1. The problem asks about the nature of $\sqrt{n}$ when $n$ is not a perfect square. 2. A perfect square is an integer that can be expressed as $k^2$ where $k$ is an integer. 3. If $n$ is not a perfect square, then $\sqrt{n}$ cannot be an integer because integers squared give perfect squares. 4. $\sqrt{n}$ is also not a natural number for the same reason, since natural numbers are positive integers. 5. $\sqrt{n}$ cannot be rational unless $n$ is a perfect square, because the square root of a non-perfect square integer is irrational. 6. Therefore, if $n$ is not a perfect square, $\sqrt{n}$ is irrational. 7. In summary, for $n$ not a perfect square, $\sqrt{n}$ is irrational.