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Groups > sci.physics.relativity > #612974 > unrolled thread
| Started by | patdolan <patdolan@comcast.net> |
|---|---|
| First post | 2023-06-15 08:14 -0700 |
| Last post | 2023-06-15 23:26 -0700 |
| Articles | 20 on this page of 174 — 20 participants |
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Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-15 08:14 -0700
Re: Einstein and Big Ben The Starmaker <starmaker@ix.netcom.com> - 2023-06-15 08:23 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-15 09:35 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-15 16:17 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-15 17:57 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-15 19:08 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-15 22:45 -0700
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-15 23:01 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 01:21 -0700
Re: Einstein and Big Ben "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2023-06-16 15:18 +0200
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-16 06:24 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 07:54 -0700
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-16 17:15 +0000
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 10:29 -0700
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-16 20:22 +0000
Re: Einstein and Big Ben Athel Cornish-Bowden <athel.cb@gmail.com> - 2023-06-17 17:39 +0200
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-17 09:27 -0700
Re: Einstein and Big Ben "Dono." <eggy20011951@gmail.com> - 2023-06-17 09:40 -0700
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-17 17:38 +0000
Re: Einstein and Big Ben Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2023-06-19 13:24 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-19 16:32 -0700
Re: Einstein and Big Ben "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-19 16:41 -0700
Re: Einstein and Big Ben Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2023-06-19 16:50 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-19 17:06 -0700
Re: Einstein and Big Ben Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2023-06-19 18:47 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-19 21:57 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-20 03:30 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-20 06:52 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-20 08:02 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-20 10:29 -0700
Re: Einstein and Big Ben Athel Cornish-Bowden <athel.cb@gmail.com> - 2023-06-20 20:47 +0200
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-20 11:48 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-20 12:09 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-20 18:04 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-20 19:35 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-20 20:41 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-21 09:41 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-21 18:46 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-22 10:25 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-22 16:51 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-22 17:51 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-22 19:40 -0700
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-22 21:29 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-23 07:57 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-23 16:37 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-23 17:45 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-23 18:36 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 12:17 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-24 14:23 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 14:46 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-24 18:06 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-25 03:10 +0200
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 20:40 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-24 20:53 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 21:24 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-24 21:40 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 22:13 -0700
Re: Einstein and Big Ben "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 22:30 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-24 22:53 -0700
Re: Einstein and Big Ben "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 23:55 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-24 23:40 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-25 00:18 -0700
Re: Einstein and Big Ben "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 00:28 -0700
Re: Einstein and Big Ben "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 00:30 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-25 00:47 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-25 00:54 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-25 10:55 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-25 16:05 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 02:29 -0700
Re: Einsteine Miquel Niftrik <lile@iulmqnqu.ni> - 2023-06-26 13:47 +0000
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 06:56 -0700
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-26 07:20 -0700
Re: Einstein and Big Ben Volney <volney@invalid.invalid> - 2023-06-26 14:02 -0400
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 11:12 -0700
Re: Einstein and Big Ben Volney <volney@invalid.invalid> - 2023-06-26 15:33 -0400
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-26 12:52 -0700
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-26 11:36 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 09:33 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 11:46 -0700
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-26 12:11 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 12:17 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 16:24 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 18:12 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 18:37 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 18:41 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 19:22 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 19:30 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 19:48 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 19:51 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 21:11 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 21:31 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-26 21:33 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-26 22:54 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 07:53 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 17:34 +0200
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-27 16:19 +0000
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 09:41 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 18:42 +0200
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 09:54 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 19:05 +0200
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-27 11:07 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 11:52 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 12:44 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 13:00 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 13:03 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 22:07 +0200
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 13:34 -0700
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 22:44 +0200
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-27 21:38 +0000
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 23:39 +0200
