Problem Set 3 Solutions

Transcription

1 Problem Set 3 Solutions Section 3.1. For each of the following, use a counterexample to prove the statement is false. (a) For each odd natural number n, if n > 3 then 3 divides (n 1). Consider n = 9. Then, n 1 = 80 and (b) For each natural number n, (3 n + 3 n + 1) is a prime number. For n = 6, = 1651 = (c) For all real numbers x and y, x + y > xy. If x = y = 1, then we have >, which is false. (d) For each integer a, if 4 divides (a 1), then 4 divides (a 1). The statement can be shown to be false if we take a = 3, which gives that a 1 = 8 and 4 8, but a 1 = and Determine if each of the following statements is true or false. If a statement is true, then write a formal proof of that statement, and if it is false, then provide a counterexample that shows it is false. (a) For all integers a, b and c with a 0, if a b then a (bc).. Let a, b, c Z with a 0. We will show that if a b then a (bc). If a b, then k Z such that b = ka. Then, bc = (ak)c = a(kc) and hence, a bc. (b) For all integers a and b with a 0, if 6 (ab) then 6 a or 6 b. False. Consider a = and b = 3. then, 6 6, but 6 and 6 3. (c) For all integers a, b, and c with a 0, if a divides (b 1) and a divides (c 1), then a divides (bc 1). Let a, b, c Z. If a (b 1) and a (c 1) then there exists integers m, n such that am = b 1 and an = c 1. Then and since amn + m + n Z, a (bc 1). bc 1 = (am + 1)(an + 1) 1 = a mn + am + an = a(amn + m + n) (d) For each integer n, if 7 divides (n 4), then 7 divides (n ). False. Let n = 5. Then, 7 1 but 7 3. The trick is to note that n 4 = (n + )(n ) and to look for an n such that 7 (n + ) as it will therefore not be able to divide n. (e) For every integer n, 4n + 7n + 6 is an odd integer. False. If n is even, then for some k Z, we have that n = k. Then, 4n + 7n + 6 = 4(k) + 7(k) + 6 = 16k + 14k + 6 = (8k + 7k + 3) and since 8k + 7k + 3 Z, 4n + 7n + 6 is even whenever n is even.

2 (f) For every odd integer n, 4n + 7n + 6 is an odd integer. Let n be an odd integer. Then, there exists k Z such that n = k + 1. Now, 4n + 7n + 6 = 4(k + 1) + 7(k + 1) + 6 = 4(4k + 4k + 1) + 14k = 16k + 30k + 17 = (8k + 15k + 8) + 1 Since 8k + 15k + 3 Z, 4n + 7n + 6 is odd whenever n is odd. (g) For all integers a, b, and d with d 0, if d divides both a b and a + b, then d divides a. False. Let a = 3, b = 1 and d =. Then, d 4 and d but d 3. (h) For all integers a, b, and c with a 0, if a (bc) then a b or a c. False. Consider a = 6, b = 4 and c = 3. Then we have 6 1 but 6 4 and Determine if each of the following statements is true or false. If a statement is true, then write a formal proof of that statement, and if it is false, then provide a counterexample that shows it is false. (a) For each integer a, if there exists an integer n such that a divides (8n + 7) and a divides (4n + 1), then a divides 5.. If a (8n + 7) and a (4n + 1), then there exists integers k, m such that ak = 8n + 7 and am = 4n + 1. Multiplying the second equation by gives am = 8n + and subtracting the second equation from the first yields ak am = 5, or 5 = a(k m). Since k m Z, we have that a 5. (b) For each integer a, if there exists an integer n such that a divides (9n + 5) and a divides (6n + 1), then a divides 7. If a (9n + 5) and a (6n + 1) then there exists integers k, m such that ak = 9n + 5 and am = 6n + 1. Multiplying the first by and the second by 3 yields ak = 18n + 10 and 3am = 18n + 3. Subtracting the second from the first gives ak 3am = 7, or 7 = a(k 3m). Since k 3m Z, we have that a 7. (c) For each integer n, if n is odd, then 8 divides (n 4 + 4n + 11).. Suppose n is an odd integer. we will show 8 (n 4 + 4n + 11). If n is odd, then k Z such that n = k + 1. Consider n 4 + 4n + 11 = (k + 1) 4 + 4(k + 1) k 4 + 3k 3 + 4k + 8k k + 16k = 16k 4 + 3k k + 4k + 16 = 8(k 4 + 4k 3 + 5k + 3k + ) and since k 4 + 4k 3 + 5k + 3k + is an integer, 8 (n 4 + 4n + 11). (d) For each integer n, if n is odd, then 8 divides (n 4 + n + n). False. Let n = 1. Then, we have 8 4, which is not true.

