[SOLVED] Prove that the cuberoot of 2 is irrational

[Goldenwind]

1. The problem statement, all variables and given/known data
Prove that the cuberoot of 2 is irrational

3. The attempt at a solution
Assume it is rational, and find a contradiction.
2^(1/3) = a/b, where a, b are integers, where a/b is in lowest terms, and where b != 0.
2 = a^3 / b^3
2b^3 = a^3

Aaaand stuck.
Can someone give me a little push?

[NateTG]

It works just like the square root proof. What happens if a is not divisible by 2?

[shukrri]

http://72.14.205.104/search?q=cache:sMLv_NLBkOgJ:mathforum.org/library/drmath/view/51568.html+cube+root+irrational+proof&hl=en&ct=clnk&cd=1&gl=ca&client=firefox-a

[Goldenwind]

To be completely honest, neither of your answers really make sense to me :(
I mean, I'm sure they're completely valid and correct answers, but I'm having trouble grasping what you're getting at.

Can someone elaborate a little? I realize it has something to do with the definitions of a and b (Being integers, lowest terms, and b != 0), but I don't see what.

And NateTG, I have the squareroot proof in my book, in front of me... but it uses the fact that a^2 and b^2 = even. I can't use that fact here, as a^3 will be of the same polarity as a, and b^3 the same polarity as b. I can't rely on it being even, nor odd.

[jimmypoopins]

since 2b^3 = a^3 don't you know that a^3 is even? therefore making a even?

then a = 2c for some integer c, and subbing back in you get 2b^3=(2c)^3=8c^3?

does this help?

[Goldenwind]

since 2b^3 = a^3 don't you know that a^3 is even? therefore making a even?

then a = 2c for some integer c, and subbing back in you get 2b^3=(2c)^3=8c^3?

does this help?

If a^3 = 6, then a = cuberoot(6), which in theory would be irrational (However proving this would just be another huge circle).

If a^3 is even, why is a even?
Ooohh... 'cause a has to be an integer...

Yeah, I think I see now.

[Goldenwind]

And then because a and b are both even, they're both divisible by 2, which breaks the hypothesis of "lowest common terms".
Yes?

[Mathdope]

If a^3 = 6, then a = cuberoot(6), which in theory would be irrational (However proving this would just be another huge circle).

If a^3 is even, why is a even?
Ooohh... 'cause a has to be an integer...

Yeah, I think I see now.
If a is odd what's a^3, odd or even?

[cepheid]

And then because a and b are both even, they're both divisible by 2, which breaks the hypothesis of "lowest common terms".
Yes?

Yeah, which is the same proof as for square root 2.

[Goldenwind]

Yep, solved the problem now. Wanna know what was confusing me? For some reason, I thought that a^2 = Even, regardless of value of a. But then I sat there and thought... 3^2=9...wtf?

Yeah... sorry for the stupid questions, and thanks for your help everyone :)
With your help, I've not only completed this question, but blasted straight through the next one ^^;