# prove -0 =0

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prove :

-0=0

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Why? You didn't even bother to specify why you are asking this question and what your background is. I see little point in working out a proof based on group theory if you don't even know what algebra (also known as "abstract algebra" in the US ) is. And in the case that you are currently learning algebra, then showing you a proof would probably violate the "homework rules" of this forums. And if you knew algebra, you wouldn't ask such a question. And if you weren't looking for an algebra-based answer, then you seemingly forgot to add what kind of answer you are looking for.

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This would depend entirely on a) What type of object this zero is (Real number, group element, ect.) and b) how this type of object was developed. If this is a real/rational/integer number, then from every construction I have seen, -0 is not defined as negative numbers are generally defined as additive inverses or come from your formulation of order.

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If you take zero to be the identity element with respect to addition in some group, then just use the definitions of the identity element with respect to addition and of the additive inverse, and show that +0 and -0 must be the same.

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This would depend entirely on a) What type of object this zero is (Real number, group element, ect.) and b) how this type of object was developed. If this is a real/rational/integer number, then from every construction I have seen, -0 is not defined as negative numbers are generally defined as additive inverses or come from your formulation of order.

This is the real zero

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-0 = -1*0 by definition of "-"

-1*0 = 0 by definition of "0"

-0 = 0 by transitive property of "="

Q.E.D.

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-0 = -1*0 by definition of "-"

-1*0 = 0 by definition of "0"

-0 = 0 by transitive property of "="

Q.E.D.

But in real Nos zero is not defined as you mention in your second line of proof.

How do you define "-" in real Nos.

I know that "-" in real Nos is defined by the equation : x+(-y) = x-y

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But in real Nos zero is not defined as you mention in your second line of proof.
0 is defined as the number with these two properties:
• x+0=x
• x*0=0

So in fact, it is.

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$-0=0+(-0)=0$

First identity holds for 0 is the additive neutral of the real field, second holds because a+(-a)=0 for every a.

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-0 = -1*0 by definition of "-"

-1*0 = 0 by definition of "0"

-0 = 0 by transitive property of "="

Q.E.D.

I have seen a lot of mathematical proofs ,but a correct one with a wrong justification??

0 is defined as the number with these two properties:

• x+0=x
• x*0=0

So in fact, it is.

x*0=0 ,is a theoerem not a definition,it can be proved

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In the context of ring theory triclino is right. The defining axiom of the zero element is x+0 = x for all x.

Then x 0 = x (0+0) = x 0 + x 0, which implies x 0 =0. (also easy to show 0 x =0 )

So we have "patched up" the trees proof.

Also we can extend the proof to rings directly.

Edited by ajb

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Even easier: $0+0=0 \implies -0=0$ due to the uniqueness of the additive inverse.

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Even easier: $0+0=0 \implies -0=0$ due to the uniqueness of the additive inverse.

It is a completely different business in proving the uniqueness of the zero ,from the proof : -0 = 0

In the context of ring theory triclino is right. The defining axiom of the zero element is x+0 = x for all x.

Then x 0 = x (0+0) = x 0 + x 0, which implies x 0 =0. (also easy to show 0 x =0 )

So we have "patched up" the trees proof.

Also we can extend the proof to rings directly.

How x0 = x0+x0 imply x0 =0??

Also this is the real zero as kavlas said,so we have the axioms of the reals.And one of them is : x+0 =x

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prove :

-0=0

Givien all this discussion, you need to define what you mean by "0".

The symbol "0" is used in mathematics for many analagous concepts.

The most common is the integrer 0.

It sometimes used as the symbol for the additive identity in an abstract abelian group.

It is sometime used as the symbol for the additive identity of the abelian group that is part of a ring structure.

Similarly "-0" might mean the additive inverse of the abstract abelian group element "0".

It might also mean "-1 x 0" in the context of a ring.

The proofs are very similar in all cases, but it would help understanding for the context of your question to be made clear.

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It is a completely different business in proving the uniqueness of the zero ,from the proof : -0 = 0

Actually, it is related. Proving that inverses are unique implies that as long as 0 is an inverse of 0, it is the unique inverse of 0.

