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 $t\in -B$

  • $\begingroup$ 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. $\endgroup$ – M. Winter Sep 17 '18 at 11:27
  • $\begingroup$ 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 $\endgroup$ – Winther Sep 17 '18 at 11:33
  • $\begingroup$ 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 ? $\endgroup$ – Winther Sep 17 '18 at 11:41
  • $\begingroup$ 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 $t\in -B$"? $\endgroup$ – M. Winter Sep 17 '18 at 11:51
  • $\begingroup$ Note that if you multiply an inequality by $-1$, then the direction of the inequality changes. For example, $x>1 \implies -x<-1$. $\endgroup$ – celtschk Sep 17 '18 at 12:21

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.

  • $\begingroup$ 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) $\endgroup$ – Eric Stucky Oct 10 '18 at 5:01

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