1. ## Rational numbers

If z is a real number, then $\ {z^n}$ is defined recursively for non-negative integers

$\ n: {z^0}=1$

And, for $n\geq 0$

$\ {z^{n+1}} = z ({z^n})$.

If z is a positive real number, and b a natural number, then we may define the positive bth root

$\ y= {z^{1/b}$ to be the positive real number y such that

$\ {y^b} = z$.

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

2. ## Re: Rational numbers

suppose that x,y > 0 with xn = yn.

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/yn, can you do this?).