math 140a

# math 140a - 140A Midterm 1 Solutions Fall 2009 Problem 1...

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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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## This note was uploaded on 09/30/2011 for the course MATH 140a taught by Professor Staff during the Fall '08 term at UCSD.

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math 140a - 140A Midterm 1 Solutions Fall 2009 Problem 1...

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