We want to proove that:
If p1,...,pn are positive, distinct, prime numbers then
√p1⋯pn∉Q.
Let's assume that √p1⋯pn∈Q. Then, ∃(a,b)∈Z∗×Z∗:√p1⋯pn=ab with gcd 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.
Answer
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.
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