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December 6th, 2017, 07:15 AM   #41
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Quote:
 Originally Posted by Conway51 I have specified the class of things (-0) belongs to. I stated clearly that (-0) is an integer. I stated that it belonged to the zero element set... I thank you for your time.
But that standard arithmetic is a ring. And your construction is not a ring. So there is an inconsistency. That's enough of my time.

December 6th, 2017, 07:17 AM   #42
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Quote:
 Originally Posted by JeffM1 But that standard arithmetic is a ring. And your construction is not a ring. So there is an inconsistency. That's enough of my time.
I believe it to be a ring.....and a field....but neither is the exercise....thank you for your time.

December 6th, 2017, 08:25 AM   #43
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 Originally Posted by Conway51 Perhaps you would wish to take up the debate with jebm1....as this person allows there is NO proof showing there is only one unique additive identity. I however wish to remain on topic.
If $I_1$ and $I_1$ are both identity elements then
\begin{align*}
I_1 &= I_1 + I_2 &\text{because $I_2$ is an identity element} \\
I_2 &= I_1 + I_2 &\text{because $I_1$ is an identity element} \\
Thus \\
I_1 &= I_2 &\text{by transitivity}\end{align*}
So there is only one identity element for each operation. We can't even escape by doing away with commutativity. The only way to get out of this is for $I_1 + I_2$ to take more than one value, but then the questions, for any given scenario, are which and why?

Last edited by v8archie; December 6th, 2017 at 08:30 AM.

December 6th, 2017, 08:54 AM   #44
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Quote:
 Originally Posted by JeffM1 That's enough of my time.
Thanks Jeff.

Last edited by skipjack; December 6th, 2017 at 10:45 AM.

December 6th, 2017, 09:12 AM   #45
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Quote:
 Originally Posted by v8archie If $I_1$ and $I_1$ are both identity elements then \begin{align*} I_1 &= I_1 + I_2 &\text{because $I_2$ is an identity element} \\ I_2 &= I_1 + I_2 &\text{because $I_1$ is an identity element} \\ Thus \\ I_1 &= I_2 &\text{by transitivity}\end{align*} So there is only one identity element for each operation. We can't even escape by doing away with commutativity. The only way to get out of this is for $I_1 + I_2$ to take more than one value, but then the questions, for any given scenario, are which and why?
I_1 =/= I_2

Nothing says that two additive identity elements Must be equal
TRANSITIVITY does not work for the additive identity elements/but remains the same elsewhere
I_1 can NOT replace I_2
This does NOT mean addition doesn't work for the additive identity elements

Last edited by Conway51; December 6th, 2017 at 09:14 AM.

December 6th, 2017, 09:13 AM   #46
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Originally Posted by studiot
Quote:
 Originally Posted by JeffM1 That's enough of my time.
Thanks Jeff.

Last edited by skipjack; December 6th, 2017 at 10:49 AM.

December 6th, 2017, 09:23 AM   #47
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Quote:
 Originally Posted by Conway51 TRANSITIVITY does not work for the additive identity elements/but remains the same elsewhere I_1 can NOT replace I_2 This does NOT mean addition doesn't work for the additive identity elements
Transitivity is a property of relations, not of elements. Also, if $I_1 \ne I_2$ then $n \ne n$ for all numbers $n$ because $n=n+I_1$ and $n=n+I_2$.

Your system appears to be degenerating into incoherence. Given any number, nobody can tell which version of the number it is.

Last edited by greg1313; December 6th, 2017 at 09:45 AM.

December 6th, 2017, 09:43 AM   #48
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Quote:
...and yours is no better. Thread closed.

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