Chapter 4 More on Directed Proof and Proof by Contrapositive 4.1 Proofs Involving Divisibility of Integers 4.2 Proofs Involving Congruence of Integers 4.3 Proofs Involving Real Numbers 4.4 Proofs Involving sets 4.5* Fundamental Properties of Set Operations 4.6* Proofs Involving Cartesian Products of Sets

Section 4.1 Proofs Involving Divisibility of Integers In general, for integers a and b with a≠0, we say that a divides b if there is an integer c such that b=ac. In this case, we write a | b. If a | b, then we also say that b is a multiple of a and that a is a divisor of b. If a does not divide b, then we write a b.  Result: Let a, b, and c be integers with a≠0 and b ≠0. If a|b and b|c, then a|c. Proof. Assume that a|b and b|c. Then b=ax and c=by, where x, y  Z. Therefore, c=by=(ax)y=a(xy). Since xy  Z, a|c.

Examples Result: Let a, b, c, x, y  Z, where a≠0. If a|b and a|c, then a|(bx+cy). Exercise. Result: Let x, y  Z, If 3 xy, then 3 x and 3 y. Proof: Assume that 3 | x or 3 | y. WLOG, assume that 3 divides x. Then x=3z for some integer z. Hence xy=(3z)y=3(zy). Since zy is an integer, 3 | xy. #

Examples Let x  Z. If 3 (x 2 -1), then 3 | x. Proof. Assume that 3 x. Then either x=3q+1 for some integer q, or x=3q +2 for some integer q. We consider these two cases. Case 1. x =3q+1 for some integer q. Then x 2 -1=(3q+1) 2 -1=3(3q 2 +2q). Since 3q 2 +2q is an integer, 3 | x Case 2. x=3q+2 for some integer q. Then x 2 -1=(3q+2) 2 -1=3(3q 2 +4q+1). Since 3q 2 +4q+1 is an integer, 3 | x #

Section 4.2 Proofs Involving Congruence of Integers For integers a, b, and n≥2, we say that a is congruent to b modulo n, written a  b (mod n), if n | (a-b). For example: 15  7 (mod 4) since 4 | (15-7), but 14 4 (mod 6) since 6 (14-4).  Note that: since every integer can be expressed as x=2q or as x=2q+1 for some integer q, it follows that either 2|(x-0) or 2|(x-1); that is, x  0 (mod 2) or x  1 (mod 2). Similarly, we have x  0(mod 3), x  1(mod 3), or x  2(mod 3). Etc.

Examples Result: Let a, b, k, and b be integers, where n≥2. If a  b (mod n), then ka  kb (mod n). Proof: Assume that a  b(mod n). Then n | (a-b). Hence a-b =nx for some integer x. Therefore, ka-kb=k(a-b)=k(nx)=n(kx). Since kx is an integer, n | (ka-kb) and so ka  kb (mod n). #

Examples Result: Let a, b, c, d, n  Z, where n ≥2. If a  b (mod n) and c  d (mod n), the ac  bd (mod n). Proof: Exercise.

Examples Let n  Z. If n 2 n (mod 3), then n 0(mod 3) and n 1(mod 3). Proof. Let n be an integer such that n  0 (mod 3) or n  1(mod 3). We consider these two cases. Case 1. n  0(mod 3). Then n=3k for some integer k. Hence n 2 -n=(3k) 2 -(3k)=3(3k 2 -k). Since 3k 2 -k is an integer, 3 | n 2 -n. Thus n 2  n (mod 3). Case 2. n  1(mod 3). Then n=3x+1 for some integer x. Hence n 2 -n=(3x+1) 2 -(3x+1)=3(3x 2 +x). Since 3x 2 -x is an integer, 3 | n 2 -n and so n 2  n (mod 3). #

Section 4.3 Proofs Involving Real Numbers Some facts about real numbers that can be used without justification. a 2 ≥0 for every real number a. a n ≥0 for every real number a if n is a positive even integer. If a<0 and n is a positive odd integer, then a n <0. If the product of two real numbers is positive if and only if both numbers are positive or both are negative. If the product of two real numbers is 0, then at least one of these numbers is 0. Let a, b, c  R. If a ≥b and c ≥0, then ac ≥ bc. Indeed, if c>0, then a/c ≥b/c. If a>b and c>0, then ac>bc and a/c>b/c. If a>b and c<0, then ac<bc and a/c<b/c.

Theorem Theorem: If x and y are real numbers such that xy=0, then x=0 or y=0. Proof. Assume that xy=0. We consider two cases, x=0 or x≠0. Case 1. x=0. Then we have the desired result. Case 2. x ≠0. Multiplying xy=0 by the number 1/x, we obtain 1/x(xy)=1/x(0)=0. Since 1/x(xy)= ((1/x)x)y=y, it follows that y=0. #. Result: Let x  R. if x 5 -3x 4 +2x 3 -x 2 +4x-1 ≥0, then x ≥0. Proof. Assume that x<0. Then x 5 <0, 2x 3 <0, and 4x<0. In addition, -3x 4 <0, -x 2 <0. Thus x 5 -3x 4 +2x 3 -x 2 +4x-1<0-1<0, as desired. #

Examples Result. If x, y  R, then 1/3x 2 +3/4y 2 ≥xy. Proof. Since (2x-3y) 2 ≥0, it follows that 4x 2 -12xy+9y 2 ≥0 and so 4x 2 +9y 2 ≥12xy. Dividing this inequality by 12, we obtain 1/3x 2 +3/4y 2 ≥xy. #

Section 4.4 Proofs Involving sets Recall, for set A and B contained in some universal set U, A  B={x: x  A or x  B}. A   B={x: x  A and x  B}. A - B={x: x  A and x  B}. To show the equality of two sets C and D, we can verify the two sets inclusions C  D and D  C. To establish the inclusion C  D, we show that every element of C is also an element of D; that is, if x  C then x  D.

Result. For every two sets A and B, A-B=A  Proof. First we show that A-B  A . Let x  A and x  B. Since x  B, it follows that x . Therefore, x  A and x  ; so x  A . Hence A-B  A . Next we show that A   A- B. Let y  A . Then y  A and y . Since y , we see that y  B. Now because y  A and y  B, we conclude that y  A- B. Thus, A   A-B. # Examples

Result. Let A and B be sets. Then A  B= A if and only if B  A. Proof. First we prove that if A  B= A, then B  A. We use a proof by contrapositive. Assume that B is not a subset of A. Then there must be some element x  B such that x  A. Since x  B, it follows that x  A  B. However, since x  A, we have A  B≠A. Next we prove the converse, namely, if B  A, then A  B=A. We use a direct proof here. Assume that B  A. To verify that A  B= A, we show that A  A  B and A  B  A. The set inclusion A  A  B is immediate. It remains only to show then that A  B  A. Let y  A  B. Thus y  A or y  B. If y  A, then we already have the desired result. If y  B, then since B  A, it follows that y  A. Thus A  B  A. #.