True or false
Between any two unequal irrational numbers there is a rational number.
I'm leaning towards true, but I can't come up with a convincing argument.
Hi
What you think is right, what is said is that $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}.$
Let $\displaystyle a,b,\ a<b$ be two (unequal) reals. Then $\displaystyle d=b-a>0.$ Since the sequence $\displaystyle \left(\frac{1}{k}\right)_{k\in\mathbb{N}^*}$ strictly decreases and has limit $\displaystyle 0,$ there is a $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \frac{1}{n}<d.$
Let $\displaystyle m\in\mathbb{Z}$ be the greatest integer such that $\displaystyle \frac{m}{n}\leq a.$ Then $\displaystyle a<\frac{m+1}{n}=\frac{m}{n}+\frac{1}{n}\leq a+\frac{1}{n}<a+d<b,$ therefore $\displaystyle a<\frac{m+1}{n}<b$
Since $\displaystyle m,n,\ n \neq 0$ are integers, $\displaystyle \frac{m+1}{n}$ is a rationnal; we've proved that between two unequal reals lies a rational number. In particular, there's always a rational number between two unequal irrational numbers.