• Nov 20th 2009, 08:49 AM
hebby
Prove by contradiction that 2^(1/3) is irrational.

a^2= 2^(2/3)

then 2^(2/3)= M^2 because we assume a is rational
N^2
then what do I do
• Nov 20th 2009, 08:57 AM
aman_cc
Quote:

Originally Posted by hebby
Prove by contradiction that 2^(1/3) is irrational.

a^2= 2^(2/3)

then 2^(2/3)= M^2 because we assume a is rational
N^2
then what do I do

Do you know the proof of $\sqrt(2)$ is irrational?
Go along same lines.

Let $2^{1/3} = p/q$. Where $(p,q)=1$
Thus $2q^3=p^3$
so $2|p^3$=> $2|p$
Can you complete?
• Nov 20th 2009, 09:05 AM
hebby
Thanks...I can see the proof now...Merci!..it was quite easy