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Regorous and formal proof of (-1).x =-x

  1. Jun 6, 2009 #1
    Give a rigorous and then a formal proof of the theorem :

    [tex]\forall x [ (-1).x = -x ][/tex]
     
  2. jcsd
  3. Jun 6, 2009 #2
    Hint: (-1).x = -x means that (-1).x + x = 0
    This should be in a homework forum
     
  4. Jun 6, 2009 #3
    Meaning that formal proofs are homework stuf??
     
  5. Jun 7, 2009 #4

    HallsofIvy

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    Once again, we return to the question, "What do you mean by 'formal proof'?"

    I think you are just talking about the kind of proof you would find in a math paper or calculus book- far from what, say, logicians would mean by "formal proof". A true "formal proof" of such a thing would probably require an entire book! How many pages did Russel and Whitehead require to prove "1+ 1= 2"?

    Another question: what algebraic system are you working in? The proof for an abstract ring or integral domain would be quite different than for the real numbers.

    MXSCNT's point is that the only good reason for doing such "fiddly" stuff is practise: homework.
     
    Last edited: Jun 7, 2009
  6. Jun 7, 2009 #5
    It would be something like:

    0.x = 0
    (-1+1).x = 0
    -1.x + 1.x = 0
    -1.x + x = 0
    -1.x = -x
     
  7. Jun 7, 2009 #6
    The definition was given in another thread by tgt and is the following:

    With the only difference that i may add:

    each formula could be, apart from axiom, a theorem or definition


    Go to pages 121 to 139 in ANGELO'S MARGARIS book :

    FIRST ORDER MATHEMATICAL LOGIC.There you will find many true "formal proofs" not more than half a page long

    You know how many books you need before you double integrate a function ,or write a proof in analysis??

    On the following axiomatic system i will base any formal or rigorous proofs:

    The primitive symbols are:
    = for equality
    + for addition
    . for multiplication
    - for the inverse in addition
    0 constant
    1 constant
    / for inverse in multiplication

    [tex] 1\neq 0[/tex]
    AND the axioms are:

    1) [tex]\forall a\forall b[ a+b = b+a].......\forall a\forall b[ a.b = b.a ][/tex]

    2)[tex]\forall a\forall b\forall c[ a+(b+c) = (a+b)+c]........\forall a\forall b\forall c[ a(bc)=(ab)c][/tex]


    3)[tex]\forall a[ a+0 = a]............\forall a[ 1.a =a][/tex]


    4) [tex]\forall a[ a+(-a) = 0]..........\forall a[ a\neq 0\Longrightarrow a.\frac{1}{a} = 1]
    [/tex].



    5)[tex]\forall a\forall b\forall c [ a(b+c) = ab + ac][/tex]


    AND now the rigorous proof of (-1)x = -x



    (-1)x = (-1)x + 0 =.............................................................by axiom 3 (for addition)

    =(-1)x +[ x + (-x)]=.............................................................by axiom 4 (for addition)

    =[(-1)x +x] + (-x) =.............................................................by axiom 2 (for addition)

    =[(-1)x + 1x] + (-x)=.............................................................by axiom 3 (for multiplication)


    =[x.1 + x(-1)] + (-x)=............................................................by axiom 1 ( for addition and multiplication)


    =[ x( 1 + (-1))] + (-x)=.........................................................by axiom 5


    = x.0 + (-x) = ......................................................................by axiom 4 (for addition)


    = 0 + (-x) =.......................................................................by the theorem 0.x = 0


    = -x ..............................................................................by axiom 4 (for addition)

    Next post the formal proof
     
  8. Jun 8, 2009 #7
    Sounds like homework to me.
     
  9. Jun 8, 2009 #8

    CRGreathouse

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    360. But it's complicated only if you're trying to prove that equation under R or C; if you restrict yourself to Peano arithmetic it follows from the definition of 2 and the symmetry of equality.
     
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