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December 5th, 2017, 05:06 PM   #21
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Originally Posted by Conway51 View Post
0(the element) + (-0)( the additive element) = 0

0 + (-0) =/= (-0)
You have again not answered the point that I made. Put in a different way we have $$\left.\begin{aligned}n &= n + 0 \\ (- n) + n & = 0 \end{aligned}\right\} \quad \left\{\begin{aligned}n &= n + (-0) \\ (- n) + n & = (-0) \end{aligned}\right.$$
Taken together we then have $$(-0) =(- n) + n = 0$$ as I pointed out before.

Also, if $(-0)$ is the additive identity, we get $$0 = 0+0=\big((-0)+0\big) + \big(0+(-0)\big) = (-0) + (0+0) + (-0) = (-0) + (-0) = (-0)$$
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December 5th, 2017, 05:11 PM   #22
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Without meaning to be obnoxious, is there not a theorem in group theory that proves an identity element to be unique? (Sorry: it has been literally decades since I worried about group theory, and I never studied the theory of rings or fields.)

Whether your new mathematics is a group or not, what is the point of this particular new set of axioms? I GET that one view of mathematics is that it is simply a particularly difficult game, but I do not see why I should want to play this particular variant of the game. Perhaps computers may actually view - 0 and + 0 as distinct floating point numbers, and the consequences of that distinction may be important. Without, however, any clue given as to why a different set of axioms may be worthwhile or even amusing to explore, I am a bit reluctant to drop down a rabbit hole for no particular reason.
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December 5th, 2017, 05:33 PM   #23
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Originally Posted by v8archie View Post
So you have $$n = n \times (-0) = n \times 0 = 0$$
Or is transitivity broken?
I now understand what you were asking me here. I apologize for the miscommunication here. It is along the same lines as what Romesk helped with. You pointed it out first. I failed to understand you.

Again please allow the following edits

0 =/= (-0)
|0| = |-0|

I apologize
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December 5th, 2017, 05:35 PM   #24
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Originally Posted by romsek View Post
so in your system both $(-0)$ and presumably $1$ (is this incorrect?) are multiplicative identities?

Is the use of $(-0)$ constrained somehow? It seems to be what some would call an infinitesimal. But the multiplicative identity property rules this out.
yes both (-0) and 1 are multiplicative identites

(-0) is constrained to being the divisor in division....always

Last edited by Conway51; December 5th, 2017 at 05:51 PM.
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December 5th, 2017, 05:42 PM   #25
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Quote:
Originally Posted by v8archie View Post
You have again not answered the point that I made. Put in a different way we have $$\left.\begin{aligned}n &= n + 0 \\ (- n) + n & = 0 \end{aligned}\right\} \quad \left\{\begin{aligned}n &= n + (-0) \\ (- n) + n & = (-0) \end{aligned}\right.$$
Taken together we then have $$(-0) =(- n) + n = 0$$ as I pointed out before.

Also, if $(-0)$ is the additive identity, we get $$0 = 0+0=\big((-0)+0\big) + \big(0+(-0)\big) = (-0) + (0+0) + (-0) = (-0) + (-0) = (-0)$$
Your equation is wrong. (-0) + 0 =/= (-0)
Your equation is wrong. (-0) + (-0) =/= (-0)
as stated in the op

(-n) + (-0) = -n
(-n) + (0) = -n
therefore
(-n) + (n) = 0
therfore
(-0) + 0 = 0
(-0) + (-0) = 0

Last edited by Conway51; December 5th, 2017 at 05:49 PM.
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December 5th, 2017, 05:47 PM   #26
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Originally Posted by JeffM1 View Post
Without meaning to be obnoxious, is there not a theorem in group theory that proves an identity element to be unique? (Sorry: it has been literally decades since I worried about group theory, and I never studied the theory of rings or fields.)

Whether your new mathematics is a group or not, what is the point of this particular new set of axioms? I GET that one view of mathematics is that it is simply a particularly difficult game, but I do not see why I should want to play this particular variant of the game. Perhaps computers may actually view - 0 and + 0 as distinct floating point numbers, and the consequences of that distinction may be important. Without, however, any clue given as to why a different set of axioms may be worthwhile or even amusing to explore, I am a bit reluctant to drop down a rabbit hole for no particular reason.


I am not aware of any such theorems. I would be glad to.

The last two sentences are the particular reason you may or may not want to drop down this rabbit hole
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December 5th, 2017, 06:08 PM   #27
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Quote:
Originally Posted by Conway51 View Post
I am not aware of any such theorems. I would be glad to.

The last two sentences are the particular reason you may or may not want to drop down this rabbit hole
OK. I did a bit of research. Apparently my recollection was wrong. That each element of the set defined by the group has a unique left and right identity element is an axiom used to prove that there is only one identity element in the group. Therefore, if there not a unique identity element, it is not a group. OK. No problem.

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Without, however, any clue given as to why a different set of axioms may be worthwhile or even amusing to explore, I am a bit reluctant to drop down a rabbit hole for no particular reason.
I am mystified. How does that last sentence help me to decide whether to drop down ANY rabbit hole.

The next-to-last sentence may be what you were referring to. If the imprecision of floating point numbers is the proposed motivation, my next question is why is this not satisfactorily covered in the theory of errors? In fact, when I studied numerical methods back in ancient times, the theory of errors was where the whole course started. The computer doesn't work quite right so you sort a set of numbers before adding them to reduce potential error, etc. It was a sort of revelation for me about non-Platonic mathematics. The computer is not precise; the data are always somewhat dirty, so how do you cope? I built a rather successful business career on understanding, but avoiding naive credulity about, mathematical results.
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December 5th, 2017, 06:17 PM   #28
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OK. I did a bit of research. Apparently my recollection was wrong. That each element of the set defined by the group has a unique left and right identity element is an axiom used to prove that there is only one identity element in the group. Therefore, if there not a unique identity element, it is not a group. OK. No problem.


I am mystified. How does that last sentence help me to decide whether to drop down ANY rabbit hole.

The next-to-last sentence may be what you were referring to. If the imprecision of floating point numbers is the proposed motivation, my next question is why is this not satisfactorily covered in the theory of errors? In fact, when I studied numerical methods back in ancient times, the theory of errors was where the whole course started. The computer doesn't work quite right so you sort a set of numbers before adding them to reduce potential error, etc. It was a sort of revelation for me about non-Platonic mathematics. The computer is not precise; the data are always somewhat dirty, so how do you cope? I built a rather successful business career on understanding, but avoiding naive credulity about, mathematical results.
Well there-in is much philosophy. Let me just make a statement for the entire set of readers.

"This is an exercise in theory and nothing more."

However....it is my BELIEF that computers have trouble because division by zero is undefined...

Find a way to define it....rewrite binary and computer "math" accordingly...and.....all the extra steps needed for computers to find a away around division by zero is no longer needed.



Consider the last sentence....

Therefore...math...is relative.

Last edited by Conway51; December 5th, 2017 at 06:29 PM.
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December 5th, 2017, 08:00 PM   #29
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I am not aware of any such theorems.
I demonstrated in your last thread.
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December 5th, 2017, 08:03 PM   #30
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Quote:
Originally Posted by Conway51 View Post
Your equation is wrong. (-0) + 0 =/= (-0)
I don't believe I said that it was.
Quote:
Originally Posted by Conway51 View Post
Your equation is wrong. (-0) + (-0) =/= (-0)
So it's not an additive identity element as you claimed. I'm rather afraid that all these rules turn out to be completely arbitrary and, as likely as not, contradictory.
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