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

• Aug 9th 2010, 04:32 PM
tempins
Proof of how multiplying a positive and negative constant changes the sup and inf
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• Aug 9th 2010, 05:23 PM
Vlasev
For your second question, it's more along those lines. Let $\displaystyle A = \inf(S)$ and $\displaystyle B= \sup(S)$ where $\displaystyle A < B$. Let a be a positive number. Then $\displaystyle -a$ is a negative number.

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

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

Try an example: the interval $\displaystyle (1,2)$for example. $\displaystyle A = 1$, $\displaystyle B = 2$. When you take $\displaystyle (-1)S$, you get $\displaystyle (-2,-1)$and it should be clear.
• Aug 10th 2010, 03:25 AM
HallsofIvy
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.
• Aug 10th 2010, 03:51 AM
Vlasev
The question was along the lines of:

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

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

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