《数学分析原理》英文答案.doc

MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1Generally, a “solution” is something that would be acceptable if turned in in theform presented here, although the solutions given are often close to minimal in thisrespect. A “solution (sketch)” is too sketchy to be considered a complete solutionif 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 6∈ 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 6∈ Q implies r + x 6∈ Q.Similarly, if rx ∈ Q, then x = (rx)/r ∈ Q. (Here, in addition to the ?eldproperties of R and Q, we use r6= 0.) Thus x 6∈ Q implies rx 6∈ 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 thatx2 = 12. Write x =mn in lowest terms. Then x2 = 12 implies that m2 = 12n2.Since 3 divides 12n2, it follows that 3 divides m2. Since 3 is prime (and by uniquefactorization in Z), it follows that 3 divides m. Therefore 32 divides m2 = 12n2.Since 32 does not divide 12, using again unique factorization in Z and the fact that3 is prime, it follows that 3 divides n. We have proved that 3 divides both m andn, contradicting the assumption that the fraction mnis in lowest terms.(Alternate solution (Sketch): If x ∈ Q satis?es x2 = 12, thenx2 is in Q and satis?esx￠2= 3. Now prove that there is no y ∈ Q such that y2 = 3 by repeating the2proof that√2 6∈ Q.Problem 1.5: Let A ? R be nonempty and bounded below. Set ?A = {?a : a ∈A}. Prove that inf(A) = ? sup(?A).Solution: First note that ?A is nonempty and bounded above. Indeed, A containssome element x, and then ?x ∈ A; moreover, A has a lower bound m, and ?m isan upper bound for ?A.We now know that b = sup(?A) exists. We show that ?b = inf(A). That ?b isa lower bound for A is immediate from the fact that b is an upper bound for ?A.To show that ?b is