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MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1

Generally, a “solution” is something that would be acceptable if turned in in the form presented here, although the solutions given are often close to minimal in this respect. A “solution (sketch)” is too sketchy to be considered a complete solution if turned in; varying amounts of detail would need to be ?lled in. Problem 1.1: If r ∈ Q \ {0} and x ∈ R \ Q, prove that r + x, rx ∈ Q. Solution: We prove this by contradiction. Let r ∈ Q\{0}, and suppose that r +x ∈ Q. Then, using the ?eld properties of both R and Q, we have x = (r + x) ? r ∈ Q. Thus x ∈ Q implies r + x ∈ Q. Similarly, if rx ∈ Q, then x = (rx)/r ∈ Q. (Here, in addition to the ?eld properties of R and Q, we use r = 0.) Thus x ∈ Q implies rx ∈ Q. Problem 1.2: Prove that there is no x ∈ Q such that x2 = 12. Solution: We prove this by contradiction. Suppose there is x ∈ Q such that x2 = 12. Write x = m in lowest terms. Then x2 = 12 implies that m2 = 12n2 .


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자료제목 : 월터루딘 해석학 rudin w. , solution manual of principles of mathematical analysis 솔루션
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