Copy of s1 - Solutions to homework # 1. 1. Let r be...

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Solutions to homework # 1. 1. Let r be rational and x be irrational. If r + x were rational, then x = ( r + x ) - r would be rational too, as a difference of two rational numbers. Hence r + x is irrational. Likewise, if rx were rational, then, since 1 /r exists and is rational, the number x = ( rx )(1 /r ) would be rational too as a product of two rational numbers. Thus rx must be irrational. 2. Suppose there exists a rational number whose square is 12. Write such a number in lowest terms as m/n , where m and n are integers having no factors in common. Then m 2 = 12 n 2 = 3 · 2 2 · n 2 . Since 3 appears in the right-hand side, 3 divides m 2 . But 3 is prime, so it must divide m ; hence m 2 must be divisible by 9. But then 3 divides n 2 and therefore n as well, contrary to the assumption that m and n have no common factors. Contradiction! 3. Since E is nonempty, it has at least one element, say, x . As α is a lower bound of E , we get α x . Since β is an upper bound of E , x β
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This document was uploaded on 11/17/2010.

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