# 1=2 via Continued Fractions

This is true of course that 1=\frac{2}{3-1}. Now, let's substitute this very expression for 1 in the denominator:

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

We can do that one more time:

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

And one more time to make sure there is no misunderstanding of the construction,

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

At this point I am assuming that further steps could be performed by any reader who got this far. To indicate that possibility I'll use the ellipsis:

1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}}.

Well, we also know that 2=\frac{2}{3-2}. With this as a starting point, we follow in the footsteps of the previous example. Replacing 2 in the denominator with that expression gives

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

To continue as before:

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

And finally,

2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}}.

But this is exatly the same continued fraction. By necessity we conclude that 1=2.