# Principal nth Root of a Real Number

1. Feb 18, 2008

### cepheid

Staff Emeritus
According to Wikipedia, if n is odd, then every real number A has a unique real nth root having the same sign as A and known as the principal nth root of A. It is denoted by

$$\sqrt[n]{A}$$

My question is, how do we know that this is true i.e. that $\sqrt[n]{A}$ exists for all real numbers if n is odd, and for all positive real numbers if n is even?

Note: I have studied Complex Analysis. I am not interested in the other n-1 complex roots of the number.

2. Feb 18, 2008

### cepheid

Staff Emeritus

Let y be a negative real number:

$$y = A e^{i(\pi + 2k\pi)}, \ \ \ \ \ k \in \mathbb{Z}$$

Then

$$y^{1/n} = \sqrt[n]{A} e^{i(\pi/n + 2k\pi/n)}, \ \ \ \ \ k \in \{0,...,n-1\}$$

We can prove that one of the roots must be real by equating the argument to pi so that we just get a negative real number. Solving for k, we get:

k = (n-1)/2

which is an integer in the allowable range of k values IF n is odd.

Conclusion: $\sqrt[n]{y}$ exists and is equal to -$\sqrt[n]{A}$ where A = |y| (provided n is odd)

So my question reduces to, "how do we know that every POSITIVE real number has a unique, real nth root?"

3. Feb 18, 2008

### John Creighto

How about using the infimum axiom the superimum axiom and the squeeze theorem.

4. Feb 18, 2008

### morphism

x^n is a continuous bijection on the appropriate domain and codomain.

5. Feb 18, 2008

### cepheid

Staff Emeritus
I'm not familiar with those axioms. I haven't been taught any number theory formally (if that is indeed what this is), so maybe I won't be able to understand the answer to this question.

6. Feb 18, 2008

### cepheid

Staff Emeritus
Haha...ironically, I have a problem asking me to determine whether f(x) = x^3 + 1 is a bijection, given a domain of R and a codomain of R, and THAT problem is what raised this question in my mind in the first place! I have no doubt that if the claim I was asking about is true, then it follows that what you have said is true, and vice versa. I just have no idea how to go about proving either statement.

7. Feb 18, 2008

### Hurkyl

Staff Emeritus
This real analysis; in fact, all you need is calc I. You said you took complex analysis -- well, you could use that too, can't you?

Last edited: Feb 18, 2008
8. Feb 18, 2008

### John Creighto

The axioms are necessary for the existence of the real numbers.

9. Feb 18, 2008

### morphism

A sketch of the graph of f(x) indicates that it's increasing. Can you think of how we can prove this formally? This should give us that f(x) is 1-1. To prove that f(x) is onto, we can use the intermediate value theorem.

10. Feb 18, 2008

### John Creighto

Well, at one time we believed all numbers were constructed by ratios. I suppose calc I may have a theorem that could be applied but the result won't be from first principles. Perhaps we can argue the existence of the inverse since x^n=y is one to one and onto.