Proving that there is an irrational number between any two unequal rational numbers.
Sophia Terry 
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 $b$.
$\endgroup$ 32 Answers
$\begingroup$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 $b$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1
