sol1 115A

# sol1 115A - MAT 115A University of California Fall 2010...

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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 well-ordering 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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sol1 115A - MAT 115A University of California Fall 2010...

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