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Unformatted text preview: a 2 . Since 5 is prime, 5 must divide a, so a = 5c and then we get 5b 2 = 5c 2 = 25c 2 . Divide by 5 to get b 2 = 5c 2 . Just as with a, we see that now 5 must divide b, so the fraction a/ b is NOT completely reduced. This is big trouble, so 5 cannot be rational. #30. Suppose 3 2 is rational, so that 3 2 = a b , where a and b are integers and the fraction is completely reduced. Cube both sides to get 2 = a 3 b 3 so 2b 3 = a 3 and 2 divides a 3 . Since 2 is prime, 2 must divide a, so a = 2c and then we get 2b 3 = 2c 3 = 8c 3 . Divide by 2 to get b 3 = 4c 3 . Just as with a, we see that now 2 must divide b , so the fraction a/b is NOT completely reduced. This is big trouble, so 3 2 cannot be rational....
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 Spring '08
 Irwin
 Division, Prime number, Rational number, Integral domain, Dr. Schurle Assignment

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