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I'm trying to prove that there is an irrational number between any two unequal rational numbers $a, b$. Here's a "proof" I have right now, but I'm not sure if it works.

Let $a, b$ be two unequal rational numbers and, WLOG, let $a < b$. Suppose to the contrary that there was an interval $[a, b]$, with $a, b$ rational, which contained no irrational numbers. That would imply that the interval contained only rational numbers since the reals are composed of rationals and irrational numbers. Furthermore, this interval has measure $b-a$, a contradiction since this is a subset of $\mathbb{Q}$ which has measure zero.

Does this work? Is there an easier way to go about it, perhaps through a construction?

Construction: Let $a = \frac{m}{n}$, $b = \frac{p}{q}$. WLOG $a>b$. Then $a-b = \frac{m}{n}-\frac{p}{q} = \frac{mq-np}{nq}$. Since $mq - np > 1$, we can construct an irrational number $a + \frac{1}{nq\sqrt2}$ which is between $a$ and $b$.

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    $\begingroup$ It works, but uses an awful lot of background machinery. Probably a proof is expected that uses no more than completeness of the reals. You might ty to show that between $a$ and $b$ there is a number of the form $\frac{p}{q}\sqrt{2}$, where $\frac{p}{q}$ is a non-zero rational. $\endgroup$ Commented Jan 16, 2015 at 2:15
  • $\begingroup$ @AndréNicolas Yea, I was thinking it was overkill. Do you know an alternate way that doesn't use as much background? $\endgroup$
    – MT_
    Commented Jan 16, 2015 at 2:17
  • $\begingroup$ I added a suggestion on what to try. On the right side of this page under Related you will find a solution, the question has been asked several times on MSE. $\endgroup$ Commented Jan 16, 2015 at 2:19

3 Answers 3

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Your proof looks fine, but there is a slightly more elementary way of doing this: $\sqrt{2}$ is irrational. A rational times an irrational is irrational, and the sum of a rational with an irrational is irrational. It's then easy to check that $$s = a+\frac{\sqrt{2}(b-a)}{2}$$ is an irrational number between $a$ and $b$.

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    $\begingroup$ Thanks. While you posted this I added my own construction, but I now realize that this is much simpler. Not sure why I didn't see this. $\endgroup$
    – MT_
    Commented Jan 16, 2015 at 2:25
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Your proof is fine, though its appeal to measure theory (or at least to the statement that the interval $[a,b]$ has measure $b-a$ and the rationals have measure $0$) might raise some eyebrows since those are somewhat deeper results than what you're trying to prove.

I think an insightful proof would be to notice that $f(x)=mx+b$ for rational $m\neq 0$ and $b$ has that if $x$ is rational $f(x)$ is rational, and if $x$ is irrational, $f(x)$ is irrational. Notice that any interval $[a,b]$ for rational $a$ and $b$ can be written as an image $f([0,1])$ for some $m$ and $b$. Then, if you prove that there is an irrational in $[0,1]$, you prove the desired statement. This is to say, since all intervals are isomorphic in a sense, we only need to prove it's true of one interval.

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    $\begingroup$ I see. So I suppose for any interval $[a,b]$ with $a, b$ rational, we can "morph" it into $[0,1]$ by some dilations $(m)$ / translations $(b)$. And of course there is an irrational number in $[0,1]$, e.g. $\frac{1}{\sqrt{2}}$. That makes sense, thanks! $\endgroup$
    – MT_
    Commented Jan 16, 2015 at 2:26
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(a,b) is uncountable, so if all elements of (a,b) were rational, we'd have an uncountable set of rationals. We can't have that!

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