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Unformatted text preview: Ma1a 2007 HW 1 Solutions Question 1. Prove that √ 3 √ 5 is irrational. Solution 1. Let’s assume, by way of contradiction, that √ 3 √ 5 is rational. So there exist integers a and b , with b 6 = 0, such that √ 3 √ 5 = a b . Squaring this gives 3 2 √ 15 + 5 = a 2 b 2 and rearranging this expression yields √ 15 = 8 b 2 a 2 2 b 2 . Thus we deduce that if √ 3 √ 5 is rational then we must have that √ 15 is also rational. We now show that √ 15 is irrational. So suppose, by way of contradiction, that √ 15 is rational. Then there exist integers c and d , with d 6 = 0, such that √ 15 = c d . Moreover we can assume that c and d have no common factors. Squaring this gives 15 d 2 = c 2 . Now 3 divides the left hand side and hence also the right hand side of this expression. Therefore 3 divides c 2 and since 3 is prime it must also divide c . So we can write c = 3 c 1 . But then we have 15 d 2 = 9 c 2 1 and so 5 d 2 = 3 c 2 1 . Thus we deduce that 3 divides d 2 and hence 3 divides d . But now we have that 3 divides both c and d which gives a contradiction to our assumption that c and d had no common factors. Hence we have that √ 15 is irrational and therefore √ 3 √ 5 is also irrational. Question 2. (pg. 35, #1c) Prove the following by induction: 1 3 + 2 3 + ··· + n 3 = (1 + 2 + 3 + ··· + n ) 2 . 1 Solution 2. For a positive integer n let A ( n ) denote the statement 1 3 + 2 3 + 3 3 + .. . + n 3 = (1 + 2 + 3 + .. . + n ) 2 ....
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 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra, Integers, Natural number, 1 K, 2k, 1 k, 1 2 k

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