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We want to proove that:
If $p_1,...,p_n$ are positive, distinct, prime numbers then $\sqrt{p_1\cdots p_n} \notin \Bbb{Q}$.
Let's assume that $\sqrt{p_1\cdots p_n} \in \Bbb{Q}$. Then, $\exists (a,b)\in \mathbb{Z^*\times Z^*}:\sqrt{p_1\cdots p_n}=\frac{a}{b}$ with $\gcd (a,b)=1.$ So, $a^2=p_1\cdots p_n \cdot b^2. $ But how do we continue? Is this technique right or should we follow something different?
PS: This is a part this proof, and I would like to discuss it.
Thank you.
The classic proof that $\sqrt 2 \notin \Bbb Q$ readily extends to this case, to wit:
if
$\sqrt{p_1p_2 . . . p_n} = \dfrac{a}{b} \tag 1$
with $a, b \in \Bbb N$, $\gcd(a, b) = 1$, then
$p_1p_2 \ldots p_n b^2 = a^2, \tag 2$
whence $p_1 \mid a^2$; thus $p_1 \mid a$ and so $a = p_1c$; thus
$a^2 = p_1^2c^2 = p_1p_2 \ldots p_n b^2, \tag 3$
whence
$p_1c^2 = p_2p_3 \ldots p_n b^2; \tag 4$
since the $p_i$ are distinct, we must have $p_1 \mid b^2$, whence $p_1 \mid b$, contradicting our assumption that $\gcd(a, b) = 1$.
I think the ancient roots or this proof are worthy or respect, and I also note the method readily extends to show many similar propositions hold.
I presented this answer because it seems to me that the linked proof is pretty complex for this particular problem, although it is certainly engaging in and of itself, and leads in entaging directions.
Hint: think about the prime factorizations of $a$ and $b$, and note that $\gcd(a,b)=1$ implies that they do not share any prime factors. Use this to show that $a^2 = p_1 \cdots p_n b^2$ cannot happen.
If you look at the factorization of $q^2$ for some $q \in \mathbb Q$, you will see that all the prime powers are indeed even. In your case, they're not, hence there is no rational square root.
A ready consequence of Fundamental theorem of arithematic is that A perfect square must have even powers of primes. Since all the primes are coprime with each other so there is no chance for thing in you square root to be a perfect square since every prime will have an odd power $(1)$
A somewhat different result is that if $n$ is not a perfect square then $\sqrt{n}$ is irrational.
Here is one of the many proofs: Follow-up Question: Proof of Irrationality of $\sqrt{3}$
Of course you then have to show that $\prod p_i$ is not a perfect square.
It is worth mentioning Rational root theorem (RRT). From $$\sqrt{\prod\limits_{i=1}^{n} p_i}=x \Rightarrow x^2 - \prod\limits_{i=1}^{n} p_i=0 \tag{1}$$ which is a polynomial with integer coefficients, leading coefficient is $1$, thus $x$, the solution, can only be integer or irrational. Let's assume it's an integer, then (from the same RRT) it should divide the constant term $\prod\limits_{i=1}^{n} p_i$ or $\gcd\left(x,\prod\limits_{i=1}^{n} p_i \right)=x$. Then either $x$ is a prime $p_k$ or divisible by such a prime, in both cases $p_k \mid x$ and $p_k \mid \prod\limits_{i=1}^{n} p_i$. Using Euclid's lemma we will conclude that $p_k \in \{p_1,p_2,...,p_n\}$ and from $(1)$ $$p_k \mid \prod\limits_{i=1,i\ne k}^{n} p_i$$ which, from the same Euclid's lemma, is not possible.
Correct me if wrong:
gcd(a,b)=1:
Assume :
$ a^2= p_1 p_2p_3 .....p_n b^2$.
$\rightarrow$ $p_1$ divides $a^2$;
$\rightarrow$ $p_1$ divides $a,$ by the Theorem below
$ \rightarrow$ $ (p_1)^2$ divides $a^2$.
Likewise
$(p_2)^2,(p_3)^2.,.....,(p_n)^2$ divide $a^2$.
Hence:
$a^2 =$
$(p_1p_2p_3......p_n)^2 q^2 =$
$p_1p_2p_3.....p_n b^2$;
$\rightarrow$ $ b^2 = p_1p_2p_3...p_n q^2$
$\rightarrow$ $p_1,p_2,p_3,....,p_ n$ divide $b^2$;
$\rightarrow$ $p_1,p_2,p_3,....,p_n$ divide $ b$ by the Theorem below.
Hence $p_1,p_2,p_3,.....,p_n$ are common divisors of $a$ and $b$,
in contradiction to the assumption
gcd$(a,b) =1$.
$ $
Used: Theorem of Number Theory:
If $p$ is a prime and $p$ | $ab$, then
$p$ | $a$ or $p$ | $b$.
It follows: If $p$ | $a^2$ then $p$ | $a$.
