Irrational number "test"?

Question

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 $n$ ?

Answer 1

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 $(u_n,v_n)=1$.)

Answer 2

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.

Answer 3

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