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  1. #1
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    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 ??
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  2. #2
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    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).
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    Re: difficult problem

    Quote Originally Posted by awkward View Post
    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
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    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 ....
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    Re: difficult problem

    Quote Originally Posted by awkward View Post
    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
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    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.
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    Re: difficult problem

    Quote Originally Posted by awkward View Post
    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
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    Re: difficult problem

    Quote Originally Posted by psolaki View Post
    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.
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    Re: difficult problem

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

    Quote Originally Posted by psolaki View Post
    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.)
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    Re: difficult problem

    Quote Originally Posted by awkward View Post
    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
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  12. #12
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    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.
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  13. #13
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    Re: difficult problem

    Quote Originally Posted by psolaki View Post
    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 ??
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    Re: difficult problem

    Quote Originally Posted by Deveno View Post
    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
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    Re: difficult problem

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