# Continued fraction fallacy: $1=2$

It's easy to check that for any natural $n$ $$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$

Now,

$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots =\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$

$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots =\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$

Since the right hand sides are the same, hence $1=2$.

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they arent the same. One right hand side is a continued fraction that converges to 2, one converges to 1. By using the dots, you cut away information. – CBenni Jun 11 '13 at 10:10
Just because $\frac{n}{n+1}$ tends to 1 as $n$ tends to $\infty$ doesn't mean that when you take the whole limit, it is equal to the first expression where you just have $2-1$ in the denominator of the repeating stuff. Also, the fact that you get $1=2$ should immediately tell you that you have went wrong somewhere. – Andrew D Jun 11 '13 at 10:25
@MherSafaryan no you dont. Dots imply a repitition of a given structure; They are not a rigorous mathematical notation. – CBenni Jun 11 '13 at 10:58
Is there an implied question here? E.g. "what's wrong with this fake proof"? – LarsH Jun 11 '13 at 14:38
Uh ya...what's the question? – nzifnab Jun 11 '13 at 16:27

Another example where dots are misleading: $$1= \frac{ 1 \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{red}{4} \cdots}{ 2 \cdot \color{blue}{3} \cdot \color{green}{4} \cdot \color{red}{5} \cdots} \leq \frac{1}{2}$$

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The dots are not misleading here, it is clear in both interpretations what the dots are replacing, unlike the question and @Didi's example. – jwg Jun 27 '13 at 8:25
Maybe just rephrasing @jwg's comment, I fail to see what explains the $=$ sign here. – Did Aug 13 '13 at 9:06
I didn't want to make my comment dismissive; in fact, I don't really understand the problem. – Seirios Aug 13 '13 at 9:48
Seirios: this fallacy is based on the order of taking limits (like many others). The crux of it is that $\lim\left(\frac{n!}{n!}\right)$ is not the same as $\frac{\lim(n!)}{\lim(n!)}$. Mine and @Did's comments are trying to say that this is not the same as 'hiding' two different expressions behind an ellipsis. – jwg Aug 13 '13 at 10:11
Would you prefer it if I deduced $1 \leq \frac{1}{2}$ from $1= \frac{1}{1}= \frac{1 \cdot 2}{1 \cdot 2} = \dots= \frac{1 \cdot 2 \cdot 3 \cdots}{1 \cdot 2 \cdot 3 \cdots}$ and $\frac{1}{2}= \frac{1}{1 \cdot 2} \geq \frac{1 \cdot 2}{1 \cdot 2 \cdot 3} \geq \frac{1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} \geq \dots \geq \frac{1 \cdot 2 \cdot 3 \cdots}{1 \cdot 2 \cdot 3 \cdots}$? – Seirios Aug 13 '13 at 11:01

A variant: note that $$\color{red}{\mathbf 1}=0+\color{red}{\mathbf 1}=0+0+\color{red}{\mathbf 1}=0+0+\cdots+0+\color{red}{\mathbf 1}=0+0+0+\cdots$$ and $$\color{green}{\mathbf 2}=0+\color{green}{\mathbf 2}=0+0+\color{green}{\mathbf 2}=0+0+\cdots+0+\color{\mathbf green}{2}=0+0+0+\cdots$$ "Since the right hand sides are the same", this proves that $\color{red}{\mathbf 1}=\color{green}{\mathbf 2}$.

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Damn, this is a simple, straightforward and elegant as one can expect! +1 – DonAntonio Jun 11 '13 at 11:27
@DonAntonio I think it is invalid. It should be written as $$\Large\color{blue}{1}=0+0+0+\cdots+\Large\color{blue}{1}$$ and $$\Large\color{red}{2}=0+0+0+\cdots+\Large\color{red}{2}$$ – Tunk-Fey Apr 13 '14 at 9:40
@Tunk-Fey You seem to seriously think that some people on this page wish to prove that 1=2. Then indeed you might be missing the point... – Did Apr 13 '14 at 11:19
@Did Well, I just think instead of showing other fallacies of $1=2$, why didn't show a valid proof that the argument in the question is wrong. – Tunk-Fey Apr 13 '14 at 11:35
@Tunk-Fey Because explanations by analogy may be valuable (as a long tradition shows). But you still did not explain why, in your first comment, you saw fit to throw around the word "invalid"... – Did Apr 13 '14 at 11:38

