Is the proof in my book number theory correct? Why?

[Steven G]

Below is a proof in my Number Theory book.
I have a problem accepting that the summation in the statement of the theorem equals f(n) as sometimes that summation equals 0. Shouldn't it be necessary that f(n) = 0 as well when the summation is 0? There is no reason (that I see) why f(n)=0 when the summation = 0. I do see that if the summation is not 0, then it equals f(n).

In case you need to know the definition of the mu function, I defined it at the end of my attachment.

Thanks for your time!

 

[blamocur]

There is no reason (that I see) why f(n)=0 when the summation = 0.

Why? I haven't checked the proof, but why cannot f(n) be 0 ?

 

[Steven G]

Why? I haven't checked the proof, but why cannot f(n) be 0 ?

Sure f(n) can be 0. My question is why is f(n)=0 when the summation is 0, that is where in the proof does it show that to be the case?

 

[blamocur]

I find it difficult to read hand-written math (especially my own), so I've typeset it in [imath]\LaTeX[/imath] the best I could. I think I mostly understand the text, but I still don't understand your question (probably because my math is better than my English . Can you point to a specific line relevant to your question?

Thank you.


Theorem: Let [imath]F[/imath] and [imath]f[/imath] be two number theoretic functions related by the formula
[math]F(n) = \sum_{d|n} f(d)[/math]Then
[math]f(n) = \sum_{d|n} \mu(d) F\left(\frac{n}{d} \right)[/math]where
[math]\mu(n) = \left\{ \begin{array}{ccl} 1 & \text{if } & n = 1\\ 0 & \text{if } & p^2 | n \text{ for some prime } p\\ (-1)^r &\text{if }& n = p_1 p_2...p_r \\ \end{array} \right.[/math]
Lemma 1: [imath]d | n \wedge c | (n/d) \Leftrightarrow c|n \wedge d|(n/c)[/imath]

Lemma 2: If [imath]m \neq 1[/imath] then [imath]\sum_{d|m)} \mu(d) = 0[/imath].

Proof of the theorem:
[math]\sum_{d|n} \mu(d)F\left(\frac{n}{d} \right) = \sum_{d|n}\left( \mu(d) \sum_{c|(n/d)} f(c) \right) =[/math][math]= \sum_{d|n}\left( \sum_{c|(n/d)} f(c)\mu(d) \right)[/math][math]= \sum_{c|n}\left( \sum_{d|(n/c)} f(c)\mu(d)\right) = \sum_{c|n}\left( f(c) \sum_{d|(n/c)} \mu(d)\right)[/math]
Using Lemma 2:

[math]\sum_{c|n}\left( f(c) \sum_{d|(n/c)} \mu(d)\right) = \sum_{c=n}\left( f(c) \sum_{d|1} \mu(d)\right) = f(n)[/math]

 

[Dr.Peterson]

I still don't understand your question (probably because my math is better than my English

I don't follow his English either. Here is why:

I have a problem accepting that the summation in the statement of the theorem equals f(n) as sometimes that summation equals 0.

I think this means that you think the theorem itself is false, because you think it is false when the summation is 0.

And I think the summation your refer to is the RHS of [imath]f(n) = \sum_{d|n} \mu(d) F\left(\frac{n}{d} \right)[/imath].

Shouldn't it be necessary that f(n) = 0 as well when the summation is 0?

I think you are saying that in order for the theorem to be true in the case where the summation is zero, f(n) must be zero; that's obviously an implication of the theorem. Apparently you think the latter is not necessarily true. What you haven't done is to show that. Why do you think that?

There is no reason (that I see) why f(n)=0 when the summation = 0.

Are you saying that the proof doesn't apply when the sum is zero? Where do you think it fails?

I do see that if the summation is not 0, then it equals f(n).

I think you're saying that you agree with the proof except when [imath]\sum_{d|n} \mu(d) F\left(\frac{n}{d} \right)=0[/imath]. What is the difference in that case?

My question is why is f(n)=0 when the summation is 0, that is where in the proof does it show that to be the case?

What in the proof are you saying doesn't apply in this case? You need to point out what detail you are referring to.

When we understand what you are claiming, it will be easier to look at the proof, or the statement of the theorem, to see if we agree. I haven't tried yet, because I don't see what you are saying.

 

[Steven G]

OK, I see my problem now. I thought that n/c was a constant, but now I see otherwise. n/c takes on (possibly) many values one of which is always 1. I got it fine now.
Thanks for your help.