Is $(x^n)$ decreasing if and only if $x\in(0,1)$? I was reading something where one of the steps said $x\leqslant-1\implies x^{4}\geqslant x^{2}$. I stopped and thought on this part about $x^{4}$ being greater than or equal to $x^{2}$. In the case of $x>1$, I could believe that powers of this number would get bigger and bigger, while, for example, if it's something like $\frac{1}{2}$, then powers of it take halves and make the number smaller. I reckoned then that these $x^{n}$s only got smaller if $-1<x<1$.
So I thought maybe we can claim $(x^n)$ is decreasing if and only if $x\in(0,1)$, but wasn't totally sure how I would prove this.
 A: The sequence $(x^n)$ is decreasing for $(0,1)$. You can prove this by extending $n$ to a continuous variable $y$, i.e. $$f(y) = x^y,\qquad  x > 0 $$ and note that $f'(y) = x^y \ln x$  is negative for $x\in (0,1)$.
For $x < 0$ you can always prove that the sequence is alternating, since $x^n < x^{n + 1}$ implies $x^{n+1} > x^{n + 2}$ by multiplying by $x$.
A: If $0 < x < 1$ then use the axiom:  If $a < b$ and $c > 0$ then $ac < bc$.
So $0< x < 1$ so $0*x < x*x < 1*x$ so $0 < x^2 < x< 1$ and by induction if you have $0 < x^n < x^{n-1}$ you have $0*x < x^n*x <x^{n-1}x $ so $0< x^{n+1}< x^n$.
So $x^n$ is decreasing.  (which doesn't prove $x^n\to 0$ FWIW.)
Now if $x \in (0,1)$ we have.
$x < 0$:  Well then $x^n > 0$ if $n$ is even and $x^n < 0$ if $n$ is odd.  So that's not decreasing.
$x = 0$:  Well then $x^n = 0^n = 0$ which constant; not decreasing.
$x =1$: Well then $x^n =1^n=1$ which ...ditto...
$x> 1$ then by similar induction $1 < x$ so $1*x < x*x$ and if $1 < x^n$ then $1 < x \implies x^n < x*x^n=x^{n+1}$ so $x^n$ is increasing.
