Prove that -0 = 0

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• Aug 4th 2011, 04:08 AM
Siron
Re: Prove that -0 = 0
I think you're step 2 is incorrect, if you're working with ring axioms, you can say in general for a number $\displaystyle a\in R$ (where R is a ring), there exists a neutral element $\displaystyle e$ (and for the addition: 0) wherefore: $\displaystyle a+e=a=e+a$, that means in this case (for the addition): $\displaystyle -0+0=-0$.

I would prove it this way:
$\displaystyle 0+0=0$ (this is an axiom of the existence of a neutral element in a commutative group in a ring)
$\displaystyle \Leftrightarrow 0=0-0$ (putting 0 to the other side)
$\displaystyle \Leftrightarrow 0=-0+0$ (in a ring the addition is commutative)
Because of the existence of the neutral element $\displaystyle 0$ for the addition, like I stated earlier: $\displaystyle -0+0=-0$ therefore:
$\displaystyle 0=-0+0 \Leftrightarrow 0=-0$
• Aug 4th 2011, 10:29 AM
LoblawsLawBlog
Re: Prove that -0 = 0
Quote:

Originally Posted by Alexrey
So my proof was wrong?

Yes. You started with -0=0, which is what you're trying to prove in the first place, and you end up with the true statement 0=0. This isn't a valid form of proof unless you can show that each step is reversible, so we have to start from the bottom and see if that's true.

0=0. So far so good...

-0+0=0+0. This is true because -0=0, but unfortunately that's what you're trying to prove, so you can't use this step.

I still say that any proof over one line is unnecessarily long. 0+0=0 is all you need. Think about what it says. If you add 0 to itself you get 0, and isn't 0 the additive identity in a ring?
• Aug 5th 2011, 08:29 AM
topspin1617
Re: Prove that -0 = 0
Quote:

Originally Posted by Alexrey
So my proof was wrong?

Well... yeah. You started off by assuming what you want to prove. You can't prove a statement P by starting like "assume P is true".

All you need to do is say 0+0=0. By definition, this means -0=0.
• Aug 7th 2011, 05:48 AM
Alexrey
Re: Prove that -0 = 0
How does 0+0=0 imply -0=0? I would really appreciate it if that statement could be explained further.
• Aug 7th 2011, 06:08 AM
LoblawsLawBlog
Re: Prove that -0 = 0
What is, say, -7? It's 7 reflected across 0 on the number line, it's -1 times 7, but more fundamentally, it's the solution to 7+x=0. The solution to -2+x=0 is 2 and so by definition, -(-2)=2. The solution to 0+x=0 is 0, so by definition, -0=0.
• Aug 7th 2011, 07:28 AM
TheChaz
Re: Prove that -0 = 0
Quote:

Originally Posted by Alexrey
How does 0+0=0 imply -0=0? I would really appreciate it if that statement could be explained further.

a + b = 0 implies that b is the (additive) inverse of a.
That b is the inverse of a means, exactly, that b = -a.
So if
a + a = 0
Then a (the one on the right, if you will) = -a (i.e. the inverse of a, the one on the left)
Replace a with 0.
If you don't understand, read it again. Otherwise, it just looks like the old "dumb troll" routine.
• Aug 7th 2011, 11:32 PM
Alexrey
Re: Prove that -0 = 0
Awesome, thanks a lot guys.
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