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Let $p\in \mathbb{Q}$ such that $p^2<2$. Find $\varepsilon>0$ such that $p+\varepsilon\in \mathbb{Q}$ and $(p+\varepsilon)^2<2$.

I need the explicit form of $\varepsilon$. Can anyone show how to find $\varepsilon$?

RFZ
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    What have you tried? Do you know any classical methods for getting approximate values for square roots? – lulu Jul 08 '15 at 17:04
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    $(p+\epsilon)^2-2=(p^2-2)+2p\epsilon+\epsilon^2$. – David C. Ullrich Jul 08 '15 at 17:06
  • I wrote the expression in this form but what it gives? I can't understand – RFZ Jul 08 '15 at 17:10
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    Take a look at, say, the Babylonian method for getting square roots: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots. (special case of Newton's method). That gives you a good way to improve any estimate you have and it happens that if your estimate is rational, so is the better approximation produced by the method. Warning: on its own the method will give you a value greater than #\sqrt{2} but it is not difficult to get what you want out of it. – lulu Jul 08 '15 at 17:14
  • It's all great but I thought that explicit form of $\varepsilon$ can be found more simple. Maybe somebody know this. – RFZ Jul 08 '15 at 17:26
  • @Pacman I'm pretty sure David is suggesting that you now use the quadratic formula, solve $$A\varepsilon ^2+B\varepsilon+C<0$$ where $A = 1$, $B = 2p$ and $C = p^2-2$. This will yield two solutions for $\varepsilon$, so take the one that is negative (if it exists) – graydad Jul 08 '15 at 18:09
  • Any choice of rational $\epsilon \in (0, \sqrt2-|p|)$ will work. – Macavity Jul 08 '15 at 18:11
  • cf. this question – J. W. Tanner May 05 '21 at 02:35

2 Answers2

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If $p<0$ then we can take $\varepsilon = -p >0$. So let's assume that $p\ge0$.

$(p + \varepsilon) ^ 2 = p^2 + 2p\varepsilon +\varepsilon^2$ and so it is enough to take $\varepsilon$ such that $p^2 + 2p\varepsilon +\varepsilon^2< 2$.

We shall look for $\varepsilon < 1$. Then $\varepsilon^2 < \varepsilon$ and so it is enough to take $\varepsilon$ such that $p^2 + 2p\varepsilon +\varepsilon\le 2$. The largest solution is $$ \varepsilon = \frac{2-p^2}{2p+1} $$ If $p$ is rational, then so is $\varepsilon$. Moreover, $p^2 < 2$ implies $\varepsilon >0$.

However, this $\varepsilon$ may not satisfy $\varepsilon <1$. So we take $$ \varepsilon = \min\big(1,\frac{2-p^2}{2p+1}\big) $$

lhf
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  • How you got $\varepsilon=\dfrac{2-p^2}{2p+1}$? – RFZ Jul 08 '15 at 18:23
  • Why the cases $p<0$ and $p\geqslant 0$ are so important? Where did you use them? – RFZ Jul 08 '15 at 18:29
  • See also http://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1. – lhf Jul 08 '15 at 18:35
  • Dear lhf, why the cases $p<0$ and $p\geqslant 0$ are so important? I can't find this in source that you gave – RFZ Jul 08 '15 at 18:44
  • @Pacman, they simplify the analysis. – lhf Jul 08 '15 at 18:44
  • Dear lhf. If $\varepsilon=\dfrac{2-p^2}{2p+1}$ then how to prove that $(p+\varepsilon)^2<2$. Your post is very hard for me and I can't understand some moments :( – RFZ Jul 09 '15 at 07:12
  • @Pacman, the last value I gave is the one that works. – lhf Jul 09 '15 at 11:43
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Let

$$\epsilon = \frac{1}{\lceil \frac{1}{\sqrt{2} - \lvert p \rvert} \rceil + 1} \in \mathbb{Q}$$

Because

$$0 < \frac{1}{\sqrt{2} - \lvert p \rvert} < \lceil \frac{1}{\sqrt{2} - \lvert p \rvert} \rceil + 1$$

We have

$$0 <\epsilon < \sqrt{2} - \lvert p \rvert \leq \sqrt{2} - p $$ $$ -\sqrt{2} < p + \epsilon< \sqrt{2}$$ $$(p + \epsilon) ^ 2 < 2$$

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