How x0 = x0+x0 imply x0 =0??

x*0 = x*(0 + 0) = (x*0) + (x*0)

By well-definedness of addition,

(x*0) + -(x*0) = ((x*0) + (x*0)) + -(x*0)

By associativity of addition,

(x*0) + -(x*0) = x*0 + ((x*0) + -(x*0))

By definition of inverse,

0 = x*0 + 0

By definition of additive identity,

0 = x*0

Some steps were left out by ajb, but it is clear what they meant.

=Uncool-

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It is a completely different business in proving the uniqueness of the zero ,from the proof : -0 = 0

What are you talking about?

$0+0=0$ is obviously true. From this it follows that 0 is the additive inverse to itself. Since -0 is defined to be the additive inverse to 0, and additive inverses are unique in $(\mathbb{R},+)$, it follows that $-0=0$.

The uniqueness of inverses is sometimes even included in the definition of a group, yet it follows directly from the group axioms.

Edited by MathIsCool

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prove :

-0=0

the question is moot.

from Wiki

As a number

0 is the integer immediately preceding 1. In most cultures, 0 was identified before the idea of negative things (quantities) that go lower than zero was accepted. Zero is an even number,[22] because it is divisible by 2. 0 is neither positive nor negative. By most definitions[23] 0 is a natural number, and then the only natural number not to be positive. Zero is a number which quantifies a count or an amount of null size.

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I tend to think of zero as being both positive and negative. So when I end up with $a = -a$ I know that $a=0$, in the context of ($Z_{2}$-graded) rings.

But do note DrRockets very valid point, 0 can mean different things.

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Actually, it is related. Proving that inverses are unique implies that as long as 0 is an inverse of 0, it is the unique inverse of 0.

x*0 = x*(0 + 0) = (x*0) + (x*0)

By well-definedness of addition,

(x*0) + -(x*0) = ((x*0) + (x*0)) + -(x*0)

By associativity of addition,

(x*0) + -(x*0) = x*0 + ((x*0) + -(x*0))

By definition of inverse,

0 = x*0 + 0

By definition of additive identity,

0 = x*0

Some steps were left out by ajb, but it is clear what they meant.

=Uncool-

You converted a two line proof into a long,long story:

We have : 0x =(0+0)x=0x+0x,by distributive property,and the axiom : x+0=x

But : ox +0 =0x (by x+0=x again).

Hence : ox =0 ,by cancellation law

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You converted a two line proof into a long,long story:

We have : 0x =(0+0)x=0x+0x,by distributive property,and the axiom : x+0=x

But : ox +0 =0x (by x+0=x again).

Hence : ox =0 ,by cancellation law

The last 3 steps are effectively an explanation of why the cancellation law works. Cancellation is not an axiom; it's a result of the axioms that I named.

=Uncool-

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I could imagine that the number system has two groups, positive numbers and negative numbers, and that they intersect at Zero

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I could imagine that the number system has two groups, positive numbers and negative numbers, and that they intersect at Zero

Interestingly enough, that is essentially how the IEEE floating point format works. You are most likely using IEEE floating point numbers whenever you use some product like Excel or whenever you do floating point computations in some common programming language. Positive 0.0 and negative 0.0 are different things in that format. They are not "equal". It would nonetheless be nice (very, very nice) if comparing +0 against -0 for mathematical equality yielded true. In fact, it's not just very nice; this is required behavior of a compliant implementation. The underlying machinery has to go out of its way to ensure that the result of such a comparison is true even though -0 and +0 have different representations.

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I'm not sure of this, but I think that:

By the definition of $\mathbb{R}^*$:

$\mathbb{R}^* = (\mathbb{R}^+ \cup \mathbb{R}^-) \setminus 0$

$\mathbb{R}^* = (\mathbb{R}^+ \cup \mathbb{R}^-) \setminus (\mathbb{R}^+ \cap \mathbb{R}^-)$

Since only 0 satisfy: $0 \in \mathbb{R}^+$ and $0 \in \mathbb{R}^-$ we get $0 \in (\mathbb{R}^+ \cap \mathbb{R}^-)$

Where $(\mathbb{R}^+ \cap \mathbb{R}^-) = (\mathbb{R}^+ \cup \mathbb{R}^-) \setminus \mathbb{R}^*$

Edited by khaled

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0=-0

Ad 0 to both sides

0+0=

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