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Math Help - function proof

  1. #1
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    function proof

    Challenge problem


    Given, the following definition of a function from a set A to a set B
    denoted as: :

    1)

    2) for all , ,there exists a unique ,such that

    Prove that the following f:

    f={(1,a),(2,b)} is a function from A to B,where :

    A = {1,2} and

    B = {a,b}

    Moderator approved CB
    Last edited by CaptainBlack; July 3rd 2010 at 10:14 AM. Reason: correction
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  2. #2
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    Quote Originally Posted by alexandros View Post
    Challenge problem


    Given, the following definition of a function from a set A to a set B
    denoted as: :

    1)

    2) for all , ,there exists a unique ,such that

    Prove that the following f:

    f={(1,a),(2,b)} is a function from A to B,where :

    A = {1,2} and

    B = {a,b}

    Moderator approved CB
    I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple (A,B,F) with F \subseteq (A\times B) as you defined above.

    Anyway it's easy to check that {(1,a),(2,b)} \subseteq(A\times B)={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

    (Fixed wording.)
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  3. #3
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    Quote Originally Posted by undefined View Post
    I don't understand why this is a challenge problem; it's just a simple definition. Usually the formal definition is that f is an ordered triple (A,B,F) with F \subseteq (A\times B) as you defined above.

    Anyway it's easy to check that {(1,a),(2,b)} \subseteq(A\times B)={(1,a),(1,b),(2,a),(2,b)} and that for x=1, the unique y is a, and for x=2, the unique y is b.

    (Fixed wording.)
    I think there is a difference between ""check" and "prove",don't you?
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  4. #4
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    Quote Originally Posted by alexandros View Post
    I think there is a difference between ""check" and "prove",don't you?
    There is no difference in any finite case.
    I absolutely agree, I cannot see why this is a challenge problem.
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  5. #5
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    If you want me to reveal my proof so you can see the difference ,i can do so.
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  6. #6
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    Quote Originally Posted by alexandros View Post
    If you want me to reveal my proof so you can see the difference ,i can do so.

    Please do. Without putting any shame on you, I think that including questions like this one in the challenge problems section demerit and defile the intention of it.

    Tonio
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  7. #7
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    proof:

    Let : \vee denote "or" ,and

    let : \wedge denote "and"



    1)Let (x,y)εf =>

    [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B


    2) Let xεΑ => χ=1v x=2.

    For x=1

    put y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B

    Also


    (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f



    Thus there exists y such that, y\in B and (x,y)\in f

    So much for the existence part.

    Now for the uniqueness part.For that we must prove:

    for all ,y, z: [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]

    So let :

    (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]



    BUT 1\neq 2 (an assumption needed in the initial hypothesis for f to be a function),thus 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)

    ,since x=1.

    In the same way . (x,z)\neq (2,b)

    Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

    So:

    for x= 1 there exists a unique y\in B ,such that (x,y)\in f

    in the same way:

    for x=2 there exists a unique y\in B ,such that (x,y)\in f

    Therefor.:

    For all ,x: x\in A there exists a unique y\in B ,such that (x,y)\in f .

    Thus : f:A\to B
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  8. #8
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    Quote Originally Posted by alexandros View Post
    proof:

    Let : \vee denote "or" ,and

    let : \wedge denote "and"



    1)Let (x,y)εf =>

    [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B


    2) Let xεΑ => χ=1v x=2.

    For x=1

    put y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B

    Also


    (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f



    Thus there exists y such that, y\in B and (x,y)\in f

    So much for the existence part.

    Now for the uniqueness part.For that we must prove:

    for all ,y, z: [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]

    So let :

    (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]



    BUT 1\neq 2 (an assumption needed in the initial hypothesis for f to be a function),thus 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)

    ,since x=1.

    In the same way . (x,z)\neq (2,b)

    Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

    So:

    for x= 1 there exists a unique y\in B ,such that (x,y)\in f

    in the same way:

    for x=2 there exists a unique y\in B ,such that (x,y)\in f

    Therefor.:

    For all ,x: x\in A there exists a unique y\in B ,such that (x,y)\in f .

    Thus : f:A\to B

    With all due respect the above is completely absurd and if I may add, even ridiculous: nobody not out of his mind would ever present such a cumbersome, long and ridiculously messy proof of such a basic, easy-to-check and straighforward fact in such a basic case, UNLESS he's been said to do so in order to check his/her command in this or that logic-related or definition-related matters.

    Anyway, and as has already been pointed out by some other members of the forum, unless some really interesting, unexpected and/or brilliant insight is produced I cannot understand how this question has been aproved as a challenge one, and even less can I understand how the above horrible, awful proof could possibly be accepted by any teacher/lecturer or professor in any university.

    Tonio
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  9. #9
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