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This question already has an answer here:

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 $1=2.$

by 24.0.94.225 at 20110927 via 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:

  1. I could undersatnd it (I studied Maths at university)

  2. 16 year olds (with no formal knowledge of analysis or the language of convergence) would understand it?

Thank you.

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marked as duplicate by Simply Beautiful Art, Martin R, Lord Shark the Unknown, Derek Elkins, miracle173 Jun 27 '17 at 7:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ We generally prefer that questions stand on their own, without making everyone follow your link to find out what is going on. $\endgroup$ – Teepeemm Jun 26 '17 at 19:37
  • $\begingroup$ Sorry for not spotting the duplication of an earlier question! $\endgroup$ – Jezza Judge Jun 27 '17 at 18:45
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If you look very carefully at each and every step, the "paradox" ends up being a "sleight of hand". When it says "I'll use the ellipsis", it means to indicate that any number of finite steps can be performed. The second time it does not even say "ellipsis", but it means the same. But then, at the end, it says "this is exa[c]tly the same continued fraction". It is not. They are not continued fractions: they are expressions with finite steps. Continued fractions (as you can see in the other answers) are infinite expressions with some careful definition of convergence.

(The sleight of hand is that they look the same as the usual form for continued fractions, by using a meaning for ellipsis that is different from the meaning it has in continued fractions.)

[You should really make your question stand alone, by the way.]

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When we evaluate a continued fraction, we're trying to solve an equation: in this case, $$x={2\over 3-x}.$$

This turns into $3x-x^2=2$, or $x^2-3x+2=0$; or, better yet, $(x-2)(x-1)=0$. This has two solutions, namely $x=2$ and $x=1$.

The "contradiction" here comes from the implicit assumption that a continued fraction always describes a single number - in the same way that a decimal expansion picks out a specific real - as opposed to being a description that could apply to several things. Basically, the point is that $$x={2\over 3-{2\over 3-{2\over 3-...}}}$$ is really just a tricky way of saying $$\mbox{"$x$ has the property that $x^2-3x+2=0$,"}$$ which of course doesn't refer uniquely. The expression $${2\over 3-{2\over 3-{2\over 3-...}}}$$ looks like something that refers to a specific number, but actually doesn't.

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  • $\begingroup$ Thank you! Useful to point out the similarity with quadratic equations - both cases where you might think there is only one solution. $\endgroup$ – Jezza Judge Jun 27 '17 at 18:49
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Fundamentally, this is an error where you see something that "looks like" a calculation, and you assume it actually means something. But it doesn't. The expression:

$$\dfrac{2}{3-\dfrac{2}{3-\frac{2}{3-\dots}}}$$ looks like something, but it isn't well defined. Is it the limit of:

$$\frac{2}{3},\dfrac{2}{3-\frac{2}{3}},\dfrac{2}{3-\dfrac{2}{3-\frac{2}{3}}},\dots$$

or is it the limit of:

$$2,\dfrac{2}{3-2},\dfrac{2}{3-\frac{2}{3-2}},\dots?$$

The first sequence converges to $1$, the second sequence converges trivially to $2$ (since every value is $2$.)

It's worth noting that if $x_0$ is defined as any value, and $x_{n+1}=\frac{2}{3-x_n}$, this sequence actually only converges to $2$ if $x_0=2$. This is because:

$$2-\frac{2}{3-x}=\frac{4-2x}{3-x}=\frac{2}{3-x}(2-x)$$

so $|x_{n+1}-2|>|x_n-2|$ when when $x_n>1$ and $x_n\neq 2$. So the only way for it to converge to $2$ is if it is always equal to $2$.


You get this problem, too, if you just take a simple equation:

$$x = 1+\frac{1}{1+x}$$

which has solutions $\pm \sqrt{2}$. That seems to mean that the continued fraction:

$$1+\frac{1}{2+\frac{1}{2+\dots}}$$

is likewise ambiguous. The only reason we don't see it as a paradox is that intuitively, we know this is only the positive value.

So the real problem is in extrapolating from the equation $\gamma = a+\frac{b}{c+\gamma}$ to some infinite form, which may ore may not have meaning.


A similar error can be seen more simply if you start with:

$$2=1+(-1)+2$$

You would never conclude that $$2=1+(-1)+1+(-1)+\cdots$$

The finite sums always equal $2$, but the $2$ is no longer present in the infinite sum, and you the meaning you ascribe to the infinite sum depends on definitions.

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  • $\begingroup$ In your example, the first limit/sequence should start with $1$ so that it trivially goes to $1$ in the same manner the second one does. :-) $\endgroup$ – Simply Beautiful Art Jun 26 '17 at 19:14
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    $\begingroup$ Yes, but it also converges to one of $x_0=0$ or $x_0=1.9999$. The point is that we get way more cases converging to $1$. $\endgroup$ – Thomas Andrews Jun 26 '17 at 19:31
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    $\begingroup$ Ah, okay. :D My bad $\endgroup$ – Simply Beautiful Art Jun 26 '17 at 19:38
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Channeling my inner child for an extremely simple explanation:

The last step states "But this is exatly the same continued fraction." [sic]

They are not the same. One has a 1 in the bottom right. One has a 2 in the bottom right. The author is using ellipses to elide different things, then claiming the results are equal because they now look the same.

This is like saying

1/3 = 0.33333...

1/3 + 1/3000000 = 0.33333...

therefore 1/3000000 = 0.

from this and the following:

n * 0 = 0

1 * n / n = 0 for n != 0

1 * n = n * 1

we have

3000000 * 1/3000000 = 0

therefore 3000000 * 1/3000000 = 1 * 3000000 / 3000000

therefore 1 * 1 = 0

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  • $\begingroup$ "1 * n / n = 0 for n != 0" ??? I understand that you are trying to do a 'fake' proof, so something should be wrong, but in a proper fake proof, the mistake should be hidden so the casual reader misses it... $\endgroup$ – Pakk Jun 27 '17 at 7:03

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