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Getting into the depths of a proof

  1. Nov 21, 2009 #1
    For the following proof:

    [tex](a+b)^2 = (a+b)(a+b) = (a+b)a +(a+b)b =a^2 +b^2 +2ab[/tex].

    Write:

    a) a list of theorems and definition involved in the proof.

    b) a list of the laws of logic involved in the proof by transforming the theorems and the definitions of the proof into the statements of the proof
     
  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi evagelos! :smile:

    You first! :wink:

    Start with (a+b)2 = (a+b)(a+b) :smile:
     
  4. Nov 22, 2009 #3
    well, the main thing you use here is distributivity of multiplication over addition and the fact that "=" is a symmetrical relation, and rules of general calculation.

    for example

    the first thing you do is:

    [tex]{(a+b)}^2={(a+b)(a+b)}[/tex] because that's how multiplication of a number with itself a a finite amount of times is defined: [tex]{x*x*x*...*x \ n \ times \ is} \ x^n[/tex] and because "=" is a symmetrical relation (meaning that if x=y then y=x)

    than you use the fact that multiplication is distributive over addition (and again that "=" is symmetrical) to expand the expression and get (a+b)(a+b)=(a+b)a+(a+b)b and so on and so forth... that's for a) and for b) you list the rules of deduction that "allow" those computation to be made (such as if [tex]{(P\rightarrow S)and(P)[/tex] is true, then S is true).

    hope you get the idea. :)
     
    Last edited: Nov 22, 2009
  5. Nov 22, 2009 #4
    Should we say that the definition involved is:

    for all ,nεN , xεR : [tex] x^n = xx^{n-1}[/tex] ?
     
  6. Nov 22, 2009 #5
    you must say that for any set [tex]S[/tex] closed under an associative algebraic operation "[tex]\ast[/tex]" i.e. ([tex]S , \ast[/tex]) is a semigroup,

    [tex]\forall x \in S[/tex] , [tex]n \in \mathbb{N}[/tex] , [tex]x^n = \underbrace{x \ast x \ast ... \ast x}_{n \ times}[/tex] .

    that the algebraic operation is associative is very important (the definition doesn't hold if the operation is not associative)...
    but x needn't necessarily be a real number. for example the definition "works" for ([tex]\mathbb{Z}_n , \cdot[/tex]) as well.
     
    Last edited: Nov 22, 2009
  7. Nov 22, 2009 #6

    tiny-tim

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    erm :redface:

    actually all i was expecting you to say was that it's a definition … x2 = xx :wink:

    ok … now (a+b)(a+b) = (a+b)a + (a+b)b ?​
     
  8. Nov 22, 2009 #7

    should we say now for that statement responsible is the axiom : for all ,x,y,z : z(x+y) = zx + zy??
     
  9. Nov 22, 2009 #8

    tiny-tim

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    Yes (well, the other way round :wink:).

    (and that's called the "…" law ? :smile:)

    ok, now (a+b)a + (a+b)b = a2 + b2 + 2ab ? :smile:
     
  10. Nov 22, 2009 #9
    No .the other way round ,because if you put ; z= a+b , x=a ,y = b we have:

    z(x+y) = (a+b)a + (a+b)b and to use the same distributive law we must use commutativity and change ,(a+b)a + (a+b)b to a(a+b) + b(a+b) ,and using again distributivity we have:

    [tex] a^2 + ab + ab + b^2 [/tex]

    Now we must prove that : ab + ab = 2ab
     
  11. Nov 22, 2009 #10

    tiny-tim

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    That's easy … ab + ab = ab(1) + ab(1) :wink:

    ok, now part (b)

    (if I were you, I'd do that tomorrow :zzz:)
     
  12. Nov 22, 2009 #11
    O.K we curry on tomorrow
     
  13. Nov 23, 2009 #12
    O.K ,you start first now.
     
  14. Nov 23, 2009 #13

    tiny-tim

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    hmm … logic isn't my strong point. :rolleyes:

    ok, well I suppose it must involve the principle that things which are equal to a third thing are equal to each other?

    phew! :redface:

    your turn! :smile:
     
  15. Nov 23, 2009 #14
    By applying two times the law of Universal Elimination on the definition:

    for all, nεN ,xεR : [tex] x^n = xx^{n-1}[/tex] ,where we put n=2 and x = a+b

    we have the 1st statement of the proof;

    [tex](a+b)^2 =(a+b)(a+b)[/tex]

    Now what theorem or axiom and what law, the application of which will give us the 2nd statement of the proof i.e

    (a+b)(a+b) = (a+b)a + (a+b)b,
    do we have to use??
     
  16. Nov 23, 2009 #15
    I'm not exactly sure how things work around here (if only homework helpers should post or something like that... if it wasn't appropriate for me to post I apologize but...).

    Now, I think part b) asks for logical rules of inference (such as modus ponens, reduction ad absurdum, introduction rules etc) used when proving the statement...

    I noticed that the depths of this proof "oughtn't" be too rigorous (silly me:confused:) but "things which are equal to a third thing are equal to each other" while correct (it follows from the fact that "=" is an equivalence relation -the identity relation-, or since we're talking about logic, from the fact that the equality predicate is transitive), is not a rule of inference.

    anyway, I think I intruded enough here, so I'll take my leave. :smile:
     
    Last edited: Nov 23, 2009
  17. Nov 23, 2009 #16
    You are right it is a theorem concerning the predicate of equality
     
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