Mathematics 220 Midterm Practice problems from old exams Page 1 of 8

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1 Mathematics 220 Midterm Practice problems from old exams Page 1 of 8 1. (a) Write the converse, contrapositive and negation of the following statement: For every integer n, if n is divisible by 3 then n 2 is divisible by 3. Converse: For every integer n, if n 2 is divisible by 3, then n is divisible by 3. Contrapositive: For every integer n, if n 2 is not divisible by 3 then n is not divisible by 3. Negation: There exists an integer n such that n is divisible by 3 and n 2 is not divisible by 3. (b) Let A n be the interval [ n, n 3 ) for n N. Find n N required). n N A n = [ 1, 2] n N A n = (, 6) A n and n N A n (no proof is (c) Write R N as a union of an indexed collection of sets where each set is an interval. If we let I := N {0}, let A 0 := (, 1) and, for every n N, let A n := (n, n + 1), then: R N = i I A i. 2. Consider the following two statements: 1. n N, z Z such that z = n 2. z Z such that n N, z = n One of the statements is true, and the other is false. Determine which is which and prove both of your answers. The first statement is true. Proof. Since N Z, given any n N we can simply choose z to be equal to n. The second statement is false.

2 Mathematics 220 Midterm Practice problems from old exams Page 2 of 8 Proof. We prove that its negation: z Z, n N such that z n is true. Given any z Z we can simply choose n := z Since z N and since z z for any integer z (why not?) this completes the proof. 3. (a) Give the definition of the power set P(A) of a set A. See textbook. (b) Let A = {1, 2, {1, 2}}. Determine whether the following statements are True or False (and provide a brief explanation why). (a) {1, 2} A. (b) {1, 2} P(A). (c) {1, 2} A. (d) {1, 2} P(A). (a) True because 1 A and 2 A. (b) False because 1 / P(A) and 2 / P(A). (c) True because it is listed as an element of A. (d) True because (a) was true. 4. Let A := {n N : z Z such that n = 2z + 1} and let B := {n N : k N such that n = 2k}. Determine the following: 1. A B, 2. A B, 3. A B, and 4. B A. Notice that A coincides with the odd natural numbers while B coincides with the even natural numbers. As such: 1. A B =, 2. A B = N, 3. A B = A, and 4. B A = B.

3 Mathematics 220 Midterm Practice problems from old exams Page 3 of 8 5. True or False: You do not have to give explanations. (a) N Z. T (b) [2, 3] P(R). T (c) 2 P(R). F (d) n Z (n, n + 2) = R. T (e) Q ( 2, ) = Q [ 2, ). T 6. Using any method you like, prove that the following statements are logically equivalent: Statement 1: Q (R S) Statement 2: (Q R) S. Q (R S) ( Q) (R S) ( Q) (( R) S) On the other hand (Q R) S ( (Q R)) S (( Q) ( R)) S The two right hand sides are logically equivalent by Associative law. 7. Let n Z. Prove that n 3 5n is odd. Proof. Let n Z and consider the following two (exhaustive) cases:

4 Mathematics 220 Midterm Practice problems from old exams Page 4 of 8 If n is even, then there is some k Z such that n = 2k. In this case, n 3 5n = (2k) 3 5(2k) + 13 = 8k 3 10k = 2(4k 3 5k + 6) + 1 and since 4k 3 5k + 6 Z we conclude that n 3 5n is odd. On the other hand, if n is odd, then there is some k Z such that n = 2k + 1. In this case, n 3 5n = (2k + 1) 3 5(2k + 1) + 13 = 8k k 2 + 6k k = 8k k 2 4k = 2(4k 3 + 6k 2 2k + 4) + 1 and since 4k 3 + 6k 2 2k + 4 Z we conclude that n 3 5n is odd. 8. (a) Write the negation of the following statement: For every positive ɛ there exists a positive δ such that if x < δ then f(x) < ɛ. We start negating this sentence, and get: There exists a positive ɛ such that for every positive δ, x < δ does not imply that f(x) < ɛ. If we recall the rules for negating the implication, it is tempting to write: There exists a positive ɛ such that for every positive δ, x < δ and f(x) ɛ. Note, however, that this doesn t mean the same thing as the sentence just above (and actually, unlike the above sentence, it is not a statement at all, but is an open sentence with x being a variable). This happens because when we say x < δ implies f(x) < ɛ, we mean that this happens for all x that satisfy x < δ. Thus, when we say x < δ does not imply that f(x) < ɛ, we mean that there exists x such that x < δ and f(x) ɛ. Putting it all together we get: There exists a positive ɛ such that for every positive δ, there exists x such that x < δ and f(x) ɛ. (b) Is the number 0.7(π ) rational or irrational? (Include a short proof; you can assume without proof that π is irrational). It is irrational. Two ways to prove this: could refer to the results from class: the sum/difference of a rational and irrational number is irrational; the product of a non-zero rational number and an irrational number is irrational. Alternatively, you could just use proof by contradiction (this is how we obtained those results, anyway): Let a = 0.7(π ). Suppose this number was rational. Note that the numbers 0.7 and are both rational. Then we have: π = a would have been rational as well a contradiction (we 7 are given that π is irrational). 9. Prove that if n and m are odd integers, then n 2 m 2 0 (mod 8).