Re: Einstein and Big Ben Richard Hachel <r.hachel@tiscali.fr> - 2023-06-27 21:51 +0000
Re: Einstein and Big Ben Python <python@invalid.org> - 2023-06-27 23:54 +0200
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 14:52 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 16:33 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 17:49 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 20:43 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 20:55 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-27 21:00 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 01:10 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 01:26 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 06:38 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 07:55 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 08:10 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 08:58 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 08:14 -0700
Re: Einstein and Big Ben Paul Alsing <pnalsing@gmail.com> - 2023-06-28 09:21 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 09:50 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 09:52 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 10:32 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 10:46 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 11:10 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 11:16 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 11:52 -0700
Re: Einstein and Big Ben Athel Cornish-Bowden <athel.cb@gmail.com> - 2023-06-28 20:57 +0200
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 16:05 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 17:24 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-28 19:06 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-28 22:57 -0700
Re: Einstein and Big Ben Paul Alsing <pnalsing@gmail.com> - 2023-06-28 12:01 -0700
Re: Einstein and Big Ben Volney <volney@invalid.invalid> - 2023-06-28 14:34 -0400
Re: Einstein and Big Ben Braxton Ramakers <ssor@ekrattrt.mr> - 2023-06-28 21:31 +0000
Re: Einstein and Big Ben Woodrow Aalst <soas@ooaowlat.ow> - 2023-06-28 21:41 +0000
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-28 22:29 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-27 13:06 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-25 09:44 -0700
Re: Einstein and Big Ben whodat <whodaat@void.nowgre.com> - 2023-06-22 20:48 -0500
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-22 19:06 -0700
Re: Einstein and Big Ben whodat <whodaat@void.nowgre.com> - 2023-06-23 05:19 -0500
Re: Einstein and Big Ben Athel Cornish-Bowden <athel.cb@gmail.com> - 2023-06-23 11:13 +0200
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-23 02:22 -0700
Re: Einstein and Big Ben Thomas Heger <ttt_heg@web.de> - 2023-06-25 07:25 +0200
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-16 12:13 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 13:01 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-16 16:33 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 17:22 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-16 18:17 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-16 19:02 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-16 22:15 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-17 09:24 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-17 10:44 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-17 11:11 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-17 12:12 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-17 19:07 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-18 14:34 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-19 11:13 -0700
Re: Einstein and Big Ben Trevor Lange <trevorlange97@gmail.com> - 2023-06-19 12:04 -0700
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-19 12:20 -0700
Re: Einstein and Big Ben Tom Roberts <tjoberts137@sbcglobal.net> - 2023-06-19 14:40 -0500
Re: Einstein and Big Ben patdolan <patdolan@comcast.net> - 2023-06-19 16:04 -0700
Re: Einstein and Big Ben Tom Roberts <tjoberts137@sbcglobal.net> - 2023-06-19 19:34 -0500
Re: Einstein and Big Ben Maciej Wozniak <maluwozniak@gmail.com> - 2023-06-19 23:29 -0700
Re: Einstein and Big Ben The Starmaker <starmaker@ix.netcom.com> - 2023-06-20 23:31 -0700
Re: Einstein and Big Ben "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-18 19:58 -0700
Re: Einstein and Big Ben The Starmaker <starmaker@ix.netcom.com> - 2023-06-15 23:26 -0700
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 06:38 -0700 |
| Message-ID | <ffe83375-0344-4be8-be9a-aff2412cc988n@googlegroups.com> |
| In reply to | #613917 |
On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > > Let's be clear about this. Your grand argument is > > > > > > > > i^4 = i^8 .................... equivalent to (-1)^2 = (1)^2 and to 1=1 > > > > sqrt[ i^4 ] = sqrt[ i^8 ] ......... equivalent to sqrt[(-1)^2] = sqrt[(1)^2] > > > > +/- ( i^2 ) = +/- ( i^4 ) ...... equivalent to -1 = 1 (and to 1 = -1) > > > > > > > > Like every clueless newbie, you claim that by taking the square > > > > root of both side of (-1)^2 = (1)^2 we get -1 = 1. Sheesh. > > > > > > I didn't take the sqrt of both sides of (-1)^2 = (1)^2. I took the sqrt of > > > both sides of i^4 = i^8. > > > > Note: i^4 = (i^2)^2 = (-1)^2, and i^8 = (((i^2)^2)^2 = (1)^2. Making these substitutions, your grand argument is > > > > (-1)^2 = (1)^2 > > sqrt[(-1)^2] = sqrt[(1)^2] > > -1 = 1 > > > > Thus you are indeed committing the standard 3th grade fallacy beginning with (-1)^2 = (1)^2, and then taking the square root of both sides to give -1 = 1. To simplify still more, since i^2 = -1, i^4 = 1 and i^8 = 1, your grand argument is > > > > 1 = 1 > > sqrt(1) = sqrt(1) > > -1 = 1 > > > > You see, as explained before, your second claimed "equation" is not really an equation, because the expressions represent two roots, and then your third claimed "equation" is equating the negative root on the left with the positive root on the right. LOL If I had never seen your ideas on relativity, I would have said this is the dumbest thing I've ever seen. > > iiii = iiiiiiii > sqrt[ iiii ] = sqrt[ iiiiiiii ] Those are equivalent to 1=1 sqrt(1)=sqrt(1) the first of which is a tautology, and the second of which, without some further stipulation, is not an equality because the square root of a number has two values. > Principle, positive branch root: > -1 = 1 LOL. No, the principal value of an analytic multivalued function is simply a particular value chosen *by convention* to be returned for a given argument. This is sometimes designated differently, such as by capatilizing Log instead of log, or Sqrt instead of sqrt, or placing "pv" in front, such as "pv sqrt(z)", just to remind the newbie that some specific convention has been adopted. Now, for positive real arguments, one common convnetion is to select the positive root as the principal value, in which case we have Sqrt(1)=Sqrt(1) represents the tautology 1=1. In general, any complex number z can be expressed essentially uniquely in the form as r e^(iq) where r is a positive real number and q is a real phase angle, and by convenetion the principle value of the square root is Sqrt(r) e^[i(q/2)] where q is stipulated to be in the range -PI < q =< PI. Hence your conclusion is 1=1. Duh. > [You claimed] sqrt[ -1 x -1 ] = sqrt[ -1 ]sqrt[ -1 ]. Again, that is a lie. What I carefully stated repeatedly is that "The expression sqrt(1) represents either of two values, +1 and -1, and the expressions sqrt[(-1)(-1)] and sqrt(-1)sqrt(-1) also each represent either of two values, +1 and -1.", and I repeatedly informed you that what you have typed there is invalid, and why it is invalid. Remember?
