2026/01/08

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2026-01-08 23:28:19 +0100 <Leary> Yes.
2026-01-08 23:27:38 +0100 <TMA> Leary: ax = a(x+0) = ax + a0; now add -ax to both sides: -ax + ax = -ax + ax + a0 ==> 0 = 0 + a0 = a0, do I understand you correctly?
2026-01-08 23:27:34 +0100Digit(~user@user/digit) (Remote host closed the connection)
2026-01-08 23:26:52 +0100merijn(~merijn@host-cl.cgnat-g.v4.dfn.nl) merijn
2026-01-08 23:26:51 +0100 <tomsmeding> nice
2026-01-08 23:25:38 +0100 <ncf> 0 = a * 0 - a * 0 = a * (0 + 0) - a * 0 = a * 0 + a * 0 - a * 0 = a * 0
2026-01-08 23:24:38 +0100AlexNoo(~AlexNoo@178.34.163.50)
2026-01-08 23:24:29 +0100 <monochrom> :)
2026-01-08 23:24:16 +0100 <Leary> Actually the -0 could have been anything, should have just used 0.
2026-01-08 23:23:11 +0100 <tomsmeding> so fully: 0 = -(a * -0) + a * -0 = -(a * -0) + a * (-0 + 0) = -(a * -0) + a * -0 + a * 0 = 0 + a * 0 = a * 0
2026-01-08 23:22:53 +0100 <ncf> Leary++
2026-01-08 23:21:59 +0100 <Leary> a * -0 = a * (-0 + 0) = a * -0 + a * 0 ==> a * 0 = 0 (by cancellation)
2026-01-08 23:20:17 +0100 <TMA> the other order of operands need showing (-1)*a = a*(-1)
2026-01-08 23:19:49 +0100peterbecich(~Thunderbi@71.84.33.135) (Ping timeout: 264 seconds)
2026-01-08 23:19:47 +0100 <Leary> That still relies on `a * -1 = -a`.
2026-01-08 23:19:38 +0100 <tomsmeding> oh also TMA :)
2026-01-08 23:19:32 +0100 <tomsmeding> ncf: thank you
2026-01-08 23:19:02 +0100 <TMA> 0*a = (x-x)*a = xa - xa = 0 for any x
2026-01-08 23:18:47 +0100 <ncf> a * 0 = a * (1 - 1) = a - a = 0 ?
2026-01-08 23:17:44 +0100 <tomsmeding> right, I just proved it in a long-winded way (see 6 minutes ago), but surely there's a more direct way
2026-01-08 23:17:16 +0100 <geekosaur> but I think that's derived, not amxiom
2026-01-08 23:17:06 +0100 <tomsmeding> geekosaur: no, because you can prove it so it need not be an axiom :p
2026-01-08 23:16:51 +0100 <monochrom> Oh oops, sorry! Delete everything I said.
2026-01-08 23:16:49 +0100 <geekosaur> 0 is required to be an annihilating element in multiplication