This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View Full
Document
 Fall '08
 COSTELLO
 Logic, Vector Space

Click to edit the document details