Proof by Contraposition - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

Download Proof by Contraposition - Discrete Mathematics - Lecture Slides and more Slides Discrete Mathematics in PDF only on Docsity!

CSE115/ENGR160 Discrete Mathematics 02/01/

1.6 Direct proofs of p→q

• First assume p is true
• Then show q must be true (using axioms, definitions, and previously proven theorems)
• So the combination of p is true and q is false never occurs
• Thus p→q is true
• Straightforward
• But sometimes tricky and require some insight

Example

• Note the theorem states
• By definition of odd integer, n=2k+1, where k is some integer
• n^2 =(2k+1)^2 =4k^2 +4k+1=2(2k^2 +2k)+
• By definition of odd integer, we conclude n^2 is an odd integer
• Consequently, we prove that if n is an odd integer, then n^2 is odd

4

∀ n ( p ( n )→ q ( n ))

Example

• “If m and n are both perfect squares, then nm is also a perfect square (an integer a is a perfect square if there is an integer b such that a=b^2 )
• By definition, there are integers s and t such that m=s^2 , and n=t^2
• Thus, mn=s^2 t^2 =(st)^2 (using commutativity and associativity of multiplication)
• We conclude mn is also a perfect square

Example

• Show that “if n is an integer and 3n+2 is odd, then n is odd”
• Use
• Proof by contraposition:
  • Assume n is even, i.e., n=2k, for some k
  • It follows 3n+2=3(2k)+2=6k+2=2(3k+1)
  • Thus 3n+2 is even

7

p → q ≡ ¬ q →¬ p

Example

• Prove that if n=ab, where a and b are positive integers, then

8

a ≤ n or b ≤ n

ab n

ab n n n

a n b n

a n b n

≠

⋅ =

∧ >

Assume¬( ≤ ∨ ≤ )

p → q ≡ ¬ q →¬ p

Trivial proof

• Trivial proof : a proof of p→q that uses the fact q is true - Often important when special cases are proved
• Let p(n) be “If a and b are positive integers with a≥b, then an^ ≥bn^ where the domain consists all integers
• The proposition p(0) is “If a≥b, then a 0 ≥b^0 ”. a^0 ≥b^0 is true, hence the conditional statement p(0) is true

Example

• Definition: the real number r is rational if there exist integers p and q with q≠0 such that r=p/q
• A real number that is not rational is irrational
• Prove that the sum of two rational numbers is rational (i.e., “For every real number r and every real number s, if r and s are rational numbers, then r+s is rational”)
• Direct proof? Proof by contraposition?

Example

• Prove that if n is an integer and n^2 is odd, then n is odd
• Direct proof? Proof by contraposition?

Proof by contradiction

• Suppose we want to prove a statement p
• Further assume that we can find a contradiction q such that ┐p→q is true
• Since q is false, but ┐p→q is true, we can conclude ┐p is false, which means p is true
• The statement ┐r˄r is contradiction , we can prove that p is true if we can show that ┐p→(┐r˄r), i.e., if p is not true, then there is a contradiction

Example

• Prove that is irrational by giving a proof by contradiction
• Let p be the proposition “ is irrational”
• ┐p: is rational, and thus where a and b have no common factors
• Thus 2=a 2 /b^2 , 2b^2 =a 2 , and thus a^2 is even
• a 2 is even and so a is even. Let a=2c for some integer c, 2b^2 =4c^2 , and thus b^2 =2c^2 , and b^2 is even 16

2

2 2 2 = a^ / b

Example

• Use the fact that if the square of an integer is even, then the integer itself is even, we conclude b must be even
• ┐p leads to where and b have no common factors, and both a and b are even (and thus a common factor), a contradiction
• That is, the statement “ is irrational” is true

17

2 = a / b

2

Proof by contradiction

• Can rewrite a proof by contraposition of a conditional statement p→q as proof by contradiction
• Proof by contraposition: show if ┐q then ┐p
• Proof by contradiction: assume p and ┐q are both true
• Then use steps of ┐q→┐p to show ┐p is true
• This leads to p˄┐p, a contradiction

Example

• Proof by contradiction “If 3n+2 is odd, then n is odd”
• Let p be “3n+2 is odd” and q be “n is odd”
• To construct a proof by contradiction, assume both p and ┐q are both true
• Since n is even, let n=2k, then 3n+2=6k+2= 2(3k+1). So 3n+2 is even, i.e. ┐p,
• Both p and ┐p are true, so we have a contradiction