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Unformatted text preview: Ma 1a  Fall 2005 Homework 1 Solutions Problem 1. Prove that √ 3 √ 6 is irrational. Solution 1. This will be a proof by contradiction. Assume that √ 3 √ 6 is rational; that is, √ 3 √ 6 = p q , where p and q are integers. Squaring both sides, we obtain 3 6 √ 2 + 6 = p 2 q 2 , so ( √ 3 √ 6) 2 is also a rational number. Subtracting 3 and 6 from the left and rewriting, we obtain 6 √ 2 = p 2 q 2 3 6 = p 2 3 q 2 6 q 2 q 2 , a rational number once again. Dividing by 6, we see that √ 2 = p 2 3 q 2 6 q 2 6 q 2 . Note that the numerator and the denominator are both integers, so √ 2 is here written as a rational number. Of course, this contradicts our previously established fact (from in class) that √ 2 is irrational. Thus, we are forced to conclude that our initial assumption that √ 3 √ 6 is rational is false; thus, √ 3 √ 6 is irrational. Problem 2. (Apostol p. 36, # 6) Let A ( n ) denote the statement that 1 + 2 + 3 + ... + n = 1 8 (2 n + 1) 2 . Solution 2. a) Prove that if A ( k ) is true for an integer k , then A ( k + 1) is true. We wish to show that 1 + 2 + 3 + ... + k + ( k + 1) = 1 8 (2( k + 1) + 1) 2 knowing that 1+2+3+ ... + k = 1 8 (2 k +1) 2 . So, 1+2+3+ ... + k +( k +1) = (1+2+3+ ... + k )+( k +1). By our assumption, we may substitute the first k terms to obtain 1 8 (2 k + 1) 2 + ( k + 1) = 1 8 (4 k 2 + 4 k + 1) + ( k + 1) = 1 8 (4 k 2 + 4 k + 1) + 1 8 (8 k + 8) = 1 1 8 (4 k 2 + 8 k + 4) + (4 k + 4) + 1 = 1...
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This note was uploaded on 03/30/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.
 Fall '08
 Borodin,A
 Integers

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