Are the square roots of all non-perfect squares irrational? I was asked to define a non-perfect square. Now obviously, the first definition that comes to mind is a square that has a root that is not an integer. However, in the examples, 0.25 was considered a perfect square. And the square itself + its root were both not integers.
Is it that all non-perfect squares have irrational roots, e.g. $\sqrt{2}$?
 A: It seems you want to extend the term "perfect square" from integers to rationals. then we want it to mean exactly those rationals which are squares of rationals. That is, whose square roots are rational. 
Now we know that a rational number can be written as $\frac p q$ with with $p,q$ coprime (or $p=0, q = 1$). It is easy to check that this is the square of some rational exactly when $p$ and $q$ are both squares of integers. In other words, $p=m^2, q=n^2$ for some integers $m,n$.
A: In the integers, a perfect square is one that has an integral square root, like $0,1,4,9,16,\dots$  The square root of all other positive integers is irrational.  In the rational numbers, a perfect square is one of the form $\frac ab$ in lowest terms where $a$ and $b$ are both perfect squares in the integers.  So $0.25=\frac 14$ is a perfect square in the rationals because both $1$ and $4$ are perfect squares in the integers.  Any rational that has a reduced form where one of the numerator and denominator is not a perfect square in the integers is not a perfect square.  For example, $\frac 12$ is not a perfect square in the rationals.  $1$ is a perfect square in the integers, but $2$ is not, and there is no rational that can be squared to give $\frac 12$
