# 3.3 Proofs Involving Quantifiers

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1 3.3 Proofs Involving Quantifiers 1. In exercise 6 of Section 2.2 you use logical equivalences to show that x(p (x) Q(x)) is equivalent to xp (x) xq(x). Now use the methods of this section to prove that if x(p (x) Q(x)) is true, then xp (x) xq(x) is true. (Note: The other direction of the equivalence is quite a bit harder to prove. See exercise 12 of Section 3.5.) Proof. Suppose x(p (x) Q(x)). Suppose xp (x). Since x(p (x) Q(x)) there exists some object x 0 such that P (x) Q(x) is true. In other words, P (x 0 ) Q(x 0 ). In order for this statement to be true, one of the following conditions must be met: P (x 0 ) and Q(x 0 ) are true, P (x 0 ) is false and Q(x 0 ) is true, or P (x 0 ) and Q(x 0 ) are false. But then, since xp (x), P (x 0 ) is true. Then it follows that Q(x 0 ) is true. Therefore, we have shown xp (x) xq(x). 2. Prove that if A and B\C are disjoint, then A B C. Proof. Suppose A and B\C are disjoint, and suppose x A B. Then since A and B\C are disjoint, if x A, then x / B\C. By the definition of /, x / B\C is equivalent to x B\C. By the definition of \, this is equivalent to ((x B) (x / C)). By DeMorgan s law, this is equivalent to (x B) (x / C). By the definition of, this is equivalent to x / B x C. By the conditional law, this is equivalent to x B x C. 1

2 Therefore, since x B x C and x A B, x C. Since x was an arbitrary element of A, we can conclude that A B C. 3. Suppose A P(A). Prove that P(A) P(P(A)). Proof. Suppose A P(A). Let x be an arbitrary element of A. Then since A P(A) and x A, x P(A). Then there exists y such that y(y x y A). Since y A and A P(A), y P(A). Thus, since x P(A) and y P(A), it follows that x P(P(A)). Therefore, we have just shown P(A) P(P(A)) as required. 4. The hypothesis of the theorem proven in exercise 3 is A P(A). (a) Can you think of a set A for which this hypothesis is true? Definition Suppose A is a set. The power set of A, denoted P(A), is the set whose elements are all the subsets of A. In otherwords, P(A) = {x x A}. Therefore, if A has an element, since A is a subset of itself, the hypothesis is true. If A is an empty set, since the empty set is a subset of every set, the hypothesis is true. Therefore, any set will do (such as R, Q, Z...). 2

3 (b) Can you think of another? There are countless examples of sets where the hypothesis is true. 5. Suppose x is a real number. (a) Prove that if x 1 then there is a real number y such that y+1 y 2 = x. Proof. Let x be an arbitrary real number, and suppose x 1. Let y = 2x + 1 x 1 which is defined since x 1. Then, 2x+1 y + 1 y 2 = + 1 x 1 2x+1 2 = x 1 (2x+1)+(x 1) x 1 (2x+1) 2(x 1) x 1 = 3x x 1 3 x 1 = 3x 3 = x. (b) Prove that if there is a real number y such that y+1 = x, then y 2 x 1. Proof. Let x = y+1 which is defined if y 2. Solving this y 2 equation for y, we have x = y+1 y 2 x(y 2) = y + 1 xy 2x = y + 1 xy y = 2x + 1 y(x 1) = 2x + 1 y = 2x+1. x 1 Therefore, in order for y to be defined, x must not be 1. 3

4 6. Prove that for every real number x, if x > 2 then there is a real number y such that y + 1 y = x. Proof. Let x be an arbitrary number, and suppose x > 2. Let y = x + x which is defined since x > 2. Then, y + 1 y = x + x x + x 2 4 = (x + x 2 4) (x + x 2 4) = (x2 + 2x x (x 2 4)) + 4 2(x + x 2 4) = 2x2 + 2x x 2 4 2(x + x 2 4) = 2x(x + x 2 4) 2(x + x 2 4) = x. 7. Prove that if F is a family of sets and A F, then A F. Proof. Suppose A F. Let x be an armitrary element of F. Since A F and x A, x F. Therefore, clearly, x F. But x was an arbitrary element of F, so this shows that A F. 4

