# Find the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}\leq \frac{1}{100}$

Find the smallest positive integer $n$ such that $\sqrt n-\sqrt{n-1}\le \frac{1}{100}$.

First I multiplied by the conjugate and got $$\frac 1{\sqrt n + \sqrt{n-1}} \le \frac{1}{100},$$ or $\sqrt n+ \sqrt{n-1} \ge 100$. Now I squared both sides: $$2n-1+2 \sqrt n \sqrt{n-1}\ge 100.$$ So $$\sqrt n (\sqrt n +\sqrt{n-1})\ge \frac {101}2.$$ However, we know that $\sqrt n+\sqrt{n-1}\ge 100$, so I substituted that: $$\sqrt n (100) \ge \frac {101}{2}.$$ However, if I solve for $n$ I get around $.25$, and there is no lower positive integer. What did I do wrong? Thanks!

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The latex commands "\leq" for $\leq$ and "\geq" for $\geq$ will do the trick. – ThisIsNotAnId Jun 2 '13 at 5:25
Yea thanks I really need to learn latex – Ovi Jun 2 '13 at 5:27
Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. You can also see how I've edited your post here. – Zev Chonoles Jun 2 '13 at 5:27
Also, please choose more descriptive, informative titles in the future. – Zev Chonoles Jun 2 '13 at 5:28
When you square both sides it is useful to square both sides. – André Nicolas Jun 2 '13 at 5:29

Let us stop temporarily at $$\sqrt{n}+\sqrt{n-1}\ge 100.\tag{1}$$ Maybe the algebraic manipulations can stop here. Note that (1) will certainly be true if $2\sqrt{n-1}\ge 100$, that is, if $n-1 \ge 2500$, But it is conceivable that (1) also holds at $n=2500$. It doesn't.

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Thanks. And by the way, Ted said in the comments that my substitution was not valid. What do you think about that? I think it's valid since we want to minimize $n$, meaning we want to minimize $sqrt(n)+sqrt(n-1)$, so it makes sense to substitute in the lowest possible value of $sqrt(n)+sqrt(n-1)$ – Ovi Jun 2 '13 at 5:43
@Ovi: Sure, you are right. The inequality you started with is equivalent to the more tractable $\sqrt{n}+\sqrt{n-1}\ge 100$. – André Nicolas Jun 2 '13 at 5:48
Ok thank you very much! – Ovi Jun 2 '13 at 5:49

$\sqrt n-\sqrt{n-1}\le\frac1{100}\implies \sqrt{n-1}\ge \sqrt n-\frac1{100}$

$$\text{On Squaring , }n-1\ge n+\frac1{100^2}-\frac{2\sqrt n}{100}$$

$$\implies \frac {\sqrt n}{50}\ge\frac1{100^2}+1$$

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