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3 Hrs MAT246H1S - Concepts in Abstract Mathematics Examiners: D

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3 Hrs MAT246H1S - Concepts in Abstract Mathematics Examiners: D University of Toronto Faculty of Arts and Science Solutions to Final Examination, April 2019 Duration: 3 hrs MAT246H1S - Concepts in Abstract Mathematics Examiners: D. Burbulla, S. Homayouni Time: 9 AM-12 Noon, April 22, 2019. Aids permitted: None. General Comments: • Since it was only required to do 12 questions, there were many pages left blank. In particular, many students ended up skipping one or two of Questions 3, 5, 6, 7, 8, 9, 10, 11, 12 and 14. The breakdown of the number of zero's by question is: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 3 0 28 4 99 44 86 49 55 31 35 98 5 130 • On the whole, the proof questions were done very poorly. Note that Question 12 was actually a very simple question, but many students didn't even try it. • There were only six questions with passing averages: Questions 1, 2, 3, 9, 10 and 13. For the most part these questions were either definitions or computations. It seems as if proofs are still a challenge to most students. Breakdown of Results: 216 students wrote this exam. The marks ranged from 2/140 to 129/140, and the average was 67.2/140. However, since we ended up taking your score on the exam as a mark out of 100, the average on the exam is in effect 67.2%. (It turned out that fifteen students had a mark greater than 100.) A histogram of the results by decade is to the right: 1. [course avg: 7.33/10] Define the following. Your definitions must be concise, accurate, complete and a definition, not an equivalent condition. (a) [2 marks] A multiplicative inverse of the natural number a; modulo m: Solution: a multiplicative inverse of the natural number a; modulo m; is an integer (accept natural number) x such that a x ≡ 1 (mod m): (b) [3 marks] A tower of fields. Solution: a finite sequence F ; F ;:::; F of subfields of such that F = ; and for each i; 0 1 n Rp 0 Q p from 1 to n; there is a positive number ri 2 Fi−1 such that ri 2= Fi−1 but Fi = Fi−1( ri); is called a tower of fields. We have F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn−1 ⊂ Fn: (c) [1 mark] A surd. Solution: a surd is a number that is in some field that is in a tower of fields. (d) [1 mark] An algebraic number. Solution: a real number is algebraic if it is a root of a polynomial with integer coefficients. (e) [1 mark] The power set of a set S: Solution: the power set of S; denoted by P(S); is the set of all subsets of S: p (f) [1 mark] Q( 7) p p Solution: Q( 7) = fa + b 7 j a; b 2 Qg (g) [1 mark] The characteristic function fT of a subset T of the set S: Solution: the characteristic function fT : S −! f0; 1g is the function defined by 1; if s 2 T f (s) = T 0; if s2 = T 2 2. [course avg: 8.3/10] 2.(a) [3 marks] Use the Euclidean algorithm to show that 60 and 43 are relatively prime. Solution: use the Euclidean algorithm. 60 = 1 · 43 + 17 (1) 43 = 2 · 17 + 9 (2) 17 = 1 · 9 + 8 (3) 9 = 1 · 8 + 1 (4) 8 = 8 · 1 + 0 (5) Thus gcd(60; 43) = 1: 2.(b) [4 marks] Find a solution to the congruence 43 x ≡ 11 (mod 60) Solution: use part (a) to find a multiplicative inverse of 43 modulo 60: (4) ) 1 = 9 − 8; (3) ) 1 = 9 − (17 − 9) = 2 · 9 − 17; (2) ) 1 = 2(43 − 2 · 17) − 17 = 2 · 43 − 5 · 17; (1) ) 1 = 2 · 43 − 5(60 − 43) = 7 · 43 − 5 · 60: Thus 7 · 43 ≡ 1 (mod 60); and so 43 x ≡ 11 (mod 60) ) 7 · 43 x ≡ 7 · 11 (mod 60) ) x ≡ 77 ≡ 17 (mod 60); and x = 17 is one solution to the congruence. 2.(c) [3 marks] Prove that if p is a prime number that does not divide a, then ap2 ≡ ap (mod p2): Proof: since p does not divide a; p2 and a are relatively prime. So Euler's Theorem applies. 2 φ(p2) = p2 − p ) ap −p ≡ 1 (mod p2) 2 ) ap · ap −p ≡ ap (mod p2) 2 ) ap ≡ ap (mod p2) 3 3. [course avg: 6.88/10] 3.