5 8. Prove that if F is a family of sets and A F, then F A. Proof. Suppose A F. Let x be an arbitrary element of F. Then by the definition , A(A F x A). Since A F and A(A F x A), it follows that x A. But x was an arbitrary element of F, so this shows that F A, as required. 9. Suppose F and G are families of sets. Prove that if F G then F G. Proof. Suppose F G. Let x be an arbitrary element of F. By definition , x F means A(A F x A). Then, since A F and F G, A G. Therefore, it follows that A(A G x A). But x was an arbitrary element of F, so this shows that F G, as required. 10. Suppose F and G are nonempty families of sets. Prove that if F G then G F. Proof. Suppose F G. Let A be an arbitrary element of F. Then since F G and A F, A G. Now, let x be an arbitrary element of G, which is defined since G is nonempty. Then, by the definition , A(A G x A). Thus, 5

6 x A. Since A was an arbitrary element of F and x A, it follows that x F, which is defined since F is nonempty. But x was an arbitrary element of G, so this shows that G F. 11. Suppose {A i i I} is an indexed family of sets. Prove that i I P(A i ) P( i I A i ). (Hint: First make sure you know what all the notation means!) Proof. Suppose {A i i I} is an indexed family of sets. Let x be an arbitrary element of i I P(A i ). Let y be an arbitrary element of x. Then, by the alternative notation of union of an indexed family of sets, i I( y(y x y A i )). In other words, there exists some i 0 I such that every element of x is an element of A i0. Therefore, y i I A i. Since y was an arbirary element of x and y i I A i, we can conclude that x P( i I A i ). But x was an arbitrary element of i I P(A i ), so this shows that i I P(A i ) P( i I A i ), as required. 12. Prove the converse of the statement proven in Example In other words, prove that if F P(B) then F B. Proof. Suppose F P(B). Let x be an arbitrary element of F. Then by definition , A(A F x A). In other words, there exists some set A such that A F and x A. Since A F and F P(B), A P(B). Then, by the definition of power set, all the element of A is an element of B. Since x A, we can conclude x B. But x was an arbitrary element of F, so this shows that F B, as required. 6

7 13. Suppose F and G are nonempty families of sets, and every element of F is a subset of every element of G. Prove that F G. Proof. Suppose F and G are nonempty families of sets, and every element of F is a subset of every element of G. Let x be an arbitrary element of F. Then, by definition , A(A F x A). In other words, there exists some set A such that A F and x A. Since A F and every element of F is a subset of every element of G, A G. Therefore, since A G and x A, by the definition , x G, which is defined since G is not an emptyset. But x was an arbitrary element of F, so this shows that F G, as required. 14. In this problem all variables range over Z, the set of all integers. (a) Prove that if a b and a c, then a (b + c). Proof. Let a, b, and c be arbitrary integers and suppose a b and a c. Since a b, we can choose some integer m such that ma = b. Similarly, since a c, we can choose an integer n such that na = c. Therefore b + c = ma + na = (m + n)a, so since m + n is an integer, a (b + c). (b) Prove that if ac bc and c 0, then a b. Proof. Let a, b, and c be arbitrary integers and suppose ac bc and c 0. Since ac bc, we can choose some integer m such that mac = bc. Since c 0, we can devide both sides by c. Therefore ma = b, so since m is an integer, a b. 7

8 15. Consider the following theorem: Theorem. For every real number x, x 2 0. What s wrong with the following proof of the theorem? Proof. Suppose not. Then for every real number x, x 2 < 0. In particular, plugging in x = 3 we would get 9 < 0, which is clearly false. This contradiction shows that for every number x, x 2 0. We cannot let one example, x = 3 represent the whole set of real numbers for the statement to a contradiction. Proof. There are three cases to be considered: x < 0, x = 0, and x > 0. Case 1. x < 0. Since x < 0, multiplying both sides by x, we would get x x > 0 x, which is equivalent to x 2 > 0. Case 2. x = 0. Multiplying both sides by x, we would get x x = 0 x, which is equivalent to x 2 = 0. Case 3. x > 0. Since x > 0, multiplying both sides by x, we would get x x > 0 x, which is equivalent to x 2 > 0. Therefore, we have shown that for every real number x, x 2 0 (the equality holds only when x = 0). 8

