whole roots and powers

[georg gill]

(a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{ 2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

(a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p} })^n

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:



\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}}

to become right side by only using roths and powers of integers.

[HallsofIvy]

Certainly a^{1/2n}a^{1/2n}= a^{1/2n+ 1/2n}= a^{2/2n}= a^{1/n}.

And \left(a^{1/n}\right)^m= a^{(1/n)m}= a^{m/n}. Those are well known properties of the exponentials.

[D H]

If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.

[georg gill]

If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.


As far as I can see or from what I get from the proof at least

(xy)^m)^{\frac{1}{n}}=(y)^m)^{\frac{1}{n}}(x)^m)^{ \frac{1}{n}}

I want to prove that

(xy)^m)^{\frac{1}{n}}=y)^{\frac{m}{n}}x^{\frac{m}{ n}}

What I have managed to show is that

(xy^{\frac{1}{np}})^{np}

(a^{\frac{1}{np}})^{np}=a

((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac {1}{n}})^{\frac{1}{p}})^n)^p

since n and p are integers to put p outside as own power goues without proving (could of course prove this
and we get

((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac {1}{n}})^{\frac{1}{p}})^n)^p=a

so

xy^{\frac{1}{np}}=(a^{\frac{1}{n}})^{\frac{1}{p}}