The question is
Prove that if $0 \leq x < y$ then $x^n < y^n$, $n = 1, 2, 3, \ldots $
Spivak's solution was using (if $0 \leq x < y$ then $y^2 < x^2$) and generalising it for all n. Spivak does acknowledge the fact that the proof isn't rigorous
My method was: as $0 \leq x$, I used cases
First taking the case $x = 0$. We have $x= 0$ and $0 < y$, thus any power of zero will be greater than any power of $y$.
Then taking the case $0 < x < y$,
I use the fact that $(y - x)$ is in p.
Now, $y^n - x^n = (y - x) (y^{(n-1)} + y^{(n-2)}x \cdots + yx^{(n-2)} + x^{(n-1)})$
$(y - x)$ is positive.
The other term in the product is also positive due to closure under addition because we're adding only positive terms to positive terms.
Now as both the terms are positive, due to closure under multiplication, $y^n -x^n$ is also positive.
i.e, $y^n - x^n$ is in P. Thus $y^n > x^n$.
I'm confident I have used only those results that I have proved during earlier exercises and not assuming anything. Is my method consistent or flawed? Thank you very much for your time. :-)
I don't see anything wrong in your proof using $$y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+\ldots +x^{n-1}).$$ But you can also use a simple number line argument, noting that $x$ and $y$ are both nonnegative such that $y>x$. Then $y^n-x^n$ is also nonnegative.