MAT246H1S - Concepts In Abstract Mathematics. Solutions to Term Test 1

MAT246H1S - Concepts In Abstract Mathematics Solutions to Term Test 1 - February 1, 2018 Time allotted: 110 minutes. Aids permitted: None. Comments: Statements of Definitions, Principles or Theorems should be spot on. Anything else is unacceptable. You must make your arguments logical and complete. We are not responsible for figuring out what you are doing; you are supposed to make it completely clear what you are doing. When using the Well-Ordering Principle or any of the Principles of Mathematical Induction, the definition of the set to which you will apply WOP or Induction must be clearly given. Breakdown of Results: 352 students wrote this test. The marks ranged from 12% to 93%, and the average was 53%. Some statistics on grade distribution are in the table on the left, and a histogram of the marks (by decade) is on the right. Grade % Decade % 90-100% 2.2% A 8.2% 80-89% 6.0% B 10.8% 70-79% 10.8% C 16.5% 60-69% 16.5% D 18.7% 50-59% 18.7% F 45.8% 40-49% 21.6% 30-39% 14.8% 20-29% 7.1% 10-19% 2.3% 0-9% 0.0%

1. [10 marks; avg: 7.1] Let a, p, m and n be natural numbers. (a) [6 marks; 2 for each part] Define the following: (i) a is a divisor of n Solution: a is a divisor of n if there is a natural number k such that n = k a. (ii) p is a prime number Solution: a natural number p greater than 1 is a prime number if the only natural number divisors of p are 1 and itself. (Symbolically: p > 1 and a p a = 1 or a = p) (iii) m is a composite number Solution: a natural number m greater than 1 is a composite number if it is not a prime number. (That s the definition in the book.) Or: a natural number m greater than 1 is a composite number if there are natural numbers a, b with 1 < a, b < m and m = a b. (b) [4 marks; 2 for each part] Determine if the following numbers are prime or composite. Solution: observe that neither of the two numbers is even, so 2 is not a divisor; neither of the two numbers is divisible by 3, since the sum of their digits is not a multiple of 3; and neither of the two number is divisible by 5, since the last digit of each number is neither 0 nor 5. The next step is to divide each number by 7... (i) 119 is composite since 119 = 7 17. (ii) 179 is prime since neither 7, nor 11, nor 13 divide it, 179 = 7 25 + 4 179 = 11 16 + 3 179 = 13 13 + 10, and the next prime, 17, is too large to bother with since 17 > 179. 2

2. [10 marks; avg: 5.7] 2.(a) [3 marks] State the Well-Ordering Principle. Solution: every non-empty set of natural numbers contains a least element. OR: if S N and S, then there is an m S such that m n for all n S. 2.(b) [7 marks] Use the Well-Ordering Principle to prove that every natural number other than 1 has a prime divisor. Proof: let S = {n N n > 1 and n has no prime divisor}. Suppose S. By the Well-Ordering Principle, S has a least element, call it m. Since m S, m has no prime divisor. But m divides itself, so m cannot be a prime. Therefore m is a composite number and there are numbers a, b such that m = a b and 1 < a, b < m. In particular a < m, so a / S. Then a has a prime divisor, call it p. Thus p a and a m. This means p divides m, contradicting the fact that m S. Thus S =. That is, every natural number other than 1 has a prime divisor.

3. [10 marks; avg: 7.3] 3.(a) [3 marks] State the Principle of Mathematical Induction. Solution: if S is any set of natural numbers such that 1 is in S, and k + 1 is in S whenever k is in S, then S = N. 3.(b) [7 marks] Use the Principle of Mathematical Induction to prove that for every natural number n. Proof: let S = {n N ( ) is true for n}. 1 S since 1 < 2 = 2 1. 1 + 1 2 + 1 3 + + 1 n < 2 n ( ) Now assume k S. Then ( ) holds for n = k and 1 + 1 2 + 1 3 + + 1 k + 1 k + 1 < 2 k + 1 k + 1 ( using ab a + b 2 ; so k k + 1 k + k + 1 ) 2 = 2 k k + 1 + 1 k + 1 k + k + 1 + 1 k + 1 = 2(k + 1) k + 1 = 2 k + 1 Thus ( ) holds for n = k + 1 as well. By the Principle of Mathematical Induction S = N and the formula is true for all n. OR: if you don t know the inequality ab a + b, establish the induction step using more algebra: 2 2 k + 1 k + 1 < 2 k + 1 2 k k + 1 + 1 < 2(k + 1) 2 k(k + 1) < 2k + 1 4k(k + 1) < (2k + 1) 2 4k 2 + 4k < 4k 2 + 4k + 1 0 < 1

