Prove (2^(1/2)) + (2^(1/3)) is irrational

This is what I have

2^(1/2) + 2^(1/3) = r with r being a rational number

2^(1/2) = r - 2^(1/3) then square both sides

2 = r^(2) - 2r(2)^(1/2) + 4^(1/3)

2r(2)^(1/2) = r^(2) + 4^(1/3) -2

2^(1/2) = [r^2 + 4^1/3 -2 ]/2r. Here is my problem, how can I make the RHS rational?

Re: Prove (2^(1/2)) + (2^(1/3)) is irrational

Hey dave52.

Square both sides and show that 4^(1/3) is irrational and since r^2 - 2 is also rational and irrational + rational = rational, then you are done.

Re: Prove (2^(1/2)) + (2^(1/3)) is irrational

Quote:

Originally Posted by

**dave52** This is what I have

2^(1/2) + 2^(1/3) = r with r being a rational number

2^(1/2) = r - 2^(1/3) then square both sides

2 = r^(2) - 2r(2)^(1/2) + 4^(1/3)

2r(2)^(1/2) = r^(2) + 4^(1/3) -2

2^(1/2) = [r^2 + 4^1/3 -2 ]/2r. Here is my problem, how can I make the RHS rational?

You have a typo in the third line. You want

2 = r^(2) - 2r(2)^**(1/3)** + 4^(1/3)

I would suggest however that you put the $\displaystyle 2^{\frac13}$ on the left-hand side and cube it:

$\displaystyle 2^{\frac13}=r-2^{\frac12}$

$\displaystyle 2=r^3+6r-(3r^2+2)\sqrt2$

Now it’s easy to see the required contradiction that $\displaystyle \sqrt2$ is rational.