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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.

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marked as duplicate by t.b., Sasha, Martin Sleziak, Arturo Magidin, David Mitra Dec 2 '11 at 14:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Consider the rational number between a and b. Add the last infinity digits of pi to it. – Jacob Wakem May 24 at 20:32

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 $\frac{m}{n} = 0$.

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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.

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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.

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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.

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By scaling and translating you only need to show there is an irrational number between, say, $0$ an $1$. Just show that $\alpha$ irrational $=> r \alpha $ irrational and $n + \alpha$ irrational, $\forall r, n $ with $ r $ rational, $n$ integer.

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