# Zero Element Equivalency

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Zero Element Equivalency

Can this be considered a field?

Can this be considered a solution for division by zero?

Can this sufficiently create varying amounts of zero?

Allow that there exists an integer zero element ( -0 ).

0 =/= (-0)

|0| = |-0|

0 |=| (-0)

Where |=| is defined as “Zero Element Equivalency”, where any two unique or similar additive identities are considered equal because they share the same absolute value and cardinality but may or may not possess different multiplicative properties.

Allow that :

0: possess the additive identity property and possess the multiplicative property of zero.

(-0): possess the additive identity property and possess the multiplicative identity property.

The addition of any two additive identities is not expressible as a sum, except with |=|.

0 + 0 =/= 0

0 + 0 |=| 0

0 + ( -0 ) |=| 0

( -0 ) + ( -0 ) |=| 0

Where n =/= 0:

n + 0 = n = 0 + n

n + ( -0 ) = n = ( -0 ) + n

Multiplication of any two additive identities is not expressible as a product, except with |=|.

0 * 0 =/= 0

0 * 0 |=| 0

0 * ( -0 ) |=| 0

( -0 ) * ( -0 ) |=| 0

Where n =/= 0:

n * 0 = 0 = 0 * n

n * ( -0 ) = n = ( -0 ) * n

1 * 0 = 0 = 0 * 1

1 * ( -0 ) = 1 = ( -0 ) * 1

The division of any two zero elements is not expressible as a quotient, except with |=|.

0 / ( -0 ) =/= 0

0 / ( -0 ) |=| 0

( -0 ) / 0 |=| 0

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

Where n =/= 0:

0 / n = 0

( -0 ) / n = ( -0 )

n / 0 = n

n / ( -0 ) = n

Therefore the multiplicative inverse of 1 is defined as ( -0 )

1 * ( -0 ) = 1

1/( -0 ) * ( -0 )/1 ) = 1

0 remains without a multiplicative inverse.

Examples containing the distributive property:

a( b + c) = a * b + a * c

Where: a=1, b= 0, c=0

1( 0 + 0) = 1* 0  + 1* 0

1 * 0 = 1 * 0 + 1 * 0

Where: a=1, b=0, c=( -0 )

1( 0 + ( -0 ) = 1 * 0 + 1 * ( -0 )

0 + 1 = 0 + 1

Where: a=1 b=( -0 ) , c=( -0 )

1( ( -0 ) + ( -0 ) ) = 1 * ( -0 ) + 1 * ( -0 )

1 + 1 = 1 + 1

Therefore, non-zero elements divided by zero elements are defined.

Therefore, the product of non-zero elements multiplied by zero elements is relative to which integer zero element is used in the binary expression of multiplication.

The rules for exponents and logarithms exist without change. It continues that multiplication of any zero elements by any zero elements is not expressible as a product except with |=|.

n^0 = 1

n^(-0) = 1

0^0 = 1

( -0 )^0 = 1

0^( -0 ) = 1

( -0 )^( -0 ) = 1

0^n = 0

( -0 )^n = 1

0^(-n) = 1

( -0 )^(-n) = 1

log0 |=| 0

log( -0 ) |=| 0

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If 0 + 0 is not equal to 0, then 0 doesn't have the property of being an additive identity (with respect to that equivalence relation).

1 already has a multiplicative inverse: 1.

If n/(-0) = n, then it's not hard to deduce that (-0) = 1, unless you are assuming that = has very few properties in common with what we usually think of as equality.

In short, your idea takes away many of the things that make fields useful in the first place.

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This is more towards your second question regarding division by zero .

ive seen one documentary that is recent by a few years claiming Indian cultures had reasoned division by zero was infinity .

I cannot claim this, but it does seem the closer you get to zero when dividing,  the closer the output gets to infinity.

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1 hour ago, uncool said:

If 0 + 0 is not equal to 0, then 0 doesn't have the property of being an additive identity (with respect to that equivalence relation).

1 already has a multiplicative inverse: 1.

If n/(-0) = n, then it's not hard to deduce that (-0) = 1, unless you are assuming that = has very few properties in common with what we usually think of as equality.

In short, your idea takes away many of the things that make fields useful in the first place.

0 + 0 |=| 0.........it still equals zero.....just under a new form of equivalencey

There is nothing saying a given number can not have more than 1 multiplicative inverse....both (-0) and 1 both fit the order of defintion for such

Please deduce any equations using the rules presented to show how (-0) = 1....it shouldn't be hard as you claim

In short...I do not understand how anything useful has been taken away only added....

