# What are some classic fallacious proofs?

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
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
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
So 1+1 can be equal to 5 for very large values of 2? –  Cole Johnson Apr 2 '13 at 1:07

Wikipedia has a long list of these:

http://en.wikipedia.org/wiki/Mathematical_fallacy

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Let's see what happens if we allow 2=1 to be an OK result. What does it mean? That where the number 2 is first introduced, namely in the second proposition, it is to be replaced by 1 as a kind of correction. So:

1. a = b

2. a^1 = ab

3. a^1 = a

4. a = ab

5. a/a = ab/a

6. Cancelling, 1 = b

7. 1 = a

8. 1 = 1

Back to normal again! I call this revolutionary new method of mine by a term cobbled together from medieval Latin: Modus Corrigens - the corrective mode, to be compared with the more familiar Modus Tollens, the destructive mode. It can be applied in other situations, such as when deriving an apparently absurd 2=1 type result on attempting to solve simultaneous equations.

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This doesn't answer the question asked. –  Daniel Rust Feb 19 at 10:21
Strictly right Daniel. However I did deal with the question in detail, and suggested a new approach to apparently absurd results, by using the specific information they contain to go back and redefine, redirect the task. I think this is more positive and less pedantic, and may even break new ground in mathematical logic. If you still want a fallacy then Jon's own answer, that you can't divide out (a-b), is pretty good. Lastly, I have myself already been a keen fallacy hound on Stack Exchange - see my answers to "finding the fallacy in this broken proof", and to question 427635 on Grandi. –  Chris Goldfrap Feb 19 at 14:07
If you agree that this does not answer the question, then why have you not deleted it yet? If you want to advertise your 'revolutionary new methods' then write a blog. –  Daniel Rust Feb 19 at 14:12
"Revolutionary" was a bit strong, though only half-serious. But I'm also going to retract my admission that I didn't answer the question. In a deeper sense I think I did in that Jon was really asking how we should deal with these classic fallacious proofs that end in the equality of two different numbers. His coming up with an answer implies he wanted more. My answer, and his question, probably belong under a different tag, like "logic" or "philosophy of maths" where slightly wider discussion is (hopefully) permissible. It's an issue going back at least to Aristotle. As for a blog - good idea. –  Chris Goldfrap Feb 19 at 19:27

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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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.