# 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 generalise 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! :-)