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 07:55 -0700 |
| Message-ID | <ffcaf0de-b915-4af4-8a32-e45416b3096dn@googlegroups.com> |
| In reply to | #613925 |
On Wednesday, June 28, 2023 at 6:38:45 AM UTC-7, Trevor Lange wrote: > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > > > Let's be clear about this. Your grand argument is > > > > > > > > > > i^4 = i^8 .................... equivalent to (-1)^2 = (1)^2 and to 1=1 > > > > > sqrt[ i^4 ] = sqrt[ i^8 ] ......... equivalent to sqrt[(-1)^2] = sqrt[(1)^2] > > > > > +/- ( i^2 ) = +/- ( i^4 ) ...... equivalent to -1 = 1 (and to 1 = -1) > > > > > > > > > > Like every clueless newbie, you claim that by taking the square > > > > > root of both side of (-1)^2 = (1)^2 we get -1 = 1. Sheesh. > > > > > > > > I didn't take the sqrt of both sides of (-1)^2 = (1)^2. I took the sqrt of > > > > both sides of i^4 = i^8. > > > > > > Note: i^4 = (i^2)^2 = (-1)^2, and i^8 = (((i^2)^2)^2 = (1)^2. Making these substitutions, your grand argument is > > > > > > (-1)^2 = (1)^2 > > > sqrt[(-1)^2] = sqrt[(1)^2] > > > -1 = 1 > > > > > > Thus you are indeed committing the standard 3th grade fallacy beginning with (-1)^2 = (1)^2, and then taking the square root of both sides to give -1 = 1. To simplify still more, since i^2 = -1, i^4 = 1 and i^8 = 1, your grand argument is > > > > > > 1 = 1 > > > sqrt(1) = sqrt(1) > > > -1 = 1 > > > > > > You see, as explained before, your second claimed "equation" is not really an equation, because the expressions represent two roots, and then your third claimed "equation" is equating the negative root on the left with the positive root on the right. LOL If I had never seen your ideas on relativity, I would have said this is the dumbest thing I've ever seen. > > > > iiii = iiiiiiii > > sqrt[ iiii ] = sqrt[ iiiiiiii ] > Those are equivalent to > 1=1 > sqrt(1)=sqrt(1) > the first of which is a tautology, and the second of which, without some further stipulation, is not an equality because the square root of a number has two values. > > > Principle, positive branch root: > > -1 = 1 > > LOL. No, the principal value of an analytic multivalued function is simply a particular value chosen *by convention* By convention!!!! Show this forum the minutes of that convention or cite a reference or admit that you just made it up. to be returned for a given argument. This is sometimes designated differently, such as by capatilizing Log instead of log, or Sqrt instead of sqrt, or placing "pv" in front, such as "pv sqrt(z)", just to remind the newbie that some specific convention has been adopted. Now, for positive real arguments, one common convnetion is to select the positive root as the principal value, in which case we have Sqrt(1)=Sqrt(1) represents the tautology 1=1. > > In general, any complex number z can be expressed essentially uniquely in the form as r e^(iq) where r is a positive real number and q is a real phase angle, and by convenetion the principle value of the square root is Sqrt(r) e^[i(q/2)] where q is stipulated to be in the range -PI < q =< PI. Hence your conclusion is 1=1. Duh. > > > [You claimed] sqrt[ -1 x -1 ] = sqrt[ -1 ]sqrt[ -1 ]. > > Again, that is a lie. What I carefully stated repeatedly is that "The expression sqrt(1) represents either of two values, +1 and -1, and the expressions sqrt[(-1)(-1)] and sqrt(-1)sqrt(-1) also each represent either of two values, +1 and -1.", and I repeatedly informed you that what you have typed there is invalid, and why it is invalid. Remember? What you and Python are failing to understand is that the mechanics of arithmetic and its representations are ambiguous when i is included in the set of number over which arithmetic operates. You want it to be pure and true. It isn't. Just like most everything else in life. You and Python rejected the world in all its beauty and ugliness and instead chose the life of mathematical monks believing you had truth by the tail. Foolish little Legion.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 08:10 -0700 |
| Message-ID | <692197e4-0b88-4f49-a76d-c17b70fa9464n@googlegroups.com> |
| In reply to | #613930 |
On Wednesday, June 28, 2023 at 7:55:09 AM UTC-7, patdolan wrote: > On Wednesday, June 28, 2023 at 6:38:45 AM UTC-7, Trevor Lange wrote: > > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > > > > Let's be clear about this. Your grand argument is > > > > > > > > > > > > i^4 = i^8 .................... equivalent to (-1)^2 = (1)^2 and to 1=1 > > > > > > sqrt[ i^4 ] = sqrt[ i^8 ] ......... equivalent to sqrt[(-1)^2] = sqrt[(1)^2] > > > > > > +/- ( i^2 ) = +/- ( i^4 ) ...... equivalent to -1 = 1 (and to 1 = -1) > > > > > > > > > > > > Like every clueless newbie, you claim that by taking the square > > > > > > root of both side of (-1)^2 = (1)^2 we get -1 = 1. Sheesh. > > > > > > > > > > I didn't take the sqrt of both sides of (-1)^2 = (1)^2. I took the sqrt of > > > > > both sides of i^4 = i^8. > > > > > > > > Note: i^4 = (i^2)^2 = (-1)^2, and i^8 = (((i^2)^2)^2 = (1)^2. Making these substitutions, your grand argument is > > > > > > > > (-1)^2 = (1)^2 > > > > sqrt[(-1)^2] = sqrt[(1)^2] > > > > -1 = 1 > > > > > > > > Thus you are indeed committing the standard 3th grade fallacy beginning with (-1)^2 = (1)^2, and then taking the square root of both sides to give -1 = 1. To simplify still more, since i^2 = -1, i^4 = 1 and i^8 = 1, your grand argument is > > > > > > > > 1 = 1 > > > > sqrt(1) = sqrt(1) > > > > -1 = 1 > > > > > > > > You see, as explained before, your second claimed "equation" is not really an equation, because the expressions represent two roots, and then your third claimed "equation" is equating the