Analysis2.1 - Real Analysis Chapter 2.1 David Joseph Stith Page 29 Number 4 Exercise If a R satises a a = a prove that either a = 0 or a = 1 Suppose a R
Analysis2.1 - Real Analysis Chapter 2.1 David Joseph Stith...
Real Analysis: Chapter 2.1David Joseph StithPage 29, Number 4. Exercise.Ifa∈Rsatisfiesa·a=a, prove that eithera= 0ora= 1.Supposea∈Randa·a=a. Then eithera= 0 oranegationslash= 0.Supposea= 0. Thena·a=ais satisfied since we know that 0·0 = 0.Now supposeanegationslash= 0. Thena=a·1=a·(a·1a)= (a·a)·1a=a·1asincea·a=a= 1Therefore ifa∈Rsatisfiesa·a=a, then eithera= 0 ora= 1.Q.E.D.Page 30, Number 6.Exercise.Use the argument in the proof of Theorem 2.1.4 toshow that there does not exist a rational numberssuch thats2= 6.Supposepandqare integers such that (pq)2= 6. Without loss of generality we mayassume thatpandqare positive and have no common integer factors other than 1. Sincep2= 6q2, we see that 6|p2.We claim that 6|p2=⇒6|p. To show this we will show that the contrapositive istrue:6negationslash |p=⇒6negationslash |p2Suppose 6negationslash |p.Then either 2negationslash |por 3negationslash |p.We know thatp=ax11ax22. . . axkkfor primefactorsa1. . . akand powersx1. . . xk∈Nis a unique representation ofp.Hence eitherannegationslash= 2 orannegationslash= 3 for alln∈N,n≤k. Now,p2=a2x11a2x22. . . a2xkkhas the same sameprime factors aspraised to twice the powers of each. Therefore 6negationslash |p=⇒6negationslash |p2. Therefore6|p2=⇒6|p.
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Fall '08
Akhmedov,A
Integers, Rational number, David Joseph Stith, ab < b2