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Math Help - Simple proof -1<0<-1 using axioms

  1. #1
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    Simple proof -1<0<-1 using axioms

    This is probably a 2 or 3 line proof but just can't seem to get it. (mistyped in the title)

    Prove -1<0<1 using Axioms 1-10 and a>0 => -a<0

    Thanks!
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  2. #2
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    Quote Originally Posted by economanc View Post
    Prove -1<0<1 using
    Axioms 1-10 and a>0 => -a<0
    If you want any help you must give us the list of Axioms 1-10.
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  3. #3
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    Quote Originally Posted by Plato View Post
    If you want any help you must give us the list of Axioms 1-10.
    oops ok:
    A1: Addition is associative
    A2: Addition is commutative
    A3: every integer has an identity with respect to addition
    A4: every integer has an inverse with respect to addition
    A5: Multiplication is associative
    A6: Multiplication is commutative
    A7: every integer has an identity with respect to multiplication
    A8: Distributive Law: x(y+z)=xy+xz
    A9: Closure Property: a set A is closed with respect to addition and multiplication if x+y is in A and xy is in A
    A10: Trichotomy Law: For every integer x, exactly one of the following is true: x is either in the set of negative integers, set of positive integers, or x=0

    You can also use the following properties that I have already proved:

    P1: a+b=a+c => b=c
    P2: a0=0a=0
    P3: (-a)b=a(-b)=-(ab)
    P4: -(-a)=a
    P5: (-a)(-b)=ab
    P6: a(b-c)=ab-ac
    P7: (-1)a=-a
    P8: (-1)(-1)=1
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  4. #4
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    There seems to be a missing definition.
    You should have: a>b if and only if a-b is a positive integer.

    Having that we get 1-0=1 which is a positive integer so 1>0.

    By the given you get a > 0\; \Rightarrow \; - a < 0.
    There you have it.
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  5. #5
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    you are correct, left off that definition.

    Thank you! So simple, I should have known
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