# Comments on Product of Negatives

## Prunthaban said:

Nice proof. So now we should be able to talk in a very generic sense.
The additive inverse of \( a \) is \( -a. \) From your proof it looks
like in a *field* the multiplication of two values is equal to the
multiplication of their additive inverses. That makes your proof
independent of real numbers and assumes only that the system in question
is a field (assuming your proof is not using anything other than the 9
field axioms).

## Susam Pal said:

Prunthaban,

That is an interesting way to look at it. We can generalize it further.
We neither need commutativity of multiplication in our proof nor do we
need existence of multiplicative inverse (two of the field axioms), so
our proof holds good for the elements of a *ring* as well.

## Sunita Rajamani said:

Nice post. I really understood something this time! :-)