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

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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.
  • #1
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[tex]e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in N\rbrace[/tex]
where [tex]n=p_r^{e_r}[/tex]


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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  • #2
if [tex]0 \in N[/tex] then I get [tex]e_r+1[/tex]. If [tex]N[/tex] starts with 1, then OK.

Example: How many powers of 2 divide [tex]8 = 2^3[/tex] ... Answer: 4 of them, namely 1,2,4,8.
 
  • #3
The problem is not the first element of [itex]\mathbb N[/itex] to be 0 or 1; rather, is that 1 is always a divisor of any integer, so the set will always have [itex]e_r + 1[/itex] elements.
 
  • #4
I think g_edgar is right. If [tex]0\notin \mathbb N[/tex] then the set does not contain 1.

We were given

[tex]\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}[/tex]
Prove that [tex]\Lambda(n)= \sum_{d\mid n}{\mu\left(\frac{n}{d}\right) \log d}[/tex]
Hint: Calculate [tex]\sum_{d\mid n}{\Lambda (d)}[/tex] 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:

[tex]\sum_{d\mid n}{\Lambda (d)}=\log n[/tex]

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 [tex]e_1=0,\;\;\;k\in\mathbb N\;\;\;[/tex]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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  • #5
To make the first statement in my last post clearer...

[tex]1\notin \mathbb P[/tex]
[tex]\therefore d=1\Leftrightarrow k=0[/tex]EDIT:
So I guess I should write
[tex]e_r=\#\lbrace d:d\mid n\;\;\; \textup{and}\;\;\; d=p_r^k, k\in \mathbb N^*\rbrace[/tex]
where [tex]n=p_r^{e_r}[/tex]
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 [tex]\mathbb N[/tex] 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.

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

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