University of Toronto Faculty of Arts and Science Solutions to Final Examination, April 2017 MAT246H1S - Concepts in Abstract Mathematics

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1 University of Toronto Faculty of Arts and Science Solutions to Final Examination, April 2017 MAT246H1S - Concepts in Abstract Mathematics Examiners: D. Burbulla, P. Glynn-Adey, S. Homayouni Time: 7-10 PM, April 12, Aids permitted: None. General Comments: 1. Questions 2(e) and 2(f) were right out of the exercises from Chapter 7; yet they were poorly done. 2. The statement in 8(e) is False; no part marks were awarded for proving it is True. 3. The overall average for Question 2 was 16.9/30; for Question 8, it was 15.8/ The proofs in Questions 4(b) and 6 were right out of the book. Breakdown of Results: 264 students wrote this test. The marks ranged from 8 to 114 (out of 120), and the average was out of 120. A histogram of the marks (by decade) is below:

2 1.(a) [2 marks] Define the cardinal numbers ℵ 0 and c. Solution: ℵ 0 = N and c = R 1.(b) [8 marks; 1 for each part] Indicate if the cardinality of the following sets is ℵ 0, c, or neither. (No justification is required.) Z Answer: ℵ 0 Q Answer: ℵ 0 the open interval (0, 1) Answer: c the set of algebraic numbers Answer: ℵ 0 the set of transcendental numbers Answer: c the set of constructible numbers Answer: ℵ 0 the set of all subsets of N Answer: c the set of all subsets of R Answer: neither Average this page: 7.83/10 2

3 2.(a) [6 marks; 2 for each part] State the following theorems: (i) Fermat s Theorem Solution: if p is a prime number and a is any natural number not divisible by p, then a p 1 1 (mod p). (ii) Wilson s Theorem Solution: if p is a prime number, then (p 1)! (mod p). (iii) Euler s Theorem Solution: if m is a natural number greater than 1 and a is a natural number that is relatively prime to m, then a φ(m) 1 (mod m). 2.(b) [4 marks] Prove: if p is a prime number that does not divide a, then a p2 a p (mod p 2 ). Solution: since a and p are relatively prime, so are a and p 2. Euler s Theorem a φ(p2) 1 (mod p 2 ) a p2 p 1 (mod p 2 ) a p2 p a p 1 a p (mod p 2 ) a p2 a p (mod p 2 ) Average this page: 7.63/10 3

4 2.(c) [4 marks] Use Wilson s Theorem to prove: if p is a prime number greater than 3, then 2(p 3)! 1 (mod p). Solution: p > 3, so (p 1)! = (p 1) (p 2) (p 3)!. Thus Wilson s Theorem (p 1)! 1 (mod p) (p 1) (p 2) (p 3)! 1 (mod p), by above observation (0 1) (0 2) (p 3)! 1 (mod p), since p 0 (mod p) ( 1) ( 2) (p 3)! 1 (mod p) 2 (p 3)! 1 (mod p) 2.(d) [6 marks] Find the remainder when (34! ) 39 is divided by 37. Solution: there are many ways to calculate this, but you will need to use one or all of Wilson s Theorem, Fermat s Theorem and part (c). Also, you will need to use the fact that 37 is prime; 75 1 (mod 37) (mod 37) 1 (mod 37); and, if a and 37 are relatively prime, then a 36 1 (mod 37) a 39 a 3 (mod 37). Method 1: from part (c) we know 2 34! 1 (mod 37). So 34! (34!+1) (34! 36!) 34!( ) 34!(1+35) 34! ! ( mod 37) (34! ) 39 ( 18) 3 ( ) (mod 37) Method 2: 2 34! 1 36 (mod 37) 34! 18 (mod 37). Then (34! ) 39 (18 + 1) ( 18) 3 (mod 37) and proceed as before. Method 3: work with 19 instead of 18 : (34! ) (mod 37) (mod 37) (mod 37), since 19 2 = (mod 37) (mod 37) (mod 37) 14 1 (mod 37) 14 (mod 37) Average this page: 7.01/10 4

