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$a_i$, $b_i$, $i=0,\ldots,\infty$, are two integer sequences with $\gcd(a_i,b_i)=1$ for all $i$.

Is then the limit $\frac{a_\infty}{b\infty}$ irrational if not $\frac{a_\infty}{b\infty}=\frac{A}{B}$ for some $\gcd(A,B)=1$ (assuming the limit exists at all)?

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I think this can't quite be what you're asking? It looks like you're asking, "if the limit exists at all, is it irrational if it's not rational?" (See Zev Chonoles' answer.) – Ben Blum-Smith Feb 19 '12 at 15:45
You could get an irrational limit, for example consecutive Fibonacci numbers ($a_n=b_{n+1}=\frac{\phi^n-\psi^n}{\phi-\psi}$ for $\phi=\frac{1+\sqrt{5}}2$, $\psi=\frac{1-\sqrt{5}}2$) approaching the golden mean, but there are way too many possible such sequences to make such a statement. – bgins Feb 19 '12 at 16:26
up vote 6 down vote accepted

No - for example, let $a_i=2$ and $b_i=3$ for all $i$. Then $\gcd(a_i,b_i)=\gcd(2,3)=1$ and $$\lim_{i\to\infty}\frac{a_i}{b_i}=\frac{2}{3}$$ is rational.

Your question, as edited, is trivially true because any rational number can be expressed as $\frac{A}{B}$ where $\gcd(A,B)=1$; so if the limit cannot be expressed that way, it will be irrational. That the number occurs as a limit of anything is irrelevant.

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No. For example $(n+1)/n \to 1$, and $\operatorname{pgcd}(n+1,n)=1$.

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Excuse my ignorance: what is a $pgcd$? I know $gcd$, but I have not heard of $pgcd$. – user2468 Feb 19 '12 at 15:47
PGCD (plus grand commun diviseur) is French for GCD (greatest common divisor). – Did Feb 19 '12 at 16:27

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