# Thread: Sum of nth powers

1. ## Sum of nth powers

Hello friends.

I was just wondering if there is an easy proof for the following:

For ever $n>1$, $\sum_{j=1}^{n}\sqrt[j]{j}\notin\mathbb{Q}$
I may have found a horrid proof using the rational root theorem, but I have not checked the details.

Any input is appreciated!

2. You can probably try to show that the set $\{1, \sqrt{2}, \sqrt[3]{3},\ldots, \sqrt[n]{n}\}$ is linearly independent over the field $\mathbb{Q}$. The result will follow immediately.

It seems to be fairly straightforward if you know some field theory. (But I haven't actually tried it, so I could be wrong.)

3. Originally Posted by roninpro
You can probably try to show that the set $\{1, \sqrt{2}, \sqrt[3]{3},\ldots, \sqrt[n]{n}\}$ is linearly independent over the field $\mathbb{Q}$. The result will follow immediately.

It seems to be fairly straightforward if you know some field theory. (But I haven't actually tried it, so I could be wrong.)
I am sorry, it was my bad for not saying. I was aware there was an algebraic proof, but you see my cousin asked me and he's in calc III, I was hoping there may be an "elementary" number theoretic proof. Thanks anyways, I appreciate it!

4. Wouldn't it be possible to somehow get rid of the roots by taking the inverses of the exponents while keeping the sum equivalent ? Dealing with powers seems simpler somehow.

This is just an idea, I guess it would be possible by messing around with logs and using their irrational properties.

5. Originally Posted by Bacterius
Wouldn't it be possible to somehow get rid of the roots by taking the inverses of the exponents while keeping the sum equivalent ? Dealing with powers seems simpler somehow.

This is just an idea, I guess it would be possible by messing around with logs and using their irrational properties.
I'm sorry, I'm not quite sure what you mean.

$\sqrt[y]{x} = b^\frac{\log_b{(x)}}{y}$ - or even better - $\sqrt[j]{j} = b^\frac{\log_b{(j)}}{j}$
Seems easier than dealing with $\sqrt[j]{j}$ somehow.