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Proof using the Axioms of Addition and Multiplication

  1. Feb 21, 2006 #1
    So does everybody know the Axioms of Addition and Multiplication?
    They are too long to type, but they are listed:
    A1, A2, A3, A4, A5, M1, M2, M3, M4, M5 and the distributive law, DL.

    anyways, I want to prove:

    1. (-x)y = -(xy) and 2. (-x)(-y) = (xy) using ONLY the axioms of additon and multiplication.

    Can somebody please help start me off, ive spent an hour on this and am getting now where :grumpy:

    1. seems quite elementry to me.
    (-x)y = (-1)(x)y by M4
    (-x)y = (-1)(xy) by M3
    (-x)y = -(xy) by M3 again

    how does that sound?
    =
     
  2. jcsd
  3. Feb 21, 2006 #2

    arildno

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    Hmm.. I didn't know that (-x)=(-1)x was an axiom.
    You should prove that one, and all your troubles vanish.
     
  4. Feb 21, 2006 #3

    honestrosewater

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    Well, (-x)y + xy = (-x + x)y = ...
     
  5. Feb 21, 2006 #4

    quasar987

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    I don't know what you call M1,2,3,4, etc but here's how I'd do it:

    1° Prove that -x = (-1)(x)

    It follows that

    2° (-x)y = [(-1)(x)]y = (-1)[(x)(y)] (associativity) = -(xy) (by 1° again)

    edit: I like honestrose's way better, but it's good to prove once and for all that -x = (-1)(x).
     
    Last edited: Feb 21, 2006
  6. Feb 21, 2006 #5

    Prove that -x = (-1)(x) : That is a theorm in the textbook, and the proof is their with it.

    x + (-1)x = x + x(-1)by M2
    =x1 + x(-1) by M4
    x[1+(-1)]
    x0

    about honestrose's way,(-x)y + xy = (-x + x)y = ... (0)y = 0. Is this what he was trying to show as the proof?

    Well, (-x)y + xy = (-x + x)y = ...
     
  7. Feb 21, 2006 #6

    quasar987

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    (-x)y + xy = (-x + x)y (distributivity) = (0)y = 0. Hence (-x)y is the additive inverse of xy, which we note -(xy).
     
  8. Feb 21, 2006 #7

    honestrosewater

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    Right, and if a + b = 0, then b = ?

    I'm a she, by the bye. :smile:
     
  9. Feb 21, 2006 #8
    if a + b = 0 then b = -a
     
  10. Feb 21, 2006 #9

    honestrosewater

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    So if (xy) + (-x)y = 0, then (-x)y = ?

    What do you have for (2)?
     
  11. Feb 21, 2006 #10
    if (xy) + (-x)y = 0, then (-x)y = -(xy) as required.

    For (2), i did it a similar way

    2. Prove that (-x)(-y) = (xy) using only the axioms....

    (-x)(-y) - (xy) = (-x + x)(-y) = 0(-y) = 0
    thus, (-x)(-y) - (xy) = 0
    therefore, (-x)(-y) = (xy)

    how does that sound!? i have a feeling its good!
     
    Last edited: Feb 21, 2006
  12. Feb 21, 2006 #11

    quasar987

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    The general idea is good, and there is nothing you've written that is false.

    But it's not detailed enough. You have to justify each and everyone of your = signs by a statement such as "by definition of ..." or "by axiom ..." or "according to theorem ..." or "by the result found in (1)". That's the whole point of proving using the axioms.
     
    Last edited: Feb 21, 2006
  13. Feb 22, 2006 #12
    that you quasar987, In my answer, I would write "by M1, by M2, by A4... etc" (refering to the specific axiom.)

    I came accross another Proving question in the book, (in the same section as the axiom one.)

    Prove: If [itex]x \geq 0[/itex] and [itex]x \leq \varepsilon[/itex] for all [itex] \varepsilon > 0[/itex] then [itex]x = 0[/itex]

    Firstly, I decided to write it up more mathematically.

    Prove: If [itex]0 \leq x \leq \varepsilon[/itex] for all [itex] \varepsilon > 0[/itex] then [itex]x = 0[/itex]

    Now, this statement dosn't even look true to me. Suppose x = 9 and [itex]\varepsilon = 11[/itex], then dosn't the statement hold? But since it says Prove, ill just accept the fact that its true.

    So is it possible to prove this using ONLY the axioms? i took the contrapositive and that didnt help :grumpy:
     
  14. Feb 22, 2006 #13

    quasar987

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    The statement is that if [itex]0\leq x \leq \epsilon[/itex] FOR ALL [itex]\epsilon>0[/itex]. So if you say x =9, while it is true for e = 11, it is not true for e = 8, hence it is not true for all e>0.
     
  15. Feb 22, 2006 #14

    quasar987

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    Try proof by contradiction. Suppose x is not zero. Can you find a number epsilon smaller than x?
     
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