In the proof of theorem 16.4 in Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, the authors state without proof that if $a \ge 1$ then {$a^{1/n}$} is a decreasing real sequence. As an exercise I decided to prove this by induction. But I'm stuck trying to prove that $S(n) \Rightarrow S(n+1)$. Question: how does one prove the inductive step?
$n \in \mathbb{P}$
$a,r,q,z,y,w \in \mathbb{R}$
$S(n)$ is $(z^n = y^{n+1} = a \ge 1) \Rightarrow (z \ge y)$
Suppose $w^{n+2} = a$,
$S(n+1)$ is $(y^{n+1} = w^{n+2} = a \ge 1) \Rightarrow (y \ge w)$
Suppose $r^1 = q^2 = a$,
$S(1)$ is $(r = q^2 \ge 1) \Rightarrow (r \ge q)$
$0 < q < 1 \Rightarrow q^2 < 1$, therefore $q \ge 1$ and $r = q^2 \ge q$
So the base case is true...