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.