Is there a formula to find a rational number between $x$ (where $x$ is rational and $x^2>2$) and $\sqrt{2}$? I'm not looking for an existence proof, but a formula that can give the result.
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3 Answers
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Let $x$ be a positive rational with $x^2>2$. Then $(2/x)^2<2$ and the average $y=(1/2)(x+2/x)$ of $x$ and $2/x$ should be nearer $\sqrt2$ than either. By AM/GM actually $y>\sqrt2$. As $2/x<x$ then $y<x$ too.
answered May 3, 2018 at 17:56
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Take $n\in\mathbb N$ such that $\left(x-\frac1n\right)^2>2$. Such a $n$ exists, since$$\left(x-\frac1n\right)^2>2\iff x^2-\frac2n+\frac1{n^2}>2\iff x^2-2>\frac2n-\frac1{n^2}.$$Then, take the number $x-\frac1n$.
answered May 3, 2018 at 18:01
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just take $x-f(x)$ where $f:(\sqrt{2},\infty)\rightarrow \mathbb R$ is a function that is small and always rational if $x$ is rational.
How small? Smaller than $(x-\sqrt{2})$
We propose $f(x)=\frac{(-(x-\sqrt2)(x+\sqrt2))}{(x+2)\times(\text{favorite big number that makes us confident)}}$