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 $b$.
Best Answer
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 $b$.