Indirect Argument: Contradiction and Contraposition Lecture 15 Section 3.6 Mon, Feb 7, 2005

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.

Example Theorem: Let a, b, c be integers. If a | b, but a does not divide c, then b does not divide c. Proof: Suppose that a | b. Now suppose further that b | c. Then it follows that a | c, which is a contradiction. Therefore, a | b and a does not divide c, then b does not divide c.

Contraposition vs. Contradiction Was that last proof an example of contraposition or contradiction? The form of the theorem was (p  q)  r. This is logically equivalent to p  (q  r). In the proof we demonstrated that p  (r  q).

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 Is this proof by contradiction or is it proof by contraposition? Proof by contraposition is preferred.

A Lemma Lemma: Let f(x) be a continuous function, [a, b] an interval, and c a real number. If f(x)  c for all x  [a, b], then [a, b] f(x) dx  c(b – a). Proof: Use Riemann sums.

A Lemma f(x) c x a b

A Lemma f(x) c c(b – a) x a b

A Lemma f(x) c [a, b] f(x) dx x a b

Proof by Contradiction? Theorem: Let f(x) be a continuous function and [a, b] an interval and suppose that f(x)  0 for all x  [a, b]. Then [a, b] f(x) dx = 0 if and only if f(x) = 0 for all x  [a, b].

Proof by Contradiction? Suppose f(x) = 0 for all x  [a, b]. Then it is clear from the definition of the integral that [a, b] f(x) dx = 0 Proof: () Suppose that [a, b] f(x) dx = 0. Suppose that f(x) > 0 for some x  [a, b], say f(c) =  > 0, for some c  [a, b].

Proof by Contradiction? Then there is an interval [c – , c + ] over which f(x)  /2, for some  > 0. It follows from the lemma that [c – , c + ] f(x) dx  (/2)(2) =  > 0. But then [a, b] f(x) dx > 0, which is a contradiction. Therefore, f(x) = 0 for all x  [a, b].

Proof by Contradiction? The last part of that proof was really proof by contraposition. It assumed that the statement x  [a, b], f(x) = 0 was false, and proved that the statement [a, b] f(x) dx = 0 was false. That is, it proved that q  p.