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Unformatted text preview: Abstract Vector Spaces (Christopher Heil)  MATH 2406 Fall 2011 Jonathan Johnson January 11, 2012 2 Contents 3 4 Chapter 1 Chapter 1 : Logic and Proofs 1.1 Playing with Proofs Theorem 1. √ 2 is irrational Proof. (We prove by contradiction). Suppose that √ 2 is a rational number. We first must define a rational number. Definition 1. A real number r is rational if for some integers m,n that r = m n , where n ≠ . So then, by definition, this means that there are integers p and q such that, √ 2 = p q (1.1) We can then assume that p and q have no integer factors other than 1 in common (if they do, then we just divide both p and q by that factor  we cancel it out, in other words). Another way to say this is that we can assume that the fraction p q is in lowest terms Squaring both sides of the equation, we see that 2 = p 2 q 2 which implies that p 2 = 2 q 2 (1.2) Hence, p 2 is an even number. Definition 2. n is even if n = 2 k for some integer k . Definition 3. n is odd if n = 2 k + 1 for some integer k 5 Lemma 1. If n 2 is even, then n is even. Proof. n is either even or odd. If n is odd, then n = 2 k + 1, and then n 2 = ( 2 k + 1 ) 2 = 4 k 2 + 4 k + 1. But we know that n 2 is even, not odd. Q . E . D . From this we can know that p = 2 n , for some integer n . Then this tells us, p 2 = 4 n 2 . From equation 1.2 we know that p 2 = 2 q 2 . So then, 2 q 2 = p 2 = 4 n 2 From which it follows that q 2 = 2 n 2 . So then we know q 2 is a multiple of 2, so q 2 is even....
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 Fall '08
 COSTELLO
 Logic, Vector Space, Prime number, upper bound, Q.E .D

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