4. [10 marks; avg: 5.6] 4.(a) [4 marks] State the Generalized Principle of Complete Mathematical Induction. Solution: If S is any set of natural numbers with the properties that m is in S, and k + 1 is in S whenever k is a natural number and all the natural numbers from m through k are in S, then S = {n N n m}. 4.(b) [6 marks] Use the Generalized Principle of Complete Mathematical Induction to prove that every natural number other than 1 is a product of primes. Solution: let S be the set of natural numbers greater than 1 that can be written as a product of primes. 2 S, since 2 is a prime, it is a product of a single factor, itself. Let k > 2 and suppose that all natural numbers from 2 through k can be written as a product of primes, that is all the numbers from 2 through k are in S. Consider the number k + 1. There are two cases: 1. If k + 1 is prime, then it is a product of a single factor, itself, so k + 1 S. 2. If k + 1 is composite then there are numbers a, b such that 1 < a, b < k + 1 and k + 1 = a b. Since 2 a, b k, the inductive hypothesis implies that both a and b can be written as a product of primes. Suppose in particular that a = p 1 p 2 p m and b = q 1 q 2 q n for primes p i and q j. Then k + 1 = a b = p 1 p 2 p m q 1 q 2 q n, which means that k + 1 is also a product of primes. Hence k + 1 S. By the Generalized Principle of Complete Mathematical Induction, S = {n N n 2.}

5. [10 marks; avg: 4.8] Let n be a natural number. Prove that if n! + n 2 + n + 1 is prime then n 2 + n + 1 is prime. Proof: by contraposition. If n 2 + n + 1 is composite, then n 4 and n 2 + n + 1 = ab, with a, b N and 1 < a, b < n 2 + n + 1. One of a, b must be less than or equal to n because otherwise a > n and b > n a n + 1 and b n + 1 ab (n + 1) 2 = n 2 + 2n + 1 > n 2 + n + 1 = ab, which is a contradiction. So, assume without loss of generality that 1 < a n. Then a divides n!. Since a already divides n 2 + n + 1 we can conclude that a divides n! + n 2 + n + 1, which means that n! + n 2 + n + 1 is also composite.

6. [10 marks; avg: 7.0] 6.(a) [3 marks] Let n be a natural number other than 1. Define the canonical factorization of n into a product of primes. Solution: if n > 1 then the canonical factorization of n is n = p α 1 1 pα 2 2 pαr r, where each p i is a prime, p i < p i+1, and α i is a natural number. 6.(b) [7 marks] Find the canonical factorization of 1605240. Solution: 1605240 is even, ends in a zero, and the sum of its digits is divisible by 3; so primes 2, 3 and 5 all divide it. We have 8 1605240 since 1605240 = 8 200655 5 200655 since 200655 = 5 40131 9 40131 since 40131 = 9 4459 Is 4459 divisible by 7? Yes: 4459 = 7 637 = 7 2 91 = 7 3 13. Thus the canonical factorization of 1605240 is 1605240 = 2 3 3 2 5 7 3 13. The increasing order of the primes is part of the correct answer.

7. [10 marks; avg: 2.7] 7.(a) [6 marks] Let a and b be integers, let p and q be distinct primes. Prove that if a b (mod p) and a b (mod q), then a b (mod p q). Proof: a b (mod p) p a b a b = p m, for some integer m. Also, a b (mod q) q a b q p m. Since q is prime, q p or q m. Since p and q are distinct primes, we must have q m. Thus m = q n, for some integer n, and which means a b (mod p q). a b = p m = p q n, 7.(b) [4 marks] Prove that if a is any integer such that a 1 (mod 30), then a 1 (mod 2) or a 1 (mod 3) or a 1 (mod 5). Proof: by contraposition. Suppose a 1 (mod 2) and a 1 (mod 3) and a 1 (mod 5). By part (a), a 1 (mod 6). So a 1 = 6k, for some integer k. But also a 1 (mod 5). Therefore 5 divides a 1 = 6k. Since 5 doesn t divide 6, 5 k, and so k = 5j, for some integer j. Consequently a 1 = 6k = 6 5 j = 30 j, which means a 1 (mod 30).

