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

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I wanted to check the error:

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

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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 $\sqrt{n}-\sqrt{n-1}<0.01$.

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