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Unformatted text preview: MAT 137Y 20082009 Winter Session, Solutions to Problem Set 3 1 (a) Suppose there is a largest negative real number x < 0. Then x 2 > x . The existence of x 2 contradicts the assumption that x is the largest negative real number, so there does not exist a largest negative real number. (b) Suppose 3 is rational. Then a / b = 3 = a 2 = 3 b 2 , where a and b are positive integers and a / b is in lowest terms. Since a 2 and 3 b 2 are integers, and 3 b 2 is divisible by 3, then a 2 must also be divisible by 3. It must follow as a result that a is divisible by 3 (suppose not; then a = 3 k + 1 or a = 3 k + 2 for some integer k ; but ( 3 k + 1 ) 2 = 9 k 2 + 9 k + 1 and ( 3 k + 2 ) 2 = 9 k 2 + 12 k + 4, and both are not divisible by 3, a contradiction). Hence a = 3 m for some integer m , so ( 3 m ) 2 = 3 b 2 = 9 m 2 = 3 b 2 = b 2 = 3 m 2 . This implies that b 2 is divisible by 3, and hence, b is divisible by 3. Since a and b are both divisible by 3, then a / b is not in lowest terms, a contradiction. (c) (SHE 1.2 #74) If a = b = 2 then a + b = 2 2, which is irrational. If a = 2 and b = 2, then a + b = 0, which is rational. If a = 2 and b = 1 + 2, then ab = 2 + 2 is irrational (see 1.2 #71). However, if a = b = 2, then ab = 2, which is rational. (d) Suppose 2 + 3 = p / q , where p and q are nonzero integers. 3 = p q 2 = 3 = p 2 q 2 2 p 2 q + 2 = 2 p 2 q = p q 2 1 = 2 = p 2 q 2 2 pq , but this implies that 2 is rational, a contradiction. Therefore 2 + 3 is irrational. 2 (i) (SHE 1.8 #14) Computing the product for various choices of n , we have 1 1 2 2 = 3 4 , 1 1 2 2 1 1 3 2 = 2 3 = 4 6 , 1 1 2 2 1 1 3 2 1 1 4 2 = 5 8 , 1 1 2 2 1 1 3 2 1 1 4 2 1 1 5 2 = 3 5 = 6 10 , 1 1 2 2 1 1 3 2 1 1 4 2 1 1 5 2 1 1 6 2 = 7 12 ....
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 Spring '08
 UPPAL

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