Comment on Why don't we define “imaginary” numbers for every “impossibility”? (2012)parentComments−poizan423yIt'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))
Comments
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))