5 2.(e) [5 marks] Suppose that a and m are relatively prime numbers and k is the smallest natural number such that a k 1 (mod m). Prove that k divides φ(m). Solution: by Euler s Theorem, a φ(m) 1 (mod m). Thus k φ(m). Use the division algorithm to write φ(m) = q k + r, for some non-negative integers q, r with r {0, 1, 2,... k 1}. If r = 0 we are done. Suppose r > 0. Then a φ(m) 1 (mod m) a q k+r 1 (mod m) (a k ) q a r 1 (mod m) a r 1 (mod m), since a k 1 (mod m) Since r < k, this contradicts the assumption that k is the smallest number such that a k 1 ( mod m). Thus r = 0 and k divides φ(m). Alternate Solution: there are integers x, y such that gcd(k, φ(m)) = xk + yφ(m); so a gcd(k,φ(m) = (a k ) x (a φ(m) ) y 1 (mod m) k gcd(k, φ(m)) k gcd(k, φ(m)) = k k φ(m) 2.(f) [5 marks] Show that the number m is prime if there is a number a such that a m 1 1 ( mod m) and a k is not congruent to 1 (mod m) for every natural number k less than m 1. Solution: first note that a and m must be relatively prime: a m 1 1 (mod m) implies there is an integer x such that a m 1 = 1 + xm a m 1 xm = 1. So if p is a prime that divides both a and m then p must divide 1; which is impossible. Then there are at least two ways to proceed: Directly: we have that k = m 1 is the smallest natural number such that a k 1 (mod m). Since a and m are relatively prime, part(e) implies that m 1 divides φ(m). Therefore m 1 φ(m). On the other hand, by definition of the Euler function, φ(m) m 1. Combining these two inequalities gives φ(m) = m 1; that is, every number from 1 to m 1 is relatively prime to m. Thus the only divisors of m are 1 and itself; so m must be prime. By Contradiction: suppose m is not prime. Then there is a divisor of m in the set {2, 3,..., m 1} and consequently φ(m) < m 1. But since a and m are relatively prime, Euler s Theorem implies that a φ(m) 1 ( mod m). Thus for k = φ(m) we have a k 1 ( mod m) and k < m 1. Contradiction. Average this page: 2.25/10 5

6 3. [2 marks for each part] Decide if the following statements are True or False. If the statement is True, explain why; if it is False, give a counter-example. (a) The sum of two rational numbers is rational. True: if r 1 = m 1 n 1 and r 2 = m 2 n 2, then r 1 + r 2 = m 1n 2 + m 2 n 1 n 1 n 2. Or simply say, the rational numbers are a field, so they are closed under addition. (b) The sum of two irrational numbers is irrational. False: (3 2) + (2 + 2) = 5 (c) If x and y are irrational numbers then x y is irrational. False: you have to use logic on this one. Consider the two irrational numbers 2 and 3. If ( 3) 2 is rational, then we have a counter-example. If ( 3) 2 is irrational, then and we have a counter-example. ( ( 3) 2 ) 2 = ( 3) 2 = 3, (d) There are uncountably many irrational numbers. True: the rational numbers are countable. If the irrational numbers were also countable, then the real numbers would be the union of two countable sets. This would imply that the real numbers are countable, which is a contradiction. (e) Every irrational number is a surd. False: e, π, 3 2 are all examples of irrational numbers that are not surds. Average this page: 6.97/10 6

7 4. Let S and T be sets, not necessarily finite. (a) [4 marks; 2 for each part] Define the following: (i) S < T Solution: S T but S T (ii) P(S), the power set of S Solution: the set of all subsets of S (b) [6 marks] Prove: S < P(S). Solution: reproduce the (classic) proof in the book. First show that S P(S) : Define f : S P(S) by f(s) = {s}. f is one-to-one: f(s 1 ) = f(s 2 ) {s 1 } = {s 2 } s 1 = s 2. Therefore S P(S). Second, show that S P(S) : Suppose that g : S P(S) is any one-to-one function. Suppose it is also onto. Consider the subset T of S defined by T = {s S s / g(s)}. Because g is onto, and T P(S), there is a t S such that T = g(t). Is t g(t)? If the answer is yes, then t g(t) = T. But by definition of T this means t / g(t), and we have a contradiction. It the answer is no, then t / g(t). Then, by the definition of T, we have t g(t), and again we have a contradiction. Thus the function g cannot be onto. And so S P(S). Therefore S < P(S). Average this page: 5.15/10 7

