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Proof checking

  1. Mar 5, 2005 #1
    I'm trying to show the following:

    Let [tex]a^n+b^n=c^n[/tex] where [tex]a, b, c[/tex] are not equal to zero. There is no real solution for [tex]a, b, c[/tex] as [tex]n\rightarrow\infty[/tex].

    Proof:

    Assume the contrary. If there is a real solution for [tex]a, b, c[/tex] as [tex]n\rightarrow\infty[/tex], then

    [tex]lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=1[/tex]

    must also be true.

    We can use properties of limits and algebra to obtain, from the previous equation

    [tex]lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1[/tex]

    It can be shown that if [tex]a^n+b^n=c^n[/tex] (where [tex]a, b, c[/tex] are not equal to zero) is true, then [tex]c>a[/tex] and [tex]c>b[/tex]

    Therefore

    [tex]0<\frac{a}{c}<1[/tex] and [tex]0<\frac{b}{c}<1[/tex]

    Using properties of limits, we can state that

    [tex]lim_{n\rightarrow\infty} \frac{a^n}{c^n} = lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 0[/tex]

    and therefore

    [tex]lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=0[/tex]

    Which is not 1, and we have thus shown that the contrary to the first statement cannot be true.

    p.s. I hate tex
     
    Last edited by a moderator: Mar 5, 2005
  2. jcsd
  3. Mar 5, 2005 #2
    Icebreaker,
    [tex]lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1[/tex]

    for starters where did you get this?

    Before we get too caught up in that, I have to ask, What does a limit have to do with it?
     
    Last edited: Mar 5, 2005
  4. Mar 5, 2005 #3
    [tex]lim_{n\rightarrow\infty}\frac{a^n+b^n}{c^n}
    =lim_{n\rightarrow\infty}\frac{a^n}{c^n}+\frac{b^n}{c^n}
    =lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}
    =1[/tex]

    If you have a way to prove it without limits, be my guest.
     
    Last edited by a moderator: Mar 5, 2005
  5. Mar 5, 2005 #4

    Curious3141

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    First of all, I don't like the problem "statement" at all, it makes little mathematical sense. Did you get that out of a textbook, or did you make it up ?

    The ("Fermat-like") equation [tex]a^n + b^n = c^n[/tex] has infinitely many real solution sets (a,b,c) for every positive integral value of n. Asking what solutions exist when n is infinite, or even "n tends to infinity" is meaningless. This is not evaluating the limit of an expression, this is gobbledegook.
     
  6. Mar 5, 2005 #5
    Why is asking what solutions exist when n increases without bound meaningless?
     
  7. Mar 5, 2005 #6

    Curious3141

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    Think about what you're asking.

    Do real solution sets (a,b,c) exist for n = 1000 ? YES

    Do real solution sets (a,b,c) exist for n = 100,000 ? YES

    Do real solution sets (a,b,c) exist for n = 10^(10^(10^10) ? YES

    Is n "tending" to infinity in the above ? I suppose so.

    Do real solutions exist all the way regardless of the finite value of n, no matter how large ? Obviously.

    Does it make sense to ask the question for an infinite value of n ? NO.

    ***

    Do you see now ?
     
  8. Mar 5, 2005 #7
    Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.
     
  9. Mar 5, 2005 #8

    Curious3141

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    As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

    If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

    When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large. Note that even if the limit exists, the expression can NEVER actually equal the limit. It can only approach it.

    Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.
     
  10. Mar 5, 2005 #9
    Oh? [tex]lim_{n\rightarrow\infty}\frac{1}{n}=0[/tex]

    Of course there's a solution for every finite n. So?

    So I'm investigating the value of the expression as a variable is increasing without bounds. And?

    I never said n EQUALS infinity.
     
  11. Mar 5, 2005 #10
    Congratulations, you've proven that there are at least 3 solutions for the equation.

    Why do you think I set up the limit? Because I don't simply "suppose so".

