# Prove that -0 = 0

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• Aug 3rd 2011, 11:52 AM
Alexrey
Prove that -0 = 0
I know that this is probably trivial for all of you, but I'm having a little trouble figuring it out. This is what I wrote down, but I'm not too sure that it's correct:

-0 = 0
-0 + 0 = 0 + 0
0 = 0
Therefore -0 = 0

Is this correct or am I missing something?
• Aug 3rd 2011, 11:59 AM
anonimnystefy
Re: Prove that -0 = 0
"0 is neither positive nor negative."

from http://en.wikipedia.org/wiki/0_(number)
• Aug 3rd 2011, 12:00 PM
Siron
Re: Prove that -0 = 0
Hmm, I don't think you've to prove that, 0 is neither negative neither positive.
Why are you proving this? ...
• Aug 3rd 2011, 12:27 PM
TheChaz
Re: Prove that -0 = 0
-0 = 0
Add 0 to both sides
0 + (-0) = 0 + 0
0 = 2(0)
0 = 0
QED

edit... I guess this actually demonstrates, using ring axioms (ish), that
"If -a = a, then a = 0"
• Aug 3rd 2011, 12:28 PM
anonimnystefy
Re: Prove that -0 = 0
ok prove that 2(0)=0
• Aug 3rd 2011, 12:32 PM
TheChaz
Re: Prove that -0 = 0
Quote:

Originally Posted by anonimnystefy
ok prove that 2(0)=0

Surprised you couldn't look this one up on your own!
PlanetMath: zero times an element is zero in a ring
• Aug 3rd 2011, 12:35 PM
anonimnystefy
Re: Prove that -0 = 0
but that proof is based on the fact that a-a=0,and a+0=a.
• Aug 3rd 2011, 12:38 PM
TheChaz
Re: Prove that -0 = 0
Quote:

Originally Posted by anonimnystefy
but that proof is based on the fact that a-a=0,and a+0=a.

... and??? Those are axioms. Is there some other punchline?
• Aug 3rd 2011, 12:40 PM
anonimnystefy
Re: Prove that -0 = 0
they could be axioms but you can prove them just by defining addition how you want,but let not go any deeper into this.
• Aug 3rd 2011, 01:42 PM
topspin1617
Re: Prove that -0 = 0
Quote:

Originally Posted by anonimnystefy
they could be axioms but you can prove them just by defining addition how you want,but let not go any deeper into this.

Of course if you are allowed to redefine things the way you want, you can prove ANYTHING.

Those ARE ring axioms, which means they are facts you can use whenever working with rings.

Axioms are axioms for a reason. Generally, if they could be proven as theorems from the other axioms, then they wouldn't BE axioms.
• Aug 3rd 2011, 01:44 PM
topspin1617
Re: Prove that -0 = 0
Quote:

Originally Posted by TheChaz
edit... I guess this actually demonstrates, using ring axioms (ish), that
"If -a = a, then a = 0"

That isn't even true. A ring may certainly have a nonzero element that is equal to its own additive inverse. The simplest example probably being $1=-1$ in $\mathbb{Z}_2$.
• Aug 3rd 2011, 02:17 PM
SammyS
Re: Prove that -0 = 0
Quote:

Originally Posted by Alexrey
I know that this is probably trivial for all of you, but I'm having a little trouble figuring it out. This is what I wrote down, but I'm not too sure that it's correct:

-0 = 0
-0 + 0 = 0 + 0
0 = 0
Therefore -0 = 0

Is this correct or am I missing something?

In my view the first problem you have is that you assumed that -0 = 0 is true at the beginning of your proof.

Another way to word this problem is:
Prove that the additive inverse of zero is zero.
• Aug 3rd 2011, 07:17 PM
LoblawsLawBlog
Re: Prove that -0 = 0
0+0=0

That's the entire proof. Can you justify it from the group axioms? You should also try to prove that inverses are unique if you haven't done so already.
• Aug 3rd 2011, 09:07 PM
TheChaz
Re: Prove that -0 = 0
Quote:

Originally Posted by topspin1617
That isn't even true. A ring may certainly have a nonzero element that is equal to its own additive inverse. The simplest example probably being $1=-1$ in $\mathbb{Z}_2$.

My edit/amendment was unto your point, not a claim of veracity. I proved the converse!
• Aug 4th 2011, 03:45 AM
Alexrey
Re: Prove that -0 = 0
So my proof was wrong?
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