3 ms·
It's not making a multiplicative inverse of 0 exist though, it just defines a '/' operator that is slightly different from our usual one (i.e. a/b = a*b^(-1))
by poizan42 4y ago
It's not making a multiplicative inverse of 0 exist though, it just defines a '/' operator that is slightly different from our usual one (i.e. a/b = a*b^(-1))