# Find the smallest positive integer $n$

The smallest positive integer $$n$$ for which the difference $$\sqrt {n} - \sqrt {n-1}$$ becomes less than $$0.01$$

I wanted to check the error:

I made using differentials, which gives an approximation to the root

The answer $$2501$$ is right, but check out this method.
1) $$(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})=n-(n-1)=1$$ by difference of squares.
2) You can't have $$\sqrt{n}+\sqrt{n-1}>100$$ until you have $$\sqrt{n}>50$$. On the other hand you surely have $$\sqrt{n}+\sqrt{n-1}>100$$ when $$\sqrt{n-1}\ge 50$$.
So from (2) the minimum $$n$$ for $$\sqrt{n}+\sqrt{n-1}>100$$ is $$50^2+1=2501$$. And then from (1) that is also the minimum for $$\sqrt{n}-\sqrt{n-1}<0.01$$.