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(a,b)=1. So, a2=p1⋯pn⋅b2. 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 √2∉Q readily extends to this case, to wit:
if
√p1p2...pn=ab
with a,b∈N, gcd(a,b)=1, then
p1p2…pnb2=a2,
whence p1∣a2; thus p1∣a and so a=p1c; thus
a2=p21c2=p1p2…pnb2,
whence
p1c2=p2p3…pnb2;
since the pi are distinct, we must have p1∣b2, whence p1∣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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