Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y<z<x$.

ban1ka1u
2022-07-02
Answered

Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y<z<x$.

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furniranizq

Answered 2022-07-03
Author has **20** answers

$x-y>0$ so $(x-y{)}^{2}>0$. By the Archimedean property, there is a positive integer $m$ such that $\frac{1}{m}<(x-y{)}^{2}$. By increasing $m$ if necessary, you can assume that $m=2{n}^{2}$ so that $\frac{1}{2{n}^{2}}<(x-y{)}^{2}$. This means

$\frac{1}{\sqrt{2}n}<x-y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}1+\sqrt{2}ny<\sqrt{2}nx.$

Now let $k$ be the largest integer that is less than or equal to $\sqrt{2}ny$. This implies

$1+k\le 1+\sqrt{2}ny<\sqrt{2}nx.$

But we also have $\sqrt{2}ny<k+1$, so

$\sqrt{2}ny<k+1<\sqrt{2}nx\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y<\frac{k+1}{\sqrt{2}n}<x.$

$\frac{1}{\sqrt{2}n}<x-y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}1+\sqrt{2}ny<\sqrt{2}nx.$

Now let $k$ be the largest integer that is less than or equal to $\sqrt{2}ny$. This implies

$1+k\le 1+\sqrt{2}ny<\sqrt{2}nx.$

But we also have $\sqrt{2}ny<k+1$, so

$\sqrt{2}ny<k+1<\sqrt{2}nx\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y<\frac{k+1}{\sqrt{2}n}<x.$

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