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Beyond + and -

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Beyond + and -
JmsNxn Offline
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#2
08/17/2012, 01:25 AM (This post was last modified: 08/17/2012, 01:26 AM by JmsNxn.)
Your operator is very simple actually:

x§y = \sgn(x) (|x|-y)

Nothing really special about it at all


0 § y is absolutely nothing because \sgn(0) is inexistent

And the solution to your algebraic equation (-X+Y=X+Y) is X = 0
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Messages In This Thread
Beyond + and - - by Benny - 08/16/2012, 05:35 PM
RE: Beyond + and - - by JmsNxn - 08/17/2012, 01:25 AM
RE: Beyond + and - - by Benny - 08/17/2012, 11:58 AM
RE: Beyond + and - - by tommy1729 - 08/17/2012, 03:18 PM
RE: Beyond + and - - by Benny - 08/19/2012, 05:24 PM
RE: Beyond + and - - by tommy1729 - 08/21/2012, 11:04 PM
RE: Beyond + and - - by tommy1729 - 08/22/2012, 11:04 AM
RE: Beyond + and - - by Benny - 08/22/2012, 12:09 PM
RE: Beyond + and - - by tommy1729 - 08/22/2012, 02:05 PM
RE: Beyond + and - - by Benny - 08/22/2012, 06:34 PM
RE: Beyond + and - - by tommy1729 - 08/22/2012, 10:07 PM
RE: Beyond + and - - by tommy1729 - 08/22/2012, 10:20 PM
RE: Beyond + and - - by Benny - 08/23/2012, 12:08 PM
RE: Beyond + and - - by tommy1729 - 08/23/2012, 02:12 PM
RE: Beyond + and - - by Benny - 08/23/2012, 03:33 PM
RE: Beyond + and - - by tommy1729 - 08/23/2012, 04:12 PM
RE: Beyond + and - - by Benny - 08/23/2012, 07:41 PM
RE: Beyond + and - - by hixidom - 05/28/2014, 01:40 AM
RE: Beyond + and - - by JmsNxn - 05/28/2014, 04:33 PM
RE: Beyond + and - - by hixidom - 05/28/2014, 04:46 PM
RE: Beyond + and - - by tommy1729 - 05/28/2014, 10:44 PM



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