Chapter 4 More on Directed Proof and Proof by Contrapositive

1. 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

2. 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.

3. 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. #

4. Examples Let x Z. If 3 (x2-1), 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 x2-1=(3q+1)2-1=3(3q2+2q). Since 3q2+2q is an integer, 3 | x2-1. Case 2. x=3q+2 for some integer q. Then x2-1=(3q+2)2-1=3(3q2+4q+1). Since 3q2+4q+1 is an integer, 3 | x2-1. #

5. 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.

6. 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). #

7. 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.

8. Examples Let n  Z. If n2 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 n2-n=(3k)2-(3k)=3(3k2-k). Since 3k2-k is an integer, 3 | n2-n. Thus n2  n (mod 3). Case 2. n  1(mod 3). Then n=3x+1 for some integer x. Hence n2-n=(3x+1)2-(3x+1)=3(3x2+x). Since 3x2-x is an integer, 3 | n2-n and so n2  n (mod 3). #

9. Section 4.3 Proofs Involving Real Numbers Some facts about real numbers that can be used without justification. • a2≥0 for every real number a. • an≥0 for every real number a if n is a positive even integer. • If a<0 and n is a positive odd integer, then an<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.

10. 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 x5-3x4+2x3-x2+4x-1 ≥0, then x ≥0. Proof. Assume that x<0. Then x5<0, 2x3<0, and 4x<0. In addition, -3x4<0, -x2<0. Thus x5-3x4+2x3-x2+4x-1<0-1<0, as desired. #

11. Examples Result. If x, y R, then 1/3x2+3/4y2 ≥xy. Proof. Since (2x-3y)2 ≥0, it follows that 4x2-12xy+9y2 ≥0 and so 4x2+9y2 ≥12xy. Dividing this inequality by 12, we obtain 1/3x2+3/4y2 ≥xy. #

12. 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.

13. Examples 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. #

14. 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. #.