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Suppose we have a finite quantity $a$, which we would like to prove to be irrational, supposing that it is indeed irrational.

Then, would it be enough to show that $$a=\lim_{n\to\infty}\frac{u_n}{v_n},$$ for some positive integers $u_n,v_n$, where $u_n,v_n\to\infty$ as $n\to\infty$. If so, then would there have to be some divisor properties between the denominator and numerator so that cancellation does not produce an integer or rational number as $n\to\infty$, e.g. suppose $(u_n,v_n)=1$ for all $n$ ?

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    $\begingroup$ As the rational numbers $\mathbb Q$ are dense in the real numbers $\mathbb R$, any real number (either rational or irrational) is the limit of a sequence of rational numbers. So, no, the answer is no (any $a$ can be written in that form for suitable integers $u_n$ and $v_n$) $\endgroup$ – AndreasT Aug 15 '14 at 15:24
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    $\begingroup$ The condition that $v_n \nmid u_n$ is extremely ineffective, it only rules out $a$ being an integer. Don't you mean that the gcd of $v_n$ and $u_n$ should be 1? $\endgroup$ – Erick Wong Aug 15 '14 at 15:48
  • $\begingroup$ @Erick yes you're right. I've updated the question. $\endgroup$ – Pixel Aug 15 '14 at 15:49
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    $\begingroup$ All rational numbers have finite continued-fraction representations. So if you can prove a number has an infinite continued fraction representation it must also be irrational. $\endgroup$ – NovaDenizen Aug 15 '14 at 15:53
  • $\begingroup$ Given a number, it is undecidable in general whether it is real or not (because there are more real numbers than there are Turing machines). $\endgroup$ – user1729 Aug 15 '14 at 15:55
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Here's a similar condition that is sufficient: there exists a sequence of integers $u_n, v_n \to\infty$ such that $(u_n, v_n) = 1$ and $$\lim_{n\to\infty} v_n a - u_n = 0.$$ (Alternatively, we could just require that $a$ not be exactly equal to any $u_n/v_n$ rather than $(u_n,v_n)=1$.)

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  • $\begingroup$ Thanks Erick. Do you have a reference for this or a resource for further details? I'd like to study more. $\endgroup$ – Pixel Aug 15 '14 at 15:53
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    $\begingroup$ I'd be shocked if the idea isn't in any book on irrationality, but I don't know a reference offhand (maybe try something basic like Niven's "Numbers: Rational and Irrational"?). The proof is very simple: if $a=p/q$, and $a \ne u_n/v_n$, then $v_n a - u_n$ is a non-zero rational with denominator (at most) $q$. Therefore it is at least $1/q$ in absolute value, and can't go to $0$. $\endgroup$ – Erick Wong Aug 15 '14 at 15:56
  • $\begingroup$ @ Erick but isn't this just the same as my original question, since $$\lim_{n\to\infty}v_n a-u_n=0 \iff \lim_{n\to\infty}a=\lim_{n\to\infty}\frac{u_n}{v_n}\iff a=\lim_{n\to\infty}\frac{u_n}{v_n},$$ and $u_n,v_n\to\infty$ as $n\to\infty$, $(u_n,v_n)=1$ ? $\endgroup$ – Pixel Aug 19 '14 at 8:53
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    $\begingroup$ @pbs Absolutely not. Compare your first iff against $a=1$, $u_n = n$, $v_n = n+1$. The $\Leftarrow$ is very much false. $\endgroup$ – Erick Wong Aug 19 '14 at 14:30
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    $\begingroup$ @ Erick wow - I see that now: while $$\lim_{n\to\infty}\frac{u_n}{v_n} = \lim_{n\to\infty}\frac{n}{n+1}=1,$$ we have $$\lim_{n\to\infty}(n+1)\times 1-n=\lim_{n\to\infty} 1 = 1\neq 0.$$ $\endgroup$ – Pixel Aug 19 '14 at 15:35
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Consider

$$u_n = \frac{10^n - 1}{3} \hspace{1cm}v_n = 10^{n}$$

Then:

$$\frac{u_n}{v_n} = 0.\underbrace{3333\cdots 3}_{n \text{ times}}$$

What means that

$$\lim_{n\to\infty} \frac{u_n}{v_n} = 0.333\cdots = \frac{1}{3}$$

So no, that condition is not sufficient.

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This will not suffice. Let $a$ be an irrational number then there is some sequence of rational numbers $(q_{n})_{n=1}^{\infty}$ that converges to $a$. Denote $q_{n}=\frac{u_{n}'}{v_{n}'}$ where $u_{n}',v_{n}'$ are coprime. Let $u_{n}=nu_{n}',v_{n}=nv_{n}'$ then $$ \lim_{n\to\infty}\frac{u_{n}}{v_{n}}=\lim_{n\to\infty}\frac{nu_{n}'}{nv_{n}'}=\lim_{n\to\infty}\frac{u_{n}'}{v_{n}'}=\lim_{n\to\infty}q_{n}=a $$

So every irrational number satisfy your property.

But similarly if $a$ is rational take $q_{n}\equiv\frac{na}{n\cdot1}$ which will show that every rational number also satisfy this test. So in fact all real numbers satisfy the test and so we can't get information about a number that satisfies this test - they all do

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