This was insipired by Treadstone71.
Proof that for $\displaystyle n\in Z^+$
Then, $\displaystyle \sqrt{n}$ is rational or irrational.
It suffices to prove that the square of a non-integer rational number cannot be an integer.
$\displaystyle n,m\in \mathbb{Z}$
Let $\displaystyle \frac{n}{m}$ non-integer and let it be in its lowest terms. That is, the intersection of the set of prime factors of n and m is empty. Let n/m be squared. However, since the set of prime factors of n and m did not change, the interesction is still empty, and therefore no cancellation can be made. $\displaystyle \frac{n^2}{m^2}$ is therefore a rational number in its lowest terms with its denominator not equal to 1, and is therefore not an integer. QED.
Here I thought of a proof,
If,
$\displaystyle \sqrt{x}=\frac{p}{q}$
then,
$\displaystyle x=\frac{p^2}{q^2}$
Let, $\displaystyle d=\gcd(p,q)$ then $\displaystyle \gcd(\frac{p}{d},\frac{q}{d})=1$
Express $\displaystyle x$ as,
$\displaystyle x=\frac{d^2\left(\frac{p}{d}\right)^2}{d^2\left( \frac{q}{d}\right)^2}$
Thus,
$\displaystyle \left(\frac{q}{d}\right)^2x=\left(\frac{p}{d} \right)^2$
But,
$\displaystyle \gcd(\left(\frac{p}{d}\right)^2,\left(\frac{q}{d} \right)^2)=1$
and,
$\displaystyle \left(\frac{p}{d} \right)^2\left|\left(\frac{q}{d} \right)^2x \right$
Thus,
$\displaystyle \left(\frac{p}{d} \right)^2 \left|x\right$
Thus,
$\displaystyle k\left(\frac{p}{d} \right)^2=x$ for some $\displaystyle k\in\mathbb{Z}$
Substituting that into the equation we have,
$\displaystyle k\left(\frac{q}{d}\right)^2 \left(\frac{p}{d} \right)^2=\left(\frac{p}{d} \right)^2$
Thus,
$\displaystyle k=1$
Thus,
$\displaystyle \left(\frac{p}{d} \right)^2=x$
Thus, $\displaystyle x$ is a square of some integer.
Q.E.D.