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Show that there is a rational number and an irrational number between any two real numbers.

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marked as duplicate by Kugelblitz, Fabio Somenzi, Antonios-Alexandros Robotis, Shailesh, Leucippus May 2 '17 at 3:41

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    $\begingroup$ Are you familiar with the archimedian principal? $\endgroup$ – fleablood May 2 '17 at 2:45
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    $\begingroup$ @fleablood: "principle" ;) To the OP: the answer to this question will depend heavily on what properties of the reals you are taking as given. $\endgroup$ – Martin Argerami May 2 '17 at 2:47
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We know that $\mathbb{Q}$ is $dense$ in $\mathbb{R}$. This means for any $x\in\mathbb{R}$ and for any $r>0$ , $\mathbb{Q} \cap B(x;r)\not=\phi $ where $B(x;r)=\lbrace y\in\mathbb{R}:|x-y|<r \rbrace$.

Let $a, b\in\mathbb{R}$ be such that $a<b$. Then we have $\mathbb{Q}\cap B(\frac{a+b}{2};\frac{b-a}{3})\not=\phi$ by above statement. This simply means there is at least one rational number between $a$ and $b$.

Similarly, $\mathbb{R-Q}$ is also $dense$ in $\mathbb{R}$. By the similar process as above , one can easily derive that there is at least one irrational number between any two distinct real numbers.

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  • $\begingroup$ Note that, usually, the way to show that $\mathbb Q$ is dense is precisely to show that between any two reals there is a rational. It depends on how one defines/constructs the reals. $\endgroup$ – Martin Argerami May 2 '17 at 4:18

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