For $n \ge 1$, $\sqrt{n^{2} + 1}$ lies between

For $n \ge 1$, $\sqrt{n^{2} + 1}$ lies between
  1. $n$ and $n+1$
  2. $n-1$ and $n$
  3. $n+1$ and $n+2$
  4. $n+0.5$ and $n+1.5$

Solution

$n^{2} < n^{2} + 1 < (n+1)^{2} = n^{2} + 2n + 1$ for $n \ge 1$, so $n < \sqrt{n^{2}+1} < n + 1$.

Asked in: IMO

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