# Sum of nth powers

• May 8th 2010, 08:25 PM
Drexel28
Sum of nth powers
Hello friends.

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

Quote:

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!
• May 8th 2010, 08:57 PM
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.)
• May 8th 2010, 09:01 PM
Drexel28
Quote:

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!
• May 8th 2010, 11:50 PM
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. (Wondering)
• May 8th 2010, 11:57 PM
Drexel28
Quote:

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. (Wondering)

I'm sorry, I'm not quite sure what you mean. (Worried)
• May 9th 2010, 12:05 AM
Bacterius
$\sqrt[y]{x} = b^\frac{\log_b{(x)}}{y}$ - or even better - $\sqrt[j]{j} = b^\frac{\log_b{(j)}}{j}$
Seems easier than dealing with $\sqrt[j]{j}$ somehow.