# Order Preserving

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.

Hints:

1. $a\leq b$ is the same as $0\leq b-a$.

2. Can you factor $f(b)-f(a)$?

3. Can you say anything about the signs of the factors of $f(b)-f(a)$?

• I'm not 100% sure what you mean by can hint #2. Like I know that $f(b)$ and f(a) would both be $x^2$. But I do know that $0 \leq f(b)-f(a)$ – hockeynl Jan 31 '17 at 18:03
• Wait no it would be $b^2$ and $a^2$ – hockeynl Jan 31 '17 at 18:18
• Correct, it would be $f(b)-f(a)=b^2-a^2$. – Michael Burr Jan 31 '17 at 22:12