definitionelementary-number-theoryterminology
According to Wikipedia: …a square number or perfect square is an integer that is the square of an integer…
Is the statement strictly true? And if it is, why are only Integers considered to be square numbers?
For example, if I have a square in the real world, with all sides being 1.5 units. Why is 2.25 not considered a square number? As: $(1.5)^2$ = 2.25
Consider the square root of 2.25. As the result is a Rational number.
$\sqrt{2.25}$ = 1.5
A nice generalization of the fundamental theorem of arithmetic is that every rational number is uniquely represented as a product of primes raised to integer powers. For example:
$$\frac{4}{9} = 2^{2}*3^{-2}$$
This is the natural generalization of factoring integers to rational numbers. Positive powers are part of the numerator, negative powers part of the denominator (since $a^{-b} = \frac{1}{a^b}$).
When you take the $n$th root, you divide each power by $n$:
$$\sqrt[n]{2^{p_2}*3^{p_3}*5^{p_5}...} = 2^{p_2/n}*3^{p_3/n}*5^{p_5/n}...$$
For example:
$$\sqrt{\frac{4}{9}} = 2^{2/2}*3^{-2/2} = \frac{2}{3}$$
In order for the powers to continue being integers when we divide (and thus the result a rational number), they must be multiples of $n$. In the case where $n$ is $2$, that means the numerator and denominator, in their reduced form, are squares. (And for $n=3$, cubes, and so on...)
In your example, when you multiply the numerator and denominator by the same number, they continue to be the same rational number, just represented differently.
$$\frac{2*4}{2*9} = 2^{2+1-1}*3^{-2} = 2^{2}*3^{-2}$$
You correctly recognize the important of factoring, though you don't really want to use it in your answer. But the most natural way to test if the fraction produced by dividing $a$ by $b$ has a rational $n$th root, is to factor $a/b$ and look at the powers. Or, equivalently, reduce the fraction and determine if the numerator and denominator are integers raised to the power of $n$.
We use the complex plane because it provides a very helpful visualization of complex numbers. The reason we can use it is because each complex number $z = a + bi$ is completely determined by the real numbers $a$ and $b$. In a way, thinking about a complex number as a point in $\mathbb{R}^2$ is just a different way of notating that number. There's no danger of confusion or ambiguity, because everyone can agree that a given point $(a,b)$ corresponds to the complex number $a + bi$, and vice versa. This is explained also in Atmos's answer.
I think it's worth mentioning that there is actually more than just this correspondence between the elements of each set, however.
We can also think about different operations that can be done with complex numbers and vectors in $\mathbb{R}^2$, and we can compare these. We know that complex numbers can be added together, as can vectors in $\mathbb{R}^2$. Complex numbers and vectors in $\mathbb{R}^2$ both add in the same way. Compare the following, for $a,b, c, d \in \mathbb{R}$
$$ (a + bi) + (c + di) = (a + c) + (b + d)i \\ (a,b) + (c,d) = (a + c, b + d) $$
You can see that to add two complex numbers you can just think of each complex number as a point in $\mathbb{R}^2$ and add these points as you normally would. The result is the point in $\mathbb{R}^2$ corresponding to the sum of the complex numbers.
We can also scale complex numbers (multiply them by a real constant), and also scale vectors in $\mathbb{R}^2$. Again, this scaling behaves in the same way in $\mathbb{C}$ as it does in $\mathbb{R}^2$. Let $a,b,k \in \mathbb{R}$ and compare again
$$ k(a + bi) = ka + (kb)i \\ k(a,b) = (ka, kb) $$
We see that to scale a complex number by a real constant, we can just think of the complex number as a point in $\mathbb{R}^2$, scale that point as we normally would, and the result is the point in $\mathbb{R}^2$ corresponding to the scaled complex number. If you'd like, the precise way of saying all of this is to say that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic as $\mathbb{R}$-vector spaces. So you can be confident that the things you do with vectors (i.e. scalar multiplication, and vector addition) are the same in $\mathbb{C}$ and $\mathbb{R}^2$, so we are permitted to use this helpful visualization.
Edit: Here's an explanation of some extra behavior exhibited by $\mathbb{C}$ that is not compatible with the behavior of vectors in $\mathbb{R}^2$.
Above I did not explain the whole story of the multiplication operation. I explained how scaling by a real number works, but I did not mention what happens when you scale by an arbitrary complex number. Multiplication of complex numbers has a nice geometric interpretation when thinking of the elements of $\mathbb{C}$ as points on the plane. Each complex number $z = a + bi$ has associated to it a magnitude and an angle as measured counterclockwise from the positive $x$ axis. When multiplying complex numbers $z$ and $w$, the result is a new complex number with magnitude $|z|\cdot|w|$ and an angle that is the sum of the angles of $z$ and $w$. That is, multiplication of complex numbers corresponds to scaling and rotating on the complex plane. I explained a special case of this above when I talked about scaling by a real number. A real number has an angle of zero (it is on the $x$ axis), so multiplying by a real number only scales and does no rotation.
I mentioned that this operation is not compatible with a corresponding operation on $\mathbb{R}^2$. On one hand, this is the case because we are thinking of $\mathbb{R}^2$ as a vector space. And in a vector space, multiplication of two vectors need not be defined. So, this operation of multiplication is some sort of extra property possessed by the complex numbers that is not possessed by the vector space $\mathbb{R}^2$. This doesn't really change anything that I've already said, however. We can still think of complex numbers as points in $\mathbb{R}^2$. We just have to remember this extra rule for what happens to two points when we multiply them together.
And finally, some more detail if you still happen to be interested. We can define multiplication of elements in $\mathbb{R}^2$, and the natural way to do this is to say that for $a,b,c,d \in \mathbb{R}$
$$ (a,b) \cdot (c,d) = (ac, bc) $$
This makes $\mathbb{R}^2$ into a ring. This way of multiplying elements, however, is not the same as the way we multiply elements in $\mathbb{C}$ because in $\mathbb{C}$ we have the relation that $i^2 = -1$, which changes things slightly. The precise way to state this difference is to say that $\mathbb{R}^2$ and $\mathbb{C}$ are not isomorphic as rings. Again though, this often doesn't matter to us, because in the situations where we want to think of $\mathbb{C}$ as $\mathbb{R}^2$ we are thinking of their compatibility as vector spaces, not as rings.
Best Answer
Non-Integer numbers can be considered squares, in certain contexts. Although a square number is usually meant to be an Integer. Especially at elementary level, and when it is stated to be a "perfect square".
There is no rigid source of definitions for math terminology.
These statements are based on my understanding, of the comments by other users, on the question above.