A = 2C where C is a positive integer. Substituting in gives 14B 2 = 4C 2. Dividing by 2 yields 7B 2 = 2C 2. The right hand side is divisible by 2, so the left hand side must be too. Since 2 does not divide 7, it must divide B 2 and hence 2 divides B since 2 is prime. So, we can write B = 2D where D is a positive integer. However, note that therefore, both A and B have a common divisor of 2 which contradicts the fact that gcd(a, B) = 1. So, our assumption is wrong and in fact, 14 is irrational. Before we continue to the second part of the problem, note that if we add, multiply, subtract, or divide two rational numbers, we again get a rational number. The second part of the problem requires us to show that 2+ 7 is irrational. Again, we proceed with a proof by contradiction. Assume 2 + 7 is rational. So, write 2 + 7 = A B where A, B Z and B 0 (we could again insist that gcd(a, B) = 1, but it actually will not be necessary). Then B( 2 + 7) = A, which after squaring yields B 2 (2 + 2 14 + 7) = A 2. We now want to isolate the 14, so subtracting 9B 2 and dividing by 2B 2 (which is nonzero because B 0) gives 14 = A 2 9B 2 2B 2. Note that the numerator is an integer as A, B and 9 are. Moreover, the denominator is a non-zero integer since B 0. So, we have written 14 as a ratio of an integer with a non-zero integer. So we conclude 14 is rational which contradicts the first part of this problem where we proved 14 is actually irrational. So, we conclude that our assumption was false and indeed it follows that 2 + 7 is irrational. Problem 5. The proposition we are going to prove is the following: P(n): Given polynomial f of degree n and real number b, there exists a polyno- 4

mial g of degree n 1 and a real number c such that f(x) = (x b)g(x) + c. In this form, we can prove this using induction on n, the degree of our polynomial. One must start with the base case, which entails proving the statement for n = 1. Base Case: We want to show that if f(x) = a 1 X + a 0, with a 0, a 1 R, and b R, we can find a degree n 1 = 0 polynomial (i.e. a constant, say g R), and c R, such that f(x) = (x b)g + c. We can actually just solve this by equating polynomials: a 1 x + a 0 = gx bg + c. This yields g(x) = g = a 1 and c = a 0 + bg = a 0 + ba 1. Inductive Step: To complete this step, one must prove that the statement is true for n = k + 1 assuming it is true for all n k. First, we assume that for all n k, for any degree n polynomial f and real number b, there is a degree n 1 degree polynomial g and real number c such that f(x) = (x b)g(x) + c. (1) Next, we look at an arbitrary k + 1 degree polynomial F (x) = α k+1 x k+1 + α k x k + + α 0. Now for arbitrary b R, we can observe that: F (x) = x(α k+1 x k + α k x k 1 + + α 1 ) + α 0. (2) Notice that inside the brackets in Equation 2, is the degree k polynomial: α k+1 x k + α k x k 1 + + α 1. So, we can apply Equation 1, which tells us that there is a degree k 1 polynomial g(x) and c R such that α k+1 x k + α k x k 1 + + α 1 = (x b)g(x) + c. (3) Substituting Equation 3 into Equation 2 yields F (x) = x((x b)g(x) + c) + α 0. (4) We are not done as Equation 4 is not in the form of the proposition. We are looking for a degree k polynomial G and a real number C such that F (x) = (x b)g(x) + C. (5) Like in the base case, we attempt to solve for G and C simply using polynomial equality. In particular, using Equations 4 and 5 gives us (x b)g(x) + C = x((x b)g(x)) + c) + α 0 = (x b)(xg(x)) + cx + α 0 = (x b)(xg(x) + c) + cb + α 0. So, setting G(x) = xg(x) + c and C = cb + α 0 gives the required G and C and the induction is complete. 5

Hence, by the principle of mathematical induction, given polynomial f of degree n and real number b, there exists a polynomial g of degree n 1 and a real number c such that f(x) = (x b)g(x) + c. The second part of the problem requires us to show that b is a root of f iff c = 0. This is now much easier after having completed the induction. (only if direction): Suppose b is a root of of f. Then f(b) = 0, so (b b)g(b)+ c = 0 which simplifies to c = 0. (if direction): Suppose c = 0, then f(x) = (x b)g(x). Then f(b) = (b b)g(b) = 0, so b is a root of f. Variations on the inductive step in Problem 5 There are a couple of variations to the proof of the induction step worth mentioning. First of all, one could eliminate the top term by writing F (X) = α k+1 x k (x b) + (α k + bα k+1 )x k + α k 1 x k 1 + + α 1 + α 0. Then notice that (α k + bα k+1 )x k + α k 1 x k 1 + + α 1 + α 0 is of degree at most k, so we can apply the inductive step here: (α k + bα k+1 )x k + α k 1 x k 1 + + α 1 + α 0 = (x b)g(x) + c. Then, F (x) = α k+1 x k (x b)+(x b)g(x)+c = (x b)(g(x)+α k+1 x k )+c. So, G(x) = g(x) + α k+1 x k and C = c satisfy the requirements. Another variation is to first prove that for single term polynomials of the form x n, for any b R, there exists some degree n 1 polynomial, h(x), and c R such that x n = (x b)h(b)+c. The induction step in this proof can be done by writing x k+1 = x x k = x(x b)h(x)+cx = (x b)(xh(x)+c)+bc. Next, we can observe that the sum of degree k polynomial and degree < k polynomials is also a degree k polynomial. Also, that a nonzero scalar product of a degree n polynomial is also degree n. Now, the induction step of the original problem can be proven by applying the first step. Denote F 1 (x) = α n x n and F 2 (x) = α n 1 x n 1 + + α 0 (which is not necessarily a degree n 1 polynomial as α n 1 could be 0). Then F (x) = F 1 (x) + F 2 (x) = (x b)h(x) + c 1 + (x b)g(x) + c 2, where h(x) is a degree n 1 polynomial, and g(x) is of degree at most n 2, hence g(x) + h(x) is a degree n 1 polynomial. 6