# Intuition behind irrationality of $\pi$

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.

• Do you mean the numbers that appear in the expansion, or the fact that the expansion is infinite? – Andrés E. Caicedo Aug 14 '16 at 2:00
• ...do you know that there are arbitrarily large terms in the continued fraction expansion of $\pi$? – Steven Stadnicki Aug 14 '16 at 2:01
• I don't know that there are arbitrarily large terms in the continued fraction expansion of $\pi$. – pdmclean Aug 14 '16 at 2:41
• 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. – Steven Stadnicki Aug 14 '16 at 2:51
• 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. – Deepak Aug 14 '16 at 3:19

But any proof that the simple continued fraction expansion of $\pi$ does not terminate must require some work.
• 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. – Mario Carneiro Aug 14 '16 at 4:17
• Because if $a>b$ then the remainder when $a$ is divided by $b$ is less than $b$, and a fortiori is less than $a$. $\qquad$ – Michael Hardy Aug 14 '16 at 4:19
• @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. $\qquad$ – Michael Hardy Aug 14 '16 at 5:04
• 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. – Mario Carneiro Aug 14 '16 at 5:09