negative root on the left with the positive root on the right. LOL If I had never seen your ideas on relativity, I would have said this is the dumbest thing I've ever seen. > > > > > > iiii = iiiiiiii > > > sqrt[ iiii ] = sqrt[ iiiiiiii ] > > Those are equivalent to > > 1=1 > > sqrt(1)=sqrt(1) > > the first of which is a tautology, and the second of which, without some further stipulation, is not an equality because the square root of a number has two values. > > > > > Principle, positive branch root: > > > -1 = 1 > > > > LOL. No, the principal value of an analytic multivalued function is simply a particular value chosen *by convention* > By convention!!!! Show this forum the minutes of that convention or cite a reference or admit that you just made it up. > to be returned for a given argument. This is sometimes designated differently, such as by capatilizing Log instead of log, or Sqrt instead of sqrt, or placing "pv" in front, such as "pv sqrt(z)", just to remind the newbie that some specific convention has been adopted. Now, for positive real arguments, one common convnetion is to select the positive root as the principal value, in which case we have Sqrt(1)=Sqrt(1) represents the tautology 1=1. > > > > In general, any complex number z can be expressed essentially uniquely in the form as r e^(iq) where r is a positive real number and q is a real phase angle, and by convenetion the principle value of the square root is Sqrt(r) e^[i(q/2)] where q is stipulated to be in the range -PI < q =< PI. Hence your conclusion is 1=1. Duh. > > > > > [You claimed] sqrt[ -1 x -1 ] = sqrt[ -1 ]sqrt[ -1 ]. > > > > Again, that is a lie. What I carefully stated repeatedly is that "The expression sqrt(1) represents either of two values, +1 and -1, and the expressions sqrt[(-1)(-1)] and sqrt(-1)sqrt(-1) also each represent either of two values, +1 and -1.", and I repeatedly informed you that what you have typed there is invalid, and why it is invalid. Remember? > What you and Python are failing to understand is that the mechanics of arithmetic and its representations are ambiguous when i is included in the set of number over which arithmetic operates. You want it to be pure and true. It isn't. Just like most everything else in life. You and Python rejected the world in all its beauty and ugliness and instead chose the life of mathematical monks believing you had truth by the tail. Foolish little Legion. We now come to the heart of the matter. If arithmetic and its sqrt operator cannot be trusted to unfailingly produce well formed and valid strings of arithmetic from other well formed, valid strings, then can we trust the validity of Lorentz transforms? I have shown elsewhere that the assumption that the "v" in the LTs is equivalent for all observers in all inertial frames is indeed a fallacious assumption.
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 08:58 -0700 |
| Message-ID | <19c6d4be-e152-458f-815a-b5b8da757f79n@googlegroups.com> |
| In reply to | #613930 |
On Wednesday, June 28, 2023 at 7:55:09 AM UTC-7, patdolan wrote: > > > > > > Let's be clear about this. Your grand argument is > > > > > > > > > > > > i^4 = i^8 .................... equivalent to (-1)^2 = (1)^2 and to 1=1 > > > > > > sqrt[ i^4 ] = sqrt[ i^8 ] ......... equivalent to sqrt[(-1)^2] = sqrt[(1)^2] > > > > > > +/- ( i^2 ) = +/- ( i^4 ) ...... equivalent to -1 = 1 (and to 1 = -1) > > > > > > > > > > > > Like every clueless newbie, you claim that by taking the square > > > > > > root of both side of (-1)^2 = (1)^2 we get -1 = 1. Sheesh. > > > > > > > > > > I didn't take the sqrt of both sides of (-1)^2 = (1)^2. I took the sqrt of > > > > > both sides of i^4 = i^8. > > > > > > > > Note: i^4 = (i^2)^2 = (-1)^2, and i^8 = (((i^2)^2)^2 = (1)^2. Making these substitutions, your grand argument is > > > > > > > > (-1)^2 = (1)^2 > > > > sqrt[(-1)^2] = sqrt[(1)^2] > > > > -1 = 1 > > > > > > > > Thus you are indeed committing the standard 3th grade fallacy beginning with (-1)^2 = (1)^2, and then taking the square root of both sides to give -1 = 1. To simplify still more, since i^2 = -1, i^4 = 1 and i^8 = 1, your grand argument is > > > > > > > > 1 = 1 > > > > sqrt(1) = sqrt(1) > > > > -1 = 1 > > > > > > > > You see, as explained before, your second claimed "equation" is not really an equation, because the expressions represent two roots, and then your third claimed "equation" is equating the negative root on the left with the positive root on the right. LOL If I had never seen your ideas on relativity, I would have said this is the dumbest thing I've ever seen. > > > > > > iiii = iiiiiiii > > > sqrt[ iiii ] = sqrt[ iiiiiiii ] > > Those are equivalent to > > 1=1 > > sqrt(1)=sqrt(1) > > the first of which is a tautology, and the second of which, without some further stipulation, is not an equality because the square root of a number has two values. > > > > > Principle, positive branch root: > > > -1 = 1 > > > > LOL. No, the principal value of an analytic multivalued function is simply a particular value chosen *by convention* > > By convention!!!! Show this forum the minutes of that convention... Was that an attempt at humor, or do you really not know the meaning of "convention"? > or cite a reference or admit that you just made it up. LOL. See any number of standard definitions of "principal value", or simply note for yourself that it is self-evidently conventional. Sheesh. Look, another (equivalent) way of explaining the standard newbie fallaicy to which you've succumbed is to say that you think the principal value of a product equals the product of the principal values. That is obviously false. Do you understand this?
