# Are the results of all non-perfect squares irrational? [duplicate]

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)?

Thanks!

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## marked as duplicate by Zev Chonoles, Sasha, William, iostream007, mrfAug 16 '13 at 5:53

By definition, yes. –  MathApprentice Aug 16 '13 at 4:31
A non-perfect square is a number such that there is no rational number p/q such that (p/q)^2 = n (where n is a perfect square). –  MathApprentice Aug 16 '13 at 4:33
@ronno, both can be expressed using fractions so it's whether a non perfect square, is by definition, not able to be expressed as a fraction. –  user115422 Aug 16 '13 at 4:33

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 $\frac 12$

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Thanks, that makes sense. Thanks for explaining it out in detail! I'll mark it as answer. –  user115422 Aug 16 '13 at 4:43
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 $m,n$.