9
$\begingroup$

This is Problem 4.26 on p.58 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).

Prove that $\left|\sqrt2-\dfrac ab\right|\geq\dfrac1{3b^2}$ for all positive integers $a,b$.

Here's what I've done so far. Split into cases:

  • (i) $\dfrac ab<\sqrt2$
  • (ii) $\dfrac ab>\sqrt2+\dfrac13$
  • (iii) $\sqrt2<\dfrac ab<\sqrt2+\dfrac13$.

(i) I followed the method used by the authors in proving a similar result in that chapter. Let $f(x)=x^2-2$; this is the minimum polynomial of $\sqrt2$ in $\mathbb Q$ so $f\left(\sqrt2\right)=0$. By the mean-value theorem: $$\frac{f\left(\sqrt2\right)-f\left(\dfrac ab\right)}{\sqrt2-\dfrac ab}\ =\ f'(\xi)$$ for some $\dfrac ab<\xi<\sqrt2$. We have $f'(\xi)=2\xi<2\sqrt2<3$, and so $$\left|f\left(\frac ab\right)\right|\ <\ 3\left|\sqrt2-\frac ab\right|$$ But $$\left|f\left(\frac ab\right)\right|=\left|\frac{a^2}{b^2}-2\right|=\frac{|a^2-2b^2|}{b^2}\geq\frac1{b^2}$$ Hence $$\left|\sqrt2-\frac ab\right|\ \geq\ \frac1{3b^2}$$ as required.

(ii) Done immediately as $$\left|\sqrt2-\frac ab\right|=\frac ab-\sqrt2\,> \frac13\geq\frac1{3b^2}$$

(iii) This is where I'm stuck.

So I've basically done much of the hard work above and the final piece is all I need to complete the jigsaw. I would be grateful for some help. Thanks.

$\endgroup$
3

3 Answers 3

3
$\begingroup$

Do the case distinction with $1/12$ instead of $1/3$. This does not affect (i) and (iii) works in the same way as (i), since you still have $2 \xi < 3$ as $2(\sqrt{2} +1/12)<3$.

Of course, you need to reconsider (ii) then. But it is hardly more difficult, as you have $\frac{a}{b} - \sqrt{2} \ge \frac{1}{12}$ and $\frac{1}{12} = \frac{1}{3 \cdot 4^2} \ge \frac{1}{3b^2}$, except if $b=1$. Yet then if $b=1$, you have that $a/b$ is an integer so that in this case $a/b \ge 2$ and the difference is at least $1/3$ in this case too.

There may be a shorter method, but this was what came to mind.

$\endgroup$
2
$\begingroup$

Prove that $$\left|\sqrt2-\dfrac ab\right|>\dfrac1{3b^2}$$ for integers $a$ and $b$ with $b\neq 0$.

Without loss of generality, assume that $b>0$. Note that $$\left|\sqrt{2}-\frac{a}{b}\right|=\frac{\left|a^2-2b^2\right|}{b(a+\sqrt{2}b)}\,.\tag{*}$$ If $a<0$, then $$\left|\sqrt{2}-\frac{a}{b}\right|>\sqrt{2}>\frac{1}{3b^2}\,.$$ If $0\leq a\leq b$, then, using (*), $$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{b^2}{b(b+\sqrt{2}b)}\geq \frac{1}{1+\sqrt{2}}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$ If $a\geq \frac{3}{2}b$ and $b>1$, then $$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{3}{2}-\sqrt{2}>\frac{1}{12}\geq\frac{1}{{3b^2}}\,.$$ If $a>1$ and $b=1$, then $$\left|\sqrt{2}-\frac{a}{b}\right|\geq 2-\sqrt{2}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$ If $b<a<\frac{3}{2}b$, then, using (*), $$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{b\left(a+\sqrt{2}b\right)}>\frac{1}{b\left(\frac{3}{2}b+\sqrt{2}b\right)}=\frac{1}{\left(\frac{3}{2}+\sqrt{2}\right)b^2}>\frac{1}{3b^2}\,.$$

Indeed, the smallest constant $\gamma>0$ such that $$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{\gamma\,b^2}$$ for all integers $a$ and $b$ with $b\neq 0$ is $\gamma=\frac{3}{2}+\sqrt{2}$. The equality holds if and only if $a=3$ and $b=2$.