5 Mathematics 220 Midterm Practice problems from old exams Page 5 of 8 The intended solution. The easiest way to solve this problem was by cases: consider the possible remainders mod 8. If n is odd, it can be congruent to 1, 3, 5 or 7 mod 8. Then n 2 is congruent mod 8 to, respectively, 1 2 = 1, or 3 2 = 9 1 mod 8, or 5 2 = 25 1 mod 8, or 7 2 = 49 1 mod 8. Thus we see that in all the cases, if n is odd, then n 2 1 mod 8. This also shows that m 2 1 mod 8, since m is odd as well. Thus, n 2 m = 0 mod 8. Alternative solution (more popular): Since n and m are odd, we can write n = 2k + 1, m = 2l + 1 for some integers k, l. By definition of congruence, we need to prove that n 2 m 2 is divisible by 8. Then n 2 m 2 = (2k + 1) 2 (2l + 1) 2 = 4(k 2 l 2 + k l). Note that one cannot stop here: this clearly shows that n 2 m 2 is divisible by 4, but more work is needed to show that n 2 m 2 is divisible by 8: namely, we need to prove that the integer k 2 l 2 + k l is even. So let us prove the lemma: Lemma. For any two integers k, l, the number k 2 l 2 + k l is even. Proof of the Lemma: either by cases (you need to consider 3 cases: k, l both even, both odd, and one is even one is odd), or better, note that for any k, k 2 k is even (this can be proved by cases: k even/odd, or by considering congruence mod 2). Then k 2 l 2 + k l = (k 2 k) + (l 2 l) is even. Now, by the Lemma, we have that there exists an integer q such that k 2 l 2 +k l = 2q, and putting it all together, we get that n 2 m 2 = 4 2q = 8q. 10. (a) Prove that 10 is irrational. Proof by contradiction. Suppose 10 was rational. Then there exist integers a and b with no common factors, such that 10 = a b. Then a = b 10, and therefore a 2 = 10b 2 ; in particular, a 2 is even. We proved in class that a 2 is even if and only if a is even, so we get that a is even, that is, a = 2k for some integer k. Let us plug it in. We get: 4k 2 = 10b 2, and thus 2k 2 = 5b 2. Then 5b 2 is even, and therefore b 2 is even, since the product of two odd numbers is odd. Then b is even, and we arrive at a contradiction, since we were assuming that a and b had no common factors, but we arrived at the conclusion that they are both even. Note that the same argument with using divisibility by 10 rather than by 2 does not work so well, because it is harder to prove the lemma that 10 a 2 if and only if 10 a. (b) Prove that the following statement is False: If x, y are both irrational, then x y is irrational. To show that this statement is false, all we need is a counterexample. Take, for example, x = y = 2. Then x y = 0 is rational, while x and y are both irrational.

6 Mathematics 220 Midterm Practice problems from old exams Page 6 of 8 (c) Prove that 5 2 is irrational. We prove this by contradiction. Suppose 5 2 = q is a rational number. Then, squaring both sides, we get = q 2, and therefore 7 q 2 10 =. 2 Now, note that if q is a rational number, then q 2 is rational, 7 q 2 is rational, and 7 q2 2 is rational. We get a contradiction with part (a). 11. (a) Write the negation of the following statement: For every (a, b) N N, if a > b then (a + b) 2 (a b) 2. (a, b) N N s.t. (a > b) and ((a + b) 2 < (a b) 2 ). (b) Write the converse and contrapositive of the following statement: If it is raining outside then this is Vancouver. Converse If this is Vancouver then it is raining outside. Contrapositive If it is not Vancouver then it is not raining outside. (c) Give a precise mathematical definitions of the following sets B = α I S α C = α I S α α I S α = {x α I s.t. x S α } α I S α = {x α I, x S α } (d) For any n N, let A n = ( 1 n 1, n 2 ]. Simplify the following sets B = n N A n C = n N A n Union is B = ( 1, 4]. Intersection is C = (0, 3]. (e) Let A, B be sets in some universal set U. Suppose that Ā = {3, 8, 9}, A B = {1, 2}, B A = {8} and A B = {5, 7} Determine A, B, U. It really helps to note that A = (A B) (A B).

