I was watching a Mathologer video on a proof that $\pi$ is irrational that used continued fractions. At one point in the video, the person presenting the proof challenged the viewers to prove that
$\frac{2}{3-\frac{2}{3-\frac{2}{3-...}}}=1$. I attempted this the following way:
Let $x=3-\frac{2}{3-\frac{2}{3-...}}$.
We can then see that $x=3-\frac{2}{x}$. We can then solve for $x$, arriving at $x^2-3x+2=0$, which we can solve with the quadratic formula, yielding $x=1$ or $x=2$. Now, $1-3+2=0$; $3-\frac{2}{1}=1$, and the coefficient of $x^2\neq0$, so $1$ should be a valid solution for $x$, however in that case $\frac{2}{x}=2\neq1$ (which we were trying to prove), which is a bizarre result, seeing as a continuing fraction should only have one result (or so I think...).
I'm still only a high school student, and my formal math education is around Algebra I and Geometry level, so it's possible I may be missing something obvious. Clearly, my reasoning is flawed. Can someone please help me see where I'm wrong here, and what I missed? Is my approach completely wrong?
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BouncyKnight
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We don't have $x=3-\dfrac 2x$; we have $x=\dfrac 2{3-x}$ – Prasun Biswas Oct 23 '20 at 10:56
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@PrasunBiswas $$x = \frac2{3-x} \leadsto 3-x = \frac2x \leadsto x=3-\frac2x$$ both lead to the quadratic $x^2-3x+2 = 0$. – player3236 Oct 23 '20 at 11:00
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1See ... https://math.stackexchange.com/questions/2337058/continued-fractions-paradox – Donald Splutterwit Oct 23 '20 at 11:07