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!

$\sqrt n-\sqrt{n-1}\le\frac1{100}\implies \sqrt{n-1}\ge \sqrt n-\frac1{100}$
$$\text{On Squaring , }n-1\ge n+\frac1{100^2}-\frac{2\sqrt n}{100}$$
$$\implies \frac {\sqrt n}{50}\ge\frac1{100^2}+1$$