7 Mathematics 220 Midterm Practice problems from old exams Page 7 of 8 Then A = {1, 2} {5, 7} = {1, 2, 5, 7} Similarly B = {8} {5, 7} = {5, 7, 8}. Finally U = A Ā = {1, 2, 5, 7} {3, 8, 9} = {1, 2, 3, 5, 7, 8, 9}. 12. (a) Prove or disprove the following statement Let a, b, c, d R. If ab cd then a c and b d. The statement is false. Let a = b = 1 and c = d = 0. Then ab = 1 > 0 = cd, but a < c and b < d. (b) Prove or disprove the following Let a, b, c, n Z so that n 3. If a + b 1 mod n and b + c 1 mod n then a + c 2 mod n. This is false. The negation of the statement is There are some a, b, c, n Z so that n 3, so that a + b 1 mod n and b + c 1 mod n but a + c 2 mod n. Consider n = 3, a = 2, b = 2, c = 2. Then a + b = b + c = 4 1 mod 3, but a + c = 4 1 mod 3, not 2. (c) Repeat part (b), but when n = 2. This is now true. Proof. Assume a+b 1 mod 2 and b+c 1 mod 2. Then there exist k, l Z so that a + b 1 = 2k b + c = 2l hencea + 2b + c 2 = 2k + 2l a + c 2 = 2(k + l b) Since k + l b Z, it follows that a + c 2 mod 2 as required. 13. Let n Z. Prove that n is odd if and only if 7n + 3 is odd. Proof. The number n is either even or odd and so we consider the two cases. Assume n is even, so n = 2k for some k Z. Then n 2 +1 = 4k 2 +1 = 2(2k 2 )+1 and so is odd. Similarly 7n + 3 = 14k + 3 = 2(7k + 1) + 1 and so is odd.

8 Mathematics 220 Midterm Practice problems from old exams Page 8 of 8 Now assume n is odd, so n = 2k+1 for some k Z. Then n 2 +1 = 4k 2 +4k+2 = 2(2k 2 + 2k + 1) and so is even. Similarly 7n + 3 = 14k + 10 = 2(7k + 5) and so is even. Thus we see that either n and 7n + 3 are both even or both odd. Note that you can also solve this using a couple of lemmas. Namely If n is odd then n is even. and If 7n + 3 is odd then n is even. Both of which are easy to prove using their contrapositives. The proof of the main result then becomes Proof. We must prove both the forward and backward implications. Assume that n is odd, then by the previous lemma n is even and hence n = 2k for some k Z. Then 7n+3 = 14k+3 = 2(7k+1)+1. Since 7k+1 Z, it follows that 7n + 3 is odd. Now assume that 7n + 3 is odd, then by the above lemma, n is even and hence n = 2l. Then n = 4l = 2(2l 2 ) + 1. Since 2l 2 Z, it follows that n is odd. 14. Determine whether the following four statements are true or false explain your answers ( true or false is not sufficient). (i) x R, y R, if (xy 0) then (x + y 0). (ii) x R, y R s.t. if (xy 0) then (x + y 0). (iii) x R s.t. y R s.t. if (xy 0) then (x + y 0). (i) False negation is x s.t. y s.t. (xy > 0) and (x + y 0). This is true, Consider x = y = 2, then xy = 4 > 0 and x + y = 4 < 0. Since negation is true, original is false. (ii) True Pick any x and set y = x. Now if x = 0 then y = 0 and xy = x+y = 0 and so all is true. On the other hand, if x 0 then xy = x 2 < 0 and so the hypothesis is false, and the implication is therefore true. In either case the implication is true. (iii) True pick (say) x = 11, y = 3 then xy = 3 > 0 and x+y = 4 > 0 as required.

Mathematics 220 Homework 4 - Solutions. Solution: We must prove the two statements: (1) if A = B, then A B = A B, and (2) if A B = A B, then A = B.

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