Upon square roots' difference

Question

I have to prove this statement: $$\forall a\in\mathbb R\ \exists m,n\in\mathbb N^*:a<\sqrt{m}-\sqrt{n}<a+0.01$$ For example, given $a=0$, $m=50^2+2$ and $n=50^2+1$ satisfy these inequalities since $$m>n\rightarrow \sqrt m-\sqrt n>0$$ $$\sqrt m-\sqrt n=\frac 1{\sqrt m+\sqrt n}<\frac 1{2\sqrt n}<\frac1{100}$$ but the proof for the general situation eludes me... Ideas?

Answer 1

Let's have it in a more convinient form $$a<\sqrt{m}-\sqrt{n}<a+\varepsilon.\tag{1}$$ Then we extend the inequality to $$a-\varepsilon<\sqrt{m}-\sqrt{n}<a+\varepsilon.\tag{2}$$ Indeed, we can always obtain (1)-form from (2)-form with $(a'-\varepsilon',a'+\varepsilon')\to (a,a+\varepsilon)$ (it's a linear transform $a'=\frac\varepsilon 2+a, \,\varepsilon'=\frac\varepsilon 2$).

$$a-\varepsilon+\sqrt{n}<\sqrt{m}<a+\varepsilon+\sqrt{n}\tag{3}$$
$$(a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n<m<(a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n\tag{4}$$ Now we want $$(a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n- \left((a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n\right)>1\tag{6}$$ so there at least one integer $m$ within the interval $$\left( (a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n, (a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n \right)$$ or even ti's better to have at least $2$ integers for the case when one of them is equal to $n$. I'll show for $1$, for $2$ it's quite similar. (6) is equivalent to $$4a\varepsilon+4\varepsilon\sqrt{n}>1\tag{7}$$ $$a+\sqrt{n}>\frac{1}{4\varepsilon}\tag{8}$$ So it's sufficient to take $$n>\left(\frac{1}{4\varepsilon}-a\right)^2\tag{9}$$ Now, we can obtain, for example, an approximation for $\pi$ with $10^{-9}$ precision: $$\sqrt{62500000000000001}-\sqrt{62499998429203684}\approx \color{blue}{3.141592653}7392087789$$

Answer 2

If you accept a non-constructive proof, the following will do, without doing many calculations and the idea is extendable to similar problems:

From your own initial calculation, we see that

$$\forall n \ge N_0=50^2+1: \sqrt{n+1}-\sqrt{n} < 0.01$$

Set $N=\max (N_0,\lceil a^2\rceil)$.

Now consider the sequence

$$x_n:=\sqrt{n}-\sqrt{4N}$$

for $n=1,2,\ldots$.

Obviously $x_n$ is increasing, $\lim_{n\to\infty}x_n=\infty$, and we have $x_N=-\sqrt{N} \le a$. The latter is obvious for nonnegative $a$, for negative $a$ the above choice for $N$ makes sure it is true.

In addition, from above we know that $x_{n+1}-x_n = \sqrt{n+1}-\sqrt{n} < 0.01$ for $n \ge N$.

So considering the sequence $x_n$ from index $N$ onward, it starts out at or below $a$, increases from one index to the next by less then $0.01$ but still tends to infinity. It's clear that one member of the sequence must fall into the interval $(a,a+0.01)$.