Proving the Theorem: p!/[(p-i)! * i] = 1/p for Prime Number p and Integer i

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Homework Help Overview

The discussion revolves around proving a theorem related to prime numbers and combinatorial expressions, specifically the equation p!/[(p−i)! * i] = 1/p for a prime number p and an integer i, under the condition that 0 < i < p.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants express uncertainty about how to approach the proof, with some attempting to clarify the theorem's statement and others suggesting the use of induction as a potential method. Questions arise regarding the completeness of the original statement and the specific properties being proved.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the theorem's statement and exploring different approaches to proving it. Some guidance has been offered regarding the use of induction, but there is no explicit consensus on the method or interpretation of the theorem.

Contextual Notes

There are indications of confusion regarding the theorem's formulation and its relation to known properties of prime numbers and combinatorial coefficients. Participants note the importance of adhering to the forum's template for clarity.

hlzombi
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Prove the following theorem:
Theorem For a prime number p and integer i,
if 0 < i < p then p!/[(p− i)! * i] * 1/p

Not sure how to go about this. I wanted to do a direct proof and this is what I've got so far.
let i = p-n
then p!/[(p-n)!*(p-n)] but that doesn't exactly prove much.
 
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hlzombi,
First, please note that you should use the template and this seems like it might be better to place in the math forum.
hlzombi said:
Prove the following theorem:
Theorem For a prime number p and integer i,
if 0 < i < p then p!/[(p− i)! * i] * 1/p

Not sure how to go about this. I wanted to do a direct proof and this is what I've got so far.
let i = p-n
then p!/[(p-n)!*(p-n)] but that doesn't exactly prove much.
##\frac{p!}{(p-n)!*(p-n)}=\frac{p*(p-1)*...*(p-n+1)}{p-n}##
Finally, I am not sure what you are asked to prove...is there some equality or property here?
You need to be more clear.
 
Apologies, I'm new here. I tried to follow the template as best as I could.

To clarify, I'm trying to prove the theorem p!/[(p− i)! * i] * 1/p where 0<i<p when p is a prime number and i is an integer.
 
hlzombi said:
Apologies, I'm new here. I tried to follow the template as best as I could.

To clarify, I'm trying to prove the theorem p!/[(p− i)! * i] * 1/p where 0<i<p when p is a prime number and i is an integer.

Your statement is incomplete. What are you trying to prove about \frac{p!}{(p-i)!i} \times \frac 1p when p is prime and i is an integer?
 
Im trying to verify the theorem under those conditions
 
You did not write the theorem.
 
http://classes.soe.ucsc.edu/cmpe016/Fall14/hw/hw6.pdf
number 5
 
That says p divides p choose i. That is not what you wrote above.
 
To demonstrate this, you can use induction.
Show that for a base case (i=1) ##p\left| \left( \begin{array}{c} p\\i\end{array}\right) \right. ##
Assume for some n < p-1, the statement holds.
Show that ##p\left| \left( \begin{array}{c} p\\n+1 \end{array}\right) \right. ##
 
  • #10
hlzombi said:
Apologies, I'm new here. I tried to follow the template as best as I could.
When you use the template, don't delete its three parts.

Also, I moved this thread, as it was better suited in one of the math sections.
 

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