Proof of how multiplying a positive and negative constant changes the sup and inf

**tempins** · August 9th 2010, 05:32 PM

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**Vlasev** · August 9th 2010, 06:23 PM

For your second question, it's more along those lines. Let $A = \inf(S)$ and $B= \sup(S)$ where $A < B$ . Let a be a positive number. Then $-a$ is a negative number.

$-aA > -aB$ by the rules of ">".

Hence now you have that $\inf(S) = -aB$ and $\sup(S) = -aA$

Try an example: the interval $(1,2)$ for example. $A = 1$ , $B = 2$ . When you take [LaTeX ERROR: Convert failed] , you get $(-2,-1)$ and it should be clear.

**HallsofIvy** · August 10th 2010, 04:25 AM

Please do NOT erase a post just because you do not need it any more. New people looking at the thread can still learn something from it.

**Vlasev** · August 10th 2010, 04:51 AM

The question was along the lines of:

If $S$ is a set and $S \subset \mathbb{R}$ and if $a \in \mathbb{R}$ is a constant, then let $aS$ denote the set $aS = \{ ax \mid x \in S\}$ .

Then if $A = \inf(S)$ and $B = \sup(S)$ , prove:

If $a > 0$ then $\inf(aS) = a\inf(S)$ and $\sup(aS) = a\sup(S)$ .
If $a < 0$ then $\inf(aS) = a\sup(S)$ and $\sup(aS) = a\inf(S)$ .

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Proof of how multiplying a positive and negative constant changes the sup and inf

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