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Would the existence of arbitrarily large terms in the continued fraction expansion of $\pi$ imply its irrationality?

Edit: I completely changed the question to remove speculation on my part.

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  • $\begingroup$ Do you mean the numbers that appear in the expansion, or the fact that the expansion is infinite? $\endgroup$ – Andrés E. Caicedo Aug 14 '16 at 2:00
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    $\begingroup$ ...do you know that there are arbitrarily large terms in the continued fraction expansion of $\pi$? $\endgroup$ – Steven Stadnicki Aug 14 '16 at 2:01
  • $\begingroup$ I don't know that there are arbitrarily large terms in the continued fraction expansion of $\pi$. $\endgroup$ – pdmclean Aug 14 '16 at 2:41
  • $\begingroup$ Given that AFAIK it's not actually known that there are arbitrarily large terms, I don't think you can really use the word 'because' there since the supposed cause isn't even known for certain but $\pi$'s irrationality is. $\endgroup$ – Steven Stadnicki Aug 14 '16 at 2:51
  • $\begingroup$ I've deleted my answer because I'm not sure if it addresses the question. To a large extent, that's because the question statement itself has issues. However, if the asker found my reformulation in terms of periodicity useful, let me know and I'll restore the answer. $\endgroup$ – Deepak Aug 14 '16 at 3:19
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The existence of infinitely many terms in the simple continued fraction is enough to imply a number is irrational, if if they're not unbounded. But they can be unbounded only if there are infinitely many.

To see that a number must be irrational if it continued fraction expansion has infinitely many terms, consider this: \begin{align} & \frac{175}{81} = 2 + \cfrac{13}{81} & & \text{The numerator 13 is smaller than the numerator 175.} \\[10pt] = {} & 2 + \frac 1 {\left( \dfrac{81}{13} \right)} = 2 + \cfrac 1 {6+ \cfrac 3 {13}} & & \text{The numerator 3 is smaller than the numerator 81.} \\[10pt] = {} & 2 + \cfrac 1 {6 + \cfrac 1 {\left( \cfrac {13} 3 \right)} } = 2 + \cfrac 1 {6 + \cfrac 1 {4 + \cfrac 1 3}} & & \text{The numerator 1 is smaller than the numerator 13.} \end{align} You cannot keep getting smaller positive integers forever. Therefore the simple continued fraction expansion of a rational number terminates.

But any proof that the simple continued fraction expansion of $\pi$ does not terminate must require some work.

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  • $\begingroup$ The argument seems a bit confused. What is the strictly decreasing sequence of integers in question? I see $175>13$, $81>3$, $13>1$ but I don't know if you are going for $(175,13,1)$ or $(175,81,13)$ or $(13,3,1)$ or something else. $\endgroup$ – Mario Carneiro Aug 14 '16 at 4:17
  • $\begingroup$ Because if $a>b$ then the remainder when $a$ is divided by $b$ is less than $b$, and a fortiori is less than $a$. $\qquad$ $\endgroup$ – Michael Hardy Aug 14 '16 at 4:19
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    $\begingroup$ @MarioCarneiro : $$(175,81) \mapsto (13,81) \mapsto (13,3) \mapsto (1,3)$$ At each step the larger of the two numbers is replaced by a number that is smaller than the smaller of the two numbers. $\qquad$ $\endgroup$ – Michael Hardy Aug 14 '16 at 5:04
  • $\begingroup$ You could simplify that to $81>13>3>1$ if you just take the denominators on the right in each step. (Oh wait, I guess you are trying to capture the whole state with those pairs...) Also, the Euclidean algorithm is relevant. $\endgroup$ – Mario Carneiro Aug 14 '16 at 5:09

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