Chapter 4 More on Directed Proof and Proof by Contrapositive

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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

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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 bac. 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 ab and bc, then ac.
• Proof. Assume that ab and bc. Then bax and
  cby, where x, y?Z. Therefore, cby(ax)ya(xy).
  Since xy ?Z, ac.

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Examples

• Result Let a, b, c, x, y ?Z, where a?0. If ab
  and ac, then a(bxcy).
• 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 x3z for some integer z.
  Hence xy(3z)y3(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 x3q1 for
  some integer q, or x3q 2 for some integer q. We
  consider these two cases.
• Case 1. x 3q1 for some integer q. Then
• x2-1(3q1)2-13(3q22q).
• Since 3q22q is an integer, 3 x2-1.
• Case 2. x3q2 for some integer q. Then
• x2-1(3q2)2-13(3q24q1).
• Since 3q24q1 is an integer, 3 x2-1.
• 
  

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Section 4.2 Proofs Involving Congruence of
Integers

• For integers a, b, and n2, 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 x2q or as x2q1 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.

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Examples

• Result Let a, b , k, and b be integers, where
  n2. 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-kbk(a-b)k(nx)n(kx).
• Since kx is an integer, n (ka-kb) and so ka ?
  kb (mod n).
• 
  
  

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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.

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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 n3k 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 n3x1 for some
  integer x. Hence
• n2-n(3x1)2-(3x1)3(3x2x).
• Since 3x2-x is an integer, 3 n2-n and so n2 ? n
  (mod 3).
• 
  
  

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Section 4.3 Proofs Involving Real Numbers

• Some facts about real numbers that can be used
  without justification.
• a20 for every real number a.
• an0 for every real number a if n is a positive
  even integer.
• If alt0 and n is a positive odd integer, then
  anlt0.
• 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 cgt0, then a/c b/c.
• If agtb and cgt0, then acgtbc and a/cgtb/c.
• If agtb and clt0, then acltbc and a/cltb/c.

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Theorem

• Theorem If x and y are real numbers such that
  xy0, then x0 or y0.
• Proof. Assume that xy0. We consider two cases,
  x0 or x?0.
• Case 1. x0. Then we have the desired result.
• Case 2. x ?0. Multiplying xy0 by the number 1/x,
  we obtain
• 1/x(xy)1/x(0)0. Since 1/x(xy) ((1/x)x)yy, it
  follows that y0.
• 
  .
• Result Let x ?R. if x5-3x42x3-x24x-1 0, then
  x 0.
• Proof. Assume that xlt0. Then x5lt0, 2x3lt0, and
  4xlt0. In addition,
• -3x4lt0, -x2lt0. Thus x5-3x42x3-x24x-1lt0-1lt0, as
  desired.
• 
  

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Examples

• Result. If x, y ?R, then 1/3x23/4y2 xy.
• Proof. Since (2x-3y)2 0, it follows that
  4x2-12xy9y2 0 and so
• 4x29y2 12xy.
• Dividing this inequality by 12, we obtain
  1/3x23/4y2 xy.
• 
  
  

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Section 4.4 Proofs Involving sets

• Recall, for set A and B contained in some
  universal set U,
• A?? Bx x ? A
  or x ? B.
• A? ? Bx x ? A and x ? B.
• A - Bx 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.

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Examples

• Result. For every two sets A and B, A-BA ?
• 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 ?BA. 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.
• 
  
  .