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

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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

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 $m,n$.

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Sorry, the answer is pretty complicated, is it possible for you to elaborate? I am still in grade school! –  user115422 Aug 16 '13 at 4:35
This answer says that every fraction which is a perfect square must have a numerator which is a perfect square and a denominator which is a perfect square. It also says that each fraction has been reduced so that there are no common factors which can be divided out from the numerator and denominator. –  abiessu Aug 16 '13 at 4:41
A number like 100 is a perfect square because its square root is an integer, 10. A number like .25 = 1/4 is a perfect square because its square root is 1/2 which is rational. You can find a rational square root of a number like 4/9, because it will be 2/3. But you can't find a rational square root of 63 because it is not the square of any integer; or of 5/9 because the 5 is not the square of any integer. So 63 and 5/9 are not perfect squares. It is true that their roots will be irrational. –  Betty Mock Aug 16 '13 at 4:42
@abiessu thanks, that makes sense –  user115422 Aug 16 '13 at 4:42