Principal nth Root of a Real Number

  • Thread starter Thread starter cepheid
  • Start date Start date
  • Tags Tags
    Root
AI Thread Summary
The discussion centers on the existence of the principal nth root of real numbers, particularly emphasizing that for odd n, every real number has a unique real nth root with the same sign. The inquiry focuses on how to prove that every positive real number has a unique real nth root, suggesting the use of the infimum and supremum axioms along with the squeeze theorem. Participants note that the function x^n is a continuous bijection on its domain, which supports the existence of the nth root. There is also a mention of a related problem involving the function f(x) = x^3 + 1, which raises questions about its bijection properties. The conversation highlights the need for foundational understanding in real analysis to tackle these concepts effectively.
cepheid
Staff Emeritus
Science Advisor
Gold Member
Messages
5,197
Reaction score
38
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.
 
Mathematics news on Phys.org
I had a discovery:

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?"
 
How about using the infimum axiom the superimum axiom and the squeeze theorem.
 
x^n is a continuous bijection on the appropriate domain and codomain.
 
John Creighto said:
How about using the infimum axiom the superimum axiom and the squeeze theorem.

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

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.
 
cepheid said:
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.
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:
cepheid said:
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.

The axioms are necessary for the existence of the real numbers.
 
cepheid said:
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.
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
Hurkyl said:
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?

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.
 

Similar threads

Back
Top