I am preparing to teach an enrichment session to some 16 year olds, about Continued Fractions. I am confused about the following paradox.

## 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 $1=2.$

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by24.0.94.225at20110927via CTK Wiki Math

What is the hole in the argument? Presumably it has something to do with lack of convergence, but would anyone be able to explain it so that:

I could undersatnd it (I studied Maths at university)

16 year olds (with no formal knowledge of analysis or the language of convergence) would understand it?

Thank you.