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Unformatted text preview: 140A Midterm 1 Solutions  Fall 2009 November 20, 2009 Problem 1. Prove that the cube root of 12 is an irrational number. Proof. Suppose that 12 1 / 3 is rational, i.e. there exist relatively prime integers a and b such that 12 1 / 3 = a b . Then a 3 = 12 b 3 = 2 2 3 b 3 . So 3 divides a 3 . Since 3 is prime, 3 divides a and so 3 3 divides a 3 . Then 3 3 divides 2 2 3 b 3 . Since 3 is prime, 3 divides b contradicting that a and b are relatively prime. Problem 2. Describe an explicit method for constructing a bijection between the set of rational numbers and the set of positive integers. Proof. The key here is to define some function f : N Q that hits every element of Q exactly once. We construct a diagram similar to the one on page 29 of Rudin in the proof of theorem 2.12. 1 1 2 2 3 3 ... 2 1 2 1 2 2 2 2 2 3 2 3 2 ... 3 1 3 1 3 2 3 2 3 3 3 3 3 ... 4 1 4 1 4 2 4 2 4 3 4 3 4 ......
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 Fall '08
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 Math, Integers

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