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 fi 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 Q0, and suppose that r+x Q. Then, using the fi 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 fi eld properties of R and Q, we use r 6= 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 that x2= 12. Write x = m n 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 unique factorization in Z), it follows that 3 divides m. Therefore 32divides m2= 12n2. Since 32does not divide 12, using again unique factorization in Z and the fact that 3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and n, contradicting the assumption that the fraction m n is in lowest terms. Alternate solution (Sketch): If x Q satisfi es x2= 12, then x 2 is in Q and satisfi es x 2 2 = 3. Now prove that there is no y Q such that y2= 3 by repeating the proof 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 contains some element x, and then x A; moreover, A has a lower bound m, and m is an upper bound for A. We now know that b = sup(A) exists. We show that b = inf(A). That b is a lower bound for A is immediate from the fact that b is an upper bound for A. To show that b is the greatest lower bound, we let c b and prove that c is not a lower bound for A. Now c c. Then x A and x 1, fi xed throughout the problem. Comment: We will assume known that the function n 7 bn, from Z to R, is strictly increasing, that is, that for m, n Z, we have bm 0. We will also assume that the usual laws of exponents are known to hold when the exponents are integers. We cant assume anything about fractional exponents, except for Theorem 1.21 of the book and its corollary, because the context makes it clear that we are to assume fractional powers have not yet been defi ned. (a) Let m, n, p, q Z, with n 0 and q 0. Prove that if m n = p q, then (bm)1/n= (bp)1/q. Solution: By the uniqueness part of Theorem 1.21 of the book, applied to the positive integer nq, it suffi ces to show that h (bm)1/n inq = h (bp)1/q inq . Now the defi nition in Theorem 1.21 implies that h (bm)1/n in = bmand h (bp)1/q iq = bp. Therefore, using the laws of integer exponents and the equation mq = np, we get h (bm)1/n inq = hh (bm)1/n iniq = (bm)q= bmq = bnp= (bp)n= hh (bp)1/q iqin = h (bp)1/q inq , as desired. By Part (a), it makes sense to defi ne bm/n= (bm)1/nfor m, n Z with n 0. This defi nes brfor all r Q. (b) Prove that br+s= brbsfor r, s Q. Solution: Choose m, n, p, q Z, with n 0 and q 0, such that r = m n and s = p q. Then r + s = mq+np nq . By the uniqueness part of Theorem 1.21 of the book, applied to the positive integer nq, it suffi ces to show that h b(mq+np)/(nq) inq = h (bm)1/n(bp)1/q inq . Directly from the defi nitions, we can write h b(mq+np)/(nq) inq = h b(mq+np) i1/(nq)nq = b(mq+np). Using the laws of integer exponents and the defi nitions for rational exponents, we can rewrite the right hand side as h (bm)1/n(bp)1/q inq = hh (bm)1/n iniqhh (bp)1/q iqin = (bm)q(bp)n= b(mq+np). This proves the required equation, and hence the result. (c) For x R, defi ne B(x) = br: r Q (,x. Prove that if r Q, then br= sup(B(r). Solution: The main point is to show that if r, s Q with r 0 and q 0, such that r = m n and s = p q. Then MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 13 also r = mq nq and s = np nq, with nq 0, so br= (bmq)1/(nq)andbs= (bnp)1/(nq). Now mq x. If r Q (,x, then br B(k) so that br bkby Part (c). Thus bkis an upper bound for B(x). This shows that the defi nition makes sense, and Part (c) shows it is consistent with our earlier defi nition when r Q. (d) Prove that bx+y= bxbyfor all x, y R. Solution: In order to do this, we are going to need to replace the set B(x) above by the set B0(x) = br: r Q (,x) (that is, we require r 1 and N = 1,2,3,.) Proof: Clearly 1 is a lower bound. (Indeed, (b1/n)n= b 1 = 1n, so b1/n 1.) We show that 1+x is not a lower bound when x 0. If 1+x were a lower bound, then 1 + x b1/nwould imply (1 + x)n (b1/n)n= b for all n N. By Lemma 1, we would get 1 + nx b for all n N, which contradicts the Archimedean property when x 0.