November 9th, 2017, 06:28 PM  #11 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff. 
You know, I gave you the suggestion of using $\displaystyle \left ( \begin{matrix} z_1 \\ z_2 \end{matrix} \right ) $ to find away around the notations 0.z1 and 0.z2, but you still have the clumsy notation going. I'd set $\displaystyle 0.z1 = \left ( \begin{matrix} 0 \\ z_1 \end{matrix} \right ) $ and $\displaystyle 0.z2 = \left ( \begin{matrix} 0 \\ z_2 \end{matrix} \right ) $ Both have a "value" of 0 and you can set the space of the z's to be and act in any way you like. Dan 
November 9th, 2017, 06:30 PM  #12  
Member Joined: Aug 2012 Posts: 67 Thanks: 0  Quote:
"I don't know what they would be for zero" something like that....then made no reply....thanks by the way... still credit where it's due... I take you "taking" your credit as a sign of positive progress for this idea...  
November 9th, 2017, 06:36 PM  #13  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,042 Thanks: 2344 Math Focus: Mainly analysis and algebra 
There's no need to be so aggressive. Nobody has said anything remotely inflammatory. Quote:
$$b=Ib=(av)b = avb = a(vb) =aI = a$$  
November 9th, 2017, 06:38 PM  #14  
Member Joined: Aug 2012 Posts: 67 Thanks: 0  Quote:
These same equations can be extended to include representations of the field axioms. Also the distributive property is addressed in the op. (with all combinations of 0) Last edited by Conway51; November 9th, 2017 at 06:43 PM.  
November 9th, 2017, 07:06 PM  #15  
Member Joined: Aug 2012 Posts: 67 Thanks: 0  Quote:
I apologize....I should not ask you to read another post... a = 1 , b = 0, v = 1, I = 0 1 is the inverse of both 1 and 0....see post #7 b = Ib = (av)b = avb = a(vb) = aI = a 0 = 0 * (0.z1) = ( 1 * 1 ) * 0.z2 = 1 * 1 * 0.z2 = 1 ( 1 * 0.z2 ) = 1 * 0.z2 = 1  
November 9th, 2017, 07:34 PM  #16  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
0.z1 has a value of 0 and a space z1. Why do they need to be entangled the way you are doing? The way you define 0.z1 seems to suggest a vector and I gave you a way to make it so Mathematically. And I'm not taking any credit... I made a simple suggestion. This is your baby. Dan  
November 9th, 2017, 07:40 PM  #17  
Member Joined: Aug 2012 Posts: 67 Thanks: 0  Quote:
Do you or do you not then see some progress here?  
November 9th, 2017, 08:37 PM  #18  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
My best guess for one of the things you are trying to do with the z's is to generalize the concept of a physical unit? I've messed with this kind of thing before and didn't find any result that was worth the effort. Or am I off track? Dan  
November 9th, 2017, 08:51 PM  #19  
Member Joined: Aug 2012 Posts: 67 Thanks: 0  Quote:
Then consider in your words the following... 3 = 3 values in 3 spaces classic 3 = (1,1,1)values + ( _ , _ , _ )spaces = ( 1 , 1 , 1 ) = 3 new 2 = 2 values in 2 spaces classic 2 = (1,1)values + ( _ , _ )spaces = ( 1 , 1 ) = 2 new 1 = 1 value in 1 space classic 1 = (1)value + ( _ )space = ( 1 ) = 1 new 0 = 0 value in 1 space classic 0 = (0)value + ( _ )space = ( 0 ) = 0 new 0 = (0.z1, 0.z2) multiplication is "taking" the VALUE from one number and "putting" them in the SPACE of "another" number...then adding all values in all spaces. division follows but inverse  
November 10th, 2017, 03:25 AM  #20  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,042 Thanks: 2344 Math Focus: Mainly analysis and algebra  Quote:
 

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