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Thread: Rational and irrational numbers

  1. #1
    Oct 2008

    Rational and irrational numbers

    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.
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  2. #2
    Senior Member
    Nov 2008

    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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