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, $\displaystyle (-a)b=-ab=a(-b)$

The hint says use: $\displaystyle a+(-a)=0$

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

Multiple by b:

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

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

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

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Originally Posted by

**dwsmith** Prove that for all a,b in a ring R, $\displaystyle (-a)b=-ab=a(-b)$

The hint says use: $\displaystyle a+(-a)=0$

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

Multiple by b:

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

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

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

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

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

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

I hope this makes sense.

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, $\displaystyle (-a)b=-ab=a(-b)$

The hint says use: $\displaystyle a+(-a)=0$

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

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:

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

Have you proved that $\displaystyle b.0 = 0$ for all elements b in the ring?

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

Your approach is right.