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Logic confusion - formally clear but practically not so.

  1. Sep 9, 2011 #1
    Hello everyone!
    This is a pretty strange question, perhaps.
    I've studied logic and know very well the difference between A --> B and B --> A.
    However, there's a specific problem that I often encounter in proofs in maths that I find strange.
    For example: you get a certain P.D.E. and want to prove that there exists a solution and that it is unique. Then you prove: "if f(x) is a solution, then f(x) must be of a certain form (for example, f(x) = ex)". However, you also need to show that f(x) is a solution.

    What I can't really imagine, is a situation in which I prove that if f(x) is a solution, then f(x) = ex, but then to find out that ex isn't a solution.

    The only examples I could think of are "empty cases". For example, if you get the equation f'(x) +f(x) + 1 = f'(x) +f(x) + 2, I could say: if f(x) is a solution then f(x) = ex. Since f(x) is never a solution (cause there is no solution) we get a claim of the form F --> T/F, which is always "T".
    But are there other examples, other than empty cases, in which a solution must have the form f(x), but f(x) isn't a solution?
    Confused :-)
     
  2. jcsd
  3. Sep 12, 2011 #2

    Stephen Tashi

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    You could say that, but what are you claiming as the justification for that statement?

    What does that symbolism mean? You said that you've studied logic, but you aren't precisely translating verbal content into symbolic logic.

    The situation you are describing involves the existence of solutions. Have you studied the "existential quantifier", which is denoted by "[itex] \exists [/itex]" ? Do you understand the difference between a "statement" and a "predicate" ?

    If the question you have concerns logic and not the properties of PDEs, it might be simpler to look at an examples from algebra.
     
  4. Sep 13, 2011 #3
    Well, my logics volcabulary is a little rotten, but I recall that the statement "P --> Q" is TRUE when either P = Q = TRUE, or when P = FALSE (and Q can be either TRUE or FALSE).

    So I can say "if I live on the sun, then my name is Plato", and that would be true, because I don't live on the sun.
    Similarly, "If f(x) is a solution to that equation, then f(x) = ex" would be true, since the equation has no solution, and therefore the first statement must be FALSE, which would mean the whole statement is true.

    By F --> T/F I meant claims of the form P --> Q, where P is FALSE and Q is either TRUE or FALSE. I thought it was clear, sorry.

    Of course I know the existential quantifier. However, the difference between predicates and statements I'm not able to recall. I was asking the question rather loosely.
    It doesn't involve the properties of PDE's, but it's the only example I could think of that I'm encountering once in a while and somehow bothers me - as I've originally said, I can't imagine proving that if f(x) is a solution, f(x) must have a certain form, but then to find out that it isn't a solution!
     
  5. Sep 13, 2011 #4

    Stephen Tashi

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    The standard "modus tollens" argument is

    If A then B
    not B
    --------
    Therefore not A

    ( This amounts to a argument using the "contrapositive" of "If A then B".)

    If the premises of the argument are true, then A is false, so the "If A then B" is always "vacuuous"

    A "predicate" is an expression with variables that becomes a specific statement when its variables take specific values. For example, "x < 1" is a predicate since it doesn't become true or false unless x is given a specific value. When a quantifier is applied to a predicate, it can form a statement, such as "There exists an x such that x < 1" or "For each x, x < 1".

    For an example from algebra, you can consider solving the equation

    [itex] \frac{1}{x-2} + 1 = \frac{(x-1)^2}{x-2} [/itex]
     
  6. Sep 14, 2011 #5
    "f'(x) +f(x) + 1 = f'(x) +f(x) + 2"

    This is a false statement. If a false statement is used as the premise of an implication, it is called a vacuous implication. It's true, but doesn't tell us anything about the model.

    [itex](f'(x) +f(x) + 1 = f'(x) +f(x) + 2) \rightarrow f(x)=e^x[/itex]

    is therefore true, but not helpful.

    An example that you might find interesting (although it doesn't have to with differential equations) is one in fixed-point theory and the convergence of sequences. Suppose that we have some sequence defined by

    [itex]a_0=a[/itex]
    [itex]a_n=f(a_{a-1})[/itex] if n>1.

    where f is some continuous function. (Continuity is important here.) There is a theorem that states:

    if the sequence converges to a limit L, the limit must be a fixed point of f. That is, f(L)=L

    This doesn't tell us whether the sequence converges, but if the set of fixed points is finit and small (or infinite and just happens to have a convenient form) then we can check each and every one of them and find out. This is very similar to your example. The implication may be vacuous, but it is either vacuously true, or meaningful and true; it's never false.
     
    Last edited: Sep 14, 2011
  7. Sep 14, 2011 #6
    I'm not sure you've got to the bottom of what bothers me, so I'll get to a specific example:

    A theorem from P.D.E's states the following:
    u(x,t) is a solution to the wave equation [itex]\frac{\partial^{2}u}{\partial t^{2}} = c^{2}\nabla^{2}u[/itex] with the initial conditions u(x,0) = f(x) and ut(x,0) if and only if u(x,t) is given by:

    u(x,t) = [itex]\frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c}\int^{x+ct}_{x-ct}g(s)ds[/itex]

    In the proof, one first assumes that u(x,t) is a solution, and then gets the form above.
    However, they then write "We shall now have to verify that u(x,t) is indeed a solution to the equation", which they of course leave as an exercise. The thing is, taking this form, that we've just proved to be the form of a given solution, and starting to differentiate it and place it in the equation, feels like doing an almost trivial work. It just feels like there's no way it wouldn't be a solution, after proving that a solution has to have this form. While I understand this claim does not logically apply, I was wondering if I could have the importance of this process (checking that it is a solution) sharpened out. Could it have been that I would find that the solution must be of the form xyz, but then using this form would not work in the equation, other then some strange vacuous case, that I tried to portray and not sure I succeeded? :)

    Thanks a lot for the answers.
     
  8. Sep 14, 2011 #7
    Yes it absolutely could be the case. In fact, for every equation of any kind, if that equation has no solutions, then we can prove that "every solution has the form *" for any form *. It's always possible to prove that, which is why it's totally necessary to check that at least one solution actually exists.
     
  9. Sep 14, 2011 #8
    Alright, I'll just accept that. Thanks :-)
     
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