Inverses in the zero ring (poll)

Niels Möller nisse at
Thu Jan 2 18:19:08 UTC 2014

Torbjorn Granlund <tg at> writes:

> The argument is that an inverse of a is definied to exist is ab = 1 for
> some b.  In this ring 0 = 1 and thus are all integers congrent with 0.
> We have ab = 0 = 1 for any a and b.  Thus any integer is invertible in
> this ring.

And in addition, it agrees with the rule that a is invertible modulo m
if and only if gcd(a, m) = 1, which in the m = 1 case is always true,
gcd(a, 1) = 1 for all a.


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