SOLUTION SET FOR THE HOMEWORK PROBLEMSPage 5. Problem 8. Prove that if x and y are real numbers, then2xy ≤ x2+ y2.Proof. First we prove that if x is a real number, then x2≥ 0. The productof two positive numbers is always positive, i.e., if x ≥ 0 and y ≥ 0, thenxy ≥ 0. In particular if x ≥ 0 then x2= x · x ≥ 0. If x is negative, then −xis positive, hence (−x)2≥ 0. But we can conduct the following computationby the associativity and the commutativity of the product of real numbers:0 ≥ (−x)2= (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x= (((−1)(x(−1)))x = (((−1)(−1))x )x = (1x)x = xx = x2.The above change in bracketting can be done in many ways. At any rate,this shows that the square of any real number is non-negaitive. Now if x andy are real numbers, then so is the difference, x − y which is defined to bex + (−y). Therefore we conclude that 0 ≤ (x + (−y))2and compute:0 ≤ (x + (−y))2= (x + (−y))(x + (−y)) = x(x + (−y)) + (−y)(x + (−y))= x2+ x(−y) + (−y)x + (−y)2= x2+ y2+ (−xy) + (−xy)= x2+ y2+ 2(−xy);adding 2xy to the both sides,2xy = 0 + 2xy ≤ (x2+ y2+ 2(−xy)) + 2xy = (x2+ y2) + (2(−xy) + 2xy)= (x2+ y2) + 0 = x2+ y2.Therefore, we conclude the inequality:2xy ≤ x2+ y2for every pair of real numbers x and y. ♥12SOLUTION SET FOR THE HOMEWORK PROBLEMSPage 5. Problem 11. If a and b are real numbers with a < b, then thereexists a pair of integers m and n such thata <mn< b, n 6= 0.Proof. The assumption a < b is equivalent to the inequality 0 < b − a. Bythe Archimedian property of the real number field, R, there exists a positiveinteger n such thatn(b − a) > 1.Of course, n 6= 0. Observe that this n can be 1 if b − a happen to be largeenough, i.e., if b−a > 1. The inequality n(b−a) > 1 means that nb−na > 1,i.e., we can conclude thatna + 1 < nb.Let m be the smallest integer such that na < m. Does there exists such aninteger? To answer to the question, we consider the set A = {k ∈ Z : k > na}of integers. First A 6= ∅. Because if na ≥ 0 then 1 ∈ A and if na > 0 then bythe Archimedian property of R, there exists k ∈ Z such that k = k · 1 > na.Hence A 6= ∅. Choose ` ∈ A and consider the following chain:` > ` − 1 > ` −2 > ··· > ` − k, k ∈ N.This sequence eventually goes down beyond na. So let k be the first naturalnumber such that ` −k ≤ na, i.e., the natural number k such that ` − k ≤na < ` − k + 1. Set m = ` − k + 1 and observe thatna < m = ` − k ≤ +1 ≤ na + 1 < nb.Therefore, we come to the inequality na < m < nb. Since n is a positiveinteger, we devide the inequlity by n withoug changing the direction of theinequality:a =nan<mn<nbn= b.♥Page 14, Problem 6. Generate the graph of the following functions on Rand use it to determine the range of the function and whether it is onto andone-to-one:a) f (x) = x3.b) f(x) = sin x.c) f (x) = ex.d) f(x) =11+x4.SOLUTION SET FOR THE HOMEWORK PROBLEMS 3Solution. a) The function f is bi-jection since f(x) < f(y) for anypair x, y ∈ R with the relationx < y and for every real numbery ∈ R there exists a real numbex ∈ R such that y = f(x).b) The function f is neither in-jective nor surjective sincef(x + 2π) = f(x)x + π 6= x, x ∈ R, and if y > 1then there is no x ∈ R such thaty = f(x).c) The function f is injectivebecausef(x) < f(y)if x < y, x, y ∈ R, but not surjec-tive as a map from R to R, be-cause there exists no x ∈ R suchthat f(x) = −1.4 SOLUTION SET FOR THE HOMEWORK PROBLEMSd) The function f is not injectiveas f (x) = f(−x) and x 6= −x forx 6= 0, nor surjective as there isno x ∈ R such that f(x) = −1.♥Page 14, Problem 8. Let P be the set of polynomials of one real variable. If p(x) is sucha polynomial, define I(p) to be the function whose value at x isI(p)(x) ≡Zx0p(t)dt.Explain why I is a function from P to P and determine whether it is one-to-one and onto.Solution. Every element p ∈ P is of the form:p(x) = a0+ a1x + a2x2+ ··· + an−1xn−1, x ∈ R,with a0, a1, ··· , an−1real numbers. Then we haveI(p)(x) =Zx0(a0+ a1t + a2t2+ ··· + an−1tn−1)dt= a0x +a12x2+a23x3+ ··· +an−1nxn.Thus I(p) is another polynomial, i.e., an element of P. Thus I is a function from P to P.We claim that I is injective: Ifp(x) = a0+ a1x + a2x2+ ··· + am−1xm−1;q(x) = b0+ b1x + b2x2+ ··· + bn−1xn−1have I(p)(x) = I(q)(x), x ∈ R,i.e.,a0x +a12x2+a23x3+ ··· +am−1mxm= b0x +b12x2+b23x3+ ··· +bn−1nxn.Let P(x) = I(p)(x) and Q(x) = I(q)(x). Then the above equality for all x ∈ R allows us todifferentiate the both sides to obtainP0(x) = Q0(x) for every x ∈ R,SOLUTION SET FOR THE HOMEWORK PROBLEMS 5in particular a0= P0(0) = Q0(0) = b0. The second differentiation givesP00(x) = Q00(x) for every x ∈ R,in particular a1= P00(0) = Q00(0) = b1.Suppose that with k ∈ N we have P(k)(x) = Q(k)(x) for every x ∈ R. Then the differen-tiation of the both sides givesP(k+1)(x) = Q(k+1)(x) forevery x ∈ R,in particular ak+1= P(k+1)(0) = Q(k+1)(0) = bk+1. Therefore the mathematical inductiongivesa0= b0, a1= b1, ··· , am−1= bm−1and m = n,i.e., p = q. Hence the function I is injective.We claim that I is not surjective: As I(p)(0) = 0, the constant polynomial q(x) = 1cannot be of the form q(x) = I(p)(x) for any p ∈ P, i.e., there is no p ∈ P such thatI(p)(x) = 1. Hence the constant polynimial q is not in the image I(P). ♥Page 19, Problem 3. Prove that:a) The union of two finite sets is finite.b) The union of a finite sent and a countable set is countable.c) The union of two contable sets is countable.Proof. a) Let A and B be two finite sets. Set C = A ∩ B and D = A ∪ B. First, let a, band c be the total number of elements of A, B and C respectively. As C ⊂ A and C ⊂ B,we know that c ≤ a and c ≤ b. We then see that the union:D = C ∪ (A\C) ∪ (B\C)is a disjoint union, i.e., the sets C, A\C and B\C are mutually disjoint. Thus the totalnumber d of elements of D is precisely c + (a − c) + (b − c) = a + b − c which is a finitenumber, i.e., D is a finite set …