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Proving theorems

  1. Feb 2, 2005 #1
    Hi everybody,
    In each part of maths( Algebra, Geometry etc) we have set some axioms that we accept as true without proving them and, using these axioms we go on proving statements, some of which are important and are called theorems. When we prove a theorem , we usually refer to a general state and speak generally, for example, take the Bolzano theorem for the continuous functions:" if f a continuous real function defined in [a,b] and
    f(a)*f(b)<0 then there's at least one real number x with a<x<b so that f(x)=0
    This means that any function that is continuous in [a,b] where a,b two numbers and f(a)f(b)<0 then at least one number x exists with a<x<b so that f(x)=0, right? How are we sure that any function that has the properties mentioned in the theorem has one point in which its value is zero? Of course in order to prove the theorem we don't use a specific function. This is what makes us sure that it applies in every function with these properties? The fact that we prove that a function f that is continuous in a set [a,b] with a ,b real numbers that we don't know and that f(a)f(b)<0 without knowing anything else about the values of the function makes us sure that what we do in order to prove this statement can be applied in any specific function with these properties, so we generalise the statement and say that we can use it in any real function? That's why theorems are general? This is the concept behind any theorem? And one more thing: if there were some functions in which this theorem wan't true, then this should definitely come up while trying to prove it?I mean some restrictions should be taken in order to complete the proof ,and these restrictions whould show us which are the functions that this theorem doen't apply to?
    I referred to this theorem as an exampe but all these thoughts are about any theorem-statement that we prove in mathematics
     
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  3. Feb 2, 2005 #2

    matt grime

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    There really isn't much of a question in there, in fact if you remove the question marks it pretty much answers itself.

    If f is continuous and if on the interval [a,b] f(a) <0 and f(b)>0 then there is a c in [a,b] with f(c)=0. Usually called the intermediate value theorem or something like it.

    Anything satisfying the hypothesis will satisfy the conclusion, that is what a theorem (or proposition in this case) is.

    If we drop any of the hypotheses we do not know if the conclusion is still true. That is there a examples of discontinuous functions with f(a)f(b)<0 that have no zeros in the interior, and ones that do have zeroes in the interior. And there are continuous functions that have f(a)f(b)>0 that have zeroes and some that do not have zeroes in the interior.

    A proof is a series of logically valid deductions from an initial hypothesis, that is all.
     
  4. Feb 2, 2005 #3

    HallsofIvy

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    One of the things you left out: along with axioms, we also have "undefined terms".
    That's what really explains the strength of the "axiomatic method". In order to "apply" a theorem to a given situation, you only have to "fill in" the undefined terms.
    Theorems are like templates: you fill in the blanks to fit your particular need. As long as you can show that the axioms are (at least approximately) true for meanings you insert for the undefined terms, you know all theorems are also true.
     
  5. Feb 12, 2005 #4
    These "undefined terms" that we "fill in" are the ones that we refer to in the theorem generally. For example in the theorem i mentioned, these terms are the function and the set [a,b] that we in order to use the theorem must be specific? That is what you mean?
     
  6. Feb 12, 2005 #5

    mathwonk

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    your original ramble seems to me also to have contained a good understanding of the topic.

    Namely you correctly identify the lack of specificity of a theorem as a strength, since it enables it to be applied in many situations.

    Now the presence of "undefiend terms" does this as well, since it also enables the application of the theorem to various situations, in which the identity of those terms take on different forms.

    it is a little hard for me to invoke this in this example, since to me intervals are really intervals, but yes, in fact, if you identify those properties of intervals which are used in the rpoof, then the same prove could prove something else, where "intervals" actually had a different interpretation, but sharing the crucial properties.

    This is clearer to me in simpler cases like basic geometry, where with only a few aXIOMS, ONE CAN FIND MANY MODELS of lines, such as great circles, segments on a table top, or euclidean lines in R^2, etc...
     
    Last edited: Feb 12, 2005
  7. Feb 12, 2005 #6
    Thanks for your answer.In other words, what I am trying to say is that:
    generally in all theorems we don`t use specific mathematical "things", but a "random" element of a set. For example, we say " if f is a continuous function ... " so we pick randomly an element of the set "continuous functions" . That is what assures us that our results are general? If we can apply it to a random element of a set that we don't actually specify-define, this means that we can apply it in EVERY element of the set, right?

    From your answers i think that the above are correct. I just want to have a clear understanding of this really important topic
     
  8. Feb 12, 2005 #7

    Hurkyl

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    Nitpick: you mean to use the word "arbitrary element", not "random element".
     
  9. Feb 12, 2005 #8

    mathwonk

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    we don't pick anything. we just state a property (cont.) and then argue using only that property. then the argument hodls true for any case where the property holds.
     
  10. Feb 12, 2005 #9
    "For example, we say " if f is a continuous function ... " so we pick randomly an element of the set "continuous functions" ."
    I meant that we pick an arbitrary element in order to prove a theorem.
    Anyway, I got it. Thanks for your answers
     
    Last edited: Feb 12, 2005
  11. Feb 14, 2005 #10

    matt grime

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    No, we aren't picking anything at all. But I suspect this is semantics over the misuse of the word random.
     
  12. Feb 16, 2005 #11
    My English are not very good. I mean, when we want to prove a theorem, for example the one mentioned above, we say: "Let f be a continuous function in [a,b] with a,b real numbers. ..."
    That's what I meant by saying " we pick an arbitrary element" .
     
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