To Ferrum:

17 minutes ago, Ferrum said:

This is more towards your second question regarding division by zero .

ive seen one documentary that is recent by a few years claiming Indian cultures had reasoned division by zero was infinity .

I cannot claim this, but it does seem the closer you get to zero when dividing,  the closer the output gets to infinity.

perhaps...but division by zero leads to infinitesimals not infinities....thus you get into a can of worms with that one......thank you for your information and time!

Edited by Philostotle

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7 hours ago, Philostotle said:

Can this be considered a field?

Is what a field?

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5 hours ago, studiot said:

Is what a field?

Perhaps I should have said.....is this construct a valid field..?....or some other such very specific thing.....perhaps you do not think it is..... and are in agreement with uncool on this matter.  Either way .....thank you for your time.

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12 hours ago, Philostotle said:

0 + 0 |=| 0.........it still equals zero.....just under a new form of equivalencey

There is nothing saying a given number can not have more than 1 multiplicative inverse....both (-0) and 1 both fit the order of defintion for such

Please deduce any equations using the rules presented to show how (-0) = 1....it shouldn't be hard as you claim

In short...I do not understand how anything useful has been taken away only added....

Actually, the fact that multiplicative inverses are unique is a basic fact.

If n/(-0) = n, then under the usual rules of equality, we can multiply both sides by (-0): (-0) n/(-0) = n (-0).

The usual definition of division then tell us that (-0) n/(-0) = n. We therefore have that n = n(-0).

Finally, we can divide by n, and get 1 = (-0).

So for what you are saying to make sense, you have to deny either: the ability to multiply both sides of an equation by the same number, the definition of division, the ability to divide both sides of an equation by a nonzero number, or transitivity of equality.

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1 hour ago, uncool said:

Actually, the fact that multiplicative inverses are unique is a basic fact.

If n/(-0) = n, then under the usual rules of equality, we can multiply both sides by (-0): (-0) n/(-0) = n (-0).

The usual definition of division then tell us that (-0) n/(-0) = n. We therefore have that n = n(-0).

Finally, we can divide by n, and get 1 = (-0).

So for what you are saying to make sense, you have to deny either: the ability to multiply both sides of an equation by the same number, the definition of division, the ability to divide both sides of an equation by a nonzero number, or transitivity of equality.

Actually, the fact that multiplicative inverses are unique is still the case....1 and -0 both have a unique multiplicative inverses (1  just have more than one)....the term unique does not necessarily imply one and only one.....it implies only that it is different from all others.  An apple is unique....and I may have many of them....

You CAN NOT use the NORMAL rules of equality...as given in the op.......

Your math is clearly wrong......

n/(-0) = n

multiple both sides by (-0) :

(-0) * n/(-0) = n * (-0)

(-0) * n = n * (-0)

We therefore have n=n

Divide both sides by n :

n/n = n/n

1=1

*your mistake*.....n(-0) = n.....NOT......(-0)

Thank you for your time...it is very much appreciated.

Edited by Philostotle

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The term "unique" in mathematics means only one.

If you can't use the normal rules of equality, why use that version of equality at all?

Do you agree that (-0) * n/(-0) = n?

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4 hours ago, Philostotle said:

Perhaps I should have said.....is this construct a valid field..?....or some other such very specific thing.....perhaps you do not think it is..... and are in agreement with uncool on this matter.  Either way .....thank you for your time.

I'm sorry but that did not answer my question.

You need to provide some unambiguous method of specifying all the elements of field, against which I can test any candidate object as in the field or not.

The two usual ways of achieving this are by formula or by a list.

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1 hour ago, uncool said:

The term "unique" in mathematics means only one.

If you can't use the normal rules of equality, why use that version of equality at all?

Do you agree that (-0) * n/(-0) = n?

Indeed it does...my SINCEREST apologies.....allow then the following edit to the op....

All elements except for (1) have a unique multiplicative inverse....1 has two unique multiplicative inverses...itself and (-0).

I do agree with the equation you gave

(-0) * n/(-0) = n

To Studiot

No list of formula is necessary....

• There is an element 0 in R such that a + 0 = a for all a in R   (that is, 0 is the additive identity).

This is still a valid equation for all elements in R

observe

a + 0 = a : for all elements except 0 in R

a + 0 |=| a : for all zero elements in R

Thus all elements in any given field are addressed as requested.

