Why is n in $\sqrt[n]{x}$ a positive integer? Also for property 6, why that n should also be an integer instead of a rational number?
(Refer to the attachment)

Question

anonymous · Answer

Actually, the question should be why we have that condition, i.e. n is a positive integer.

anonymous · Answer

probably because you don't know what $\sqrt[\pi]{f(x)}$ would mean

anonymous · Answer

I'm sorry but I don't understand...

anonymous · Answer

they are talking about square roots, cubed roots, fourth roots, fifth roots etc

anonymous · Answer

But what happen if n is a negative integer? Or n is a rational number? Does it affect those two properties?

parthkohli · Answer

Thinking elementarily, $\sqrt[3]{64}$ yields us a number which we multiply to itself thrice to get 64. Similarly, it's not fundamentally possible to multiply a number to itself -2 number of times.

anonymous · Answer

First, for property 6, it is also stated that n is a positive integer. Why doesn't it make sense if n is a irrational number, rather than a integer?
Also, I would like to know if something like $\sqrt[\frac{2}{3}]{x}$ exist.