Last Update: 13 October 2010
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To show any proposition A:
supp. ¬A & try to derive a contradiction;
i.e., try to find B such that (B ∧ ¬B)
So we need to show: ¬Q(√2)
∴ 2 = a^{2}/b^{2}
∴ 2b^{2} = a^{2}
∴ Even(a^{2})
∴ Even(a), by lemma (Rosen, p.85: 16)
∴ ∃j[a=2j]; by EI, call it j_{1}
∴ 2b^{2} = (2j_{1})^{2} = 4j_{1}^{2}
∴ b^{2} = 2j_{1}^{2}
∴ Even(b^{2})
∴ Even(b)
But now we have Even(a) ∧ Even(b).
∴ a,b do have a common factor, namely: 2
So, we have our contradiction,
(B ∧ ¬B)
which means that our assumption for the proof by contradiction, Q(√2), was wrong.
∴ ¬Q(√2). QED
Another strategy to prove (A → B)
(another form of proof by contradiction):
Works because:
show (A_{1} → A_{2}), and
show (A_{2} → A_{3}), and
…, and
show (A_{n–1} →
A_{n}), and
show (A_{n} →
A_{1})
To prove this, compile each into the next, and the last into the first.
All dogs are mammals.
Garfield the cat is a mammal.
∴ Garfield is a dog.
proof #1:
proof #2: Show (2+2=5) → Pope(bertrand-russell) by Direct Proof:
Suppose 2+2=5 & show Pope(b-r):
1. | 2+2=5 | : assumption |
2. | (2+2=5) ∨ Pope(b-r) | : 1; Addition |
3. | ¬(2+2=5) | : theorem of arithmetic |
4. | Pope(b-r) | : 2,3; DS |