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 08:14 -0700 |
| Message-ID | <f0247d70-2883-44ef-b960-d296ee7c21b1n@googlegroups.com> |
| In reply to | #613925 |
On Wednesday, June 28, 2023 at 6:38:45 AM UTC-7, Trevor Lange wrote: > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > > > Let's be clear about this. Your grand argument is > > > > > > > > > > i^4 = i^8 .................... equivalent to (-1)^2 = (1)^2 and to 1=1 > > > > > sqrt[ i^4 ] = sqrt[ i^8 ] ......... equivalent to sqrt[(-1)^2] = sqrt[(1)^2] > > > > > +/- ( i^2 ) = +/- ( i^4 ) ...... equivalent to -1 = 1 (and to 1 = -1) > > > > > > > > > > Like every clueless newbie, you claim that by taking the square > > > > > root of both side of (-1)^2 = (1)^2 we get -1 = 1. Sheesh. > > > > > > > > I didn't take the sqrt of both sides of (-1)^2 = (1)^2. I took the sqrt of > > > > both sides of i^4 = i^8. > > > > > > Note: i^4 = (i^2)^2 = (-1)^2, and i^8 = (((i^2)^2)^2 = (1)^2. Making these substitutions, your grand argument is > > > > > > (-1)^2 = (1)^2 > > > sqrt[(-1)^2] = sqrt[(1)^2] > > > -1 = 1 > > > > > > Thus you are indeed committing the standard 3th grade fallacy beginning with (-1)^2 = (1)^2, and then taking the square root of both sides to give -1 = 1. To simplify still more, since i^2 = -1, i^4 = 1 and i^8 = 1, your grand argument is > > > > > > 1 = 1 I see that you have learned how to use Dolan's rule. So you are not completely uneducable, Legion. > > > sqrt(1) = sqrt(1) > > > -1 = 1 > > > > > > You see, as explained before, your second claimed "equation" is not really an equation, because the expressions represent two roots, and then your third claimed "equation" is equating the negative root on the left with the positive root on the right. LOL If I had never seen your ideas on relativity, I would have said this is the dumbest thing I've ever seen. > > > > iiii = iiiiiiii > > sqrt[ iiii ] = sqrt[ iiiiiiii ] > Those are equivalent to > 1=1 > sqrt(1)=sqrt(1) > the first of which is a tautology, and the second of which, without some further stipulation, is not an equality because the square root of a number has two values. > > > Principle, positive branch root: > > -1 = 1 > > LOL. No, the principal value of an analytic multivalued function is simply a particular value chosen *by convention* to be returned for a given argument. This is sometimes designated differently, such as by capatilizing Log instead of log, or Sqrt instead of sqrt, or placing "pv" in front, such as "pv sqrt(z)", just to remind the newbie that some specific convention has been adopted. Now, for positive real arguments, one common convnetion is to select the positive root as the principal value, in which case we have Sqrt(1)=Sqrt(1) represents the tautology 1=1. > > In general, any complex number z can be expressed essentially uniquely in the form as r e^(iq) where r is a positive real number and q is a real phase angle, and by convenetion the principle value of the square root is Sqrt(r) e^[i(q/2)] where q is stipulated to be in the range -PI < q =< PI. Hence your conclusion is 1=1. Duh. > > > [You claimed] sqrt[ -1 x -1 ] = sqrt[ -1 ]sqrt[ -1 ]. > > Again, that is a lie. What I carefully stated repeatedly is that "The expression sqrt(1) represents either of two values, +1 and -1, and the expressions sqrt[(-1)(-1)] and sqrt(-1)sqrt(-1) also each represent either of two values, +1 and -1.", and I repeatedly informed you that what you have typed there is invalid, and why it is invalid. Remember?
