MATH 2400 (Azoff) Fall Notes to Homework Assignments. Problems from Chapter 1 of Spivak (Assignment 1)

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1 MATH 2400 (Azoff Fall 2010 Notes to Homework Assignments Problems from Chapter 1 of Spivak (Assignment 1 1(ii. The fact that y( x (yx is not an axiom. Spivak proves it near the bottom of Page The key point is that x y 0 does not have a multiplicative inverse, so it cannot be cancelled. 4. While we will develop more efficient methods for solving inequalites later, at this point, it is important to understand the distinction between intersection and union when considering cases. When giving answers, it is best to use the symbol for union to avoid the ambiguity of the English word and which sometimes denotes union and sometimes denotes intersection. ii 5 x 2 < 8. Transposing shows this is equivalent to x 2 3 > 0. Since x 2 is always non-negative, the answer is all real numbers or (, in interval notation. iv f(x :(x 1(x 3 > 0. A shortcut to solving many inequalities is to first solve the corresponding equalities and then consider the complementary intervals one at atime. In the present problem, the corresponding equation (x 1(x 3 0 has two solutions: x 1andx 3. These split up the real line into the three intervals (, 1, (1, 3, and (3,. f(x > 0 in the first and third of these intervals, but negative in the middle interval. Thus the answer is the union of the first and third intervals, namely (, 1 (3,. An alternative approach is to base cases on the signs of the factors of f(x. Case 1. Both factors are positive. This means we must simultaneously have x>1andx>3. Thus this case contributes the intersection of the two intervals. Since the latter interval is smaller, this case contributes the interval (3,. Case 2. x 1andx 3 are similtaneously negative. In this case, the first inequlity is more restrictive, so this case contributes the interval (, 1. Since the two cases are connected by or, the answer to the problem is (, 1 (3,,asabove. vi Transposing yields x 2 x 1 > 0. Use the quadratic formula to factor the left hand side and proceed as in (iv. viii x 2 x 1> 0. Here the right hand side has no real roots. In fact, completing the square on the left yields (x > 0. Thus the answer is all real numbers. xii x 3 x < 4. This is sneaky. The right hand side increases as x increases and equals 4whenx 1. Thus the solution is (, 1. xiv f(x : x 1 > 0. As in the second approach to (iv, there are two cases: 1 both x1 factors positive and 2 both factors negative. The answer is (, 1 (1,. To use the shortcut method, you must split up the real line at 1 because f(1 0 and at 1 because f( 1 is not defined. A full explanation of why this is necessary will have to wait until we study the intermediate value theorem in Cahpter 6.

2 5(iv. The proof should be phrased in terms of the set P of positive numbers. Proposition. If a<band c>0 then ac < bc. Proof. The hypotheses mean b a P and c P. Axiom P12 then tells us that the product (b ac P, i.e., bc ac P.Thusac < bc by definition. 9(ii. The triangle inequality assures us that a b a b is always non-positive, so its sign must be changed when dropping the outer absolute value: ( a b a b ( a b a b a b a b. 11. One can analyze these using cases as in Problem 4 or apply an intuitive understanding of absolutie value. For example, saying x 3 < 8isequivalenttosayingthatx can roam up to 8 units in either direction from 3. The solution to this inequality is thus ( 5, (iv. Apply the triangle inequality: x y x ( y x y x y. 17b. x 2 3x 2y 2 4y 2 (x (y 1 2 9, The minimum value of this 4 expression occurs when the parenthetical quantities are zero. Thus the minimum value is c. Completing the square for the first two terms yields (x 2y 2 y 2 4x 6y 7. Now introduce the new variable z x 2y os that x z 2y. Substituion yields z 2 y 2 4z 2y 7(z 2 2 (y so the minimum value is 2. 19b. You will not be held responsible for this. 20. We re given x x 0 < ɛ 2 and y y 0 < ɛ 2. The key step is to use the triangle inequality: (x y (x 0 y 0 (x x 0 (y y 0 x x 0 y y 0 < ɛ 2 ɛ 2 ɛ. Problems from Chapter 2 of Spivak (Assignment 2 Be sure to explicitly define any notation you use in your proofs. If you phrase an inductive proof in terms of a set S of natural numbers, this means being precise concerning which natural numbers belong to S. Do not forget the base case, i.e., explain why 1 S. For the inductive step, be sure to assume k S. If you use the P (n notation, remember that P (n isastatement, notanumber. To avoid confusion, clearly identify P (n at the beginning of your proof. This will involve a little extra writing, but should help keep the argument clear. Also, be explicit about the logic of the argument; in particular, assume P ( at the beginning of the inductive step. 3b. Let 0 k n be natural numbers. a If 1 k n, then ( ( n1 k n ( k 1 n. b is a natural number. ( n k

