Download presentation

Presentation is loading. Please wait.

Published byMaddison Alls Modified about 1 year ago

1
Chapter 3 Elementary Number Theory and Methods of Proof

2
3.6 Indirect Argument

3
Reductio Ad Absurdum Argument by contradiction Illustration in proof of innocence – Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime. – In fact, at the time of the crime, I was meeting with 20 people far from the crime scene, as they will testify. – This contradicts the assumption that I committed the crime, since it is impossible to be in two places at one time. Hence, that assumption is false.

4
Proofs Direct Proof – start with hypothesis of a statement and make one deduction after another until the conclusion is reached. Indirect Proof (argument by contradiction) – show that a given statement is not true leads to the contradiction.

5
Example Use proof by contradiction to show that there is no greatest integer. – Starting point: Suppose not. Suppose that there is a greatest integer, N. N≥n for all integers. – To Show: This supposition leads logically to a contradiction. – Proof: Suppose not. Suppose that there is a greatest integer N. N ≥ n for every integer n. Let M = N + 1. M is an integer under the addition property of integers. Thus M > N. N is not the greatest integer; therefore a contradiction. Theorem 3.6.1

6
Even and Odd Integer Theorem – There is no integer that is both even and odd – Proof: Suppose not. That is, suppose there is an integer n that is both even and odd. n = 2a (even) and n = 2b + 1 (odd) 2a = 2b + 1, 2a – 2b = 1, 2(a – b) = 1 (a – b) = ½ Since a and b are integers then a – b should result in an integer. Thus a-b is integer and a-b is not, contradiction.

7
Sum of Rational and Irrational Theorem – The sum of any rational number and any irrational number is irrational – Proof: Suppose not. Suppose there is a rational number r and irrational number s such that r + s is rational. r = a/b (definition rational) r + s = c/d, for integers a,b,c,d with b≠0 and d≠0 a/b + s = c/d, s = c/d – a/b s = (bc – ad)/bd (integer) hence, s is quotient of integers and therefore, rational If s is rational it contradicts supposition that it is irrational

8
Argument by Contrapositive Argument by contrapositive is based on the equivalence of the statement and contrapositive. If the contrapositive is true then the statement is true.

9
Proof by Contrapositive Method of Proof 1.Express the statement to be provided in the form: ∀ x in D, if P(x) then Q(x) 2.Rewrite the statement in the contrapositive form: ∀ x in D, if Q(x) is false then P(x) is false 3.Prove the contrapositive by a direct proof. 1.Suppose x is a (particular but arbitrarily chosen) element of D such that Q(x) is false. 2.Show that P(x) is false

10
Example If the square of an integer is even, then the integer is even. – Prove that for all integers n, if n 2 is even then n is even. – Contrapositive: ∀ integers n, n is not even then n 2 is not even. – Proof: Suppose n is any odd integer. n = 2k + 1 n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 n 2 = 2r + 1 (by definition n 2 is odd) hence, the contrapositive is true therefore the statement must be true—”if n is even then n 2 is even”

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google