Kero van Gelder writes:

>> But with division,
>> 
>> 2*3 = 6
>> 6/3 = 2
>> 6*2 = 2, therefore (1/3) = 2 and 2 = 1 (since it's the identity.)

> because 6/3 = 6*1/3 and 6*2 = 6*1*2 ?
> But! you can not just mid-way-divide by 6...
> 1/6 = x mod 10, which, oh horror, does not exist!!!

Actually, the conclusion should be that (1/3) = 2 and that 6 = 1.
That was just a mistake.  The second follows from the first by
multiplying on the right by 3-inverse, while the third arises from
6*2 mod 10 = 12 mod 10 = 2.

Then, if x*y = y, then x must be the multiplicitative identity, which
is unique, so therefore y must equal 1.

The upshot is that this only works for integers mod a prime.

-- 
Johann Hibschman                           johann / physics.berkeley.edu