Prove that for x3=3 there isn't rational solution
What I did:
Suppose x=u/v is solution
(uv)3=3
Let's take third root from both sides:
uv=3√3
and 3√3 is irrationl ,
my problem is this "and 3√3 is irrationl " is it well known that this number is irrationl? same as π?
or maybe there is another why to prove it?
Answer
Just start off my assuming (u,v)=1 i.e the fraction is fully reduced. Then it follows that;
u3=3v3⇒3∣u3⇒3∣u⇒u=3M
Therefore (3M)3=27M3=3v3 and so 9∣v3 which implies 3∣v3 and so (v,u)≠1 which is a contradiction.
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