No, because neither - nor / are mentioned in your axioms.
But you might find some axioms that will do what you want anyway-- consider 4).
Given the following axioms:
For all x ,y ,z
1) x+y =y+x.............................................. .......................xy =yx
2) x+(y+z) =(x+y)+z.......................................... ...............x(yz) =(xy)z
3) x+0 =x................................................ ..........................1x = x
4) for all x there exists y : x+y = 0 .......................................for all x different than zero there exists y: xy =1
5)............................................x(y+ z) = xy +xz
Then prove that for all, a, different than zero and for all b there exitsts a unique z : az +b =0
Are we allowed here to put : z =-b/a ??
I got ahead of myself and got everything out of order-- no wonder you are confused. My apologies. OK, here's another try at it.
Using 4): Since a is non-zero, there is a c such that ac = 1. And there is a d such that b+d = 0.
Now let z = cd.
So az + b = a(cd) + b = (ac)d + b = 1d + b = b + d = 0
This shows existence. I haven't thought about showing uniqueness, but maybe you would like to have a go at it.
YES you are right and thank you for your efford.
Just as a matter of interest ,before we curry on with the existence part, how would you plot a proof for the following:
for all a ,for all ,b and for all,c : if a is different than zero then there exitst a unique z : az +b =c
OK, here we go.
There is a d such that c+d = 0.
Consider the problem az + (b+d) = 0.
By the previous part, there is a unique solution z (although we didn't show uniqueness).
Then
(az + (b+d)) + c = 0 + c = c,
and
(az + (b+d)) + c = az + ((b+d) + c) = az + (b + (d+c)) = az + (b + (c+d)) = az + (b + 0) = az + b
so
az + b = c
(Once again, I am ignoring the need to prove uniqueness.)
I think i can justify the substitution z=cd and correct me if i am wrong.
From the equation : az +b =0 we know that z = -b/a =(-b).1/a ,hence if we put :-b =d and 1/a= c we have ,z =cd,then since b +(-b) =0 we have b+d=0.
Also since a is different than zero we have a.(1/a)=1 =ac
That is a kind of backwards working
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.
I notice that this does NOT say that y is unique. That may cause a problem in proving "there exist a unique z"
5)............................................x(y+ z) = xy +xz
Then prove that for all, a, different than zero and for all b there exitsts a unique z : az +b =0
Are we allowed here to put : z =-b/a ??