Indirect Argument: Contradiction and Contraposition

1. Indirect Argument: Contradiction and Contraposition Lecture 17 Section 3.6 Mon, Feb 19, 2007

2. Form of Proof by Contraposition • Theorem: pq. • This is logically equivalent to qp. • Outline of the proof of the theorem: • Assume q. • Prove p. • Conclude that p q. • This is a direct proof of the contrapositive.

3. Benefit of Proof by Contraposition • If p and q are “negative” statements, then p and q are “positive” statements. • We may be able to give a direct proof that qp more easily that we could give a direct proof that pq.

4. Example: Proof by Contraposition • Theorem: An irrational number plus 1 is irrational. • Restate as an implication: Let r be a number. If r is irrational, then r + 1 is irrational. • Restate again using the contrapositive: Let r be a number. If r + 1 is rational, then r is rational. • Restate in simpler form: Let r be a number. If r is rational, then r – 1 is rational.

5. Form of Proof by Contradiction • Theorem: pq. • Outline of the proof of the theorem : • Assume (p  q). • This is equivalent to assuming p  q. • Derive a contradiction, i.e., conclude r  r for some statement r. • Conclude that p  q.

6. Benefit of Proof by Contradiction • The statement r may be any statement whatsoever because any contradiction r  r will suffice.

7. Proof by Contradiction • Theorem: For all integers n  0, 2 + n is irrational.

8. Example • Theorem: If x is rational and y is irrational, then x + y is irrational. • That is, if • p: x is rational • q: y is irrational • r: x + y is irrational, • Then • (p q)  r.

9. Contraposition and Compound Hypotheses • Often a theorem has the form (p q)  r. • This is logically equivalent to r  (p  q). • However, it is also equivalent to p  (q  r). so it is also equivalent to p  (r  q).

10. Contraposition and Compound Hypotheses • More generally, (p1p2 … pn) q is equivalent to (p1p2 … pn – 1)  (qpn).

11. Example • Theorem: If x is rational and y is irrational, then x + y is irrational. • Proof: • Suppose x is rational. • Suppose also that x + y is rational. • Then y = (x + y) – x is the difference between rationals, which is rational. • Thus, if y is irrational, then x + y is irrational.

12. Contradiction vs. Contraposition • Sometimes a proof by contradiction “becomes” a proof by contraposition. • Here is how it happens. • To prove: p q. • Assume (p  q), i.e.,p  q. • Using q, prove p. • Cite the contradiction p  p. • Conclude that pq.

13. Contradiction vs. Contraposition • Would this be a proof by contradiction or proof by contraposition? • Proof by contraposition is preferred.

14. Contradiction vs. Contraposition? • Theorem: If x is irrational, then –x is irrational. • Proof: • Suppose that x is irrational and that –x is rational. • Let -x = a/b, where a and b are integers. • Then x = -(a/b) = (-a)/b, which is rational. • This is a contradiction. • Therefore, if x is irrational, then -x is also irrational.