Prove that if $p_1,…,p_k$ are distinct prime numbers, then $\sqrt{p_1p_2…p_k}$ is irrational.
I do not usually prove theorems, so any hint is appreciated. I have taken a look at this and tried to repeat that argument over and over, but I messed up. Perhaps there is an easier to do it. Thanks in advance.
Edit: For the case $k=1$, suppose that $\sqrt{p}=\frac{m}{n}, n\neq0$ and $gcd(m,n)=1$. It follows that $m^2=pn^2$, so $p\mid m$. Also, $p\mid n$. Therefore, $m$ and $n$ are not relatively prime. Contradiction.
Best Answer
Assume ${p_1,...,p_k}$ are distinct primes, and assume (aiming for a contradiction) that ${\sqrt{p_1p_2...p_k}=\frac{a}{b}}$ for coprime positive integers ${a,b}$ (alternatively, you can write ${(a,b)=1}$). As before, squaring both sides and rearranging for ${a^2}$ yields
$${\Rightarrow a^2 = p_1...p_kb^2}$$
In other words,${a^2}$ contains ${p_1,...,p_k}$ as factors, and thus ${a}$ must contain ${p_1,...,p_k}$ as factors (since ${a^2}$ contains these primes as factors, this has to be the case because they are prime. This wouldn't be true for some random composite number).
Anyways, we rewrite ${a=p_1...p_ka^*}$. Plugging back in gives
$${\Rightarrow \frac{p_1...p_ka^*}{b}=\sqrt{p_1...p_k}}$$
And this implies
$${\Rightarrow \frac{p_1^2...p_k^2\left(a^*\right)^2}{b^2}=p_1...p_k}$$
You can rearrange this and get
$${b^2 = p_1...p_k\left(a^*\right)^2}$$
And we have got our desired contradiction. By the same argument as before, this would tell us ${b^2}$ has factors ${p_1...p_k}$, and because again these are primes that means ${b}$ contains factors ${p_1...p_k}$. This is a contradiction since we assumed that ${(a,b)=1}$ (that ${a,b}$ were coprime, hence could not share any common factors), and yet from assuming the rationality of our expression we have shown that ${a,b}$ both contained ${p_1,...,p_k}$ as factors!
QED. Or Quantum Electrodynamics if you are a Physicist :P