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Math Help - Proof of Preserv. of order

  1. #1
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    Proof of Preserv. of order

    I am stuck on this one!

    a>0, b>0 \implies ab>0
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  2. #2
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    Quote Originally Posted by altave86 View Post
    I am stuck on this one!

    a>0, b>0 \implies ab>0
    What axiomatic system do you consider for the order ,so that we can prove the above as a theorem??
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  3. #3
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    Quote Originally Posted by xalk View Post
    What axiomatic system do you consider for the order ,so that we can prove the above as a theorem??

    -Communtativity of addition and multip
    -distributivity of add and mult
    -associativity of add and mult
    -additive identity
    -multiplicative identity
    -additive inverse (-c and 1/c)
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  4. #4
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    Quote Originally Posted by altave86 View Post
    -Communtativity of addition and multip
    -distributivity of add and mult
    -associativity of add and mult
    -additive identity
    -multiplicative identity
    -additive inverse (-c and 1/c)
    No ,no i mean what axioms for the order.For example one such system is the following:

    1) \forall x\forall y[ x<y\Longrightarrow \neg (y<x)]

    2) \forall x\forall y\forall z[x<y\wedge y<z\Longrightarrow x<z]

    3) \forall x\forall y[ x\neq y\Longrightarrow (x<y\vee y<x)]

    4) \forall x\forall y\forall z[y<z\Longrightarrow (x+y<x+z)]

    5) \forall x\forall y\forall z[0<x\wedge y<z\Longrightarrow x.y<x.z]

    Now in a such axiomatic system, if in axiom (5) you put:

    x= a , y= 0 and z=b we have the following result:


    0< a\wedge  0< b\Longrightarrow 0.a< a.b ,which is exactly the theorem you are trying to prove ,since 0.a = 0

    However in another axiomatic system for order we can have a different proof
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  5. #5
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    Ohh sorry but I am a bit confused about that. This is just the introduction o Number theory and we are taking a lookat the real numbers, especifically R as a field and a totally ordered set. We don't look upon that stuff or symbols, just basic things. Sorry.

    Quote Originally Posted by xalk View Post
    No ,no i mean what axioms for the order.For example one such system is the following:

    1) \forall x\forall y[ x<y\Longrightarrow \neg (y<x)]

    2) \forall x\forall y\forall z[x<y\wedge y<z\Longrightarrow x<z]

    3) \forall x\forall y[ x\neq y\Longrightarrow (x<y\vee y<x)]

    4) \forall x\forall y\forall z[y<z\Longrightarrow (x+y<x+z)]

    5) \forall x\forall y\forall z[0<x\wedge y<z\Longrightarrow x.y<x.z]

    Now in a such axiomatic system, if in axiom (5) you put:

    x= a , y= 0 and z=b we have the following result:


    0< a\wedge  0< b\Longrightarrow 0.a< a.b ,which is exactly the theorem you are trying to prove ,since 0.a = 0

    However in another axiomatic system for order we can have a different proof
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  6. #6
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    Quote Originally Posted by altave86 View Post
    Ohh sorry but I am a bit confused about that. This is just the introduction o Number theory and we are taking a lookat the real numbers, especifically R as a field and a totally ordered set. We don't look upon that stuff or symbols, just basic things. Sorry.
    Here is a translation of the axioms written in a formal language.

    x<y implies \neg (y<x ) for all x,y

    x<y and y< z implies x< z for all x,y,z ( this is the transitive law)


     x\neq y implies x< y or y< x ,for all x,y ( this is another form of the trichotomy law)


    y< z implies y+x < z+x ,for all x,y,z


    0< x and y< z implies x.y < x.z ,for all x,y ,z ,
    .

    And if in this axiom we put x = a , y =0, z = b we have:

    0< a and 0< b \Longrightarrow 0< a.b .

    Which the theorem you want to prove.

    However in another axiomatic system your theorem could be considered as an axiom
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