# Is zero positive or negative ?

You don't really want to argue that "P AND Q ==> P" is false, do you?
No, this is a tautology. Please post the entire argument then we can discuss it.

Hey, I'm a bit of a rebel, I may question usefulness of certain definitions, but I would never argue against a tautology, that would be silly!

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No, this is a tautology. Please post the entire argument then we can discuss it.

0 is positive and is negative
==> 0 is positive (by the rule "P AND Q ==> P" )
==> 0 is not in the complement of positive numbers (since for each set A, it holds $A\cap A^c=\emptyset$)

0 is positive and is negative
==> 0 is positive (by the rule "P AND Q ==> P" )
==> 0 is not in the complement of positive numbers (since for each set A, it holds $A\cap A^c=\emptyset$)
line 3 contradicts line 1 in all worlds because if zero is negative you must put it in the set of negative numbers, so it would be in the complement of the positive numbers.

line 3 contradicts line 1 in all worlds because if zero is negative you must put it in the set of negative numbers, so it would be in the complement of the positive numbers.
I don't care. Just tell me where I have gone wrong.

I don't care. Just tell me where I have gone wrong.
Line 3 does not follow from line 2 if trichotomy property fails for 0.

Where you went wrong is you assumed that trichotomy property holds for zero.

-0 = 0 = +0

What is the point if you don't care?

Trichotomy property holds for all real numbers but for zero it is ill defined because one can write the above equation and it has meaning.

example: -1 = 1 = +1

clearly nonsense

-0 = 0 = +0

not clearly nonsense.

OK, so what you're saying is that the law

A∩Ac=∅

fails. Right? Because that's what I used in (2)==> (3).

So, let me prove that, tell me where I went wrong:

By contradiction: Take x in A∩Ac
==> x is in A AND x is in Ac
==> x is in A AND x is not in A
So there does not exist x∈A∩Ac. So A∩Ac=∅.

So, where did I go wrong here?

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OK, so what you're saying is that the law

$$A\cap A^c=\emptyset$$

fails. Right? Because that's what I used in (2)==> (3).

So, let me prove that, tell me where I went wrong:

By contradiction: Take x in $A\cap A^c$
==> x is in A AND x is in $A^c$
==> x is in A AND x is not in A
So there does not exist $x\in A\cap A^c$. So $A\cap A^c=\emptyset$.

So, where did I go wrong here?
I am having some trouble decoding Tex

Does the first line say 'take x in A or A complement'?

It would help if you can post it in words.

My previous post is now TeX-free.

My previous post is now TeX-free.
Thank you for Tex-free. I really mean that.

A intersection Ac = emty set

Not according to this, if x is considered the boundary

The boundary of a set is closed. The boundary of a set is the boundary of the complement of the set

Found here

http://en.m.wikipedia.org/wiki/Boundary_(topology)

http://en.m.wikipedia.org/wiki/Equivalence_relation

this last link is helpfull in analyzing -0 = 0 = +0 under reflexive, symmetric, and transitive.

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Thank you for Tex-free. I really mean that.

A intersection Ac = emty set

Not according to this, if x is considered the boundary

The boundary of a set is closed. The boundary of a set is the boundary of the complement of the set

Found here

http://en.m.wikipedia.org/wiki/Equivalence_relation
Boundary has nothing to do with this. Points in the boundary either lie in A or Ac, not in both. So boundary points are no contradiction to A intersection Ac = emptyset.

I observe that you did not point out the flaw in my proof...

Boundary has nothing to do with this. Points in the boundary either lie in A or Ac, not in both. So boundary points are no contradiction to A intersection Ac = emptyset.

I observe that you did not point out the flaw in my proof...
what does this mean in YOUR world?

The boundary of a set is closed. The boundary of a set is the boundary of the complement of the set

The boundary of a set is closed. The boundary of a set is the boundary of the complement of the set
So??

Care to explain what was wrong in my proof of A intersection Ac=empty??

Mark44
Mentor
This was the first sentence of my argument.

'Consider the real number line.'

BTW Mark44 'explanation' works ONLY if you accept the definition that zero has no sign. If, for whatever reason, one decides to question that definition, his clever argument doesn't make sense.
With all due respect, this is a specious argument. No serious mathematician believes that zero has a sign. The definition for positive in my dictionary (American Heritage Dictionary of the English Language) has this definition:
12. Mathematics. Pertaining to or designating: a. A quantity greater than zero."
The definition for negative is a "quantity less than zero."

If we can't agree on the definitions of basic terms, then further discussion is futile.
Heres what he did...

He took an apple, decided to define it as a banana, then he put it in a bag with other apples and concluded that he now had a bag full of apples and 1 banana. In fact the bag contains only apples. His definition of a particular apple is irrelevant. It is what it is.
You are mischaracterizing what I said in post #94, which I'll add here verbatim.
Mark44 said:
If I buy a bag of apples and a banana at the store, and the checker puts the banana in with the apples, that doesn't mean that the banana has somehow turned into an apple. All it means is that the bag has apples and a banana in it.
I have a bag of apples. I have a banana. The checker rings these items up, and puts the banana in with the apples. The bag contains several apples and one banana. I don't know how to say it more simply than that.

The fact still remains that

-0 = 0 = +0

For real numbers, any of the above symbols work.
The question about whether zero is positive or negative (by implication from the question, we are talking about real numbers only) has been answered long ago in this thread. There is no point in going on and on about how the terms "positive" and "negative" are defined, so I am locking the thread.