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Is this true?

  1. Mar 12, 2010 #1
    [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
    Last edited: Mar 12, 2010
  2. jcsd
  3. Mar 12, 2010 #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.
  4. Mar 12, 2010 #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.
  5. Mar 12, 2010 #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}
    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.)
    Last edited: Mar 12, 2010
  6. Mar 12, 2010 #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]

    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.
    Last edited: Mar 12, 2010
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