The law of cancellation requires that all values being cancelled have an inverse. The inverse of 0 doesn’t exist in the set of real numbers (although it does exist in the hyperreals). This doesn’t mean you can’t multiply a number by the inverse of 0, but the product doesn’t exist in real numbers, either. (Hyperreal numbers don’t cancel out the way real numbers do, however; they can leave behind a hyperreal component [ETA: Or at least that’s my understanding from the way my instructor explained why removable discontinuities couldn’t actually be removed—open to proof otherwise].)
0 doesn’t have an inverse in the hyperreal numbers either (To see why this it true, consider the first-order statement “∀x, x*0 != 1” which is true in the real numbers and therefore also true in the hyperreals by the transfer principle). From this it obviously follows that you can’t multiply a number by the inverse of 0.
Er, 1⁄0 * 0 != 1.
The law of cancellation requires that all values being cancelled have an inverse. The inverse of 0 doesn’t exist in the set of real numbers (although it does exist in the hyperreals). This doesn’t mean you can’t multiply a number by the inverse of 0, but the product doesn’t exist in real numbers, either. (Hyperreal numbers don’t cancel out the way real numbers do, however; they can leave behind a hyperreal component [ETA: Or at least that’s my understanding from the way my instructor explained why removable discontinuities couldn’t actually be removed—open to proof otherwise].)
0 doesn’t have an inverse in the hyperreal numbers either (To see why this it true, consider the first-order statement “∀x, x*0 != 1” which is true in the real numbers and therefore also true in the hyperreals by the transfer principle). From this it obviously follows that you can’t multiply a number by the inverse of 0.
Further, if you did decide to adjoin an inverse of zero to the hyperreals, the result would be the zero ring.
Going to have to investigate more, but that looks solid.