Inequalities

Re-arranging the equation gives $\sqrt{n}-0.01 < \sqrt{n-1}$ , then squared to $n-0.02\sqrt{n}+0.0001 < n-1$ $-2\sqrt{n} < -100.01$ $n > (50.005)^{2}$ $n > 2,500.500025$ Therefore the smallest positive integer is $n=2501$

Hint: Observe that $\sqrt{n}-\sqrt{n-1}=\large \frac{1}{\sqrt{n}+\sqrt{n+1}}$ .

• Press F12 to open your browser's console.
• Paste the following snippet into the console:

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let n=1;
while( Math.sqrt(n) - Math.sqrt(n++-1) >= .01 ){}
console.log(n-1);

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2501

same method