# Prove that we can choose both rational and irrational number between two distinct reals

Prove that we can choose both 1) rational and 2) irrational number between two distinct reals.

1) Let $a<b$ be real numbers. We can choose natural $n$ such that $n>\frac{1}{b-a}>0$. Now $nb>na+1$ so there is natural number $m$ between $na$ and $nb$. Thus we have $na<m<nb$, dividing both sides by $n$ we have $a<\frac{m}{n}<b$.

2) We can use 1) and the fact that sum of rational and irrational number is irrational, i.e. assume again $a<b$ are reals. It implies $a+\sqrt{2}<b+\sqrt{2}$. From 1) we can find rational number $r$ such that $a+\sqrt{2}<r<b+\sqrt{2}$ and so $a<r-\sqrt{2}<b$.

Could anyone check my reasoning please?

• Fast and fine!! – Bernard Nov 1 '15 at 14:30
• Seems perfectly legit. – Berni Waterman Nov 1 '15 at 14:31