Introduction to direct and proof by contradiction with examples on their applications in probability.

1. Kunyu (Quinn) He CAPP’20
1
Direct Proof and Proof by Contradiction in Probability
TA Kunyu (Quinn) He
Harris School of Public Policy
CAPP’20
Direct proof is a way of showing the truth or falsehood of a given statement by a
straightforward combination of established facts, without making further assumptions.
(Direct proof. In Wikipedia. Retrieved Oct. 4, 2018, from https://en.wikipedia.org/wiki/Direct_proof)
Proof by contradiction is a form of proof. It starts by assuming that the opposite
proposition is true, and then shows that such an assumption leads to a contradiction.
(Proof by contradiction. In Wikipedia. Retrieved Oct. 4, 2018, from https://en.wikipedia.org/wiki/Proof_by_contradiction)
To prove P (√2 is irrational):
1. Assume that P is false;
2. Then, ¬P is true (√2 is rational);
3. If we can derive P from ¬P; or…
If ¬P implies two mutually contradictory assertions, Q (greatest common
divisor of a and b is 1) and ¬Q (greatest common divisor of a and b ≥ 2);
(√2 =
𝑏
𝑎
, when the greatest common divisor of a and b is 1;
then
𝑏2
𝑎2 = 2, or 𝑏2
= 2𝑎2
,
𝑏2
must be even, and 𝑏 in turn must be even;
let 𝑏 = 2𝑐, then 𝑎2
= 2𝑐2
,
𝑎2
must be even, and 𝑎 in turn must be even;
if both a and b are even, there least common divisor is 2, which is not 1)
4. Since P and ¬P (or Q and ¬Q) cannot both be true, the assumption we made
in step 1 that ‘P is false’ is false, which in turn proves that P must be true.

2. Kunyu (Quinn) He CAPP’20
2
Example:
1. To prove P(∅) = 0 with direct proof:
Let S denotes the sample space, S ∪ ∅ = S, then P(S ∪ ∅) = P(S) = 1;
S ∩ ∅ = ∅, for the disjoint sequence of events S and ∅:
P(S ∪ ∅) = P(S) + 𝑃(∅)
which is:
1 = 1 + 𝑃(∅)
then P(∅) = 0.
∎
2. To prove P(∅) = 0 with proof by contradiction:
Assume that P(∅) = 0 is false, then P(∅) ≠ 0 is true;
Let S denotes the sample space, S ∪ ∅ = S, then P(S ∪ ∅) = P(S) = 1;
S ∩ ∅ = ∅, for the disjoint sequence of events S and ∅:
P(S ∪ ∅) = P(S) + 𝑃(∅)
which is:
1 = 1 + 𝑃(∅), where P(∅) ≠ 0
which is false;
then our assumption that ‘P(∅) = 0 is false’ is false, P(∅) = 0 must be true.
∎
3. To prove P(A𝑐) = 1 − 𝑃(𝐴) through direct proof:
Let S denotes the sample space, A ∪ A𝑐
= S, then P(A ∪ A𝑐
) = 𝑃(𝑆) = 1;
on the other hand, we also know that A ∩ A𝑐
= ∅, then P(A ∪ A𝑐) = P(A) + 𝑃(A𝑐
);
which means:
P(A) + 𝑃(A𝑐) = 1,
which is exactly:
𝑃(A𝑐) = 1 − P(A)
∎
4. To prove P(A𝑐) = 1 − 𝑃(𝐴) through proof by contradiction:
Assume that 𝑃(A𝑐) = 1 − P(A) is false, then P(A) + 𝑃(A𝑐) ≠ 1 is true;
we know that A ∩ A 𝑐
= ∅, then P(A) + 𝑃(A𝑐) = P(A ∪ A𝑐);
we also know that A ∪ A𝑐
= S, then P(A ∪ A𝑐
) = 𝑃(𝑆) = 1,
which means:
P(A) + 𝑃(A 𝑐) = 1
we derive P(A) + 𝑃(A𝑐) = 1 from P(A) + 𝑃(A 𝑐) ≠ 1, our assumption that ‘𝑃(A𝑐) = 1 −
P(A) is false’ is false, 𝑃(A𝑐) = 1 − P(A) must be true
∎