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Flawd logic

  1. Feb 2, 2008 #1
    Well, I'm supposed to prove 0v=0

    It is stated that I'm only allowed to use the following axioms.

    let a,b,c be vectors and V is a vector space, then
    1)a&b is in V then a+b is in V
    6)a is in V implies ka is in V
    10) 1a=a

    The book does it like this, and i think its wrong

    0v=(0+0)v=0v +0v { axioms 4&8}

    now subtract 0v from both sides { axioms ????}
    we get 0=0v

    you see the problem here? theres no justification for the subtraction step because there is no axiom allowing the step. Logically I assume that I'm only allowed to use the 10 axioms to prove this theorem.
  2. jcsd
  3. Feb 2, 2008 #2
    Actually I figured it out


    Now I just use another equation that has nothing to do with the first.
    namely, 0v+(-0v)=0
    well, this implies 0v+(-0v)=0=(0v+0v)+(-0v)
    implies 0=0v+(0v+-0v)=0v+0=0v
  4. Feb 2, 2008 #3
    But I was working on another problem in which the author uses incorrect reasoning.

    authors' proof:
    Prove: if a+c=b+c, then a=b
    (a+c)+(-c)=(b+c)+(-c) add (-c) to both sides

    you see how in this example, (-c) was actually added to both sides. There is no way to justify this according to the 10 axioms.
  5. Feb 3, 2008 #4
    you don't need a axiom to add an element to both sides of a equallity, that is how it works.

    That is if

    a = b then a+c=b+c

    this is not a axiom, this is how equality works, he proves that

    a = b if and only if a+c=b+c
  6. Feb 3, 2008 #5


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    You will do a lot better in math if you stop assuming anytime you don't understand something, the author is wrong.

    You don't have to have an axiom that says "if a= b then a+ c= b+ c", that's part of the definition of "binary operation" and is assumed whenever you have a binary operation.
  7. Feb 3, 2008 #6
    Its not a+c=b+c iff a=b. Its a+c=b+c implies a=b. If it is just a binary operation,(I presume you guys to mean it is a definition), then why can it be proved using axioms only? No assumptions are needed except the hypothesis. Having the need to create a definition merely indicates that the statement cannot be proven.

    proof: (a+c)+(-(a+c))=0 ax.5
    (a+c)+(-(b+c))=0 hypothesis
    (a+c)+((-b)+(-c)))=0 ax.7
    (a+c)+((-c)+(-b))=0 ax2
    ((a+c)+(-c))+(-b)=0 ax3
    (a+(c+(-c)))+(-b)=0 ax3
    (a+0)+ (-b) =0 ax5
    (a)+(-b)=0 ax.4
    a=b ax.5
  8. Feb 3, 2008 #7
    in your last step from

    (a)+(-b)=0 ax.4


    a=b ax.5

    what do you use there?

    yep, you are right you use what you are trying to prove (a+c=b+c implies a=b), just with
    a -> a+(-b)
    b-> 0
    c-> b

    so you haven't proved anything
  9. Feb 3, 2008 #8
    ax. 5 says



    a+(-b)= 0 => b=a
  10. Feb 3, 2008 #9
    I see. Well how about this:

    a=a+0 ax4
    a=a+(c+(-c)) ax5
    a=(a+c)+(-c) ax3
    a=(b+c)+(-c) hypothesis
    a=b+(c+(-c)) ax3
    a=b+0 ax5
    a=b ax4

    Well? Haven't I disproven that the binary operation is "merely" a definition?
  11. Feb 3, 2008 #10
    Well, in the defense of the last step my first proof,

    b+0=b=(a+(-b))+(b)=a+(b+(-b))=a+0=a ax.4,3,5&4 in order form left to right.
    therefore a=b
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