Proof of the Theorem a0=0

I'm trying to prove this a different way, but I thought I have it but noticed a problem I think that kills it and I need to make sure it's correct that I can't do this. This is what I have...

1) 0=0

2) a0=a0

3) a0=a(1-1)

4) a0=a-a

5) a0=0

The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?

Re: Proof of the Theorem a0=0

Quote:

Originally Posted by

**JSB1917** The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?

Yes, that looks like a problem.

I don't have a book with me, but I assume this is the standard way:

Let x=a0. Let y be the additive inverse of x (so that x+y=y+x=0). Since 0 is the additive identity, 0 + 0 = 0.

Now x = a0 = a(0+0) = a0 + a0 = x + x. Then adding y to both sides gives:

x + y = (x + x) + y = x + (x + y) = x + 0 = x. Since x + y = 0, have shown that 0 = x. Thus a0 = 0.

Another way:

a = a(1) = a(0+1) = a0 + a(1) = a0 + a. Then add a's additive inverse (-a, but to avoid confusion I'll call it b... thus a+b=b+a=0) to both sides to get:

a + b = ( a0 + a ) + b = a0 + ( a + b ) = a0 + 0 = a0. But then a0 = a + b = 0. Thus a0 = 0.

Re: Proof of the Theorem a0=0

Yeah, the other way i did it was 0+1=1, then multiply by a and use the additive inverse of a.

Re: Proof of the Theorem a0=0

Quote:

Originally Posted by

**JSB1917** I'm trying to prove this a different way, but I thought I have it but noticed a problem I think that kills it and I need to make sure it's correct that I can't do this. This is what I have...

1) 0=0

2) a0=a0

3) a0=a(1-1)

4) a0=a-a

5) a0=0

The problem is going from step 3 to step 4. Isn't this invoking the other theorem of (-a)b=-ab which requires the proof of a0=0?

Here is a standad proof.

You can supply the reasons(axioms).