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 November 9th, 2017, 06:28 PM #11 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,342 Thanks: 984 Math Focus: Wibbly wobbly timey-wimey 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 Thanks from Conway51 November 9th, 2017, 06:30 PM   #12
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 Originally Posted by topsquark 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
Well credit where credit is due...you did suggest to use projection operators...but you said and I paraphrase

"I don't know what they would be for zero"

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
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There's no need to be so aggressive. Nobody has said anything remotely inflammatory.
Quote:
 Originally Posted by Conway51 Addressing the multiplicative inverse of 0....the additive identity. Please note....zero stays the additive identity. Let this be our definition of multiplicative inverses...(we can change it if you like) https://www.merriam-webster.com/dict...tive%20inverse 0's multiplicative inverse is 1..................current 0 has no defined multiplicative inverse 1's multiplicative inverse is 0.......................current 1's multiplicative inverse is 1. all other multiplicative inverses work the same with out change *note it remains true that 1 is an inverse of itself as well as zero.* As it currently is an inverse of itself this is of little note at this time.
Inverses are unique (at least if commutativity and associativity are preserved). If $v$ is the inverse of both $a$ and $b$, and $I$ is the identity element, then we can form $avb$. This can be evaluated in two ways $(av)b$ and $a(vb)$ as below:
$$b=Ib=(av)b = avb = a(vb) =aI = a$$ November 9th, 2017, 06:38 PM   #14
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 Originally Posted by v8archie There's no need to be so aggressive. Nobody has said anything remotely inflammatory. Inverses are unique (at least if commutativity and associativity are preserved). If $v$ is the inverse of both $a$ and $b$, and $I$ is the identity element, then we can form $avb$. This can be evaluated in two ways $(av)b$ and $a(vb)$ as below: $$b=Ib=(av)b = avb = a(vb) =aI = a$$

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
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 Originally Posted by v8archie There's no need to be so aggressive. Nobody has said anything remotely inflammatory. Inverses are unique (at least if commutativity and associativity are preserved). If $v$ is the inverse of both $a$ and $b$, and $I$ is the identity element, then we can form $avb$. This can be evaluated in two ways $(av)b$ and $a(vb)$ as below: $$b=Ib=(av)b = avb = a(vb) =aI = a$$
V8archie

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
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 Originally Posted by Conway51 Well credit where credit is due...you did suggest to use projection operators...but you said and I paraphrase "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...
I re-read your first post here. I see a bit more of what you are aiming for and it would seem that 0.z1 has a "value" of 0 and z1 represents some kind of space. I suggested the notation to separate the two as a way to not have the two entangled (if that would be the word), not to add a level of complexity. I didn't suggest it to make things more complicated.

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
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 Originally Posted by topsquark I re-read your first post here. I see a bit more of what you are aiming for and it would seem that 0.z1 has a "value" of 0 and z1 represents some kind of space. I suggested the notation to separate the two as a way to not have the two entangled (if that would be the word), not to add a level of complexity. I didn't suggest it to make things more complicated. 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
You took credit for the use of projection operators......still not sure why then....

Do you or do you not then see some progress here? November 9th, 2017, 08:37 PM   #18
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 Originally Posted by Conway51 You took credit for the use of projection operators......still not sure why then.... Do you or do you not then see some progress here?
Progress I guess. I'm still not clear on what you are doing with the two component "numbers."

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
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 Originally Posted by topsquark Progress I guess. I'm still not clear on what you are doing with the two component "numbers." 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
Well......if it is applicable to defining division by zero as well as the other goals mentioned...

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
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 Originally Posted by Conway51 V8archie 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
Yes, so axcording to your system, 1=0. That's quite catastrophic for a n7mber system. It means that all numbers are equal. Tags approach Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post VisionaryLen Calculus 19 October 28th, 2016 12:29 PM alikim Elementary Math 6 June 10th, 2015 10:35 AM JohnA Algebra 2 February 19th, 2012 10:29 AM cmmcnamara Advanced Statistics 4 February 10th, 2010 06:49 AM

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