3 7. Let a be an integer and let n N. (a) Prove that if a 0 (mod n), then n a. If a 0 (mod n), then n (a 0), or n a. (b) Prove that if n a, then a 0 (mod n). If n a, then n (a 0) and hence a 0 (mod n). 9. Let a and b be integers. Prove that if a 7 (mod 8) and b 3 (mod 8), then (a) a + b (mod 8) If a 7 (mod 8) and b 3 (mod 8), then there exists integers k, m such that a 7 = 8k and b 3 = 8m. Then, a + b = 8k m + 3 = 8k + 8m + 8 = 8(k + m + 1) That is, a + b 0 (mod 8), or a + b (mod 8). (b) a b 5 (mod 8) If a 7 (mod 8) and b 3 (mod 8), then there exists integers k, m such that a 7 = 8k and b 3 = 8m. Then, ab 5 = (8k + 7)(8m + 3) 5 = 64km + 56m + 4k = 8(8km + 7m + 3k + ) That is, ab 5 0 (mod 8), or ab 5 (mod 8). 16. Determine if each of the following statements is true or false. Provide a counterexample for statements that are false and provide a complete proof for those that are true. (a) For all real numbers x and y, xy x+y. False. For any x, y R such that x < 0 and y < 0, we would have that xy > 0, but x+y < 0, making the inequality false. (b) For all real numbers x and y, xy ( x+y ). ( ) x + y xy xy x + xy + y 4 4xy x + xy + y 0 x xy + y 0 (x y) This is true for all real numbers x and y, so the original statement is true. (c) For all nonnegative real numbers x and y, xy x+y. Let x, y R. If xy x+y, then xy x + y. Now, xy x + y 4xy x + xy + y 0 x xy + y 0 (x y)

4 Section which is always true for nonnegative real numbers x and y. Note: we covered the failure of this if x and y are negative in (a). If only one of x and y are negative, then xy is imaginary, so the hypothesis would be invalid. (a) Write the contrapositive of the following statement: For all positive real numbers a and b, if ab a+b, then a b. For all positive real numbers a and b, if a = b, then ab = a+b. (b) Is this statement true or false? Prove the statement if it is true and provide a counterexample if it is false. a+b Suppose for positive real numbers a and b, a = b. Then, = a+a = a = a and ab = a = a. 4. Are the following statements true or false? Justify your conclusions. (a) For each a Z, if a (mod 5), then a 4 (mod 5). If a (mod 5), then k Z such that a = 5k, or a = + 5k. Consider a = ( + 5k) = 4 + 0k + 5k That is, a 4 = 5(4k + 5k ), or that a 4 (mod 5). (b) For each a Z, if a (mod 5), then a 4 (mod 5). False. There is no integer a such that a (mod 5), so the hypotheses cannot be satisfied. All squares (modulo 5) have remainders 1, 4 or 5. (c) For each a Z, a (mod 5), if and only if a 4 (mod 5). False. Part (b) shows this is false. 7. Consider the following proposition: For each integer a, a (mod 8) if and only if (a + 4a) 4 (mod 8). (a) Write the proposition as the conjunction of two conditional statements. ( a Z)([(a (mod 8)) ((a + 4a) 4 (mod 8))] [((a + 4a) 4 (mod 8)) ((a (mod 8))]) (b) Determine Determine if the two conditional statements in Part (a) are true or false. If a conditional statement is true, write a proof, and if it is false, provide a counterexample. (a (mod 8)) ((a + 4a) 4 (mod 8))

