Asked by diddy
Why does 2 equals 3
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Answered by
GPT-5 mini
AI
In ordinary arithmetic 2 does not equal 3. If someone "proves" 2 = 3 the proof must contain a hidden invalid step. The quickest way to see it's false is: if 2 = 3 then subtract 2 from both sides to get 0 = 1, which is a contradiction.
A common bogus proof and the error it hides:
- Start with a = b (so a − b = 0).
- Then a^2 = ab.
- Subtract b^2: a^2 − b^2 = ab − b^2.
- Factor: (a − b)(a + b) = b(a − b).
- Cancel (a − b) from both sides to get a + b = b.
- If a = b this gives 2b = b, and cancelling b (assuming b ≠ 0) gives 2 = 1 (or with a small variant you can get 2 = 3).
The hidden illegal move is cancelling (a − b) when a − b = 0 — that is division by zero. Division by zero is not allowed, so the cancellation step is invalid and the "proof" is wrong.
(As an aside: in some nonstandard algebraic settings you can have 2 and 3 be equal as elements — for example in the trivial ring Z/1Z all integers are equal — but not in normal arithmetic.)
A common bogus proof and the error it hides:
- Start with a = b (so a − b = 0).
- Then a^2 = ab.
- Subtract b^2: a^2 − b^2 = ab − b^2.
- Factor: (a − b)(a + b) = b(a − b).
- Cancel (a − b) from both sides to get a + b = b.
- If a = b this gives 2b = b, and cancelling b (assuming b ≠ 0) gives 2 = 1 (or with a small variant you can get 2 = 3).
The hidden illegal move is cancelling (a − b) when a − b = 0 — that is division by zero. Division by zero is not allowed, so the cancellation step is invalid and the "proof" is wrong.
(As an aside: in some nonstandard algebraic settings you can have 2 and 3 be equal as elements — for example in the trivial ring Z/1Z all integers are equal — but not in normal arithmetic.)
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