Indirect Argument: Contradiction and Contraposition Lecture 15 Section 3.6 Wed, Feb 15, 2006

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.

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.

Example: Proof by Contraposition Theorem: The sum of a rational and an irrational is irrational. Restate the theorem: Let r be a rational number and let α be a number. If α is irrational, then r + α is irrational. Restate again: Let r be a rational number and let α be a number. If r + α is rational, then α is rational.

The Proof Proof: Let r be rational and α be a number. Suppose that r + α is rational. Let s = r + α. Then α = s – r, which is rational. Therefore, if r + α is rational, then α is rational. It follows that if α is irrational, then r + α is irrational.

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.

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

Example: Euclid’s Prop. 7 Proposition 7: Given two straight lines constructed on a straight line, from its extremities, and meeting in a point, there cannot be constructed on the same straight line, from its extremities, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

Example: Euclid’s Prop. 7 B

Example: Euclid’s Prop. 7 B

Example: Euclid’s Prop. 7 B

Example: Euclid’s Prop. 7 B If C  D, then AC  AD or BC  BD

Example: Euclid’s Prop. 7 Proof: For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, BD be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it.

Example: Euclid’s Prop. 7 Proof: For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, BD be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it.

Example: Euclid’s Prop. 7 Let CD be joined. Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC. Therefore the angle ADC is greater than the angle DCB. Therefore the angle CDB is much greater than the angle DCB. Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it, which is impossible.

Example: Euclid’s Prop. 7 Let CD be joined. Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC. Therefore the angle ADC is greater than the angle DCB. Therefore the angle CDB is much greater than the angle DCB. Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it, which is impossible.

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.

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

Contradiction vs. Contraposition? Theorem: Prove that if x is irrational, then –x is also 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.

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).

Example Theorem: If x is rational and y is irrational, then x + y is irrational ((p  q)  r). Proof: Suppose x is rational (p). Suppose also that x + y is rational (r). Then y = (x + y) – x is the difference between rationals, which is rational (q). Thus, if y is irrational, then x + y is irrational (q  r).