Is square root of an integer, either an integer or an irrational number, but never (non-integer) rational? Possible Duplicate:
  $\sqrt a$ is either an integer or an irrational number.  




$\sqrt{2}$ is irrational number, but $\sqrt{9} = 3$ is an integer.
Are there such integers whose square root is a (non-integer)rational number?
Assume $x^2 =\frac{m}{n}$ where $m \in \mathbb Z$ and $ n \in \mathbb N$

this implies $x = \frac{\sqrt{m}}{\sqrt{n}}$

which means that for an integer to have a rational root it's root must be the ratio of two integer roots. For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.
Assume $x^2 =\frac{m}{n}$ where $m \in \mathbb Z$ and $ n \in \mathbb N$

this implies $x = \frac{\sqrt{m}}{\sqrt{n}}$

which means that for an integer to have a rational root it's root must be the ratio of two integer roots. For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.