5 If a (mod 8), then there exists k Z such that a = 8k, or a = 8k +. Then a + 4a 4 = (8k + ) + 4(8k + ) 4 = 64k + 3k k = 64k + 8 = 8(8k + 1) Since 8k + 1 Z, a + 4a 4 0 (mod 8), or (a + 4a) 4 (mod 8). ((a + 4a) 4 (mod 8)) ((a (mod 8)) The statement is false. Let a = 6. Then we have = 60 4 (mod 8), but 6 (mod 8). (c) Is the given proposition true or false? Explain. The given proposition is false since the statement Q P is false. 9. A real number x is defined to be a rational number provided there exists integers m and n with n 0 such that x = m n. A real number that is not rational is called an irrational number. It is known that if x is a positive rational number, then there exists positive integers m and n with n 0 such that x = m n. Is the following proposition true or false? Explain. For each positive real number x, if x is irrational, then x is irrational. The proposition is true. We will prove by contrapositive. Suppose x is rational. We will show x is rational as well. If x is rational then there exists ( ) some rational number p q with p, q Z and q 0. Then, p q = p = x, and since p and q q are integers with q 0, x is rational. 14. Are the following propositions true or false? Justify your conclusion. (a) There exists integers x and y such that 4x + 6y =. 4x + 6y = (x + 3y) and for x =, y = 1, we have that x + 3y = 1. Therefore, 4x + 6y =. (b) There exists integers x and y such that 6x + 15y =. The statement is false. 6x + 15y = 3(x + 5y). Now, x + 5y Z, so we have that 3(x + 5y) = 3n for some n Z. But, if 3n =, then n = 3, and 3 Z, so we contradict that n Z. (c) There exists integers x and y such that 6x + 15y = 9. Note that 6x + 15y = 9 is the same as x + 5y = 3. If we let x = 4 and y = 1, we see that this equation is satisfied. 17. Let a and b be rational numbers such that a = b 3. Prove each of the propositions in Parts (6a) through (6d).

6 (a) If a is even, then 4 divides a. Assume that a is even. Then, there exists k Z such that a = k. So, 4k = b 3. Since b Z and b 3 is even, b must be even as well. So, there exists m Z such that b = m. The, we have that 4k = 8m 3, or that k = m 3. Thus, k is even. Since k Z and k is even, k is even as well. So, there exists q Z such that k = q, and since a = k, we have that a = 4q, or that 4 a. (b) If 4 divides a, then 4 divides b. Since 4 a, there is an integer n Z such that a = 4n. Since a = b 3, we have that b 3 = 16n. So, b 3 is even and so b is even. Then, there exists m Z such that b = m. This implies that 8m 3 = 16n, or m 3 = n. So, since m 3 is even and m Z, m is even as well. So, there exists k Z such that m = k. But since b = m, it follows that b = 4k, or that 4 b. (c) If 4 divides b, then 8 divides a. If 4 b, then since a = b 3, we can conclude that a is even. But this implies that a is even. So, by (a), 4 a. So there exists integers s, t such that b = 4s and a = 4t. Then, 16t = 64s 3, or t = 4s 3. So, t is even and hence t must be even. Since a = 4t and t is even, it follows that 8 a. (d) If a is even, then 8 divides a. If a is even, then by (a), 4 a. If 4 a, then by (b), 4 b. And if 4 b, then by (c), 8 a. Section 3.3 (e) Give an example of natural numbers a and b such that a is even and a = b 3, but b is not divisible by 8. This is the case when a = 8 and b = 4.. Are the following statements true or false? Justify each conclusion. (a) For all integers a and b, if a is even and b is odd, then 4 does not divide (a + b ). If a is even and b is odd, then there exists s, t Z such that a = s and b = t + 1. Then, a + b = (s) + (t + 1) = 4s + 4t + 4t + 1 = 4(s + t + t) + 1 where s + t + t Z. Therefore, a + b 1 (mod 4), or 4 (a + b ). (b) For all integers a and b, if a is even and b is odd, then 6 does not divide (a + b ). From (a), we see that we can write a + b as 4(s + t + t) + 1. Notice that this is odd, so (a + b ). Since this is the case, no multiple of can divide a + b either. (c) For all integers a and b, if a is even and b is odd, then 4 does not divide (a + b ). If a is even and b is odd, then there exists s, t Z such that a = s and b = t + 1. Then, a + b = (s) + (t + 1) = 4s + 8t + 8t + = (s + 4t + 4t + 1)

