Chapter 3 Direct Proof and Proof by Contrapositive

1
Chapter 3 Direct Proof and Proof by Contrapositive

• 3.1 Trivial and Vacuous Proofs
• 3.2 Direct Proofs
• 3.3 Proof by Contrapositive
• 3.4 Proof by Cases

2
Concepts

• A true mathematical statement whose truth is
  accepted without proof is referred to as an
  axiom.
• A true mathematical statement whose true can be
  verified is referred to as a theorem.
• We will use the word theorem sparingly,
  however, primarily reserving it for true
  mathematical statements that will be used later.
  Otherwise, we will simply use the word result.
• A corollary is a mathematical result that can be
  deduced from, and is thereby a consequence of,
  some earlier result.
• A lemma is a mathematical result that is useful
  in establishing the truth of some other result.
• Most theorems (or results) are stated as
  implications.

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Section 3.1 Trivial and Vacuous Proofs

• When the quantified statement ??x?S, P(x)?Q(x) is
  expressed as a result or theorem, we often write
  such a statement as
• For x?S, if P(x)
  then Q(x).
• Or as
• Let x?S. If P(x),
  then Q(x). (3.1)
• Thus (3.1) is true if P(x)?Q(x) is a true
  statement for each x?S, while (3.1) is false if
  P(x)?Q(x) is false for at least one element x?S.
• In (3.1), if Q(x) is true for all x?S or P(x) is
  false for all x?S, then determining the truth of
  (3.1) becomes easier.

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Trivial Proof

• If Q(x) is true for all x?S (regardless the truth
  value of P(x)), then, according to the truth
  table for the implication, (3.1) is true. This
  constitutes a proof of (3.1) and is called a
  trivial proof.
• Result 3.1 Let x?R, If xlt0, then x21gt0.
• Proof Since x2??0 for each real number x, it
  follows that x21gtx2?0. Hence x21gt0.
• 
  
  
• The symbol that occurs at the end of the proof
  of Result 3.1 indicates that the proof is
  complete.

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Vacuous Proof

• Let P(x) and Q(x) be open sentences over a domain
  S. Then ??x?S, P(x)?Q(x) is a true statement if
  it can be shown that P(x) is false for all x?S
  (regardless of the truth value of Q(x)),
  according to the truth table for implication.
  Such a proof is called a vacuous proof of ??x?S,
  P(x)?Q(x) .
• Result 3.2 Let x?R. If x2 -2x2?0, then x3?8.
• Proof First observe that x2 -2x1(x-1)2 ?0.
• Therefore, x2 -2x2 (x-1)2 1 ?1gt0. Thus x2
  -2x2?0 is false for all x?R and the implication
  is true.
• 
  
  

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Section 3.2 Direct Proofs

• Let P(x) and Q(x) be open sentences over a domain
  S. If P(x) is false for some x?S, then P(x)?Q(x)
  is true for this element x. Hence we need only be
  concerned with showing that P(x)?Q(x) is true
  for all x?S for which P(x) is true.
• In a direct proof of P(x)?Q(x) for all x?S, we
  consider an arbitrary element x?S for which P(x)
  is true and show that Q(x) is true for this
  element.
• To summarize then, to give a direct proof of
  P(x)?Q(x) for all x?S, we assume that P(x) is
  true for some arbitrary element x?S and show that
  Q(x) be true as well for this element x.

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Basics for Numbers

• Lets first consider the integers and some of
  their elementary properties. We can use any of
  these properties
• The negative of every integer is an integer.
• The sum (and difference) of every two integers is
  an integer.
• The product of every two integers is an integer.
• Initially, we will use even and odd integers to
  illustrate our proof techniques. In this case,
  however, any properties of even and odd integers
  must be verified before they can be used.
• An integer is even if n2k for some integer k. An
  integer is odd if n2k1 for some integer k.
• Every integer is either even or odd.

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Examples

• Result 3.4
• If n is an odd integer, then 3n7 is an even
  integer.
• Proof Assume that n is an odd integer. Since n
  is odd, we can write n2k1 for some integer k.
  Now
• 3n73(2k1)76k376k102(3k5).
• Since 3k5 is an integer, 3n7 is even.
• 
  
  
• Result If n is an even integer, then 3n5 is an
  even integer.
• Proof Since n is aneven integer, n2k for some
  integer k. Therefore,
• 3n53(2k)53(32k5)96k52(48k5).
• Since 48k5?Z, the integer 3n5 is even.
• 
  
  

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Section 3.3 Proof by Contrapositive

• For statements P and Q, the contrapositive of the
  implication P?Q is the implication ?Q??P.
• Theorem
• For every two statement P and Q, the implication
  P?Q and its contrapositibe are logically
  equivalent that is,
• P?Q?? ?Q??P.
• A proof by contrapositive of the result
• Let x?S. If P(x), then
  Q(x).
• (or of for all x?S If P(x), then
  Q(x).)
• Is a direct proof of its contrapositive
• Let x?S. If ?Q(x), then ?
  P(x).
• (or for all x?S. If ?Q(x), then ?
  P(x).)