8 5.(a) [2 marks] State the Cantor-Bernstein Theorem. Solution: if S and T are sets such that S T and T S, then S = T. 5.(b) [8 marks; 4 for each part] Apply the Cantor-Bernstein Theorem to prove that: (i) N N N N = {(k, l, m, n) k, l, m, n N} is countable. Solution: define f : N N N N N by f(n) = (n, 1, 1, 1). Then f is one-to-one: f(m) = f(n) (m, 1, 1, 1) = (n, 1, 1, 1) m = n. Thus N N N N N. To obtain the other inequality, define g : N N N N N by g( (m 1, m 2, m 3, m 4 ) ) = 2 m1 3 m2 5 m3 7 m 4. g is one-to-one by the Fundamental Theorem of Arithmetic. Thus N N N N N. By the Cantor-Bernstein Theorem, N N N N = N, so N N N N is countable. (ii) the unit cube, [0, 1] [0, 1] [0, 1] = {(x, y, z) x, y, z [0, 1]}, has the same cardinality as [0, 1]. Solution: define f : [0, 1] [0, 1] [0, 1] [0, 1] by f(x) = (x, 0, 0). Then f is one-to-one: f(x) = f(y) (x, 0, 0) = (y, 0, 0) x = y. Thus [0, 1] [0, 1] [0, 1] [0, 1]. To obtain the other inequality, consider each element (a, b, c) [0, 1] [0, 1] [0, 1] in terms of infinite decimal expansions a = 0.a 1 a 2 a 3 a 4..., b = 0.b 1 b 2 b 3 b 4..., c = 0.c 1 c 2 c 3 c 4... and define g : [0, 1] [0, 1] [0, 1] [0, 1] by g( (a, b, c) ) = 0.a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 a 4 b 4 c 4... Then g is one-to-one since g( (x, y, z) ) = 0.a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 a 4 b 4 c 4... x = 0.a 1 a 2 a 3 a 4..., y = 0.b 1 b 2 b 3 b 4..., z = 0.c 1 c 2 c 3 c 4... Therefore [0, 1] [0, 1] [0, 1] [0, 1]. By the Cantor-Bernstein Theorem, [0, 1] [0, 1] [0, 1] = [0, 1]. Average this page: 4.43/10 8

9 6. [10 marks] Use the Principle of Mathematical Induction to prove that any finite set with n elements has 2 n subsets. Solution: n = 1. Suppose S is a set and S = 1. Then S = {s} for some element s, and the only subsets of S are the empty set,, and S itself. So S has 2 = 2 1 subsets. Now suppose n = k + 1 and that any finite set with k elements has 2 k subsets. Let S be a set with k + 1 elements: S = {s 1, s 2, s 3,..., s k, s k+1 }. Let S be the subset of S that consists of the k elements in S except for s 1. That is, S = {s 2, s 3,..., s k, s k+1 } = S \{s 1 }. Observe that S = k, so by the inductive hypothesis, S has 2 k subsets. Now consider any subset X of S. There are two possibilities: s 1 / X or s 1 X Case 1; if s 1 / X, then X S, and by induction, there are 2 k such subsets. Case 2: if s 1 X, then X = {s 1 } Y, where Y S. By induction, there are 2 k such subsets Y ; so there are 2 k subsets X in this case as well. In total, then, the number of subsets of S is 2 k + 2 k = 2 2 k = 2 k+1. By the Principle of Mathematical Induction, any finite set with n elements has 2 n subsets. Average this page: 7.51/10 9

10 7. [10 marks] Find the fourth roots of i, and plot them in the complex plane. Solution: first put i, into polar form: 8 ( ) i = i 1 2 ( ( π ) ( π )) = 16 cos + i sin 6 6 Let z = r(cos θ + i sin θ) and suppose ( ( π ) ( π )) z 4 = 16 cos + i sin 6 6 By De Moivre s Theorem, z 4 = r 4 (cos(4θ) + i sin(4θ)). Thus ( ( π ) ( π )) z 4 = 16 cos + i sin 6 6 ( ( π ) ( π )) r 4 (cos(4θ) + i sin(4θ)) = 16 cos + i sin 6 6 r 4 = 16 and 4θ = π 6 + 2πk r = 2 and θ = π 24 + πk 2 Then the four distinct solutions for z are z j = 2(cos θ j + i sin θ j ) with θ 1 = π 24, θ 2 = 13π 24, θ 3 = 25π 24, θ 4 = 37π 24. ( z 2 = 2 cos ( ) 13π + i sin 24 ( )) 13π 24 ( z 3 = 2 cos ( ) 25π + i sin 24 ( )) 25π 24 ( ( π ( π z 1 = 2 cos + i sin 24) 24)) ( z 4 = 2 cos ( ) 37π + i sin 24 ( )) 37π 24 Note: the four points are all on a circle of radius 2. Average this page: 6.66/10 10

