# Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$

• ballzac
In summary, e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in N^*\rbrace. If n is not a power of a prime, then d=1.

#### 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

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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.

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.

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.)

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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.

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## What is "Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$"?

Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$ is a mathematical proof that shows the relation between the exponent of a prime factor in a number and the number itself.

## Why is "Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$" important?

This proof is important because it helps to understand the properties of prime numbers and their exponents, which are essential in many areas of science and mathematics.

## What is the significance of the notation used in "Proof of e_r = #{d: d $\mid$ n and $d = p_r^k$, $k \in N$} for n = $p_r^{e_r}$"?

The notation used in this proof represents the set of all divisors of n that are equal to a power of p_r. This allows for a concise and clear representation of the relationship being proven.

## How is this proof related to prime factorization?

This proof is closely related to prime factorization because it shows how the exponent of a prime factor in a number can be determined by its prime factorization. In other words, it explains why the exponent of a prime factor in a number is always a natural number.

## Can this proof be applied to numbers with multiple prime factors?

Yes, this proof can be extended to numbers with multiple prime factors. The key concept remains the same – the exponent of a prime factor in a number is equal to the highest power of that prime number that divides the number.