Chapter 1Foundationsslides.en.pdf:110-139

ProofTalk

How to communicate a proof: proponent vs skeptic

Def 3.1ProofTalkDef 3.7Direct proofDef 3.4Proof by contradictionDef 3.11Proof by casesDef 3.19Proof by exhaustionDef 3.17Circular reasoningDef 3.20Proof by examples (fallacy)
Concept

ProofTalk extends MathTalk with the moves used to communicate proofs.

Proof moves (good):
- Direct proof (Def 3.7) — assume PP, deduce QQ, conclude PQP \Rightarrow Q.
- Proof by contradiction (Def 3.4) — assume ¬P\neg P, derive a contradiction.
- Proof by cases (Def 3.11) — split into exhaustive cases, prove each.
- Proof by exhaustion (Def 3.19) — enumerate all instances; usually only
works for very small sets.
- Contrapositive (Def 3.9) — to prove PQP \Rightarrow Q, prove ¬Q¬P\neg Q \Rightarrow \neg P.

Fallacies (bad):
- Proof by intimidation (Def 3.14) — "this is trivial".
- Proof by examples (Def 3.20) — verify on a few cases, claim done.
- Circular reasoning (Def 3.17) — assumes what you're trying to prove.

Proof communication (Def 3.22) — the proponent-skeptic game: the proponent
makes moves, the opponent tries to refute. A valid proof leaves the opponent
without a valid move.

Animation — prooftalk fallacies
Transcript — click a line to jump6 cues
  1. 0.0sProof by Authority is not a proof
  2. 1.2sA cyan claim box: All X are Y
  3. 2.6sCitation fades in: Professor Z said so
  4. 4.5sRed X strikes through the citation
  5. 5.7sCounter-example slides in: x0 in X but not in Y
  6. 7.9sOnly the argument matters, not who said it
Practice — score 100% to advance
Multiple choice
Q1
Which is a valid proof method?
Q2
What is circular reasoning?
Q3
Proof by contradiction: assume ¬P\neg P, then…
Q4
To prove PQP \Rightarrow Q via contrapositive, you prove…
Q5
When is proof by exhaustion practical?
Q6
Why is 'proof by intimidation' a fallacy?
Q7
In the proponent-skeptic game, who wins when a proof is valid?
Match definitions
Match each concept on the left to its definition on the right.
Order the steps
Arrange these proof steps in the correct order using the arrows.
1
Then n2=2kn^2 = 2k for some integer kk.
2
Since n2n^2 is even, nn itself must be even (lemma).
3
Assume nn is some arbitrary natural number.
4
Therefore n2n^2 is even only when nn is even. ∎
5
But n2=2kn^2 = 2k is only divisible by 2, not necessarily 4. Contradiction.
6
So n=2mn = 2m. Then n2=4m2n^2 = 4m^2, which is divisible by 4.
7
Suppose, for contradiction, that n2n^2 is even.
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