Reductio ad Absurdum
Indirect Proof
Prof Hans Georg Schaathun
Høgskolen i Ålesund
Autumn 2013 – Part 2/Session 6/Video 3
Recorded: 21st September 2015
Prof Hans Georg Schaathun
Reductio ad Absurdum
Autumn 2013 – Session 2/6 (3)
1/8
Proof by Contradiction
Principle
1
Assume s
2
Prove t (using the assumption)
3
Prove ¬t (using the assumption)
We have t ∧ ¬t — absurd
4
Conclude ¬s
Principle (Reductio ad Absurdum)
If, by assuming s, we can prove both t and ¬t, then s must be false.
Prof Hans Georg Schaathun
Reductio ad Absurdum
Autumn 2013 – Session 2/6 (3)
2/8
Proof by Contradiction
Alternative Version
Principle (Reductio ad Absurdum)
If, by assuming p ∧ ¬q, we can prove both r and ¬r , then p ⇒ q.
This is the same principle
s := p ⇒ q
¬s := p ∧ ¬q
Check the truth table. In what cases is p ⇒ q true, and in
which cases is p ∧ ¬q false?
Prof Hans Georg Schaathun
Reductio ad Absurdum
Autumn 2013 – Session 2/6 (3)
3/8
Example
Problem
Prove n2 > 4 ⇒ n > 2.
Proved before using contraposition.
Prove it now by contradiction.
Step 1 Assume that the statement is false.
I.e. n2 > 4 and n ≤ 2.
Prof Hans Georg Schaathun
Reductio ad Absurdum
Autumn 2013 – Session 2/6 (3)
4/8
Proof 1
p := n2 > 4,
(1)
q := n ≤ 2,
(2)
assume p ∧ q
Prof Hans Georg Schaathun
Reductio ad Absurdum
(3)
Autumn 2013 – Session 2/6 (3)
5/8
Proof 2
p := n2 > 4,
(4)
q := n ≤ 2,
(5)
assume p ∧ q
Prof Hans Georg Schaathun
Reductio ad Absurdum
(6)
Autumn 2013 – Session 2/6 (3)
6/8
Proof 3
Contraposition
p := n2 > 4,
(7)
q := n ≤ 2,
(8)
assume p ∧ q
Prof Hans Georg Schaathun
Reductio ad Absurdum
(9)
Autumn 2013 – Session 2/6 (3)
7/8
Exercise
Exercise
Prove (by contradiction) that for all real numbers x, x 2 − x 6= 0 implies
x 6= 0.
Prof Hans Georg Schaathun
Reductio ad Absurdum
Autumn 2013 – Session 2/6 (3)
8/8