Consider the set $ℝ^+=\{x∈ℝ|x>0\}$ together with the usual order $<$.Let $f:ℝ^+→ℝ^+$ be the function given by $f(x)=x^2$.Is $f$ order preserving?
Workings:
$f$ is order preserving if $a ≤ b$ in $P$ implies $f(a)$ ≤ $f(b)$ in $Q$
I am not entirely sure what to now so any help will be appreciated.
Another solution. Given $a,b>0$ and $a\le b$. Since $a>0$, we can multiply the given inequality by $a$, obtaining $a^2\le ab$. Since $b>0$, we can multiply the given inequality by $b$, obtaining $ab\le b^2$. Now we see that $a^2\le ab\le b^2$, as desired.