Originally Posted by

**Deveno** i would stop using the notation x/y. you have no idea what anything besides x+y and xy mean. you only know that given x, there is a y with x+y = 0, or a y with xy = 1 (if x is not 0).

no matter what b is, we can certainly find some x with:

b + x = 0 (rule 4a)

so (az + b) + x = c + x

but (az + b) + x = az + (b + x) (rule 1a), and b + x = 0, so:

az + 0 = c + x

az = c + x (rule 3a)

since a ≠ 0, there is some y with ay = 1 (rule 4b)

thus (az)y = (c + x)y

now (az)y = y(az) (rule 1b) so:

y(az) = (az)y = (c + x)y.

also y(az) = (ya)z (rule 2b).

so (ya)z = y(az) = (c + x)y.

now ya = ay (rule 1b), and ay = 1 (because that's what we defined y to be, by rule 4b), so ya = 1.

so 1z = (c + x)y.

by rule 3b, 1z = z, thus:

z = (c + x)y. this is "a" solution, we still do not know if it is unique, yet.

suppose, however, that z' is another solution, so that az' + b = c as well.

then az + b = c = az' + b.

thus az + b + x = az' + b + x (same x we used before)

az + 0 = az' + 0

az = az'

y(az) = y(az') (same y we used before)

(ya)z = (ya)z'

(ay)z = (ay)z'

1z = 1z'

z = z'

(fill in the relevant rules we have used)

which shows uniqueness.