3 Proof. For a, we work on the left-hand member of the equation. Applying the definition and finding a common denominator, we get ( n k 1 ( n n! (k 1!(n k 1! n! k!(n! k(n! (n k 1(n! (k!(n k 1! (n 1! k!(n k 1! ( n 1 k as desired. For b, we take S : {n N : ( n N for each 0 k n}. It is easy to check that 1 S. Assuming p S and 1 k p, Part a tells us that ( ( p1 k p ( k 1 p ; the ( inductive hypothesis tells us that both summands on the right are integers and thus p1 ( k N for 1 k p. Since p1 ( 0 p1 ( p1 1, we conclude p1 k N for 0 k p1, and thus p1 S by definition. Thus S is inductive and the proof is completed by invoking the Principle of Mathematical Induction. 5. For n N and r 1,wehave1r r 2 r n 1 rn1 1 r. (* a Prove this by induction. b Derive this equation without suing induction. 1 r 2 1 r For Part a, set T : {n N :( holds.}. When n 1, the right side of (* is 1r, so1 T. For the inductive step, suppose k T.Then, 1r...r k r k1 1 rk1 1 r r k1 1 rk2, 1 r so k 1 T and T is inductive. Thus we are done by PMI. For Part b, set S 1r r 2...r n.thenrs r r 2 r n r n1. Subtracting and cancelling yields (1 rs 1 r n1. Dividing by 1 r gives the desired result. 13a. 6 is irrational. Proof. Suppose for purposes of contradiction that 6 p q where p, q are integers which have no common factor above 1. Squaring and clearing fractions yields 6q 2 p 2. This forces p to be even, and we write p 2r for some integer r. Substituting and cancelling yields 3q 2 2r 2. Thus 3q 2 is even, so q 2 is even, so both p and q are even and we have reached a contradiction. 14. x : 2 3 is irrational. Proof. Note that x Thus the rationality of x would imply the rationality of 6 and we have just proved that 6 / Q. Problems from Chapter 3 of Spivak (Assignment 3, first part { 0, x Q 2. g(x x 2 and h(x. The answers to Parts i, iii, and v are 1, x R\Q giveninthetext. ii When is h(y g(y? When y Q, the inequality translates to 0 y 2 which is always true, so all rational y satisfy the the original inequality. On the other

4 hand, when y is irrational, the inequality translates to 1 y 2. Putting these together, the answer can be expressed as (, 1] {( 1, 1 Q} [1,.Other expressions are possible, but it is best to phrase your answer in terms of unions and intersections. iv g(w w. This is tantamount to solving w 2 w and the answer is [0, 1]. 5. You should distinguish between functions and their values. Thus for example, the answer to Part ii should be f s P ;nox shuold be involved. 10a. For which functions f does there exist a function g so that f g 2? The answer is iff f(x 0 for all x. The condition is necessary because squares of all numbers are non-negative. The condition is sufficient because if it is satisfied, then we can take g(x f(x for all x. Be sure you understand what this question is asking and why we have answered it successfully. 10b. For which functions b and c does there exist a function x satisfying {x(t} 2 b(tx(tc(t 0? The answer, based on the quadratic formula, is iff {b(t} 2 4c(t 0 for all t. 12. I did not grade this problem; answers can be found in the text. 21. Again, answers can be found in the text and no points were assigned to this problem. I did make some expository comments though. 23. Suppose f(g(x x for all x. a Show that if g(x g(y thenx y. (This is the contrapositive of the original statement of the problem; it means the function f is one-to-one. b Show that each b can be expressed as b f(a forsomea. (Thismeansf is onto. Proof. a Suppose g(x g(y. (This sentence must be included in the proof. Applying f to both sides of this equation, we get x f(g(x f(g(y y as desired. b Take a g(b. Then f(a f(g(b b as desired. 27b. Suppose f is a constant function. Find all functions g so that f g g f. Solution. Write c for the constant value of f. The answer is that g(c c for this particular c; this does not mean g must be the identity function. For example, if c 0,theng(x x 5 would work. 27c. Suppose f g g f for every function g. Thenf I. Proof. In view of Part b, we have f(c c for every c which makes f the identity function. Problems from Chapter 4 of Spivak (Assignment 3, second part 4. Answers are given to the odd parts of the problem. The graph for Part ii consists of 4 rays, two emanating from the point (1, 0 and the other two emanating from the point ( 1, 0. The graph for Part iv is the same as that for Part iii since 1 x x 1 for all x.

5 The graphs for Parts vi and viii each consist of two crossing lines. For Part vii, complete the square, rewriting the problem as (x 1 2 y Answers in the text; not graded. 14. Let g(x f(xc. The graph of g is always formed by moving the graph of f up c units; when c<0 this means moving the graph of f down c c units. 17b. Take f(x x [x]. For 0 x < 1, we have [x] 0 so f(x x on that interval. Since f has period 1, the rest of the graph is constructed by shifting this line segment integral amounts to the left and right. In particular, the graph jumps down at each integer.