9 16. Consider the following incorrect theorem: Incorrect Theorem. If x A(x 0) and A B then x B(x 0). (a) What s wrong with the following proof of the theorem? Proof. Let x be an arbitrary element of A. Since x A(x 0), we can conclude that x 0. Also since A B, x B. Since x B, x 0, and x was arbitrary, we can conclude that x B(x 0). (p. 105) In particular, you must not assume that x is equal to any other object already under discussion in the proof. We cannot assume that x 0 is also true for B without appropriate argument. (b) Find a counterexample to the theorem. In other words, find an example of sets A and B for which the hypotheses of the theorem are true but the conclusion is false. Let A be a set of all positive integers (A = Z + ), and B be a set of all integers (B = Z). By the definition of Z +, A does not have 0 as its element. Furthermore, all the elements in A belong to B, so A B is true. But then, by the definition of Z, 0 is an element of B, so this theorem is not true. 17. Consider the following incorrect theorem: Incorrect Theorem. x R y R(xy 2 = y x). 9

10 What s wrong with the following proofs of the theorem? Proof. Let x = y/(y 2 + 1). Then y x = y y y 2 +1 = y3 = y 2 +1 y y 2 +1 y2 = xy 2. Depending on the value of y, x will change its value. Furthermore, since the goal is to find x such that y R(xy 2 = y x), we cannot start the proof by assigning some value to x. 18. Consider the following incorrect theorem: Incorrect Theorem. Suppose F and G are families of sets. If F and G are disjoint, then so are F and G. (a) What s wrong with the following proof of the theorem? Proof. Suppose F and G are disjoint. Suppose F and G are not disjoint. Then we can choose some set A such that A F and A G. Since A F, by exercise 7, A F, so every element of A is in F. Similarly, since A G, every element of A is in G. But then every element of A is in both F and G, and this is impossible since F and G are disjoint. Thus, we have reached a contradiction, so F and G must be disjoint. If A in the proof is not an empty set, then the contradiction holds. But if A is an empty set, even though elements of other sets in F and G do not duplicate each other, the contradiction does not hold. In other words, there is an occasion such that F G = A, but ( F) ( G) =, even though none of the other sets in F and G duplicates each other. 10

11 (b) Find a counterexample to the theorem. For example, suppose F = {A, E}, where A is an empty set, and E is a set of all even numbers. Suppose G = {A, O}, where A is, again, an emptyset, and O is a set of all odd numbers. Then ( F) ( G) =, but F G = A, which is a set with an empty set as its element. 19. Prove that for every real number x there is a real number y such that for every real number z, yz = (x + z) 2 (x 2 + z 2 ). Proof. We will consider two cases z = 0 and z 0. Case 1. z = 0. yz = y 0 = 0, and (x + z) 2 (x 2 + z 2 ) = (x + 0) 2 (x ) = x 2 x 2 = 0. Therefore, no matter what y is, the equality holds. Case 2. z 0. Let x and z arbitrary real numbers, and y = 2x. Then, yz = (2x)z = 2xz = x 2 + 2xz + z 2 (x 2 + z 2 ) = (x + z) 2 (x 2 + z 2 ) By Case 1, we know that the value of y does not matter if z = 0. So we can let y = 2x for both cases. (a) Comparing the various rules for dealing with quantifiers in proofs, you should see a similarity between the rules for goals of the form xp (x) and givens of the form xp (x). What is this similarity? What about the rules for goals of the form xp (x) and givens of the form xp (x)? 11

12 To prove a goal of the form: xp (x) Let x stand for an arbitrary object, and prove P (x). (If the letter x already stands for something in the proof, you will have to use a different letter for the arbitrary object.) To use a given of the form: xp (x) Introduce a new variable, say x 0, into the proof, to stand for a particular object for which P (x 0 ) is true. The both of these rules above help us decide what fundamental properties we assign in x. To prove a goal of the form: xp (x) Find a value of x that makes P (x) true. Prove P (x) for this value of x. To use a given of the form: xp (x) You may plug in any value, say a for x, and conclude that P (a) is true. Both of these rules above are applied to add more properties to x that will help us prove the theories. (b) Can you think of a reason why these similarities might be expected? (Hint: Think about how proof by contradiction works when the goal starts with a quantifier.) Proof by contradiction works by negating the goals and adding them as givens and lead to a contradiction. If we negate our goals involving quantifiers, then by the quantifier negation law, existential quantifiers become universal, and vice versa, and, goals change into givens. Naturally, these similarities are expected. 12

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