The first expression is a continued fraction, the second isn't. A continued fraction is the limit of $$a_0, a_0 + \frac{1}{a_1}, a_0 + \frac{1}{a_1 + \frac{1}{a_2}} \ldots$$ for a fixed sequence of natural numbers $a_0, a_1, a_2 \ldots$

The second expression is the limit of fractions which look similar to these fractions, but which don't correspond to one well-defined sequence of naturals. The first lot of dots (between the equals signs) are ok, they just mean 'take the limit of this process'. This limit exists and is equal to 2, as you correctly deduce. The dots at the bottom of the final expression falsely suggest that the limit is a continued fraction, with coefficients given by the obvious sequence (eg $2, 2, 2, 2, \ldots$ implies the sequence consisting of only twos).

As @Did points out very elegantly, the same rules apply to infinite sums, and seem more obvious there - an infinite sum is not the same as the limit of an infinite sequence of sums, each with more terms in it than the one before. The common terms have to agree for any two sums.

I think this misunderstanding arises because sometime in iteration and limits we have the sense that the initial terms don't really matter, and sometimes this is the case. The first few terms of a sequence don't affect the limit, or the limit of the average, etc.

You are taking two different starting terms and iteratively applying a transformation to them. As you point out, this transformation doesn't actually change the number. This fact means though, that the starting value never becomes unimportant, and the final terms of each expression in your sequence similarly never become unimportant.

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This is of the same type as the following

$$0=(1-1)+(1-1)+(1-1)+\ldots=1+(-1+1)+(-1+1)+(-1+1)+\ldots = 1 \; .$$

Did's example is even simpler and closer to yours in spirit.

In working with an infinite number of operations, you have to be very careful about how you are performing them. Normally, one uses some kind of limit, but then what you really do is define a sequence of finite but ever increasing number of operations. Changing something in the order of these operations will change the entire limit. Or in your case, you hide away the fact that in each term of your sequence, the last operation is subtracting a different number, $1$ in the first case, $(n+2)/(n+1)$ in the second.

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This is a general feature of continued fractions. What you are doing is iterating the function $$s(x) = \frac{1}{2-x}$$ ... so, for instance, you have that $$1 = s(1) = s(s(1)) = s(s(s(1))) = \cdots$$ and $$2 = s(3/2) = s(s(4/3)) = s(s(s(5/4))) = \cdots .$$

The function $s$ has a single fixed point at 1 (first line). It is known that iterating $s$ from any starting point except $x=2$ converges to 1 (think of $s$ as a mapping of the Riemann sphere; it takes $2 \mapsto \infty$ and $\infty \mapsto 0$). For $x<1$, this is easy to check: $x<s(x)<1$.

So, if $x\neq 2$, then $$\lim_{n \to \infty} s^n(x) = 1 ,$$ where $s^n(x)=s(s(\cdots s(x)))$ is the $n$th iterate of $s$.

But if you start with some $y$, it's always possible to find an $x_n$ so that $s^n(x_n) = y$ (including $\infty$, the map is one-to-one). You've shown that $$s^n\left(\frac{n+2}{n+1}\right) = 2.$$

Another nice thing to note here is that if the numbers at the end of the fractions (the $x_n$) weren't approaching 1 from above, we'd end up with 1 -- $$\lim_n s^n(x_n) \neq 1$$ then $$x_n \searrow 1 .$$

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Today one friend of mine show a way to the fallacy $1=2$ in the given way described below.

Note that \begin{align} \log \,2 &=\log\, (1+1) \\&=1 -\dfrac 12 +\dfrac 13 -\dfrac 14 +\dfrac 15-\dfrac 16 +\dfrac 17 -\cdots \\ &= \left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)-2\times\left(\dfrac 12 +\dfrac 14 +\dfrac 16 +\dfrac 1{10}+\dfrac 1{12} +\dfrac 1{14} +\cdots\right)\\&=\left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)-\left(1 +\dfrac 12 +\dfrac 13 +\dfrac 14 +\dfrac 15+\dfrac 16 +\dfrac 17 +\cdots\right)\\&=\left(1-1\right)+\left(\dfrac12-\dfrac12\right)+\left(\dfrac13-\dfrac13\right)+\left(\dfrac14-\dfrac14\right)+\left(\dfrac15-\dfrac15\right)+\left(\dfrac16-\dfrac16\right)+ \cdots\\&=0+0+0+0+0+0+\cdots \\&=0\\&=\log\,1\end{align} This implies $1=2.$ Hence it is proved.

That's all from me.

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