# Prove that if $p_1,…,p_k$ are distinct prime numbers, then $\sqrt{p_1p_2…p_k}$ is irrational

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.

• Can you write up what you have done? Esp for the $\sqrt{p_1}$ case. The general case is very similar. – Calvin Lin Jun 29 '20 at 22:22
• it's exactly the same as in the case of two prime numbers that you mention. – alphaomega Jun 29 '20 at 22:24
• Just edited to show what I did so far. I'm not so sure that's correct. – Mauricio Mendes Jun 29 '20 at 22:38

## 3 Answers

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

First, I will prove that if $$\sqrt{p_1p_2...p_k}$$ is a rational, then it must be an integer. Suppose there are $$p, q\in\mathbb{N}$$ with $$\text{gcd}(p,q)=1$$ such that $$\dfrac{p}{q}=\sqrt{p_1p_2...p_k}.$$ Using the Euclid algorithm, we can find two (relatively prime) integers $$a, b$$ such that $$ap+bq=1.$$ Now consider $$0=(p-q\sqrt{p_1p_2...p_k})(b-a\sqrt{p_1p_2...p_k})=bp-(ap+bq)\sqrt{p_1p_2...p_k}+aq(p_1p_2...p_k).$$ Hence $$\sqrt{p_1p_2...p_k}=bp+aq(p_1p_2...p_k)\in\mathbb{Z}.$$

Hence $$p^2=p_1p_2...p_k.$$ Now from elementary number theory, you can argue that $$p_1$$ s a prime factor of $$p$$ and, it implies that $$p_1$$ is also a prime factor of $$p_2p_3...p_k,$$ which is a contradiction.

• Note that, this proof does not use any restriction on $k.$ In particular it can be used to derive $\sqrt{p}$ is irrational for any prime $p.$ – Bumblebee Jun 29 '20 at 23:27

"and tried to repeat that argument over and over"

that's all you have to do but you only have to do the argument once.

Let $$a,b$$ be integers where $$(\frac ab)^2 = p_1.....p_n$$ so

$$a^2 = bp_1......p_n$$. So for any of those prime $$p_i$$ (it doesn't matter which one) then $$p_i|a^2$$ and by Euclid's Lemma[*] the $$p_i|a$$ so $$p_i^2|a^2$$ and so $$a*\frac {a}{p_i} = b\frac {p_1..... p_n}{p_i}$$

For notation let $$\frac a{p_i} = a'$$ and let $$\frac {p_1.....p_n}{p_i} = P= p_1p_2...p_{i-1}p_{i+1}....p_n = \prod_{j=1;j\ne i}^n p_i$$.

so $$aa' = p_ia'^2 = bP$$. So $$p_i|bP$$.

So by Euclid's Lemma either $$p_i|b$$ or $$p_i|P$$. But $$P = \prod_{j=1;j\ne i}^n p_i$$ is not divisible by $$p_i$$. So $$p_i|b$$.

$$p_i|a$$ and $$p_i|b$$ so $$\frac ab$$ where not in lowest terms.

Now we could argue that we only claimed $$a,b$$ where integer, we never said they had to be in lowest terms. But we can argue if $$a,b$$ exist then we can repeat this infinitely (as $$p_i|b$$ then if $$\frac b{p_i} = b'$$ then we have $$(\frac {a'}{b'})^2 = p_1..... p_n$$ so we can repeat over and over.) This means we have an infinite series of $$a = p_ia'=p_i^2a_2=p_i^2a_3 =..... =p_k^2a_k=...$$. But as $$p_i > 1$$ this mean $$a > a' >a_2>a_3>.....$$. That's clearly impossible as there are only a finite number of natural numbers less than $$a$$[**].

So we have proven the result.

[*] Euclid's lemma: If $$p$$ is prime and $$p|ab$$ for integers $$a,b$$ then either $$p|a$$ or $$p|b$$ (or both).

Everything hinges on that. (Which presumably you have proven one way or another already.

[**] This is the well-ordered principal. Every set of natural numbers must have a minimal element. Sa $$a> a'> a_2 > a_3.....$$ then the set $$\{a,a', a_i\}$$ must have a smallest $$a_k$$ which means it only has $$k$$ elements and eventually we must get $$\frac ab$$ in lowest terms.