Negative. Let xβr. By Commutative Rings Β§ring there is wβr with x+w=0. Let w,wβ²βr with x+w=0 and x+wβ²=0. Then, by the axioms of Commutative Rings Β§ring,
w=w+0=w+(x+wβ²)=(w+x)+wβ²=(x+w)+wβ²=0+wβ²=wβ²+0=wβ².
Hence there is exactly one wβr with x+w=0.
Reciprocal. Let r be a field and let xβr with xξ =0. By Fields Β§field there is yβr with xβ
y=1, and r is a commutative ring. Let y,yβ²βr with xβ
y=1 and xβ
yβ²=1. Then, by the axioms of Commutative Rings Β§ring,
y=yβ
1=yβ
(xβ
yβ²)=(yβ
x)β
yβ²=(xβ
y)β
yβ²=1β
yβ²=yβ²β
1=yβ².
Hence there is exactly one yβr with xβ
y=1.