Edited by Philostotle

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54 minutes ago, Philostotle said:

Indeed it does...my SINCEREST apologies.....allow then the following edit to the op....

All elements except for (1) have a unique multiplicative inverse....1 has two unique multiplicative inverses...itself and (-0).

I do agree with the equation you gave

(-0) * n/(-0) = n

To Studiot

No list of formula is necessary....

• There is an element 0 in R such that a + 0 = a for all a in R   (that is, 0 is the additive identity).

This is still a valid equation for all elements in R

observe

a + 0 = a : for all elements except 0 in R

a + 0 |=| a : for all zero elements in R

Thus all elements in any given field are addressed as requested.

Pity you misread what I wrote.

I did not ask for a list of formulae.

Nor did I ask about any other field.

I asked how to determine any and all the members of your field.

This is necessary to show that your field does or does not satisfy the field axioms.

Membership of one particular field does not confer any status whatsoever in respect of a different field.

For example 1 and 0 are members of the field R.

They are also non-negative  integers (which do not constitute a field).

They are also members of what I understand to be the smallest possible field, which has exactly two members.

What I do not understand (because you haven't told me) is what are all the members of your field.

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Quote

All elements except for (1) have a unique multiplicative inverse....1 has two unique multiplicative inverses...itself and (-0).

The statement that inverses are unique is not a postulate, but a theorem. You have to break more than that statement to get things to work.

For the proof that 1 = (-0):

n = (-0) * n/(-0) = (-0) * n, right?

Edited by uncool

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7 hours ago, studiot said:

What I do not understand (because you haven't told me) is what are all the members of your field.

Perhaps I begin to see your point.  I should not be asking "is this a field"...

How about ...."Is this a valid mathematical construct?" thus: can it be used in the creation of a field?

Also it has been brought to my attention that perhaps I should work on redefining what it is to be equal: Thus have only one equality sign......I appreciate your time....

3 hours ago, uncool said:

The statement that inverses are unique is not a postulate, but a theorem. You have to break more than that statement to get things to work.

For the proof that 1 = (-0):

n = (-0) * n/(-0) = (-0) * n, right?

I agree and I am ready for you to continue with your equations....thanks!

Edited by Philostotle

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We can then divide by n:

1 = n/n = (-0)*n/n = (-0)

where the first and last equalities are from the fact that division is supposed to "undo" multiplication, and you've already agreed to the middle one.

But I just noticed something I'd consider worse: you have the equation 1*(-0) = 1. The main reason 1 is important is that it is the multiplicative identity, that it multiplied by anything is that thing. Breaking this pretty much breaks multiplication, and makes inverses all but useless.

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1 hour ago, uncool said:

We can then divide by n:

1 = n/n = (-0)*n/n = (-0)

where the first and last equalities are from the fact that division is supposed to "undo" multiplication, and you've already agreed to the middle one.

But I just noticed something I'd consider worse: you have the equation 1*(-0) = 1. The main reason 1 is important is that it is the multiplicative identity, that it multiplied by anything is that thing. Breaking this pretty much breaks multiplication, and makes inverses all but useless.

2 hours ago, Philostotle said:

n = (-0) * n/(-0) = (-0) * n, right?

n = (-0) * n/(-0) = (-0) * n

divide every element by n

n/n = 1 :    first element divided by n

(-0)/n = (-0):    second element divided by n

(n/(-0))/n = n/n = 1:      third element divided by n

(-0)/n = (-0) :     forth element divided by n

n/n = 1 :     fifth element divided by n

Therefore:

1 = (-0) * 1 = (-0) * 1

1 = 1 = 1

I do not think it breaks multiplication: as seen from the above equations it aids multiplication.......

Edited by Philostotle

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Can you explain what the point of this is?

What is the advantage of having multiple 0 values?

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You're providing more equations, but not answering the ones I have. If you dispute my equations, which of these is wrong?

Does 1 = n/n?

Does n/n = (-0)*n/n?

Does (-0)*n/n = (-0)?

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1 hour ago, uncool said:

You're providing more equations, but not answering the ones I have. If you dispute my equations, which of these is wrong?

Does 1 = n/n?

Does n/n = (-0)*n/n?

Does (-0)*n/n = (-0)?

He has been banned so ...

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Heh. I thought it was suspiciously similar, but forgot conway had been banned.

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I was just about to report him as a possible sock-puppet when he got banned.

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