1. ## difficult problem

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 ??

2. ## Re: difficult problem

No, because neither - nor / are mentioned in your axioms.

But you might find some axioms that will do what you want anyway-- consider 4).

3. ## Re: difficult problem

Originally Posted by awkward
No, because neither - nor / are mentioned in your axioms.

But you might find some axioms that will do what you want anyway-- consider 4).
Ihave been trying for two days.Trying axiom 4 it wiil not help because there is an a in front of az

4. ## Re: difficult problem

OK, here is a first step. According to 4), there is a c such that b + c = 0.
Then (az +b) + c = 0 + c. Also, 0 + c = c + 0 = c (why?) and (az + b) = c = az + (b + c) = az + 0 = az (why?).
So az = c.
The next step is ....

5. ## Re: difficult problem

Originally Posted by awkward
OK, here is a first step. According to 4), there is a c such that b + c = 0.
Then (az +b) + c = 0 + c. Also, 0 + c = c + 0 = c (why?) and (az + b) = c = az + (b + c) = az + 0 = az (why?).
So az = c.
The next step is ....
How do you get : (az+b) +c ?

Anyway if : (az+b) +c = 0 => az+(b+c) =0 => az +0 =0 =>az =0

6. ## Re: difficult problem

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.

7. ## Re: difficult problem

Originally Posted by awkward
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

8. ## Re: difficult problem

Originally Posted by psolaki
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
In the best mathematical tradition, I would try to reduce it to the previous problem.

9. ## Re: difficult problem

Originally Posted by awkward
In the best mathematical tradition, I would try to reduce it to the previous problem.
O.k ,I give up,show me

10. ## Re: difficult problem

Originally Posted by psolaki
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.)

11. ## Re: difficult problem

Originally Posted by awkward
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.
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

12. ## Re: difficult problem

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.

13. ## Re: difficult problem

Originally Posted by psolaki
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
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 ??

14. ## Re: difficult problem

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.
In the 2nd line of your proof , where the (az +b) +x =c +x equation come from

15. ## Re: difficult problem

i am still alittle confused as to which proof is correct : Deveno's or awkward's