(a) [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: Wilson's Theorem states: if p is prime then (p − 1)! + 1 ≡ 0 (modp): We have (p − 1)! + 1 ≡ 0 (modp) ) (p − 1)! ≡ −1 (modp) (for p > 3) ) (p − 1)(p − 2)(p − 3)! ≡ −1 (modp) ) (−1)(−2)(p − 3)! ≡ −1 (modp) ) 2(p − 3)! ≡ −1 (modp) 3.(b) [6 marks] Find the remainder when (38! + 6541)43 is divided by 41. Solution: work inside the brackets first. Observe that 41 is a prime. • By part (a), 2 · 38! ≡ −1 (mod 41): Then 21 · 2 · 38! ≡ −21 (mod 41) ) 38! ≡ −21 (mod 41): • Since 41 does not divide 65, Fermat's Theorem implies 6540 ≡ 1 (mod 41): Thus 6541 ≡ 6540 · 65 ≡ 65 ≡ 23 (mod 41): • Therefore 38! + 6541 ≡ (−21 + 23) ≡ 2 (mod 41): Finally, Fermat's Theorem implies 240 ≡ 1 (mod 41): Putting it all together (38! + 6541)43 ≡ 243 ≡ 23 ≡ 8 (mod 41): Thus the remainder when (38! + 6541)43 is divided by 41 is 8. 4 4. [course avg: 4.42/10] 4.(a) [4 marks] Let p be an odd prime, let n be a natural number. Prove that if x2 ≡ 1 (mod pn); then x ≡ 1 (mod pn) or x ≡ −1 (mod pn): Proof: we have pn j x2 − 1 = (x − 1)(x + 1): Since p is prime, p j x − 1 or p j x + 1: But p can't divide both x − 1 and x + 1; for then p divides the difference, p j x + 1 − x + 1 = 2; which is impossible since p is an odd prime. Thus p divides only one of the factors x − 1 and x + 1: Consequently pn j x2 − 1 ) pn j x − 1 or pn j x + 1 ) x ≡ 1 (mod pn) or x ≡ −1 (mod pn): 4.(b) [2 marks] Show that the congruence x2 ≡ 1 (mod 8) has four solutions in the set f0; 1; 2;:::; 7g: Solution: x = 1; 3; 5; 7 all satisfy x2 ≡ 1 (mod 8): 4.(c) [4 marks] Show that the congruence x2 ≡ 1 ( mod 2n) has four solutions in the set f0; 1; 2;:::; 2n −1g if n ≥ 3: Solution: the congruence x2 ≡ 1 (mod 2n) has the four solutions x = 1; 2n−1 − 1; 2n−1 + 1 or 2n − 1; since 12 = 1; (2n − 1)2 ≡ (0 − 1)2 ≡ 1 (mod 2n); and (2n−1 ± 1)2 ≡ 22n−2 ± 2 · 2n−1 + 1 ≡ 2n · 2n−2 ± 2n + 1 ≡ 0 + 0 + 1 ≡ 1 (mod 2n): 5 5. [course average: 2.67/10] 5.(a) [4 marks] Given a polynomial p(z) with real coefficients and a complex root z0 = a + bi of p(z), prove that z2 − 2az + a2 + b2 is a factor of p(z). Solution: since the coefficients of p are real, we know that z0 = a − bi is also a root of p(z): Thus (z − z0)(z − z0) = (z − (a + bi))(z − (a − bi)) = z2 − (a + bi)z − (a − bi)z + (a + bi)(a − bi) = z2 − 2a z + a2 + b2 is a factor of p(z): 5.(b) [6 marks] Explain why any non-constant polynomial with real coefficients can be factored into a product of polynomials of degrees one or two (i.e. linear or quadratic factors) with real coefficients. Solution: let p(z) be a polynomial of degree n and suppose the coefficients of p(z) are all real numbers. We know that p(z) can be factored as p(z) = c(z − z1)(z − z2) ··· (z − zn); for some real constant c and n complex numbers, z1; z2; : : : ; zn: Since the coefficients of p(z) are all real, the complex conjugates z1; z2;:::; zn are also roots of p(z): Thus the roots of p(z) are real numbers or come in pairs of complex conjugates. Suppose the roots z1; z2; : : : ; zj are all real, and the remaining non-real roots are zj+1; zj+1; zj+2; zj+2; : : : ; zj+k; zj+k; where n = j + 2k: For each pair of roots zj+m = am + bm; zj+m = am − bm i; 1 ≤ m ≤ k; part (a) implies 2 2 2 (z − zj+m)(z − zj+m) = z − 2amz + am + bm is a quadratic factor of p(z): Thus 2 2 2 2 2 2 2 2 2 p(z) = c (z − z1)(z − z2) ··· (z − zj) (z − 2a1z + a1 + b1)(z − 2a2z + a2 + b2) ··· (z − 2akz + ak + bk); | {z } | {z } j linearfactors k quadraticfactors all with real coefficients. 6 6. [course avg: 4.67/10] 6. For each of the following pairs of sets define a bijection from one set to the other, and verify that it is a bijection. (a) [5 marks] f0; 1gS and P(S) for any set S. (Recall: AB is set of all functions f : B −! A:) Solution: we can make use of the characteristic functions, fT : S −! f0; 1g; defined by 1; if s 2 T f (s) = ; T 0; if s2 = T S where T ⊂ S: For T a non-empty subset of S; define G : P(S) −! f0; 1g by G(T ) = fT ; if T = ;; define G(T ) to be the zero-map: (G(;))(s) = 0; for all s 2 S: Check that G is a bijection.
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