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| From | Paul Alsing <pnalsing@gmail.com> |
|---|---|
| Date | 2023-06-28 09:21 -0700 |
| Message-ID | <e66bb9de-352f-4995-aceb-8f040db4a40dn@googlegroups.com> |
| In reply to | #613917 |
On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > ii = iiii https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html ii = 1 iiii = 1 So yes, ii = iiii = 1, and that is pretty much all there is to it.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 09:50 -0700 |
| Message-ID | <169d39b4-1804-4530-be16-6c783f7cd520n@googlegroups.com> |
| In reply to | #613938 |
On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote: > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > ii = iiii > > https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html > > ii = 1 > iiii = 1 > > So yes, ii = iiii = 1, and that is pretty much all there is to it. Huh? ii = -1. Let's chalk that up to you not having your coffee yet. An equation is a double edged sword. "ii equals 1" also means that "1 = ii". Or do you think that 1 has greater validity and is more useful than iiii? Every expression in math contains iiii as a divisor.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 09:52 -0700 |
| Message-ID | <5c8eb7ba-f921-428d-996b-f220d54919f2n@googlegroups.com> |
| In reply to | #613945 |
On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: > On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote: > > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > ii = iiii > > > > https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html > > > > ii = 1 > > iiii = 1 > > > > So yes, ii = iiii = 1, and that is pretty much all there is to it. > Huh? ii = -1. Let's chalk that up to you not having your coffee yet. > > An equation is a double edged sword. "ii equals 1" also means that "1 = ii". Or do you think that 1 has greater validity and is more useful than iiii? Every expression in math contains iiii as a divisor. *a denominator
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 10:32 -0700 |
| Message-ID | <48089cc5-fa3c-4145-bcc9-55e72751086bn@googlegroups.com> |
| In reply to | #613945 |
On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: > On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote: > > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > ii = iiii > > > > https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html > > > > ii = 1 > > iiii = 1 > > > > So yes, ii = iiii = 1, and that is pretty much all there is to it. > Huh? ii = -1. Let's chalk that up to you not having your coffee yet. > > An equation is a double edged sword. "ii equals 1" also means that "1 = ii". Or do you think that 1 has greater validity and is more useful than iiii? Every expression in math contains iiii as a divisor. This is brilliant! I have just found the means of making arithmetic say whatever I want it to say! For any polynomial and/or transcendental function in math p&t(x) p&t(x) = sqrt[ p&t^2(x) ] = sqrt[ p&t^2(x)/1 ] = sqrt[ p&t^2(x)/iiii ] = p&t(x)/ii = -p&t(x) Now if only the Nobel committee can be convinced to start giving prizes in math.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 10:46 -0700 |
| Message-ID | <637005d9-6a2c-4c80-90e0-ffff73c873d0n@googlegroups.com> |
| In reply to | #613953 |
On Wednesday, June 28, 2023 at 10:32:39 AM UTC-7, patdolan wrote: > On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: > > On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote: > > > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > > ii = iiii > > > > > > https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html > > > > > > ii = 1 > > > iiii = 1 > > > > > > So yes, ii = iiii = 1, and that is pretty much all there is to it. > > Huh? ii = -1. Let's chalk that up to you not having your coffee yet. > > > > An equation is a double edged sword. "ii equals 1" also means that "1 = ii". Or do you think that 1 has greater validity and is more useful than iiii? Every expression in math contains iiii as a divisor. > This is brilliant! I have just found the means of making arithmetic say whatever I want it to say! > > For any polynomial and/or transcendental function in math p&t(x) > > p&t(x) = sqrt[ p&t^2(x) ] = sqrt[ p&t^2(x)/1 ] = sqrt[ p&t^2(x)/iiii ] = p&t(x)/ii = -p&t(x) > > Now if only the Nobel committee can be convinced to start giving prizes in math. Fourier and Laplace move over. Make room for the Dolan Transform! AKA the Imaginary Transform. It allows you to remove any term or expression you like from any equation or a proof.
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 11:10 -0700 |
| Message-ID | <a8bfec8c-2dff-4702-8927-7f2d7bbb4c62n@googlegroups.com> |
| In reply to | #613954 |
On Wednesday, June 28, 2023 at 10:46:44 AM UTC-7, patdolan wrote: > > The principal value of an analytic multivalued function is simply a > > particular value chosen *by convention* > > By convention!!!! Show this forum the minutes of that convention... Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > or cite a reference or admit that you just made it up. LOL. See any number of standard definitions of "principal value", or simply note for yourself that it is self-evidently conventional. Example from Wolfram MathWorld: "The principal value of an analytic multivalued function is the single value *chosen by convention* to be returned for a given argument." Another (equivalent) way of explaining the standard newbie fallacy to which your diseased brain has succumbed is to say that you think the principal value of a product equals the product of the principal values. In other words, you think that S(xy) must equal S(x)S(y), which is not generally true. Now do you finally understand?
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 11:16 -0700 |
| Message-ID | <71629d7a-544c-47ce-bc99-3913032d619an@googlegroups.com> |
| In reply to | #613957 |
On Wednesday, June 28, 2023 at 11:10:51 AM UTC-7, Trevor Lange wrote: > On Wednesday, June 28, 2023 at 10:46:44 AM UTC-7, patdolan wrote: > > > The principal value of an analytic multivalued function is simply a > > > particular value chosen *by convention* > > > > By convention!!!! Show this forum the minutes of that convention... > Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > > or cite a reference or admit that you just made it up. > LOL. See any number of standard definitions of "principal value", or simply note for yourself that it is self-evidently conventional. Example from Wolfram MathWorld: "The principal value of an analytic multivalued function is the single value *chosen by convention* to be returned for a given argument." So some math values are open to determination by consensus? Where does that leave arithmetic as a supposed pure truth, or whatever silliness you described it awhile back. > > Another (equivalent) way of explaining the standard newbie fallacy to which your diseased brain has succumbed is to say that you think the principal value of a product equals the product of the principal values. In other words, you think that S(xy) must equal S(x)S(y), which is not generally true. Now do you finally understand?
