Why the principal nth root is unique ?

[mahmoud2011]

Why the principal nth root is unique ? I mean in the definition of the princibal nth root it says:
let x be a real number and n is natural number $n \geq 1$.if n is odd , the principal nth root of x denoted $\sqrt[n]{x}$,is the unique real number satisfying $(\sqrt[n]{x})^{n} = x$.if n is even , the principal nth root of x denoted $\sqrt[n]{x}$ is defined similarly provided $x \geq 0$ and $\sqrt[n]{x} \geq 0$.

so why it is unique is there a proof .

 

[CompuChip]

The standard proof is to assume that $r_1^n = r_2^n = x$ (i.e. there are two roots) and try to show that [ itex]r_1 = r_2[/itex] (i.e. the roots must be the same).

I haven't worked out the proof completely, but I suppose it would be something along the lines of:
Consider
$$\frac{r_1^n}{r_2^n} = \left(\frac{r_1}{r_2}\right)^n = 1$$

 

[mathman]

Why the principal nth root is unique ? I mean in the definition of the princibal nth root it says:
let x be a real number and n is natural number $n \geq 1$.if n is odd , the principal nth root of x denoted $\sqrt[n]{x}$,is the unique real number satisfying $(\sqrt[n]{x})^{n} = x$.if n is even , the principal nth root of x denoted $\sqrt[n]{x}$ is defined similarly provided $x \geq 0$ and $\sqrt[n]{x} \geq 0$.

so why it is unique is there a proof .

Uniqueness is not something that requires a proof. For odd n, there is only one real root, all others are complex. For even n and x > 0, there are two real roots, one positive and the other negative, all others are complex. The unique roots are the particular roots defined to be such.

One thing that could contribute to a proof is that all the roots for the given n and x have the same magnitude, so the "unique" root must be unique.

 

[mahmoud2011]

I proved it using intermediate theorem by assuming a real number c such that 0<x<c , let f(x) = x^n then f(x) is continuous on [0,c] and 0<a<c<c^n .so here we use intermediate value theorem , where there is at least one real number m such that f(m)=a in other words m^n = a .here we proved its existence. Since f is increasing on [0,c] which can be proved easily then there must be on real number m satisfying m^n =a , a>0 (I proved that by contradiction by assuming there exist two numbers satisfying this ) .We can prove similarly that for a<0 where n is odd . I am not sure this proof is right , but this is what I can by my weak mathematical background , also before I write this proof I was stuck in the proof of if c>1 then c^n > c , n is a postive integer , this I can't prove can you help me to prove it.

 

[HallsofIvy]

I guess the real question is, what is your definition of "principle nth root"? You should find that "uniqueness" comes directly from the definition.

What definiton are you using?

 

[mahmoud2011]

Why the principal nth root is unique ? I mean in the definition of the princibal nth root it says:
let x be a real number and n is natural number $n \geq 1$.if n is odd , the principal nth root of x denoted $\sqrt[n]{x}$,is the unique real number satisfying $(\sqrt[n]{x})^{n} = x$.if n is even , the principal nth root of x denoted $\sqrt[n]{x}$ is defined similarly provided $x \geq 0$ and $\sqrt[n]{x} \geq 0$.

so why it is unique is there a proof .

This is the definition I am using.