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December 5th, 2017, 11:45 AM   #11
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Quote:
 Originally Posted by Conway51 No where in the original post does the equation.... n = n * ( -0 ) = n * 0 = 0 ONLY the equations. n * (-0) = n n * ( 0 ) = 0
you posted $(-0)=0$

first line

Equality doesn't mean what you think it means.

December 5th, 2017, 12:44 PM   #12
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Quote:
 Originally Posted by romsek you posted $(-0)=0$ first line Equality doesn't mean what you think it means.
perhaps...but if you chose to not offer examples then your time and my time is wasted...

further...

while I did claim

0 = (-0)

I did not claim anywhere.....
n = n * ( -0 ) = n * 0 = 0
neither did I claim...
n + (-n) = (-0)
it is
n + (-n) = 0
n + (-0) = 0

the addition of the element (-0) is dictated by the additive identity property...not by me.

Fortunately for me it is easy to say that....

The absolute value of 0 and of (-0) are equivalent
and that...
(0) =/= 0

BOOM....

your "perhaps" NON trivial issue set aside.

thank you

ALL else that follows in the op exists without change.

 December 5th, 2017, 01:19 PM #13 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 please express a detailed definition of your version of equality. Asshole. Thanks from Antoniomathgini
December 5th, 2017, 01:27 PM   #14
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Quote:
 Originally Posted by romsek please express a detailed definition of your version of equality. Asshole.
Lol...I gave you credit for a non trivial issue brought forth.

I said thank you.

I did NOT mean any rudeness or insult.

Though perhaps you will remember how you have treated me in the past...

0 =/= ( -0 )
0 and (-0) have equivalent absolute values.

This fixes any issue with the miss-communication of the definition of the equal sign.

NOTE

After much thought I have decided that this issue is entirely valid. Thank you Romesk. Seriously

please understand this version came from other's that were older and "foggier". As such.....well...there is no getting around it. I missed it. I was wrong.

Please allow this edit to the original post

0 =/= (-0 )
The absolute value of the zero elements is equivalent.

Again...Thank you

Last edited by Conway51; December 5th, 2017 at 02:25 PM.

 December 5th, 2017, 02:38 PM #15 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 0 =/= ( -0 ) |0|=|-0| Romesk.....you are certainly not nice....but you are helpful....perhaps I might entice you for some more of the same?
 December 5th, 2017, 02:39 PM #16 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 so you are saying that $|-0| = |0|$ now we can't just substitute one for the other. see how easy that was? Also, certain verbiage you used within the thread caused me to call you an asshole. Editing your posts after the fact and toning down such verbiage so it appears I am the one with a problem while effective is quite frowned upon as far as forum etiquette goes. Last edited by romsek; December 5th, 2017 at 02:43 PM.
December 5th, 2017, 02:41 PM   #17
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Quote:
 Originally Posted by romsek so you are saying that $|-0| = |0|$ now we can't just substitute one for the other. see how easy that was?
It WOULD have been easy if I had not had my guard up against you...because of previous interactions with you....that's on you...

NOTE*
In this version 0 and (-0) are never substituted.

Each is a unique zero element

December 5th, 2017, 02:44 PM   #18
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Quote:
 Originally Posted by Conway51 It WOULD have been easy if I had not had my guard up against you...because of previous interactions with you....that's on you... NOTE* In this version 0 and (-0) are never substituted. Each is a unique zero element
guard up.. hahaha

at any rate at least we've established the true relationship between $0$ and $-0$

 December 5th, 2017, 03:16 PM #19 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 Well said...and if I might add...that our conversations should only contain mathematics This I think will make you happy and me happy.
 December 5th, 2017, 03:58 PM #20 Senior Member     Joined: Sep 2015 From: USA Posts: 2,091 Thanks: 1087 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. Thanks from Conway51

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