# Is this true?

1. Mar 12, 2010

### ballzac

$$e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in N\rbrace$$
where $$n=p_r^{e_r}$$

I am using it in a proof (we were actually shown how to do the proof a different way, but want to see if I can complete the proof in a different way. What I have stated seems logical to me, but not sure how I would prove it. I'm not 100% on the notation I've used, so if anything is unclear I can write what I mean in words. Thanks :)

hmmm, \hspace doesn't seem to work in this environment :S

Last edited: Mar 12, 2010
2. Mar 12, 2010

### g_edgar

if $$0 \in N$$ then I get $$e_r+1$$. If $$N$$ starts with 1, then OK.

Example: How many powers of 2 divide $$8 = 2^3$$ ... Answer: 4 of them, namely 1,2,4,8.

3. Mar 12, 2010

### JSuarez

The problem is not the first element of $\mathbb N$ to be 0 or 1; rather, is that 1 is always a divisor of any integer, so the set will always have $e_r + 1$ elements.

4. Mar 12, 2010

### ballzac

I think g_edgar is right. If $$0\notin \mathbb N$$ then the set does not contain 1.

We were given

$$\Lambda(n) := \left\lbrace\begin{array}{cc} \log p & \textup{if } n \textup{ is a power of a prime } p\\ 0 & \textup{if } n=1 \textup{ or } n \textup{ is a composite number} \end{array}$$
Prove that $$\Lambda(n)= \sum_{d\mid n}{\mu\left(\frac{n}{d}\right) \log d}$$
Hint: Calculate $$\sum_{d\mid n}{\Lambda (d)}$$ and apply the Mobius Inversion Formula.

I don't think the definition really makes sense, as I thought a composite number to include powers of primes, but it seems assumed that it means 0 OTHERWISE. Using the property that I gave in the first post, I got:

$$\sum_{d\mid n}{\Lambda (d)}=\log n$$

which is what we were given in class when shown how to do this differently, so I figured it must make sense. Obviously because the powers of p do not begin with $$e_1=0,\;\;\;k\in\mathbb N\;\;\;$$that I have given, does not start at 0, but 1. Am I making sense? If I am right, how would I prove it? ('It' being the property that I gave in the first post. If I can prove that, then the rest of the problem works itself out.)

Last edited: Mar 12, 2010
5. Mar 12, 2010

### ballzac

To make the first statement in my last post clearer...

$$1\notin \mathbb P$$
$$\therefore d=1\Leftrightarrow k=0$$

EDIT:
So I guess I should write
$$e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in \mathbb N^*\rbrace$$
where $$n=p_r^{e_r}$$
I always think of the natural numbers as starting at one, but I guess it is optional and one should be explicit about what the set $$\mathbb N$$ contains. So if someone can help me figure out how to prove the statement, I would really appreciate it.

Last edited: Mar 12, 2010