Rational and irrational numbers

**noles2188** · February 11th 2009, 07:43 PM

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.

**clic-clac** · February 12th 2009, 02:42 AM

Hi

What you think is right, what is said is that $\mathbb{Q}$ is dense in $\mathbb{R}.$

Let $a,b,\ a<b$ be two (unequal) reals. Then $d=b-a>0.$ Since the sequence $\left(\frac{1}{k}\right)_{k\in\mathbb{N}^*}$ strictly decreases and has limit $0,$ there is a $n\in\mathbb{N}$ such that $\frac{1}{n}<d.$

Let $m\in\mathbb{Z}$ be the greatest integer such that $\frac{m}{n}\leq a.$ Then $a<\frac{m+1}{n}=\frac{m}{n}+\frac{1}{n}\leq a+\frac{1}{n}<a+d<b,$ therefore $a<\frac{m+1}{n}<b$

Since $m,n,\ n \neq 0$ are integers, $\frac{m+1}{n}$ 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.

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