# Prove the set -B has a minimum if and only if B has a maximum

This is a homework problem, so I would like hints and help not answers.

Let $B\subseteq\mathbb{Q}$ and define $-B=\{-b \mid b\in B\}$ which is also a subset of the rationals. Show that $-B$ has a maximum if and if only B has a maximum, and if so we have $min(-B) = -max(B)$

My thought process was to assume a $u$ so that $u=max(B)$ which would mean there would be a corresponding $-u$ in the set of $-B$ and then show that this $-u$ is minimum by showing $-u \leq t$ for $t\in -B$

• You approach is completely fine and will lead you to the result. Where do you get stuck? Try multiplying $-u\le t$ by -1 and see where this leads you to. – M. Winter Sep 17 '18 at 11:27
• I am having trouble showing my statement of $-u \leq t$ and my multiplying by -1 don't I have the exact same problem? But now I just need to prove u is maximum? Which I still have a bit of trouble with figuring out – Winther Sep 17 '18 at 11:33
• Or is the answer just to say that -t can't be a higher number than u, and therefore u=-t, and therefore follows -u=t ? – Winther Sep 17 '18 at 11:41
• Yeah, this is pretty close to a perfect reasoning. I would write that since $\max B \ge t$ for all $t\in B$, we also have $-\max B \le -t$ for all $t\in B$. Can you see how this becomes "$\min(- B)\le t$ for all $t\in -B$"? – M. Winter Sep 17 '18 at 11:51
• Note that if you multiply an inequality by $-1$, then the direction of the inequality changes. For example, $x>1 \implies -x<-1$. – celtschk Sep 17 '18 at 12:21

## 1 Answer

Assume there exists $$u=\max(B)$$; then $$-u\in -B$$. We wish to show that $$-u\leq t$$ for any $$t\in -B$$. This is equivalent to $$u\geq -t$$ for any $$t\in -B$$. Since $$u$$ is a maximum, $$u\geq b$$ for any $$b\in B$$, hence $$u\geq -(-b)$$. Because any $$t\in -B$$ is $$-b$$ for some $$b\in B$$. It follows that if $$B$$ has a maximum then $$-B$$ has a minimum.

The converse statement has an extremely similar proof.

• This answer consists mostly of the comment chain; it exists to help remove this question from the unanswered queue. Please upvote or accept this answer to complete the removal. (@ChristianWinther) – Eric Stucky Oct 10 '18 at 5:01