Valid Argument - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CSE115/ENGR160 Discrete Mathematics 01/31/

1.7 Introduction to proofs

• Proof : valid argument that establishes the truth of a mathematical statement, e.g., theorem
• A proof can use hypotheses, axioms, and previously proven theorems
• Formal proofs: can be extremely long and difficult to follow
• Informal proofs: easier to understand and some of the steps may be skipped, or axioms are not explicitly stated

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

• Definition:
  • The integer n is even if there exists an integer k such that n=2k, and
  • n is odd if there exists an integer k such that n=2k+
  • Note that an integer is either even or odd
• Show “If n is an odd integer, then n 2 is odd”

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

Proof by contraposition

• Indirect proof : sometimes direct proof leads to dead ends
• Based on
• Use ┐q as hypothesis and show ┐p must follow

8

p → q ≡ ¬ q →¬ p

Example

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

10

a ≤ n or b ≤ n

ab n

ab n n n

a n b n

a n b n

≠

⋅ =

∧ >

Assume¬( ≤ ∨ ≤ )

p → q ≡ ¬ q →¬ p

Vacuous proof

• Prove p→q is true
• Vacuous proof : If we show p is false and then claim a proof of p→q - However, often used to establish special case
• Show that p(0) is true when p(n) is “If n>1, then n 2 >n” and the domain consists of all integers
• The fact 0^2 >0 is false is irrelevant to the truth value of the conditional statement

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?

Direct proof

• Let r=p/q and s=t/u where p, q, t, u, are integers and q≠0, and u≠0.
• r+s=p/q+t/u=(pu+qt)/qu
• Since q≠0 and u≠0, qu≠
• Consequently, r+s is the ratio of two integers. Thus r+s is rational

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

• Show that at least 4 of any 22 days must fall on the same day of the week
• Let p be the proposition “at least 4 of any 22 days fall on the same day of the week”
• Suppose ┐p is true, which means at most 3 of 22 days fall on the same day of the week
• Which implies at most 21 days could have been chosen because for each of the days of the week, at most 3 of the chosen days could fall on that day
• If r is the statement that 22 days are chosen. Then, we have ┐p→(┐r˄r) Docsity.com 17

Example

• Since b^2 is even, 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

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2 = a / b

2

Proof by contradiction

• Can be used to prove conditional statements
• First assume the negation of the conclusion
• Then use premises and negation of conclusion to arrive a contradiction
• Reason: p→q≡((p˄┐q)→F)