Presentation on theme: "Chapter 3 Elementary Number Theory and Methods of Proof."— Presentation transcript:
Chapter 3 Elementary Number Theory and Methods of Proof
3.6 Indirect Argument
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.
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.
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
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.
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
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.
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
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”