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I can't seem to figure out why the proof is false. Everyone knows that $1 \ne 2$ and that $1=2$ is only proven through mathematical fallacies.

However I can't figure out where the problem is in the following fallacy:

Observe that

$$1=\frac{2}{3-1}$$

Substitute $\frac{2}{3-1}$ for the $1$ in the denominator

$$1=\frac{2}{3-\frac{2}{3-1}}$$

Substitute again $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-1}}}$$

Therefore $1$ can be written as the infinite fraction: $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$

Now notice that $$2=\frac{2}{3-2}$$

Substitute 2 similar to how 1 was substituted above

$$2=\frac{2}{3-\frac{2}{3-2}}$$

And through repeated substitution we get $$2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$

Notice that $1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$ and $2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$

Therefore, $1=2$

Just looking at this I know that it cannot be true. But they both seem to share the same infinite fraction, so what is it that I am missing?

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    $\begingroup$ math.stackexchange.com/questions/417280/… $\endgroup$ – Brenton Aug 21 '16 at 21:34
  • $\begingroup$ Whoops. I did not know that there was a post that explained this already. Thanks for showing me this. $\endgroup$ – WaveX Aug 21 '16 at 21:37
  • $\begingroup$ @WaveX Happens to us all! $\endgroup$ – Patrick Stevens Aug 21 '16 at 21:39
  • $\begingroup$ @WaveX Don't worry about it. It happens to pretty much everyone on here! :) $\endgroup$ – Brenton Aug 22 '16 at 0:50
  • $\begingroup$ Key in "seem to". They are written to look the same but they are both defined completely differently. One was defined with the number 1 which is no longer written and the other with 2. These are different things and it'd be unclear how they are defined simply by how they are written. Properly we'd define $a_0 =1; a_{n+1}=\frac 2 {3-a_n}$ and we note $\lim a_n = a_i = 1$ for all $i $. Meanwhile the other is $b_0 =2;b_{n+1}=\frac 3 {3 -b_n}; \lim b_n= b_i=2$ for all $i$. $b_n $ and $a-n $ are not at all the same because $a_0 \ne b_0$. Just because they are not written they are still there $\endgroup$ – fleablood Aug 22 '16 at 4:15
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If you look at the sequence of convergents for each continued fraction, you'll see they are different sequences. They don't exactly represent series, but you still have to be careful. Just because you can write "dot dot dot" doesn't mean that things really converge absolutely.

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