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Informal vs formal

  1. Sep 3, 2011 #1
    formal vs informal

    in proving that: [itex]lim_{n\rightarrow\infty} \frac{1}{n}\neq 1[/itex] the following proof was suggested.


    Suppose [itex]lim_{n\rightarrow\infty}\frac{1}{n} =1[/itex], but [itex]lim_{n\rightarrow\infty}\frac{1}{n} =0[/itex], hence:

    For all ε>0

    1) There exists mεN such that: [itex]n\geq m\Longrightarrow |\frac{1}{n}|<\frac{\epsilon}{2}[/itex]

    2)There exists kεN such that : [itex]n\geq k\Longrightarrow |\frac{1}{n}-1|<\frac{\epsilon}{2}[/itex]

    Choose r = max{m,k},then [itex]r\geq m,r\geq k[/itex]

    Let ,[itex]n\geq r\Longrightarrow n\geq m\wedge n\geq k[/itex].

    Hence : [itex]|\frac{1}{n}|<\frac{\epsilon}{2}[/itex] and [itex]|\frac{1}{n}-1|<\frac{\epsilon}{2}[/itex].

    Thus : [itex]|\frac{1}{n}-\frac{1}{n}+1|=1\leq |\frac{1}{n}| + |\frac{1}{n}-1|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/itex]

    ..............or 1<ε..........But since this holds for all ε>0 we put ε=1 and we have 1<1 ,a contradiction .

    Therefor [itex]lim_{n\rightarrow\infty}\frac{1}{n}\neq 1[/itex]

    Write a formal proof of the above ,thus proving that the above informal proof is wrong
    Last edited: Sep 3, 2011
  2. jcsd
  3. Sep 3, 2011 #2


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    Step 2) is wrong!
  4. Sep 3, 2011 #3


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    The proof is flawed at this point. If you're going to use a contradiction proof, you can't assume a consequence of what you're trying to prove and the contradictory statement at the same time. If you're going to prove by contradiction, only assume [itex]lim_{n\rightarrow\infty}\frac{1}{n} =1[/itex] and obtain a contradiction from that(using limit definition).
  5. Sep 3, 2011 #4
    I suppose there is an explanation for that
  6. Sep 3, 2011 #5

    I do not understand.We want to prove that: [itex]lim_{n\rightarrow\infty}\neq 1[/itex] and we assume the opposite i.e:[itex]lim_{n\rightarrow\infty}= 1[/itex] untill we get a contradiction.Meanwhile we do know (we have proved that) [itex]lim_{n\rightarrow\infty}=0[/itex] and we use that in our proof.Where is the mistake

    In logical terms we want to prove : q= ([itex]lim_{n\rightarrow\infty}\neq 1[/itex]) .We assume not q= ([itex]lim_{n\rightarrow\infty} = 1[/itex]) untill we come to a contradiction:

    s and not s.In our case s is: 1<1 and not s is: [itex]\neg 1<1[/itex] ,since we know that:

    ............[itex]\forall x(\neg x<x)[/itex].........................
  7. Sep 3, 2011 #6

    Stephen Tashi

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    You aren't making it clear whether [itex] \lim_{n \rightarrow \infty} {\frac{1}{n}} = 0 [/itex] is to be taken as a previously established theorem.

    It would much simpler to write a direct proof anyway. Let [itex] \epsilon = \frac{1}{2}[/itex]. Let M be any nonnegative integer. Let K be the max of the set {3,M+1}. Then [itex] \frac{1}{2} < 1 - \frac{1}{3} \leq 1 - \frac{1}{K} [/itex]. So [itex] | 1 - \frac{1}{K} | \geq \epsilon [/itex].
  8. Sep 3, 2011 #7


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    Let me remind people a subtle point about the definition of limit: a priori, both of the following statements could be true:

    • [tex]\lim_{x \to a} f(x) = 0[/tex]
    • [tex]\lim_{x \to a} f(x) = 1[/tex]

    If you're paying attention to the subtle details, one of the first thing you do with limits is prove that if
    • [tex]\lim_{x \to a} f(x) = L[/tex]
    • [tex]\lim_{x \to a} f(x) = M[/tex]
    are both true, then L = M. (And only once we've done this step does it make sense to ask about "the value of the limit".

    If I had to guess, evagelos is trying to replicate that argument in a special case, where he has a specific a, f, L=0, M=1, and wants to deduce that this leads to contradiction.
  9. Sep 3, 2011 #8
    Just change "but" to "but we already know that" and it's fine.
  10. Sep 4, 2011 #9

    Any No ,k, greater than M i.e M+1,M+2,M+3.....................e.t.c ,would make :

    [itex] |1-\frac{1}{k}|>\frac{1}{2}>\epsilon>0[/itex] ,you do not have to take k as the max{3,M+1}

    But the point is not finding other solutions .The point here is that the above suggested solution is wrong .

    And i proposed that the only way to find that out is by writing a formal proof for the problem.

    Unless somebody shows another way
  11. Sep 4, 2011 #10
    We have limits of sequences here,which a completely different aspect of that concerning limits of functions
  12. Sep 4, 2011 #11

    Stephen Tashi

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    Suppse M = 1 then K = M+1 doesn't make [itex] |1 - \frac{1}{k}| > \frac{1}{2} [/itex]
    Besides, a proof only has to pick a K that works. It doesn't have to pick the smallest K that works.

    How do you distinguish between "writing a formal proof for the problem" and "finding other solutions"? If you write a formal proof for the problem, you are finding another solution to proving it.
  13. Sep 4, 2011 #12


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    Limits of sequences are limits of functions. But even if you don't want to think about that and want to imagine them different, everything I've said applies to both cases.
  14. Sep 4, 2011 #13
    FOR M=1 we have M+1 =2 and [itex]|1-\frac{1}{2}|\geq\frac{1}{2}[/itex]

    I forgot the equality

    Definitely not . Formal proof is expanding on an already existing informal proof.

    For example if we want to prove that : 0x =0,we can write the following informal proof:

    1x = x => (1+0)x = x+0 => 1x +0x = x+0 => x+0x x+0 => 0x = 0

    Now if we want to write a formal proof of the above informal proof we have to justify each statement of the proof .For example (1+0)x =x+0 , where did it come from ,what laws of logic on what theorems or axioms applied to give us the desired result?

    WE do not write another proof.
  15. Sep 4, 2011 #14

    Stephen Tashi

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    Well, I'm glad that's settled. Now we can all go home.
  16. Sep 5, 2011 #15
    good by
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