A simple question about rational numbers without a simple proof?

Asked Aug 11 '15 at 00:08

Active Aug 16 '15 at 06:04

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As in this question, study the quasigroup $(Q_+,/)$ of positive rational numbers under division. There are two obvious identities:

How to decide if there are other independent identities, not possible to derive?

Could there be "mystical" identities of great complexity including several variables which are independent of 1 and 2?

The only way to go, as I can see, is to study equivalence classes of binary trees corresponding to expressions in $(Q_+,/)$.

edited Apr 13 '17 at 12:19

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asked Aug 11 '15 at 00:08

Lehs

The Knuth-Bendix algorithm can decide this. – MY USER NAME IS A LIE Aug 11 '15 at 00:17
@MYUSERNAMEISALIE: Thanks for the comment, which seems adequate. – Lehs Aug 11 '15 at 00:23
How would you derive $(ab)/(cb)=a/c$? – Akiva Weinberger Aug 11 '15 at 00:23
@columbus8myhw: The expressions are supposed to consists of variables, / and matching brackets only. No multiplication. – Lehs Aug 11 '15 at 00:26
@Lehs Can you derive $a/(b/b)=a$, then? Or $a/a=b/b$? – Akiva Weinberger Aug 11 '15 at 00:27
@columbus8myhw: I suppose not, so this might be an other identity then. What I'm wondering about is maximal sets of independent identities and the way to decide which those sets are. – Lehs Aug 11 '15 at 00:33
One of the questions I've answered singles out an element $1$ and has axioms $a/b=1$ If and only if $a=b$ and $(a/c)/(b/c)=a/b$ for all $a,b,c$, and the goal is to prove that the operation $a\cdot b=a/(1/b)$ satisfies all of the axioms for a group. – Matt Samuel Aug 11 '15 at 01:08
(the hard part is proving that $a/1=a$ for all $a$, which indeed holds.) – Matt Samuel Aug 11 '15 at 01:13
@MattSamuel, can youderive that from 1&2? – Lehs Aug 11 '15 at 01:39
@Lehs I can't tell you for sure, but looking at the axioms makes me think it's highly unlikely. In my example the identity is singled out explicitly. – Matt Samuel Aug 11 '15 at 01:41

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