$1=2$ | Continued fraction fallacy

Question

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 $1=2$.

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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 to end up with a limit above 1, we need the numbers at the end of the fractions (the $x_n$) to approach 1 from above: i.e., if for all $n$ that $$ s^n(x_n) = y > 1 $$ then $$ x_n \searrow 1 .$$

Answer 5

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.

Answer 6

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.

Answer 7

Incidentally, nobody appeared to have resolved the fallacy of the question, so I have provided an answer.

---

For all $a\in \mathbf N$, it follows $$\cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots - \cfrac 12}}}$$ such that the number of times the reciprocal in the continued fraction appears is $a$.

Proof. Note the identity $$\cfrac{1}{1+a}=1-\cfrac{1}{1+\color{red}{\cfrac 1a}}.$$ By letting $a=b-1$, it follows $$\cfrac 1b = 1-\cfrac{1}{1+\cfrac{1}{b-1}}.$$ From this we can substitute for $\color{red}{\cfrac 1a}$. $$\therefore \cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{1+\cfrac{1}{a-1}}}.$$ Clearly we can now likewise substitute for $1/(a-1)$, and the pattern will continue until for some $k\in\mathbf N$, the denominator of $1/(a-k)$ reaches $a-k=1$ since it cannot pass $0$. In consequence, we deduce as desired. (And, of course, when $a=0$, we have $1/1 = 1-0$.) This completes the proof. $\;\bigcirc$

And now, since $$\lim_{a\to\infty}\frac{1}{1+a}=0$$ then $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$