1. Proof by Contradiction
CS 270 Math Foundations of CS
Jeremy Johnson

2. Outline An example Negation Rules Indirect Proofs
Rational numbers and repeating decimals
Irrationality of 2
Negation Rules
Bottom introduction and elimination
Negation introduction and double negation
Indirect Proofs
Double negation
Law of excluded middle
DeMorgan’s Laws

3. Decimals and Fractions
3/8 = .375 =
.375
8 |3.0
2.4 60 56 40

4. Repeating Decimals 9/11 = 0.81818181 … = 0. 81 .81 11 |9.0 88 020 11 9
7/23 =
𝑥= . 81
100𝑥=81. 81
99𝑥= 81
→𝑥=9/11

5. Repeating Decimals
Theorem. A number r is rational iff it has a terminating or repeating decimal expansion
Proof
If r = a/b perform long division to compute the decimal expansion. At each step divide what is left by b, m = qb + r, 0 ≤ r < b. There are b possible remainders. If r = 0 the expansion is terminating. If r has occurred previously the expansion is repeating. After at most b steps one of these must happen.

6. Repeating Decimals
Theorem. A number r is rational iff it has a terminating or repeating decimal expansion
Proof
Without loss of generality, we can assume 0 < r < 1. If r = .a1…an, then r = a1…an/10n. If r = .b1…bk a1…an , then 10kr - b1…bk = a = . a1…an and (10n-1)a = a1…an and hence a is rational and consequently r is also rational.

7. Decimal Expansion of sqrt(2)
2 is the positive solution of 𝑥 2 −2=0.
We can approximate the solution of this equation by repeated bisection

8. Decimal Expansion of sqrt(2)
x0 := 1.0; x1 := 2.0; n := 20; for i from 1 to n do x := (x0+x1)/2; if x^2 > 2 then x1 := x; else x0 := x; end if; end do;

9. Decimal Expansion of sqrt(2)
Does the expansion terminate or repeat?
Maybe it doesn’t?
How long should I look?
Maybe it’s not rational?

10. Proof that sqrt(2) is not Rational
Proof by contradiction. Assume 2 = 𝑎 𝑏 with gcd(a,b)=1.
Then 2b2= a2.
Since an odd number squared is odd, this implies a=2p is even and 2b2= 4p2 and b2= 2p2 and b is also even.
Since a and b are both even gcd(a,b)  1 which is a contradiction
Thus we conclude that is not rational.

11. Negation Rules
Introduce the symbol (⊥ =bottom) to encode a contradiction
Bottom elimination
⊥ can prove anything
Bottom introduction ⊥
⊥ e. 
 
⊥ i ⊥

12. Negation Rules
Introduction and elimination rules (proof by contradiction)
Double negation  … ⊥
 i   … ⊥
 e 
 
  e 

13. Proof by Contradiction
Negation elimination called proof by contradiction
Assume  and derive a
a contradiction  … ⊥
PBC 

14. ex falso quodlibet Bottom (falsum) introduction (derived rule)
From falsehood, anything
Principle of explosion 1
P  P
premise 2 X
Assumption 3 P
e1 1 4 P
e2 1 5 ⊥
⊥i 3,4 6 X
¬e 2-5

15. Exercise
Prove that A  A and A  A

16. Exercise Prove that A  A 1 A premise 2 A Assumption 3 ⊥ ⊥i 1,2 4

17. Exercise Prove that A  A 1 A premise 2 A Assumption 3 ⊥ ⊥i 1,2 4

18. Law of the Excluded Middle
𝑝∨¬𝑝 [derived rule LEM] 1
 (p  p)
assumption 2 𝑝
Assumption 3
(p  p)
∨i1 3,4 4 ⊥
⊥i 3,1 5
 p
¬i 2-4 6
p  p
∨i2 3,4 7
⊥i 6,1 8
¬e 1-7

19. De Morgan’s Law (P  Q)  P  Q 1 (P  Q) premise 2 𝑃 assumption 34 ⊥
⊥i 1,3 5 P
i 2-4 6 Q 7
i2 6 8
⊥i 1,7 9 Q
i 6-8 10
P  Q
i 5,9

20. De Morgan’s Law (P  Q)  P  Q 1 P  Q premise 2 P e1 1 3 Q4
P  Q
assumption 5 P 6 ⊥
e 2,5 7 Q 8
e 3,7 9
e 4,5-6, 7-8 10
(P  Q)
i 4-9

21. Exercise
Prove (P  Q)  P  Q

22. Exercise
Prove P  Q  (P  Q)