11 8.(a) [12 marks] Define the following: (i) [4 marks] F is a subfield of R Solution: F is a set of real numbers satisfying four properites 1. 0 and 1 are in F 2. if x and y are in F, then both x + y and xy are in F 3. if x is in F, then x is in F 4. if x is in F and x 0, then 1 is in F x (ii) [2 marks] F( r), the extension of F by r Solution: if F is a subfield of R and r is a positive real number in R such that r is not in F, then F( r) = {a + b r a, b F} (iii) [4 marks] F 0 F 1 F 2 F n is a tower of fields Solution: each F i is a subfield of R such that Q = F 0, and for each i, 1 i n, F i is a field extension of F i 1 : there is a positive number r i in F i 1 such that r i is not in F i 1 and F i = F i 1 ( r i ). (iv) [2 marks] x is a surd Solution: a surd is a number that is in some field that is in a tower; that is, x is a surd if there exists a tower F 0 F 1 F 2 F n such that x F n. Average this page: 8.71/12 11

12 8.(b) [3 marks] Is Q Q( 3 2) a tower of fields? Explain your answer. Solution 1: No: prime numbers m and n such that 3 2 is not the square root of any positive number in Q. Suppose there are relatively 3 2 = m n 4 = m3 n 3 4 n3 = m 3. But then 2 m 3 2 m 8 4 n 3 2 n 3 2 n, and so m and n have a common factor 2. This contradicts our choice of m and n as relatively prime. Solution 2: No. If Q Q( 3 2) is a tower of fields, then x = 3 2 is constructible. This would say that the cubic equation x 3 2 = 0 has a constructible solution. But (by a theorem in the book), if the cubic equation x 3 2 = 0 has a constructible root, it must have a rational root. The only rational possibilities are ±1, ±2, but none of these satisfy the cubic equation x 3 2 = 0. 8.(c) [5 marks] Determine whether the number is constructible. Solution 1: Yes, it is constructible. Use the fact that if x is positive and constructible, then x is constructible. So 1 + Solution 2: Yes, 2 is constructible 2 is constructible is constructible is constructible is constructible is constructible is constructible. We have Q( 2), ( 2 Q( 2)) ( 3 2 ) 2, and x = [ ( 2 Q( ) ( 2) 3 2 )] ( ) 2. So x is a surd; and all surds are constructible. Average this page: 3.61/8 12

13 8.(d) [5 marks] Explain why the polynomial x 3 x 2 + 3x 2 has no constructible roots. Solution: make use of theorems that were proved in class (or in the book.) In particular, the cubic equation x 3 x 2 + 3x 2 = 0 has rational coefficients. So if the cubic equation has a constructible root, it must have a rational root. By the Rational Roots Theorem, the polynomial x 3 x 2 + 3x 2 has a rational root r only if r = m, with m dividing 2, and n dividing 1. n Thus r = ±1 or ± 2. But no four of these values satisfy the polynomial: ( 1) 3 ( 1) 2 + 3( 1) 2 = 7 0; (+1) 3 (+1) 2 + 3(+1) 2 = 1 0; ( 2) 3 ( 2) 2 + 3( 2) 2 = 20 0; (+2) 3 (+2) 2 + 3(+2) 2 = 8 0. Thus the polynomial x 3 x 2 + 3x 2 can have no constructible roots Average this page: 2.9/5 13

14 8.(e) [5 marks] Prove, or disprove by giving a counterexample, that any fifth degree polynomial with rational coefficients that has a constructible root must have a rational root. Solution 1: the statement is False. Consider the fifth degree polynomial (x 2 2)(x 3 x 2 + 3x 2) = x 5 x 4 + x 3 6x + 4. It has rational coefficients, and it has two constructible roots: x = ± 2. If the fifth degree polynomial had a rational root, then it would have to be a rational root of the cubic polynomial x 3 x 2 + 3x 2. But from part (d), we know that this cubic polynomial has no rational roots. Solution 2: same as Solution 1, just a different cubic part. Consider the fifth degree polynomial (x 2 2)(x 3 + x + 1) = x 5 x 3 + x 2 2x 2. It has rational coefficients, and it has two constructible roots: x = ± 2. If the fifth degree polynomial had a rational root, then it would have to be a rational root of the cubic polynomial x 3 + x + 1. But by the Rational Roots Theorem, the only rational roots could be ±1, neither of which satisfies x 3 + x + 1 = 0. Average this page: 0.61/5 14

15 This page is for rough work or for extra space to finish a previous problem. It will not be marked unless you have indicated in a previous question to look at this page. 15

16 This page is for rough work or for extra space to finish a previous problem. It will not be marked unless you have indicated in a previous question to look at this page. 16 The end.

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