Proving the Limit of a Sequence: A Formal Proof Refuting an Informal Claim

  • Thread starter evagelos
  • Start date
In summary, the conversation discusses a formal vs informal proof of the statement "lim_{n\rightarrow\infty} \frac{1}{n}\neq 1". The informal proof is flawed as it assumes both lim_{n\rightarrow\infty}\frac{1}{n} =1 and lim_{n\rightarrow\infty}\frac{1}{n} =0 at the same time. This is a contradiction and cannot be used to prove the statement. A suggested correction is to use a direct proof, choosing a specific value for \epsilon and showing that there exists a value for n that contradicts the limit definition. It is also noted that the conversation discusses limits of sequences, which is a different concept from limits
  • #1
evagelos
315
0
formal vs informal

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

Proof:

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:
Physics news on Phys.org
  • #2
Step 2) is wrong!
 
  • #3
evagelos said:
Suppose [itex]lim_{n\rightarrow\infty}\frac{1}{n} =1[/itex], but [itex]lim_{n\rightarrow\infty}\frac{1}{n} =0[/itex], hence:

...

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).
 
  • #4
mathman said:
Step 2) is wrong!

I suppose there is an explanation for that
 
  • #5
gb7nash said:
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).


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] until 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]) until 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].....
 
  • #6
evagelos,

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].
 
  • #7
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.
 
  • #8
evagelos said:
...Suppose [itex]lim_{n\rightarrow\infty}\frac{1}{n} =1[/itex], but [itex]lim_{n\rightarrow\infty}\frac{1}{n} =0[/itex]...

Just change "but" to "but we already know that" and it's fine.
 
  • #9
evagelos said:
.Meanwhile we do know (we have proved that) [itex]lim_{n\rightarrow\infty}=0[/itex] and we use that in our proof.

Stephen Tashi said:
evagelos,

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.

Well.....

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].

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
 
  • #10
Hurkyl said:
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.

We have limits of sequences here,which a completely different aspect of that concerning limits of functions
 
  • #11
evagelos said:
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}

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.

But the point is not finding other solutions .
...
the only way to find that out is by writing a formal proof for the problem.

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.
 
  • #12
evagelos said:
We have limits of sequences here,which a completely different aspect of that concerning limits of functions
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.
 
  • #13
Stephen Tashi said:
Suppse M = 1 then K = M+1 doesn't make [itex] |1 - \frac{1}{k}| > \frac{1}{2} [/itex]

FOR M=1 we have M+1 =2 and [itex]|1-\frac{1}{2}|\geq\frac{1}{2}[/itex]

I forgot the equality


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.

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.
 
  • #14
Well, I'm glad that's settled. Now we can all go home.
 
  • #15
good by
 

1. What is the difference between informal and formal?

Informal and formal are two different styles of communication. Informal communication is casual and relaxed, while formal communication is more structured and follows specific rules and conventions.

2. When is it appropriate to use informal communication?

Informal communication is typically used in personal settings, such as among friends and family, or in casual work environments. It allows for more creativity and flexibility in language and tone.

3. When is it appropriate to use formal communication?

Formal communication is usually used in professional or academic settings, such as in business meetings or academic presentations. It is necessary for conveying important information in a clear and concise manner.

4. How does the use of language differ between informal and formal communication?

In informal communication, colloquial language and slang may be used, while formal communication requires the use of proper grammar and vocabulary. Formal communication also avoids personal pronouns and contractions.

5. Can both informal and formal communication be used in the same situation?

Yes, it is possible to use both informal and formal communication in the same situation, depending on the context and audience. For example, a business meeting may start with formal introductions and then transition into more informal discussions to build rapport.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
850
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
485
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
789
  • Calculus and Beyond Homework Help
Replies
13
Views
635
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
Back
Top