8. [10 marks; avg: 2.3] 8.(a) [6 marks] Let p be a prime number, with p > 3. Prove that p = 6k ± 1, for some natural number k. Proof: every number p is congruent to one of 0, 1, 2, 3, 4 or 5, mod 6. Consider the four cases: 1. p 0 (mod 6) p = 6k + 0 6 p, so p is not prime. 2. p 2 (mod 6) p = 6k + 2 = 2(3k + 1) 2 p, so p is not prime if p > 3. 3. p 3 (mod 6) p = 6k + 3 = 3(2k + 1) 3 p, so p is not prime if p > 3. 4. p 4 (mod 6) p = 6k + 4 = 2(3k + 2) 2 p, so p is not prime if p > 3. If p > 3 is prime then it must be true that p 1 (mod 6) or p 5 1 (mod 6). So p = 6k ± 1, for some natural number k. Use this criterion to determine if 666634569 could be prime. Solution: since 666634569 = 6 111105761 + 3 3 (mod 6), the number 666634569 is not prime. 8.(b) [4 marks] Let m be a natural number other than 1. Prove that if m has the property that m divides at least one of the natural numbers a and b whenever it divides the product a b, then m is a prime. Proof: let p be a prime divisor of m. Then m = p k for some natural number k, and 1 p, k m. Since m divides itself, m divides p k. By the given property, m divides p or m divides k. Consider the two cases: 1. if m divides p, then we have p m p, which means m = p, and so m is a prime. 2. if m divides k, then as in Case 1, we have k m k, which means m = k. That implies p = 1, contradicting the fact that p is a prime. Thus Case 2 is impossible. Conclusion: m is a prime. Alternate Proof: by contradiction. Suppose m is composite. Then there are numbers a, b such that 1 < a, b < m, and m = a b. By the property, m should divide a or m should divide b. But since both a and b are less than m, m cannot divide them.

9. [10 marks; avg: 5.1] 9.(a) [4 marks] Use modular arithmetic, not calculus, to show that there is no integer solution to the equation x 3 + x 2 + x = 9001. Solution: mod 3, there are only three possibilities for any integer x: x 0, 1 or 2 (mod 3). Take each case in turn: 1. x 0 (mod 3) x 3 + x 2 + x 0 + 0 + 0 0 (mod 3). 2. x 1 (mod 3) x 3 + x 2 + x 1 + 1 + 1 0 (mod 3). 3. x 2 (mod 3) x 3 + x 2 + x 8 + 4 + 2 2 (mod 3). Thus, for all integers x, x 3 + x 2 + x 1 (mod 3). But 9001 1 (mod 3). So there is no integer solution to the equation x 3 + x 2 + x = 9001. 9.(b) [6 marks] Let p be a prime. Prove that if x 2 1 (mod p), then x 1 (mod p) or x 1 (mod p).. Proof: we use the fact that if p is prime and p divides ab, then p divides a or p divides b. x 2 1 (modp) p (x 2 1) = (x 1)(x + 1) p x 1 or p x + 1 x 1 (modp) or x 1 (modp) Give a counter example to show that the statement is not true if p is not prime Counter Example: consider p = 8. Then 1 2 = 1 1 (mod8), and 1 1 (mod8). 3 2 = 9 1 (mod8), but 3 ±1 (mod8). 5 2 = 25 1 (mod8), but 5 ±1 (mod8). 7 2 = 49 1 (mod8), and 7 1 (mod8). So the congruence x 2 1 (mod 8) has four solutions, x = 1, 3, 5, 7, but only two of them satisfy x 1 (mod 8) or x 1 (mod 8). Another Counter Example: consider p = 21 : 8 2 1 (mod 21) but 8 ±1 (mod 21).

10. [10 marks; avg: 5.1] Find the remainder when 2 4783975 7 6593 is divided by 17. Solution: consider the two expressions, 2 4783975 and 7 6593. For the first expression, we can use the fact that 2 4 = 16 1 (mod 17). Since 4783975 = 4 1195993 + 3, it follows that 2 4783975 = 2 4 1195993+3 = (16) 1195993 2 3 ( 1) 1195993 2 3 8 (mod 17). For the second expression, use 7 2 = 49 2 (mod 17). Then, similar to the above calculation, 7 6593 = 7 2 3296+1 = 49 3296 7, and so 7 6593 = 49 3296 7 ( 2) 3296 7 (2 4 ) 824 7 ( 1) 824 7 7 (mod 17). Thus (2 4783975 7 6593 ) ( 8 7) 15 2 (mod 17), and the remainder is 2. OR: reduce the second expression as follows: 7 6593 = ( 7) 6593 and observe that 7 10 ( mod 17). Since 6593 = 2 3296 + 1, it follows that 7 6593 = ( 7) 6593 10 6593 (mod 17) 10 2 3296+1 (mod 17) 100 3296 10 1 (mod 17) ( 2) 3296 10 (mod 17) 2 3296 10 (mod 17) 2 4 824 10 (mod 17) (2 4 ) 824 10 (mod 17) ( 1) 824 10 (mod 17) 1 10 (mod 17) 10 (mod 17), from which, (2 4783975 7 6593 ) ( 8 + 10) (mod 17) 2 (mod 17), as before. OR: use 7 2 2 (mod 17) 7 4 4 (mod 17) 7 8 16 1 (mod 17)

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MAT246H1S - Concepts In Abstract Mathematics. Solutions to Term Test 1 - February 1, 2018