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?

  • 2
    $\begingroup$ Fast and fine!! $\endgroup$ – Bernard Nov 1 '15 at 14:30
  • 1
    $\begingroup$ Seems perfectly legit. $\endgroup$ – Berni Waterman Nov 1 '15 at 14:31

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