1. ## Square Roots

Proof that for $n\in Z^+$
Then, $\sqrt{n}$ is rational or irrational.

2. Mmm... There's something odd about this question. Suppose n is rational, then we are done. Suppose n is not rational, then by definition it is irrational. QED?

3. Originally Posted by Treadstone 71
Mmm... There's something odd about this question. Suppose n is rational, then we are done. Suppose n is not rational, then by definition it is irrational. QED?
Probably should read $\sqrt n,\ n \epsilon \mathbb{Z}_+$ is either an integer or irrational.

RonL

4. Ha ha ha I did not realize what I said. CaptainBlack is right either an integer or irrational.

5. It suffices to prove that the square of a non-integer rational number cannot be an integer.

$n,m\in \mathbb{Z}$

Let $\frac{n}{m}$ 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. $\frac{n^2}{m^2}$ is therefore a rational number in its lowest terms with its denominator not equal to 1, and is therefore not an integer. QED.

6. Here I thought of a proof,

If,
$\sqrt{x}=\frac{p}{q}$
then,
$x=\frac{p^2}{q^2}$
Let, $d=\gcd(p,q)$ then $\gcd(\frac{p}{d},\frac{q}{d})=1$
Express $x$ as,
$x=\frac{d^2\left(\frac{p}{d}\right)^2}{d^2\left( \frac{q}{d}\right)^2}$
Thus,
$\left(\frac{q}{d}\right)^2x=\left(\frac{p}{d} \right)^2$
But,
$\gcd(\left(\frac{p}{d}\right)^2,\left(\frac{q}{d} \right)^2)=1$
and,
$\left(\frac{p}{d} \right)^2\left|\left(\frac{q}{d} \right)^2x \right$
Thus,
$\left(\frac{p}{d} \right)^2 \left|x\right$
Thus,
$k\left(\frac{p}{d} \right)^2=x$ for some $k\in\mathbb{Z}$
Substituting that into the equation we have,
$k\left(\frac{q}{d}\right)^2 \left(\frac{p}{d} \right)^2=\left(\frac{p}{d} \right)^2$
Thus,
$k=1$
Thus,
$\left(\frac{p}{d} \right)^2=x$
Thus, $x$ is a square of some integer.
Q.E.D.