
Rational numbers
If z is a real number, then $\displaystyle \ {z^n}$ is defined recursively for nonnegative integers
$\displaystyle \ n: {z^0}=1$
And, for $\displaystyle n\geq 0$
$\displaystyle \ {z^{n+1}} = z ({z^n}) $.
If z is a positive real number, and b a natural number, then we may define the positive bth root
$\displaystyle \ y= {z^{1/b}$ to be the positive real number y such that
$\displaystyle \ {y^b} = z$.
You may assume that a positive real number always has a positive bth root.
a) Given a positive real number z, is it possible for there to be two different positive bth roots of z? Justify your answer briefly.
Any help would be greatly appreciated :)

Re: Rational numbers
suppose that x,y > 0 with x^{n} = y^{n}.
show if x/y > 1, then (x/y)^{n} > 1, and similarly if x/y < 1
(you might need to prove first that (1/y)^{n} = 1/y^{n}, can you do this?).