But if x=0 then x=-0So we conclude that 0=-0Is -0 a Real Number?

[philbein]

I need help urgently asap proving that -0=0

Homework Statement

Prove that
0=-0

Homework Equations

We Can use only the following axioms for Real Numbers

(x+y)+z=x+(y+z); (xy)z=x(yz)
x+y=y+x; xy=yx
x(y+z)=(xy)+(xz)
The Additive Identity 0+x=x
The Additive Inverse for all x in the real numbers there exists -x, such that x+(-x)=0
Multipicative Identity There exists an element 1, such that x*1=x
Mult. Inverse There exists for all x an inverse (1/x), such that x(1/x)=1
If x is in the real numbers than one of the following is true
x is positive
x is 0
-x is positive

You can also add or multiply the same thing to both sides of the equation

The Attempt at a Solution

We know that x+(-x)=0
thus, we see that -(x+(-x))=-0

I'm not sure where i can go next. We can use the distributive property, but would we be allowed to use it in this situation with the (-).

 

[JG89]

Why not prove that any real number multiplied by 0 is 0, and so as a consequence of that, -0 = -1*0 = 0

 

[epsilonzero]

what about this

start out with the true statement: 0=0
using additive inverse: 0+(-0)=0
which is an additive identity for: -0=0

 

[chuy]

Another:

Let be x de inverse additive of 0 (that is -0), then by definition:

0+x=0

Since 0 is the additive identity:

x=0