# Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

• Jul 6th 2011, 08:57 AM
dwsmith
Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
Prove that for all a,b in a ring R, $(-a)b=-ab=a(-b)$
The hint says use: $a+(-a)=0$

Since any integer has a negative, we know $a+(-a)=0$

Multiple by b:

$(a+(-a))b=0(b)\Rightarrow ab+(-a)b=0$

How can I show $(-a)b=-ab=a(-b)$ from the above?
• Jul 6th 2011, 10:05 AM
Chris L T521
Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
Quote:

Originally Posted by dwsmith
Prove that for all a,b in a ring R, $(-a)b=-ab=a(-b)$
The hint says use: $a+(-a)=0$

Since any integer has a negative, we know $a+(-a)=0$

Multiple by b:

$(a+(-a))b=0(b)\Rightarrow ab+(-a)b=0$

How can I show $(-a)b=-ab=a(-b)$ from the above?

Interesting approach (I don't have a problem with it). Note that $-ab=-(ab)$. So the first part follows right away: $(-a)b=-ab$.

For the second part, you do pretty much the same thing; however, you start with $b+(-b)=0$ and then multiply by $a$ on the left:

$a(b+(-b))=0\implies\ldots$

Then because '=' is transitive, it now follows that $(-a)b=-ab=a(-b)$.

I hope this makes sense.
• Jul 6th 2011, 10:26 AM
Isomorphism
Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)
Quote:

Originally Posted by dwsmith
Prove that for all a,b in a ring R, $(-a)b=-ab=a(-b)$
The hint says use: $a+(-a)=0$

Since any integer has a negative, we know $a+(-a)=0$

integer?
I think you mean the ring is a group under addition and thus every element in the ring has an additive inverse.

Quote:

Multiple by b:

$(a+(-a))b=0(b)\Rightarrow ab+(-a)b=0$
Have you proved that $b.0 = 0$ for all elements b in the ring?

Quote:

How can I show $(-a)b=-ab=a(-b)$ from the above?