1. Chapter 1, # 1: Prove that for all sets A, B, C, the formula

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1 Homework 1 MTH 4590 Spring Chapter 1, # 1: Prove that for all sets,, C, the formula ( C) = ( ) ( C) is true. Proof : It suffices to show that ( C) ( ) ( C) and ( ) ( C) ( C). ssume that x ( C), thus either x or x C. If x, then x and x C, therefore x ( ) ( C). Now suppose that x C. Then x and x C, and so both x and x C, hence x ( ) ( C). Therefore ( C) ( ) ( C). Now we show that ( ) ( C) ( C). ssume that x ( ) ( C). Then x and x C. If x, then x ( C), so assume that x. Claim: x C Proof of claim: Suppose that x C. Then x and x C. Since x, it follows that both x and x C, a contradiction. Therefore x C. Since x C, it follows that x ( C). Therefore ( ) ( C) ( C). Therefore we have shown that ( C) = ( ) ( C), completing the proof. 2. Chapter 1, # 2: (a) Prove that ( c ) c =. Proof : Let c denote the complement of in X, i.e. c = X \. We shall show that ( c ) c and ( c ) c. Let x ( c ) c. Then x X \ c. ut X \ c = X \ (X \ ). Claim: X \ (X \ ) = Proof of claim: If z X \ (X \ ), then z is in the set of elements that are not in X \. Those elements are exactly the ones in, by definition. Conversely, if z, then z X\, and so z is not removed when forming X\(X\), showing that X \ (X \ ). Thus X \ (X \ ) =. Since x X \ (X \ ), we see from the claim that x. Thus we have shown that ( c ) c. Now let x. Then x X \. Therefore x X \ (X \ ). Since X \ (X \ ) = X \ c = ( c ) c, we see that ( c ) c. Therefore ( c ) c =, completing the proof. (b) Prove demorgan s law: Prove ( ) c = c c (1) 1

2 and derive from it the law ( ) c = c c. (2) Solution: First let us prove (1). Let x ( ) c. Then x X \ ( ). This means that x. Without loss of generality, assume that x. Then x c, and so x c c. Now let x c c. Then x c or x c. Without loss of generality, asusme that x c. Then x, so x. Thus x ( ) c. Completing the proof. To derive (2) from (1), replace with c and with c in (1) to get ( c c ) c = ( c ) c ( c ) c. (3) On the right-hand-side note that by part (a), we know ( c ) c = and ( c ) c =, so we get ( c c ) c =. Taking the complement of each side and applying part (a) to the left-hand side yields the desired result: c c = ( ) c. (c) Draw Venn diagrams to illustrate the two laws. Solution: c : c : First law: : 2

3 ( ) c = c c : Second law: : ( ) c = c c : (d) Generalize these laws to more than two sets. Solution: Suppose we have three sets,, C and we want to look at ( C) c. Write X = C, and observe ( C) c = ( X) c (1) = c X c = c ( C) c (1) = c c C c. This leads us to a conjecture: Conjecture: The following formula holds for n = 1, 2, 3,...: ( n ) c = c 1 c 2... c n. Proof : The case n = 1 holds trivially. The case n = 2 was proven in part (b). The case n = 3 was proven in the beginning of part (d). ssume the formula holds for n = N. We now prove it holds for n = N + 1: Claim: If for all sets, 1,..., N, ( 1... N ) c = c 1... c N, then for any additional set N+1, ( 1... N N+1 ) c = c 1... c N c N+1. 3

4 Proof of claim: Define N = N N+1. Then we see using the hypothesis of this claim that ( 1... N ) c = c 1... N c = c 1... ( N N+1 ) c, and so from the n = 2 case of the conjecture (proved earlier), we may conclude ( 1... N ) c = c 1... c N c N+1, completing the proof of the claim. Since this claim holds and the base cases n = 1, n = 2, and n = 3 hold, we have shown via induction that the conjecture holds. 3. Chapter 1, # 6: Why is the square of and odd integer odd and the square of an even integer even? What is the situation for higher powers? Solution: Recall that any positive natural number has a unique prime factorization n = p e1 1 pe pei i, where the p 1,..., p i are prime numbers with exponents e 1,..., e i N. Since 2 is prime and being even means being divisible by 2, a number is even if and only if one of its prime factors p 1,..., p i is equal to 2. If n is an odd integer, then when writing n = p e1 1 pe pei i, we observe that p 1 2 and p 2 2 and... and p i 2 (otherwise n would be even). Now consider the square of n: n 2 = (p e1 1 pe pei i )2 = (p e1 1 )2 (p e2 2 )2... (p ei i )2 = p 2e1 1 p 2e p 2ei i. We see that the only prime factors of n 2 are the same ones that are factors of n. Since 2 was not a factor of n, it follows that 2 is not a factor of n 2. similar argument shows that the square of an even must be even. This same argument holds for higher powers because n l = (p e1 1 pe pei i )l = p le1 1 ple p lei i. 4. In this problem, there is a known subset of R called C which has an upper bound. The set C is defined by and the set D is defined by C = {a Q: for some cut C, a } (a) Claim 1: C D is a cut. Solution: We must argue that D = Q \ C. a.) C D = Q, C, D, and C D =, 4

5 b.) if c C and d D, then c < d, and c.) C contains no largest element. y definition, C Q and C D =. To prove a.), it follows from the definition of D that C D = Q. We know that C and D because C is bounded above. To prove b.), suppose there is a c C and a d D with d c. Let c = C 1 C 2 and d = D 1 D 2. Then we may conclude that D 1 C 1. ut we cannot have D 1 = C 1, because it would follow that D 2 = C 2 and hence c = d which contradicts the fact that C D =. If D 1 C 1, then by the definition of C, d C. ut this cannot happen because D = Q \ C. Hence we have shown that D 1 C 1 and thus it is not true that d c. Therefore b.) holds. To prove c.), suppose that C does contain a largest element call it c. Since c Q we can express it as p for p, q Z with q 0. y the q definition of C, there is some cut C for which p q. Claim: = c Proof : Since p, it follows that c. Now we q show that c. Suppose that > c. Then if we write c = C D, it follows that C. ut this means there is some larger ĉ = Ĉ ˆD with C Ĉ. ut this means that ĉ C is a larger element of C than c, a contradiction to c being the largest. Therefore c and since we proved earlier that c, we may conclude that = c. So we see that may write the real number c as the cut ( c = =, p ) [ ) p Q q q, Q. From this we observe that c, a contradiction. Therefore c cannot exist, i.e. C has no largest element. Therefore a.), b.), and c.) hold and we may conclude that C D is a cut. 5. Claim 2: C D is an upper bound for C Proof : Let E F C. We will show that E F C D. Since E F C, we know by the definition of C that for all a E, a C, hence E C. Therefore E F C D. 5