7 Since s + 4t + 4t + 1 Z, (a + b ). But, notice that s + 4t + 4t + 1 = (s +t +t)+1, so this expression is odd and is therefore not divisible by. Therefore, 4 a + b. (d) For all integers a and b, if a is odd and b is odd, then 4 divides (a + 3b ). If a and b are odd, then there exists integers k and m such that a = k+1 and b = m+1. Consider a +3b = (k+1) +3(m+1) = 4k +4k+1+1m +1m+3 = 4(k +k+3m +3m+1) Since k + k 3 m + 3m + 1 Z, 4 (a + 3b ). 4. Prove that the cube root of is an irrational number. That is, prove that if r is a real number such that r 3 =, then r is an irrational number. Suppose to the contrary that 3 is rational. Then, there exists p, q Z with q 0 such that 3 = p q and p and q have no common divisors. Then, 7. = ( ) p 3 q 3 = p 3 q So, p 3 must be even and since p Z, p is even as well. Therefore there exists k Z such that p = k. Then, we have q 3 = (k) 3 q 3 = 8k 3 q 3 = 4k 3 That is, q 3 is even, and since q Z, q is even as well. But, if p and q are both even, then they share the common factor of, contradicting our assumption that they were relatively prime. Thus, 3 is irrational. (a) Give an example that shows that the sum of two irrational numbers can be a rational number. Consider and. The, + ( ) = 0, and 0 Q. (b) Now explain why the following proof that ( + 5 ) is an irrational number is not a valid proof: Since and 5 are both irrational numbers, their sum is an irrational number. Therefore, ( + 5 ) is an irrational number. The proof is invalid because we are simply stating the conclusion as fact with no justification. They assume that irrational numbers are closed under addition, but (a) points out that this is not the case. (c) Is the real number + 5 a rational number or an irrational number Justify your conclusion. + 5 is irrational. We will prove this by contradiction. So suppose + 5 is rational. Then there exists a Q such that + 5 = a. Then, we can rewrite this as 5 = a, and squaring both sides gives 5 = a a +

8 We can rewrite this to give = a 3 a Since a 3 and a are both integers, this shows that is rational, a contradiction. 17. Is the following statement true or false? Justify your conclusion. For each integer n that is greater than 1, if a is the smallest positive factor of n that is greater than 1, then a is prime. If n is prime, then it has only one prime factor greater than 1, namely n, so the smallest positive factor is indeed prime. Suppose n is composite. We will prove the assertion by contradiction. So suppose to the contrary that a is the smallest positive factor of a, but that a is composite. Then, there exists positive integers k, m such that 1 < k, m < a where a = mk. Hence, a has positive factors smaller than a, and therefore k and m are factors of n as well. But, k and m are both positive integers smaller than a, contradicting the assumption that a is the smallest positive factor of n. Therefore, if a is the smallest positive factor of an integer n, a is prime.