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Unformatted text preview: MAT 115A University of California Fall 2010 Solutions Homework 1 1. Rosen 1.1 #5, pg. 13 Use the wellordering principle to show that 3 is irrational. Solution: Suppose that 3 were rational. Then there would exist positive integers a and b with 3 = a b . Hence the set S = { k 3  k and k 3 are positive integers } is nonempty since a = b 3 is in S . Therefore, by the well ordering property, S has a smallest element, say s = t 3. Then we have s 3 s = s 3 t 3 = ( s t ) 3. Since s 3 = 3 t and s are both integers, s 3 s must also be an integer. Also, s 3 s = s ( 3 1) is positive since 3 > 1. s 3 s < s since ( s 3 s ) s = 3 t 2 3 t = ( 3 2) 3 t < 0, as 3 < 2. This contradicts the choice of s as the smallest positive integer in S . It follows that 3 is irrational. 2. Rosen 1.3 #8, pg. 27 Use induction to prove that n k =1 k 3 = n ( n +1) 2 2 ....
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 Spring '10
 anne
 Number Theory, Integers

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