Proving that there exists an irrational number in between any given real numbers 
Possible Duplicate:
Density of irrationals 

I am trying to prove that there exists an irrational number between any two real numbers a and b. I already know that a rational number between the two of them exists. My idea was to say represent a and b as $a = \sqrt{2} x, b = \sqrt{2} y$ for some $x, y \in \mathbb{R}.$ We know that there exists $\frac{m}{n}$ s.t. $x < \frac{m}{n} < y,$ so multiplying everything by the root of 2, we have $$\sqrt{2}x = a < \frac{\sqrt{2}m}{n} < \sqrt{2}y = b,$$ and so we have an irrational number in between two reals. Is this 'legit?' Are there 'better/'more elegant ways to go about this? I would really appreciate seeing alternatives proofs of this, as many as possible.
 A: If you know that between any two real numbers there is a rational, then why not just shift everything?  If $a < b$, then it follows that $\sqrt{2} + a < \sqrt{2} + b$, and so there is a rational number $r$ with $\sqrt{2} + a < r < \sqrt{2} + b$.  But then it follows that $a < r - \sqrt{2} < b$.  What sort of number is $r - \sqrt{2}$?
This is very close to your proposed solution, but look at what happens in your case when $\frac{m}{n} = 0$.
A: The interval $(a,b)$ is uncountable since $f(x) = \frac{(x-a)}{(b-a)}$ is a bijection from $(a,b)$ to $(0,1)$. 
There is an irrational number in $(a,b)$ since the rationals themselves are countable. 
A: HINT $\ $ If the non-empty interval I contains only rationals, then repeatedly shifting I left/right by a fixed positive rational less less than the length of I will cover the entire real line with rationals. Hence we conclude that every real is rational, a contradiction.
A: Here is one proof: 
Let $a,b\in \mathbb{R}~,a\lt b$. Then $(a,b)$ is uncountable. The set of rational numbers, $\mathbb{Q}$ is countable, so any subset of it is also countable.
  Now suppose $(a,b)$ contained no irrational number. Then $(a,b)$ would be an uncountable subset of $\mathbb{Q}$ which is a contradiction.
A: By scaling and translating you only need to show there is an irrational number between, say, $0$ an $1$. Just show that

If $\alpha$ is irrational then $r \alpha$ is irrational and $n + \alpha$ is irrational, for all rational numbers $r$ and integers $n$.