    I never said they didn't.

    And here is where your logic breaks down.
     
  12. Mar 5, 2005 #11
    By your logic, suppose we have [tex]y=\frac{1}{x}[/tex]. We find that there is a non-zero [tex]y[/tex] for any arbitrarily large [tex]x[/tex], and therefore makes no sense to ask whether there's a limit for [tex]y=\frac{1}{x}[/tex] as [tex]x\rightarrow\infty[/tex].
     
  13. Mar 5, 2005 #12
    I'm confused... Do you want n to equal infinity or not? :confused:

    --J
     
  14. Mar 5, 2005 #13
    No, I don't. Read the post above yours.
     
  15. Mar 5, 2005 #14
    Could you state your meaning using a more rigorous definition of limits?
     
  16. Mar 5, 2005 #15
    To Illustrate: a case where you could use the limit is the case of the Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89,144....where the series in continued by the formula: F(N-1)+F(N)=F(N+1), or we add two successive numbers to get the next one.

    Now it happens to be that this series is closely connected with the Golden Mean:
    [tex]\frac{-1+\sqrt(5)}{2}[/tex] =.618034...

    We look at: 34/55 =.618182..;55/89 =.617978...;89/144 =.618056...

    Thus we might wonder if the limit as n goes to infinity of F(n)/F(n+1) = Golden Mean...which it does!

    This limit is so good, that if divided 144 by the Golden Mean we get 232.996, which rounds off to 233 =89+144. (In fact it is good even from the second "1" giving 1/GM = 1.6180 rounds to 2. 2/GM =3.23 rounds to 3, 3/GM=4.85 rounds to 5!
     
    Last edited: Mar 5, 2005
  17. Mar 5, 2005 #16

    :confused:

    And I don't understand what the post that supposedly explains whether or not you want n to be infinity. I'm still confused on this issue.

    --J
     
  18. Mar 5, 2005 #17

    Curious3141

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    Oh-what ? [tex]\frac{1}{n}[/tex] is an EXPRESSION, not an EQUATION. And it *never* equals zero no matter what value of n you choose. Only the limit is zero. There is a fundamental distinction here you don't seem to be grasping.


    So the premise you set about to "prove" is WRONG !!



    No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.

    You CAN investigate the behavior of the EXPRESSION [tex]{(a^n + b^n)}^{\frac{1}{n}}[/tex] as n tends to infinity. If, for simplicity, we allow only for positive real values of a and b, then

    [tex]\lim_{n \rightarrow \infty}{(a^n + b^n)}^{\frac{1}{n}} = max(a,b)[/tex]

    Fair enough ?

    *If* you say [tex]c = {(a^n + b^n)}^{\frac{1}{n}}[/tex] THEN it's perfectly acceptable to say :

    [tex]\lim_{n \rightarrow \infty}c = max(a,b)[/tex]

    Here you are still investigating the behavior of the expression called "c" as n tends to infinity. It's completely fine.

    OTOH, it is not standard practice (and hence not acceptable) to do the same for an equation trying to find solutions at a limit.
     
  19. Mar 5, 2005 #18
    I tihnk your problem is that you haven't defined "solution" well enough, when you said there wasn't one.
     
  20. Mar 5, 2005 #19
    Oh? And

    [tex]\frac{a^n+b^n}{c^n}[/tex]

    is NOT an expression?

    Interesting how you can rush to that conclusion based on 3 examples where n equalled finite numbers.

    No, I am investigating the EXPRESSION

    [tex]\frac{a^n+b^n}{c^n}[/tex]

    as n increases without bound. And from what I see, there's little difference between what I am trying to do, and the example which you gave.
     
    Last edited by a moderator: Mar 5, 2005
  21. Mar 5, 2005 #20
    Perhaps. I will try and find a better definition of "solution".

    No, you read the wrong post. Read the one directly above your first reply to this thread.
     
    Last edited by a moderator: Mar 5, 2005
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