Intuition behind irrationality of $\pi$

Question

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.

Answer 1

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.