Irrational number "test"? 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 $n$ ?
 A: 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 $(u_n,v_n)=1$.)
A: 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.
A: 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
