MHB N th root of a positive number is positive ....

[issacnewton]

Hi

Let \( c>0 \) be a real number. Then I am trying to prove that \( \forall\; n\in\mathbb{N}\; (c^{1/n} >0) \).
I let \(n\) be arbitrary and then tried to use method of contradiction. But ran into difficulties. Is there another approach ?

 

[Evgeny.Makarov]

Let $x=c^{1/n}$, i.e., $x^n=c$. If n is odd, then x and c have the same signs, so x is positive. If n is a positive even number, then x can be either positive or negative. However, the principal n-th root is positive by definition, and "when one speaks of the n-th root of a positive real number b, one usually means the principal n-th root" (link above).

 

[issacnewton]

Evgeny, thanks for the reply. But can we derive these facts from, say, field axioms ? I am looking into that sort of solution. Looking at the
link given by you, it seems to me that may be this is how its defined.​ So we just take as it is.

 

[Evgeny.Makarov]

Looking at the link given by you, it seems to me that may be this is how its defined.​

Yes, $c^{1/n}$ is positive by definition for c > 0. The only thing that one may need to prove is that such positive x that $x^n=c$ exists. For this I would look at how the n-th root was defined in the textbook.

 

[issacnewton]

for completeness I am posting this from the wikipedia article on \( n^{\mbox{th}} \) root.

A real number or complex number has n roots of degree n. While the roots of 0 are not distinct (all equaling 0), the n nth roots of any other real or complex number are all distinct. If n is even and the number is real and positive, one of its nth roots is positive, one is negative, and the rest are complex but not real; if n is even and the number is real and negative, none of the nth roots are real. If n is odd and the number is real, one nth root is real and has the same sign as the number, while the other roots are not real

So one needs complex variables to appreciate all this