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 11:52 -0700 |
| Message-ID | <cb262c4b-a3c4-4a71-a33a-f56e7a1b49fbn@googlegroups.com> |
| In reply to | #613960 |
On Wednesday, June 28, 2023 at 11:16:18 AM UTC-7, patdolan wrote: > > > > The principal value of an analytic multivalued function is simply a > > > > particular value chosen *by convention* > > > > > > By convention!!!! Show this forum the minutes of that convention... > > Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > > > or cite a reference or admit that you just made it up. > > From Wolfram MathWorld: "The principal value of an analytic multivalued function is the single value *chosen by convention* to be returned for a given argument." > > So some math values are open to determination by consensus? You misunderstood. Again, we can use whatever symbols we choose to designate whatever values we want to discuss. This is called "being a grown-up". Remember, any complex number z can be expressed essentially uniquely in the form z = r e^(iq) where r is a positive real number and q is a real phase angle in the range -PI < q =< PI, and by convenetion the principle value of the square root of z (not to be confused with the square root of z) is pvsqrt(z) = is pvsqrt(r) e^[i(q/2)] . Notice that the principal value of a product need not equal the product of the principal values. In other words, S(xy) need not equal S(x)S(y). Now do you finally understand?
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| From | Athel Cornish-Bowden <athel.cb@gmail.com> |
|---|---|
| Date | 2023-06-28 20:57 +0200 |
| Message-ID | <kg3e4fF7h3nU1@mid.individual.net> |
| In reply to | #613953 |
On 2023-06-28 17:32:37 +0000, patdolan said: > On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: >> On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote:> > >> On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote:> > > >> ii = iiii> >> > >> https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html> >> >> > ii = 1> > iiii = 1> >> > So yes, ii = iiii = 1, and that is pretty >> much all there is to it. >> Huh? ii = -1. Let's chalk that up to you not having your coffee yet.>> >> An equation is a double edged sword. "ii equals 1" also means that "1 = >> ii". Or do you think that 1 has greater validity and is more useful >> than iiii? Every expression in math contains iiii as a divisor. > This is brilliant! I have just found the means of making arithmetic > say whatever I want it to say! > > For any polynomial and/or transcendental function in math p&t(x) > > p&t(x) = sqrt[ p&t^2(x) ] = sqrt[ p&t^2(x)/1 ] = sqrt[ p&t^2(x)/iiii ] > = p&t(x)/ii = -p&t(x) > > Now if only the Nobel committee can be convinced to start giving prizes > in math. You don't know about the Fields Medal? -- Athel -- French and British, living in Marseilles for 36 years; mainly in England until 1987.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 16:05 -0700 |
| Message-ID | <36adf2a1-3949-4680-9f48-b4ce187b9a94n@googlegroups.com> |
| In reply to | #613969 |
On Wednesday, June 28, 2023 at 11:57:24 AM UTC-7, Athel Cornish-Bowden wrote: > On 2023-06-28 17:32:37 +0000, patdolan said: > > > On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: > >> On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote:> > > >> On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote:> > > > >> ii = iiii> >> > > >> https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html> >> > >> > ii = 1> > iiii = 1> >> > So yes, ii = iiii = 1, and that is pretty > >> much all there is to it. > >> Huh? ii = -1. Let's chalk that up to you not having your coffee yet.>> > >> An equation is a double edged sword. "ii equals 1" also means that "1 = > >> ii". Or do you think that 1 has greater validity and is more useful > >> than iiii? Every expression in math contains iiii as a divisor. > > This is brilliant! I have just found the means of making arithmetic > > say whatever I want it to say! > > > > For any polynomial and/or transcendental function in math p&t(x) > > > > p&t(x) = sqrt[ p&t^2(x) ] = sqrt[ p&t^2(x)/1 ] = sqrt[ p&t^2(x)/iiii ] > > = p&t(x)/ii = -p&t(x) > > > > Now if only the Nobel committee can be convinced to start giving prizes > > in math. > You don't know about the Fields Medal? > > > -- > Athel -- French and British, living in Marseilles for 36 years; mainly > in England until 1987. I think I may have won that one already. Let me check my top dresser drawer and I'll let you know.
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 17:24 -0700 |
| Message-ID | <27ef5f37-fbdf-4cd7-8d82-5a57a6749067n@googlegroups.com> |
| In reply to | #613990 |
On Wednesday, June 28, 2023 at 4:05:49 PM UTC-7, patdolan wrote: > > > > The principal value of an analytic multivalued function is simply a > > > > particular value chosen *by convention* > > > > > > By convention!!!! Show this forum the minutes of that convention... > > > > Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > > > > > or cite a reference or admit that you just made it up. > > > > From Wolfram MathWorld: "The principal value of an analytic multivalued function is the single value *chosen by convention* to be returned for a given argument." > > So some math values are open to determination by consensus? As always, you misunderstood. Again, we can use whatever symbols we choose to designate whatever values we want to discuss. This is called "being a grown-up". Remember, any non-zero complex number z can be expressed uniquely in the form z = r e^(iq) where r is a positive real number and q is a real number in the range -PI < q =< PI, and by convention the principal value of the square root of z (not to be confused with the square root of z) is pvsqrt(z) = pvsqrt(r) e^[i(q/2)] . Notice that the principal value of a product of factors need not equal the product of the principal values of the factors. In other words, S(xy) need not equal S(x)S(y). For example, letting S denote the principal value (per the above conventional definition) of the square root, we have S((-1)(-1)) = 1 and S(-1)S(-1) = -1. The standard newbie fallacy you are committing is thinking those two things must be equal. Now do you finally understand?
