Prove by contradiction that 2^(1/3) is irrational.
I made a=2^(1/3).
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 $\displaystyle \sqrt(2)$ is irrational?
Go along same lines.
Let $\displaystyle 2^{1/3} = p/q$. Where $\displaystyle (p,q)=1$
Thus $\displaystyle 2q^3=p^3$
so $\displaystyle 2|p^3 $=> $\displaystyle 2|p $
Can you complete?