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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.

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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.

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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 $f(b)-f(a)$?

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  • $\begingroup$ 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 $0 \leq f(b)-f(a)$ $\endgroup$ – hockeynl Jan 31 '17 at 18:03
  • $\begingroup$ Wait no it would be $b^2$ and $a^2$ $\endgroup$ – hockeynl Jan 31 '17 at 18:18
  • $\begingroup$ Correct, it would be $f(b)-f(a)=b^2-a^2$. $\endgroup$ – Michael Burr Jan 31 '17 at 22:12

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