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2023-06-28 19:06 -0700 |
| Message-ID | <5d7a7c9e-d368-47b4-ab28-d87736d4ae76n@googlegroups.com> |
| In reply to | #613997 |
On Wednesday, June 28, 2023 at 5:24:35 PM UTC-7, Trevor Lange wrote: > On Wednesday, June 28, 2023 at 4:05:49 PM UTC-7, patdolan wrote: > > > > > The principal value of an analytic multivalued function is simply a > > > > > particular value chosen *by convention* > > > > > > > > By convention!!!! Show this forum the minutes of that convention... > > > > > > Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > > > > > > > or cite a reference or admit that you just made it up. > > > > > > From Wolfram MathWorld: "The principal value of an analytic multivalued function is the single value *chosen by convention* to be returned for a given argument." > > > > So some math values are open to determination by consensus? > As always, you misunderstood. Again, we can use whatever symbols we choose to designate whatever values we want to discuss. This is called "being a grown-up". Remember, any non-zero complex number z can be expressed uniquely in the form z = r e^(iq) where r is a positive real number and q is a real number in the range -PI < q =< PI, and by convention the principal value of the square root of z (not to be confused with the square root of z) is pvsqrt(z) = pvsqrt(r) e^[i(q/2)] . > > Notice that the principal value of a product of factors need not equal the product of the principal values of the factors. In other words, S(xy) need not equal S(x)S(y). For example, letting S denote the principal value (per the above conventional definition) of the square root, we have S((-1)(-1)) = 1 and S(-1)S(-1) = -1. The standard newbie fallacy you are committing is thinking those two things must be equal. Now do you finally understand? The problem with Legion is that he always lets me have the last word in a thread. He just disappears.
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| From | Trevor Lange <trevorlange97@gmail.com> |
|---|---|
| Date | 2023-06-28 22:57 -0700 |
| Message-ID | <7e71d56b-5801-482e-92ff-7ca4724692e7n@googlegroups.com> |
| In reply to | #614001 |
On Wednesday, June 28, 2023 at 7:06:44 PM UTC-7, patdolan wrote: > Did you just accept it ... when in college the professor drew the yellow > chalk coordinate system on top of the white chalk coordinate system and > all of a sudden you believed in relativity? The illustration of the fallacy of your claim is not a demonstration of special relativity, it's simply exploding your fallacious logic. If you want someone to teach you special relativity, you're going about it the wrong way. The point of the illustration is that, regardless of whether the trajectory of the ball is described in terms of the white coordinate system or the yellow coordinate system, it still goes into the cup. Yes, the coordinates of the ball at the end of its journey are different, but the coordinates of the cup are also different, and they match, so the ball goes into the cup. Your belief that, when described in terms of one system it goes into the cup, but when described in terms of the other system it must physically karoom into the sand trap is just bizzare. > > > > The principal value of an analytic multivalued function is simply a > > > > particular value chosen *by convention* > > > > > > By convention!!!! Show this forum the minutes of that convention... > > > > Was that an attempt at humor, or do you honestly not know the meaning of "convention"? > > > > > or cite a reference or admit that you just made it up. > > > > From Wolfram MathWorld: "The principal value of an analytic multivalued > > function is the single value *chosen by convention* to be returned for a given > > argument." > > So some math values are open to determination by consensus? As always, you misunderstood. Again, we can use whatever symbols we choose to designate whatever values we want to discuss. Remember, any non-zero complex number z can be expressed uniquely in the form z = r e^(iq) where r is a non-negative real number and q is a real number in the range -PI < q =< PI, and by convention the principal value of the square root of z is pvsqrt(z) = pvsqrt(r) e^[i(q/2)] . Notice that the principal value of a product of factors need not equal the product of the principal values of the factors. In other words, S(xy) need not equal S(x)S(y). For example, letting S denote the principal value (per the conventional definition) of the square root, we have S((-1)(-1)) = 1 and S(-1)S(-1) = -1. The standard newbie fallacy you are committing is thinking those two things must be equal. If you still don't understand any of the above answers to your questions, go ahead and ask for more clarification. If not, then, you're welcome.
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| From | Paul Alsing <pnalsing@gmail.com> |
|---|---|
| Date | 2023-06-28 12:01 -0700 |
| Message-ID | <37f7d563-2f2a-4f1b-b422-7f083d8539a4n@googlegroups.com> |
| In reply to | #613945 |
On Wednesday, June 28, 2023 at 9:51:01 AM UTC-7, patdolan wrote: > On Wednesday, June 28, 2023 at 9:21:55 AM UTC-7, Paul Alsing wrote: > > On Wednesday, June 28, 2023 at 1:10:17 AM UTC-7, patdolan wrote: > > > ii = iiii > > > > https://mathbitsnotebook.com/Algebra2/ComplexNumbers/CPPowers.html > > > > ii = 1 > > iiii = 1 > > > > So yes, ii = iiii = 1, and that is pretty much all there is to it. > Huh? ii = -1. Let's chalk that up to you not having your coffee yet. Right, no coffee... earlier you had stated... > i^4 = i^8 But i^4 = 1 and i^8 = 1, and THAT is pretty much all there is to it!
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| From | Volney <volney@invalid.invalid> |
|---|---|
| Date | 2023-06-28 14:34 -0400 |
| Message-ID | <u7hufg$1r5cg$3@dont-email.me> |
| In reply to | #613917 |
On 6/28/2023 4:10 AM, patdolan wrote: > Principle, positive branch root: > ii = iiii > -1 = 1 Your mistake here is assuming the "principle" square root is the only one valid here. > > Negative branch root: > - ( ii ) = - ( iiii ) > 1 = -1 Your mistake here is assuming the root that's NOT the "principle" square root is the only one valid here. The square root "function" is not a simple function because it has two possible values, each as valid as the other.
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