a. See the textbook for examples of proving logical equivalence using truth tables. b. There is a real number x for which f (x) < 0. (x 1) 2 > 0.

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1 For some problems, several sample proofs are given here. Problem 1. a. See the textbook for examples of proving logical equivalence using truth tables. b. There is a real number x for which f (x) < 0. Problem 2. Proof (1). Suppose that x is a positive real number with x = 1. Recall we showed in class that for every real number a = 0, we have the inequality a 2 > 0, so since x 1 = 0, we have (x 1) 2 > 0. Now, distributivity gives x 2 2x + 1 > 0, so by the addition law for inequalities (for all real numbers a, b, c, we have b > c if and only if a + b > a + c), x > 2x. Finally, recall the multiplication law for inequalities, which states that for all real numbers a, b, c with a > 0, we have b > c if and only if ab > ac. Since x > 0, use the multiplication law to arrive at our desired inequality x + 1 x > 2. Note: One quick way to prove the inequality a 2 > 0 for a = 0 is to use the multiplication law with two cases: Case 1) if a > 0, then a 2 > 0 by the multiplication law, and Case 2) if a < 0, then a 2 > 0 again by the multiplication law. 1 of

2 Proof (2). Suppose that x is a positive real number with x = 1. We will work backward. Notice this series of backward implications: x + 1 x > 2 = x > 2x by the multiplication law, since x > 0. = x 2 2x + 1 > 0 by the addition law. = (x 1) 2 > 0 by distributivity. = x 1 = 0 as shown in class. We know that x 1 = 0 since x = 1, so we are done. Note: It is critical for the proof just shown that the implications go backward because we began with what we wanted to show. You must keep this in mind while working out the proof and point out clearly to the reader what you are doing. Otherwise, the proof is simply wrong, sometimes in a catastrophic way. We often work out the ideas of a proof by starting from what we want to show, but you should take extra care if you write it out this way. Proof (3). Suppose that x is a positive real number with x = 1, and assume for contradiction that x + 1/x 2. Then x + 1 x 2 x x by the multiplication law, since x > 0. x 2 2x by the addition law. (x 1) 2 0 by distributivity. But recall from class that a 2 > 0 for all a = 0, so (x 1) 2 > 0 since x = 1. This gives a contradiction, so our assumption must have been false; instead, we must have x + 1/x > 2. Note: Notice that this proof by contradiction uses forward implications between each inequality, even though it looks very similar to the previous proof where we worked backward. There is a simple explanation for this: every implication used in this proof is the contrapositive of that used in the previous proof, so they are logically equivalent. (Remember: the contrapositive of P = Q is Q = P, which you ve learned are logically equivalent to each other.) We don t need to point out or indicate these forward implications with = because it is always understood that implications go forward unless indicated otherwise. In these two proofs, the implications between the inequalities happen to go both ways (they are ), so they are simple to deal with. However, this is not the case in general. For example, if you use an implication like x > 0 = x 2 > 0, you have to be careful because x > 0 = x 2 > 0. 2 of

3 Problem 3. Proof (1). Notice that 6m + 1n = 3(2m + n), but 3 does not divide 2, so there cannot be such m and n. Proof (2). Suppose that each of m and n is a positive integer, and assume for contradiction that 6m + 1n = 2. Then 3(2m + n) = 2. (1) Now, since m and n are both positive integers, the sum 2m + n must also be a positive integer, so 2m + n 1, 3(2m + n) 3 > 2, which is a contradiction to Equation (1), so our assumption must have been false; instead, we must have 6m + 1n = 2. Problem 4. Proof. We will prove this by induction on n. Base case: If n = 1, then n (2i + 1) = 1 (2i + 1) = (2(0) + 1) + (2(1) + 1) = 4 = (1 + 1) 2 = (n + 1) 2. Inductive step: Suppose that k (2i + 1) = (k + 1)2 for some positive integer k. Then k+1 (2i + 1) = k (2i + 1) + (2(k + 1) + 1) = (k + 1) 2 + 2(k + 1) + 1 = ((k + 1) + 1) 2. 3 of

4 Problem. a. Proof. We will prove this by strong induction on n. Base cases: We need two base cases because neither a 1 nor a 2 will be covered by our inductive step. First, if n = 1, then a n = a 1 = 1/2 < 2. Second, if n = 2, then a n = a 2 = 1 < 2. Inductive step: Suppose that for some positive integer k 2, we have a n < 2 for all positive integers n k. Then because a k < 2, we have a k 2 < 4, and furthermore a k 1 < 2 by the inductive hypothesis, so a k+1 = a k 2 + a k < = 2. b. Proof. We will prove this by strong induction on n. Base cases: Again, we need two base cases because neither a 1 nor a 2 will be covered by our inductive step. First, if n = 1, then Second, if n = 2, then a n = a 1 = 1 2 < 1 = a 2 = a n+1. a n = a 2 = 1 < /2 + 4 = a 3 = a n+1. Inductive step: Suppose that for some positive integer k 2, we have a n < a n+1 for all positive integers n k. Then because a k < a k+1, we have a 2 k < a 2 k+1, and furthermore a k 1 < a k = a (k+1) 1 by the inductive hypothesis, so a k+1 = a k 2 + a k < a k a (k+1) = a k+2. 4 of

5 Problem 6 Proof (1). We will show that the two derived sets are equal by showing that each is a subset of the other. Let each of A, B, and C be a subset of some universal set U. First, to show that A \ (B C) (A \ B) (A \ C), let x A \ (B C). Then by definition, x A and x / (B C), from which it follows that x / B and x / C. These imply that x A \ B and x A \ C, and so x (A \ B) (A \ C). Now, to show that A \ (B C) (A \ B) (A \ C), let x (A \ B) (A \ C). Then x A \ B and x A \ C, so by definition, x A and x / B, and x A and x / C. It follows that x / B C, and so x A \ (B C). Therefore, since each of the two sets in question is a subset of the other, they are equal. Note: The proof above was the kind of proof you were expected to understand and use, and you need to make sure you are able to write these proofs confidently. But the following proof using set identities was also accepted for this problem. Proof (2). Let each of A, B, and C be a subset of some universal set U. Then A \ (B C) = A (B C) = A (B C ) = (A B ) (A C ) = (A \ B) (A \ C). of