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I've read recently that any number with infinite continued fraction is irrational. See this continued fraction below

https://www.wolframalpha.com/input/?i=continued+fraction+of+the+cosine If I substitute x for $\frac{\pi}{2}$ I will get zero on the left side, but the continued fraction is infinite because its convergent ones never cancel out. Could anyone explain me this?

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    $\begingroup$ Please write down what you think you get when you "substitue $1$ for $\pi/2$". Then we may be able to help you. $\endgroup$
    – Somos
    Nov 8, 2017 at 17:45
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    $\begingroup$ @Somos You're an optimist. $\endgroup$
    – user436658
    Nov 8, 2017 at 18:42
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    $\begingroup$ What does "its convergent ones never cancel out" mean ? $\endgroup$
    – Peter
    Nov 9, 2017 at 10:30
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    $\begingroup$ What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones. $\endgroup$
    – bof
    Nov 9, 2017 at 11:01
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    $\begingroup$ Any rational number can be expanded as a simple continued fraction in two slightly different ways, and those continued fractions are finite. Any irrational number has a unique expansion as a simple continued fraction, and it's infinite. (The continued fraction is periodic if and only if the irrational number is quadratic.) $\endgroup$
    – bof
    Nov 9, 2017 at 11:05

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Answered (in general) by @bof in a comment:

[...] a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones.

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