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

1. 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?

2. Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

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.

3. Re: Prove that for all a,b in a ring R, (-a)b=-ab=a(-b)

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.

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?

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

,

,

,

,

,

,

,

prove that a(-b)=-(ab)

Click on a term to search for related topics.