Related Observation

Conjecture. Let $\gamma_b:=\max\,\left\{\frac{1}{b^2\,\left|\sqrt{2}-\frac{a}{b}\right|}\,\Big|\,a\in\mathbb{Z}\right\}$ for each $b\in\mathbb{N}$. Define $$\Gamma:=\liminf_{b\to\infty}\,b\cdot\gamma_b\,.$$ Then, $$\Gamma=\inf\,\left\{b\cdot\gamma_b\,\big|\,b\in\mathbb{N}\right\}=2\,.$$

After checking all positive integers $b\leq 1500$, I find that $$\Gamma\leq 1+\frac{29}{41}\sqrt{2}\lesssim 2.0003\,.$$ Below is a plot illustrating the credibility of this conjecture. Th horizontal axis is $b$. The fuzzy blue line shows $\gamma_b$ versus $b$, and the nice orange line is $\dfrac{2}{b}$ versus $b$ for $b=1,2,\ldots,1500$.

Plot describing $\gamma_b$ (blue) and $\frac{2}{b}$ (orange) versus $b$ for $b=1,2,\ldots,1500$.

What is very strange is $\Gamma$ seems to be equal to $2$ as well if $\sqrt{2}$ is replaced by $\sqrt{3}$, $\sqrt{5}$, $\sqrt{6}$, $\sqrt{7}$, and $\sqrt{8}$. Could $\Gamma=2$ universally hold when $\sqrt{d}$ replaces $\sqrt{2}$ for any non-square positive integer $d$? I tried replacing $\sqrt{2}$ by $\frac{1+\sqrt{5}}{2}$, $\sqrt[3]{2}$, $\pi$, $\text{e}$, and $\ln(3)$ as well, and it looks like $\Gamma=2$ still holds.

On the other hand, $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{2}\,.$$ If $\sqrt{2}$ is replaced by $\sqrt{d}$ for any non-square positive integer $d$, then $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{d}\,.$$

$\endgroup$
6
  • 1
    $\begingroup$ The notation is a bit hard to untangle for me, but fwiw it is not hard to see that every number can be approximated to $\frac{1}{2b}$ by a rational of the form $\frac{a}{b}$, as $\frac{1}{b}\mathbb{Z} + [-1/2b,1/2b]$ covers the real line, which I believe gives the $\Gamma$. $\endgroup$
    – quid
    Commented Aug 22, 2016 at 16:29
  • $\begingroup$ @quid I think you are right. I didn't try hard enough to untangle my notations (as they came directly from the code I used to experiment). The answer seems to be as simple as you have noted. $\endgroup$ Commented Aug 22, 2016 at 16:33
  • 1
    $\begingroup$ Or, it's in the other direction and the worst case, in which case my argument above is only a plausibility argument (which one could likely turn into an actual one), in that for some $b$ the $\sqrt{2}$ will be very close to a number of the form $\frac{a+1/2}{b}$ and the approx is only marginally better than $1/2b$ for that $b$. I will stop guessing around now. I should have checked before commenting. :-) $\endgroup$
    – quid
    Commented Aug 22, 2016 at 16:39
  • $\begingroup$ The last statement seemingly can be proven from my answer since there are infinitely many $(a,b)$ : $|a^2-db^2|=1$ and thus $|\sqrt d - {a \over b}| < {1 \over kb^2}$ is equivalent to $k < \sqrt d + \sqrt{d + {1 \over b^2}}$ for arbitrarily large $b$. $\endgroup$ Commented Aug 23, 2016 at 15:59
  • 1
    $\begingroup$ Yes, and it has a more general form: if $r\in\mathbb{R}$ is algebraic over $\mathbb{Q}$ whose (monic) minimal polynomial $f(X)\in\mathbb{Q}[X]$ is of degree $n$, then there exists $K>0$ such that $$\left|r-\frac{a}{b}\right|\geq \frac{1}{K\,b^n}$$ holds for all $a,b\in\mathbb{Z}$ with $b>0$. $\endgroup$ Commented Aug 23, 2016 at 16:07
0
$\begingroup$

Here is the alternate way, without case analysis.

Suppose that for some $a,b$, $|\sqrt 2 - {a \over b}| < {1 \over kb^2}$. Rewriting the inequality, $$-{1 \over kb^2} < {a \over b} - \sqrt 2 < {1 \over kb^2}\\ 2-{2\sqrt 2 \over kb^2} + {1 \over k^2b^4} < {a^2 \over b^2} < 2 + {2\sqrt 2 \over kb^2} + {1 \over k^2b^4} \\ a^2 = 2b^2 + {1 \over k^2b^2} + {\delta \over k}$$ where $|\delta| < 2 \sqrt 2$. Also, from $\sqrt 2$ being irrational we have $$|a^2 - 2b^2| \ge 1 \\ |{1 \over k^2b^2} + {\delta \over k}| \ge 1 \\ |\delta| \ge k - {1 \over kb^2} \\ k - {1 \over kb^2} < 2 \sqrt2 \\ k < \sqrt2 + \sqrt{2 + {1 \over b^2}}$$ Finally for $k=3$ $b=1$ can be excluded separately and $\sqrt 2 + \sqrt{2 + {1 \over 4}} < 3.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .