Proving Irrationality of \sqrt[3]{3}

[SpatialVacancy]

Hello all,

Prove that $$\sqrt[3]{3}$$ is irrational.

Heres what I have so far:
$$\sqrt[3]{3}=\dfrac{m}{n}$$
$$3=\dfrac{m^3}{n^3}$$
$$m^3=3n^3$$

Now, I think I am to assume that there is more than one factor of three on the right, or something, I don't know. Can someone point me in the right direction?

Thanks.

 

[dextercioby]

Check out that:https://www.physicsforums.com/showpost.php?p=449091&postcount=11

I think it's more than satisfying...

Daniel.

 

[HallsofIvy]

Any integer n is either
1) a multiple of 3: n= 3m in which case n²= 9m²= 3(3m²)

2) a multiple of 3 plus 1: n= 3m+1 in which case n²= 9m²+ 6m+ 1= 3(3m²+ 2m)+ 1

3) a multiple of 3 plus 2: n= 3m+2 in which case n²= 9m²+ 12m+ 4= 9m²+ 12m+ 3+ 1= 3(3m²+ 4m+1)+ 1.

In other words, if n² is a multiple of 3, then n must be a multiple of 3.

 

[dextercioby]

Halls,it's the 3-rd power. I've given him the link to Hurkyl's general proof in the other thread in HS homework section...

Daniel.