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What allow us to put x=a+b+c

  1. Aug 21, 2008 #1
    what allow us to put x=a+b+c


    Many times in mathematics in proving something we can put

    x=a+b+c or x= a-b^2 or x=y e.t.c.

    What axiom or theorem in mathematics allow us to do that??
  2. jcsd
  3. Aug 21, 2008 #2


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    Can you post (or link to) a specific example of where you've seen this?
  4. Aug 21, 2008 #3
    inequalities systems of equations e.t.c e.t.c for example one that comes into my mind just now is the following inequality:

    (a+b+c)/3 >= (abc)^1/3 here one way of solving the inequality is to use the substitution

    x=(a+b+c)/3 of course without using Am-Gm
  5. Aug 21, 2008 #4


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    Hmmm ... without knowing how to solve that one myself, it seems that this is a way of simplifying things by reducing the number of variables in the problem.

    What allows us to do this is the fact that, in this particular problem, "x" had not been defined as anything up to that point. So we are free to define it as (a+b+c)/3, or as anything else we wish.

    Hope that's clear enough ... perhaps others in here are familiar with this example and can elaborate a little more than I have?
  6. Aug 22, 2008 #5
    I am sorry for the example it was brought up at random ,its proof is immaterial

    i can give other examples
  7. Aug 22, 2008 #6
    can this be considered as axiom or a theorem.

    because the above is your basic argument in allowing the substitution

  8. Aug 22, 2008 #7


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    Honestly I'm not sure if that's considered an axiom or theorem ... or neither.

    This is done all the time in word problems, such as:

    Jim is twice as old as Jane was 4 years ago, and three times as old as she was 6 years ago. How old are Jim and Jane?

    And to find the answer, we might begin by saying:
    "Let x = Jim's age now, and y = Jane's age now ... "
  9. Aug 22, 2008 #8


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    This is just basic algebra: we let symbols represent quantities. I'm not sure why this would cause confusion, though.
  10. Aug 22, 2008 #9
    It comes from the axiom of the magic bag of variables.
  11. Aug 22, 2008 #10
    i think this is mathematical modeling isnt it
  12. Aug 23, 2008 #11


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    Yes, any application of mathematics is mathematical modeling.

    And we are free to define x to be a+ b, as long as we haven't already used "x" for some other meaning because we can always just assign a name to something.
  13. Aug 23, 2008 #12
    now i suppose you pushing me to open a new thread or to ask here what is the definition of the definition,because we all might not agree what a definition is.
    Because when you say we define x=a+b i don't think define is the proper word................
    for me
  14. Aug 23, 2008 #13


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    It might help if you said just where you are in your studies of math ... clearly you've already had some algebra, and I'm guessing geometry also (where theorems and axioms are introduced, at least in many USA high school curricula). Perhaps you are well beyond even these, and just looking for formal justification of what many take for granted.

    I checked a h.s. geometry book I have. It says we have four basic "rules", or types of justification, that allow us to proceed from step to step in a proof or derivation:

    Given information

    If we have only these 4 choices, I think "definition" comes closest to what you are looking for. If there is anything else outside of these, I am not aware of it.
  15. Aug 23, 2008 #14


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    Then I recommend you get a dictionary and look up the definition of "definition"!
  16. Aug 23, 2008 #15
    I'm curious as to see how you would prove AM-GM for three variables using that substitution. I know you can prove it by factoring [tex]a^3 + b^3 + c^3 -3abc[/tex] but it's not obvious to me how such a substitution helps.
  17. Aug 24, 2008 #16
    as i said in post #5 the example was picked up at random from a book the proof is not mine .The book uses the Am-Gm for two variables plus the substitution in solving
    the inequality
  18. Aug 25, 2008 #17
    But you mentioned that it doesn't use AM-GM and now you're saying it does involve AM-GM of two variables. Anyways that doesn't matter. I'm sure you can post the proof? I'm just curious that's all.
  19. Aug 26, 2008 #18
    here is a proof direct from the book:

    let x=(a+b+c)/3 = (a+b+c+x)/4 = [(a+b)/2 +(c+x)/2]/2 >= [sqroot(ab)+sqroot(cx)]/2>=

    sqroot[sqroot(ab)sqroot(cx)]= sqroot[sqroot(abcx)] = (abcx)^1/4

    hence x=(abcx)^1/4 ====> x^4>=abcx

    if x=0 then a=b=c=o.
    if x>0 then : x^3>= abc<====> x>= (abc)^1/3 <====> (a+b+c)/3>= (abc)^1/3
  20. Aug 26, 2008 #19
    That is a tricky substitution. Very nice proof, thanks!
  21. Aug 26, 2008 #20
    as i said the proof is not mine
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