811Solutions2

811Solutions2 - Homework 2 Solutions Kyle Chapman October...

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Unformatted text preview: Homework 2 Solutions Kyle Chapman October 15, 2011 18.1 a Seeking a contradiction, suppose 3 is rational. This means it can be written as p/q for some integers p and q . Without loss of generality, assume p and q have no common factors, since we can reduce to lowest terms. We then have that p 2 /q 2 = 3 so p 2 = 3 q 2 . This means that 3 divides p 2 and since 3 is prime then 3 divides p . This means p = 3 k for some integer k and so 9 k 2 = 3 q 2 so 3 k 2 = q 2 . This means tht 3 divides q 2 and so 3 divides q . This is our contradiction, however, as we had assumed that p/q had been reduced to lowest terms. Thus, 3 is not rational. b Seeking a contradiction, suppose that there exists rational numbers, r,s with the property that 3 = r + s 2. We can then square both sides and get 3 = r 2 + 2 rs 2 + 2 s 2 . If rs 6 = 0 then we can solve this to get 2 = (3- r 2- 2 s 2 ) / ( rs ) which is rational and so we have our contradiction, so suppose that rs = 0. That means either r = 0 or s = 0. If s = 0 we look at the original equation and get that 3 is rational, which we have just shown to be false. Thus, we can suppose that s 6 = 0 which leaves us with r = 0 and the equation 3 = s 2. We can write s = p/q in lowest terms and square both sides to get 3 q 2 = 2 s 2 . Since 2 divides 3 q 2...
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This note was uploaded on 12/26/2011 for the course MATH 8 taught by Professor Bigelow during the Fall '08 term at UCSB.

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811Solutions2 - Homework 2 Solutions Kyle Chapman October...

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