hatem240600 said:
Please share. I think I can handle some algebra.
##\{\,-1,+1\,\}## are the units in the ring of integers, which is the basic domain our number system with characteristic zero is built upon. The integers are the basis. In fact, it are the natural numbers and the integers are already the first step, in which we constructed a group from the monoid ##(\mathbb{N},+)## in order to reverse additions. As it turns out, ##\mathbb{Z}## is also a ring, i.e. allows a distributive multiplication. Now the units of any ring form a multiplicative group, in this case of order ##2##. There is only one neutral element in a group, which we usually write as ##1## in the multiplicative case (##0## in the additive case). This requires to have ##(-1)\cdot 1 = -1## from which all other formulas follow:
##(1 / (-1))= 1\cdot (-1)^{-1}=(-1)^{-1}= -1## for otherwise we would have
##-1\stackrel{(*)}{=}1\cdot (-1)\stackrel{(**)}{=}(-1)^{-1}\cdot (-1)\stackrel{(***)}{=}1## which cannot be in the case of characteristic zero.
##(*)## definition of unique ##1##
##(**)## assumption ##(-1)^{-1}= 1##
##(***)## definition of unique inverse
Next we have ##(-1)\cdot (-1)\stackrel{(a)}{=}(-1)\cdot (-1)^{-1}\stackrel{(b)}{=}1##
##(a)## by the previous formula
##(b)## by the definition of the inverse
The proofs that units of a ring form a multiplicative group, and that a group has a unique neutral element and unique inverse elements depend a bit on how you define a group (laws vs. solvability of equations).
