Proof that the square root of a non-perfect square is irrational Is the reasoning for this proof logically sound?
To show that the square roots of all non perfect squares are irrational, we must first know the definitions of rationality and perfect square.
A number n is said to be rational if it can be expressed as a ratio of two integers, a and b, such that a and b are a ratio of two integers in lowest terms, written as $\frac{a}{b}$. This means that the two integers must be co prime. It's worth noting that for any integer i, i is rational, as it can always be expressed as $\frac{k}{1}$ for k = i. 
A number n is said to be a perfect square, if for some integer k $\in \mathbb{Z}$, the number can be expressed such that $n = k^2$. 
Suppose we have a number n, and n is not a perfect square, but its square root is rational. Since n is not a perfect square, it cannot be expressed as $k^2$ for any integer $k \in \mathbb{Z}$. However, since $\sqrt n$ is rational, it can be expressed as $\frac{a}{b}$ for some integers a and b. 
Using this, we get $\sqrt n = \frac{a}{b}$, meaning $n = (\frac{a}{b})^2$, however this is a contradiction since we have already defined n to be a non perfect square, and thus not able to be expressed as $k^2$ for any $k \in \mathbb{Z}$. This means there is no non perfect square n that exists such that the expression $\frac{a}{b}$ in lowest terms is valid.
Therefore, the square roots of all non perfect squares are irrational.
 A: What you wrote is a "non-proof".
You say that $n=\dfrac{a^2}{b^2}$ is not possible because $n=k^2$ is not possible. Doing this, you just assume $\dfrac{a}{b}$ to be an integer, which make the whole discussion vacuous.

Consider $\sqrt2$. It is obvious that it is not a perfect square because $1^2<2$ and $2^2>2$. But what about
$$\left(\frac{577}{408}\right)^2=2.00000\cdots\ ?$$

The classical proof relies on the identity $a^2=nb^2$, showing that $a^2$ is a multiple of $b$, so that $a$ has a non-trivial factor in common with $b$.
A: //Redone Proof Based On Feedback//
Proof that the square roots of all integers that are not perfect squares are irrational.
To show that the square roots of all integers that are non perfect squares are irrational, we must first know the definitions of rationality and perfect square.
A number n is said to be rational if it can be expressed as a ratio of two integers, a and b, such that a and b are a ratio of two integers in lowest terms, written as ab. Additionally, the two integers must be co prime, meaning the only common factor between them must be the trivial factor  1. It's worth noting that for any integer $i \in \mathbb{Z}$, $i$ is rational, as it can always be expressed as $\frac{k}{1}$ for $k = i$.
An integer $n$ is said to be a perfect square, if for some integer $k \in \mathbb{Z}$, the number can be expressed such that $n=k^2$.
Suppose we have an integer n, and n is not a perfect square, but its square root is rational. Since n is not a perfect square, it cannot be expressed as $k^2$ for any integer $k \in \mathbb{Z}$. However, since $\sqrt n$ is rational, it can be expressed as $\frac{a}{b}$ for some integers $a$ and $b$.
Using this, we get $\sqrt{n} = \frac{a}{b}$, $n = \frac{a^2}{b^2}$, and $nb^2 = a^2$. However, if $a^2 = nb^2$, then $a$ and $b$ cannot be co prime, as $a$ has a non trivial factor in common with $b$.
Therefore, the square roots of all integers that are not perfect squares are irrational.