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Examples

• Result 3.10
• Let x?Z. If 3x-7 is even, then x is odd.
• Proof Assume that x is even. Then x2k for some
  integer k. So
• 5x-75(2k)-710k-710k-812(5k-4)
  1.
• Since 5k-4 ?Z, the integer 5x-7 is odd.
• 
  
  

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Examples

• Result 3.11
• Let x?Z. Then 11x-7 is even if and only if x is
  odd.
• Proof There are two implications to prove here,
• If x is odd, then 11x-7 is even, and
• If 11x-7 is even, then x is odd.
• We now prove (1). Assume that x is odd. Then x
  2k1, where k?Z. So
• 11x-711(2k1)-722k42(11k2).
• Since 11k2 ?Z, 11x-7 is even.
• We now prove (2). Assume that x is even. Then
  x2k, where k ?Z. Therefore, 11x-711(2k)-722k-7
  22k-812(11k-4)1.
• Since 11k-4 ?Z, 11x-7 is odd.
  

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Examples

• Theorem
• Let x ?Z. Then x2 is even if and only if x is
  even.
• Proof Exercise.
• Result Let x ?Z. If 5x-7 is odd, then 9x2 is
  even.
• Proof Assume that 5x-7 is odd. Then 5x-72k1
  for some integer k. Observe that
• 9x2(5x-7)(4x9)2k14x92k
  4x102(k2x5).
• Because k2x5?Z, 9x2 is even.
• 
  
  
• Note there are some other ways to prove it as
  well.

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Section 3.4 Proof by Cases

• While attempting to prove a statement concerning
  an element x in some set S, it is sometimes
  useful to observe that x possesses two or more
  properties.
• If we can verify the truth of the statement
  regardless of which properties that x may have,
  then we have a proof of the statement. This
  method is called proof by cases.
• For example, in a proof of ?n ? Z, R(n), it might
  be convenience to use a proof cases whose proof
  is divided into the two cases.
• Case 1. n is even, and Case 2. n is odd.
• or it could be divided into the three cases
• Case 1. n0, Case 2. n ? Z and nlt0, and Case 3. n
  ? Z and ngt0.
• Etc.

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Example

• Result
• If n ? Z, then n23n5 is an odd integer.
• Proof. We prove it by cases.
• Case 1. n is even. Then n2x for some x ? Z. So
• n23n5 (2x)23(2x)52(2x23x2)1.
• Since 2x23x2 ? Z, then integer n23n5 is odd.
• Case 2. n is odd. Then n2y1, where y ? Z. Thus
• n23n5 (2y1)23(2y1)54(2y25y4)1.
• Since 2y25y4 ? Z, the integer n23n5 is odd.
• 
  

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Parity

• Two integers x and y are said to be of the same
  parity if x and y are both even or are both
  odd. The integers x and y are of opposite parity
  if one of x and y is even and other is odd.
• Because the definition of two integers having the
  same (or opposite) parity requires the two
  integers to satisfy one of two properties, any
  result containing these terms is likely to be
  proved by cases.

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Theorem

• Theorem
• Let x, y ? Z. Then x and y are of the same parity
  if and only if xy is even.
• Proof. First, assume that x and y are of the same
  parity. We consider two cases.
• Case 1. x and y are even.
• Then x2a and y2b for some integers a and b. So
  xy2a2b2(ab).
• Since ab ? Z, the integer xy is even.
• Case 2. x and y are odd. Then x2a1 and y2b1,
  where a, b ? Z. Therefore, xy2(ab1).
• Since ab1 ? Z, the integer xy is even.
• For the converse, assume that x and y are of
  opposite parity. Again, we consider two cases.

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Continued Proof

• Case 1. x is even and y is odd. Then x2a and
  y2b1, where a, b ? Z. Then
  
• 
  xy2(ab)1.
• Since ab ? Z, the integer xy is odd.
• Case 2. x is odd and y is even. The proof is
  similar to the proof of the preceding case and is
  therefore omitted.
• 
  
  
• Notes there is an alternative when the converse
  is considered
• For the converse, assume that x and y are of
  opposite parity. Without loss of generality,
  assume that x is even and y is odd. Then x2a and
  y2b1, where a, b ? Z. Then
  
• 
  xy2(ab)1.
• Since ab ? Z, the integer xy is odd.

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WLOG

• We use the phrase without loss of generality
  (WLOG) to indicate that the proofs of the two
  situations are similar, so the proof of only one
  of these is needed.
• Theorem Let a and b be integers. Then ab is even
  if and only if a is even or b is even.
• Proof. First, assume that a is even or b is even.
  WLOG, let a be even. Then a 2x for some integer
  s. Thus ab2(xb). Since xb is an integer, ab is
  even.
• For the converse, assume that a is odd and be is
  odd. Then a2x1 and b2y1, where x, y ? Z.
  Hence
• ab(2x1)(2y1)2(2xyxy)1.
• Since 2xyxy is an integer, ab is odd.
  
  
  

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Note

• Note We prove the converse by contrapositive by
  assuming at it is not the case that a is even or
  b is even. By De Morgans Laws
• ?(P ? Q) ?(? P) ? (? Q).
• So the negation of a is even or b is even is a
  is odd and b is odd.