Principal nth Root of a Real Number

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In summary, according to the conversation, 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. This is denoted by \sqrt[n]{A}. However, it is questioned how we know that every positive real number has a unique, real nth root. One approach suggested is using the infimum axiom, the supremum axiom, and the squeeze theorem. Another approach is using complex analysis to prove that x^n is a continuous bijection. Finally, it is mentioned that the axioms are necessary for the existence of the real numbers.
  • #1
cepheid
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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

[tex] \sqrt[n]{A} [/tex]

My question is, how do we know that this is true i.e. that [itex] \sqrt[n]{A} [/itex] 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.
 
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  • #2
I had a discovery:

Let y be a negative real number:

[tex] y = A e^{i(\pi + 2k\pi)}, \ \ \ \ \ k \in \mathbb{Z} [/tex]

Then

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

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: [itex] \sqrt[n]{y} [/itex] exists and is equal to -[itex] \sqrt[n]{A} [/itex] 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
How about using the infimum axiom the superimum axiom and the squeeze theorem.
 
  • #4
x^n is a continuous bijection on the appropriate domain and codomain.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
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  • #8
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.
 
  • #9
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.
 

Related to Principal nth Root of a Real Number

1. What is the Principal nth Root of a Real Number?

The Principal nth Root of a Real Number is the unique real number that, when raised to the power of n, gives the original real number. It is also known as the principal root or the principal square root when n is 2.

2. How do you calculate the Principal nth Root of a Real Number?

The Principal nth Root of a Real Number can be calculated by taking the nth root of the given real number. For example, the principal cube root of 8 is 2 because 2^3 = 8.

3. Can the Principal nth Root of a Real Number be negative?

Yes, the Principal nth Root of a Real Number can be negative if the original real number is negative and n is an even number. For example, the principal square root of 25 is both 5 and -5.

4. What is the difference between the Principal nth Root and the nth Root of a Real Number?

The Principal nth Root is the unique real number that, when raised to the power of n, gives the original real number. The nth Root of a Real Number, on the other hand, can have multiple values (including complex numbers) that satisfy the given equation.

5. What are some real-life applications of the Principal nth Root of a Real Number?

The Principal nth Root of a Real Number is commonly used in financial calculations, such as compound interest and mortgage loans. It is also used in physics and engineering to calculate values such as speed, acceleration, and voltage. Additionally, it is used in computer programming and cryptography for data encryption and decryption.

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