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.