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...

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Real Analysis: Chapter 2.1 David Joseph Stith Page 29, Number 4. Exercise. If a R satisfies a · a = a , prove that either a = 0 or a = 1 . Suppose a R and a · a = a . Then either a = 0 or a negationslash = 0. Suppose a = 0. Then a · a = a is satisfied since we know that 0 · 0 = 0. Now suppose a negationslash = 0. Then a = a · 1 = a · ( a · 1 a ) = ( a · a ) · 1 a = a · 1 a since a · a = a = 1 Therefore if a R satisfies a · a = a , then either a = 0 or a = 1. Q.E.D. Page 30, Number 6. Exercise. Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s 2 = 6 . Suppose p and q are integers such that ( p q ) 2 = 6. Without loss of generality we may assume that p and q are positive and have no common integer factors other than 1. Since p 2 = 6 q 2 , we see that 6 | p 2 . We claim that 6 | p 2 = 6 | p . To show this we will show that the contrapositive is true: 6 negationslash | p = 6 negationslash | p 2 Suppose 6 negationslash | p . Then either 2 negationslash | p or 3 negationslash | p . We know that p = a x 1 1 a x 2 2 . . . a x k k for prime factors a 1 . . . a k and powers x 1 . . . x k N is a unique representation of p . Hence either a n negationslash = 2 or a n negationslash = 3 for all n N , n k . Now, p 2 = a 2 x 1 1 a 2 x 2 2 . . . a 2 x k k has the same same prime factors as p raised to twice the powers of each. Therefore 6 negationslash | p = 6 negationslash | p 2 . Therefore 6 | p 2 = 6 | p .  #### You've reached the end of your free preview.

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• Fall '08
• Akhmedov,A
• Integers, Rational number, David Joseph Stith, ab < b2
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