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The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?

Note that \begin{align} \small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt] \text{and } \\ \small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\ \text{So } & 2=1. \end{align}

Michael Hardy
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    Clever, cutting the bottom right fractions off – Heavenly96 Apr 04 '17 at 02:14
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    This pseudo-proof is cute in the way it conceals the falsehood. It recasts the problem of "proving" 1==2 into the problem of convincing the reader that the subtle difference between 3-1 and 3-2 in the context of a continued fraction expansion can be ignored; An artifice achieved by way of brazenly deleting the primary difference between the two CFRACs in their definition. – Iwillnotexist Idonotexist Apr 04 '17 at 05:46
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    Related to http://math.stackexchange.com/questions/417280/continued-fraction-fallacy-1-2 – Matthew Towers Apr 04 '17 at 09:49
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    The Facebook page is actually titled "Mathematical theorems you had no idea existed, 'cause they're false". "BestTheorems" is their Twitter username. – Akiva Weinberger Apr 04 '17 at 22:50
  • "Dot dot dot " can conceal a multitude of deceptions . It all comes down to what "dot dot dot" means, or in this case, whether it means anything at all – DanielWainfleet Apr 05 '17 at 02:50

4 Answers4

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$$x = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}$$

$$x = \frac 2 {3 - x}\\ x^2 - 3x + 2 = 0\\ (x-1)(x-2) = 0$$

$1,2$ are both solutions. However, if we consider this recurrence relation:

$$x_n = \frac 2 {3 - x_{n-1}}$$

when $x_{n-1} <1 \implies x_{n-1}<x_n<1$ And the squence converges to 1.

And

$$1<x_{n-1} < 2 \implies 1<x_n <x_{n-1}$$ and the sequence again converges.

but, $$2<x_{n-1} < 3 \implies x_{n-1}<x_n$$

and

$$x_{n-1} >3 \implies x_n< 0$$

The sequence isn't stable in a neighborhood of $2.$

and it converges to $1$ for nearly all starting conditions.

Michael Hardy
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Doug M
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  • The starting conditions where the recurrence does not converge to $1$ seem to be $x=2$ and $x=2+\frac{1}{2^n-1}=\frac{2^{n+1}-1}{2^n-1}$ for positive integer $n$ so $3, \frac{7}{3}, \frac{15}{7}, \frac{31}{15}, \ldots$ – Henry Apr 04 '17 at 17:53
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Simple, the problem is the assumption the three dots aka the ellipsis, in the first equation equal the ellipsis in the second. They are not.

Douglas
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  • This posting perhaps nicely identifies a gap in the logic, but where the gap is in the logic is easy. The point about the recurrence have one attractive fixed point and one repulsive fixed point is more interesting. – Michael Hardy Apr 04 '17 at 17:33
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For any $a$ and $b$, finding $n$ and $k$ such that

$a=\dfrac{n}{k-a}$

$b=\dfrac{n}{k-b}$

"proves" that $a=b\;\;\forall a,b\in\mathbb{N}$.

djechlin
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JMP
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    I don't understand your answer. This doesn't prove that $a=b$, as the question demonstrates (take $n=2$, $k=3$). It just proves that $a$ and $b$ are both roots of $x^2-kx+n=0$. – David Richerby Apr 04 '17 at 11:36
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    Perhaps the "proves" is sarcastic and is leveraging the proposal of the original question which "proves" $2=1$. – Ian MacDonald Apr 04 '17 at 13:46
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    Poe's law in action, I guess? But to be honest, I don't understand this answer either. – user1551 Apr 04 '17 at 14:36
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    @DavidRicherby : If $a = \dfrac n {k-a}$ then you can put $\dfrac n {k-a}$ in place of $a$ where it appears just after the minus sign, and you get $$ a = \cfrac n {k- \cfrac n {k-a}}. $$ Then you can do the same with the last $a$ that appears in that expression and get $$ a = \cfrac n {k-\cfrac n {k- \cfrac n {k-a}}} $$ and so on. You get $$ a = \cfrac n {k - \cfrac n {k- \cfrac n {k - \ddots \vphantom{\cfrac n k} }}} $$ But the same argument shows that $b$ is equal to that expression, so $a=b.$ (Of course, this argument is flawed, but I think this is what was intended.) $\qquad$ – Michael Hardy May 12 '17 at 02:09
  • @MichaelHardy Ah, yes. That makes sense. It's a shame the answerer never clarified. – David Richerby May 12 '17 at 08:10
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Suppose $$ x = 2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {\ddots}}}}}} $$ Then $$ x = 2 + \cfrac 1 {3 + \cfrac 1 x}. $$ Solving this by the usual method, one gets $$ x = \frac {3\pm\sqrt{15}} 3. $$ At this point I'd be inclined to say we obviously want the positive solution. And that that's what the fraction converges to. However, the answer from Doug M. suggests this point of view: The function $$ x \mapsto 2 + \cfrac 1 {3 + \cfrac 1 x} $$ has one attractive fixed point and one repulsive fixed point.

Michael Hardy
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