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If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:

  1. Let $a = b$.
  2. Then $a^2 = ab$
  3. $a^2 - b^2 = ab - b^2$
  4. Factor to $(a-b)(a+b) = b(a-b)$
  5. Then divide out $(a-b)$ to get $a+b = b$
  6. Since $a = b$, then $b+b = b$
  7. Therefore $2b = b$
  8. Reduce to $2 = 1$

As @jan-gorzny pointed out, in this case, line 5 is wrong since $a = b$ implies $a-b = 0$, and so you can't divide out $(a-b)$.

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Community wiki, as there is no one answer? –  FordBuchanan Jul 20 '10 at 20:59
    
Ah, I thought I checked it when I posted. Edited. –  Jon Bringhurst Jul 20 '10 at 21:00
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are you looking for more "classic fake proofs" of this type, or something along the lines of "common misconceptions in mathematics"? –  Jamie Banks Jul 20 '10 at 21:18
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The example given isn't fallacious or fake, it is perfectly acceptable apart from relying on an undefined operation (division by 0). It is however a great example of why division by 0 is left undefined, namely that it makes the arithmetic system inconsistent and therefore you can prove anything (such as 1 = 2) –  workmad3 Jul 20 '10 at 22:18
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So 1+1 can be equal to 5 for very large values of 2? –  Cole Johnson Apr 2 '13 at 1:07

4 Answers 4

The odd number $N = 198585576189 = 3^2 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 22021$ has an interesting property—it is perfect:

$$\sigma(N) = (1 + 3 + 3^2)(1 + 7 + 7^2)(1 + 11 + 11^2)(1 + 13 + 13^2)(1 + 22021) = 397171152378 = 2N$$

Now, where is the catch? (This one was found by René Descartes. It is also the only known odd number to have this property.)

We pretend that the number $22021 = 19^2 \cdot 61$ is prime.

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Well, one I've read quite some time ago (sorry, can't give a reference) is a proof that any bag that contains peas either contains only green or only yellow peas.

The proof goes by induction:

Assume the bag holds only one pea. That pea is of course either green or yellow. Therefore for only one pea, the statement is true.

Now assume we've proven it for $n$ peas, and we are given a bag of $n+1$ peas. In that case, we first take one pea out, so we now have a bag of $n$ peas, so by assumption, they all have the same colour. To find out that colour, we take another pea out, and put our first pea back in. Then by looking at the other pea, we can determine the colour of the peas in the sack, and since there are again $n$ peas in the sack, the one we had removed first also has the same colour as the others.

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Similar to this "proof", which has an attached explanation. –  WChargin Feb 8 at 23:54

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