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From Walter Rudin's Principles of Mathematical Analysis, Third Edition, page 20, Step 8:

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I want to complete the proof of (b) by proving the following claim.

Claim: For $r\in{}\mathbb{Q}^+$ and $s\in{}\mathbb{Q}^-$, $r^*\cdot{}s^*\subset{}(rs)^*$

Attempt at Proof: $q\in{}r^*\cdot{}s^*\implies{}q\in{}-[r^*\cdot{}(-s)^*]\implies{}\forall{}r'\forall{}s'(r',s'\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,r'\in{}r^*\,\,\,\wedge{}\,\,\,s'\in{}(-s)^*\implies{}\exists{}a(a\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,-q-a>r's'))$


At this point I am stuck. I need to show $rs>q$ so that $q\in{}(rs)^*$.

This post covers the question, but I think the proof is wrong at the relevant place and I'm stuck.

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1 Answer 1

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First prove for all rational $r,s$, we have:

(1) $\quad r^* < s^*$ if and only if $r < s$.

(2) $\quad (-r)^* = -(r^*)$

(3) $\quad r,s \ge 0$ implies $r^*s^* = (rs)^*$.

Then, using the definition of multiplication of reals (Rudin's Step 7), the other cases of (3) follow immediately, unless I'm missing something!

I believe the proofs of (1)–(3) are very straightforward, again unless I'm missing something.

(I'm not sure what the etiquette is on this site, but I feel like I should restrict my contribution to getting past the roadblock, not filling in every detail. Anyway, in the meantime, I have verified that the proofs of (1)–(3) are straightforward.)

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  • $\begingroup$ Jeremy is correct. If $r>0,s<0$, then $r^*\cdot{}s^*:=-[r^*\cdot{}(-s)^*]=-[(r\cdot{}-s)^*]=-[(-rs)^*]=(rs)^*$. This makes use of Jeremy's (2) and (3) above. It also means that we don't need to prove inclusion in the manner that I tried to do in the OP. That said, if anyone reads this and can prove the inclusion, I'd love to see it. $\endgroup$
    – Gary
    Commented Mar 6, 2022 at 7:51
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    $\begingroup$ It also uses (1). $\endgroup$
    – Jeremy Weissmann
    Commented Mar 6, 2022 at 12:22
  • $\begingroup$ Where? $r^*\cdot{}s^*:=-[r^*\cdot{}-(s^*)]$ (definition) = $-[r^*\cdot{}(-s)^*]$ (2) = $-[(r\cdot{}-s)^*]$ (3) = $-[(-rs)^*]$ (elementary property of rational multiplication)= $(rs)^*$ (2) $\endgroup$
    – Gary
    Commented Mar 6, 2022 at 16:42
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    $\begingroup$ Because you assumed $r > 0$ (for example) but used $r^* > 0^*$. $\endgroup$
    – Jeremy Weissmann
    Commented Mar 6, 2022 at 20:21

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