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

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The discussion focuses on proving the theorem that states for a prime number p and an integer i, if 0 < i < p, then p!/[(p−i)! * i] = 1/p. Participants express confusion about how to approach the proof, with one user attempting a direct proof by substituting i with p-n but finding it unhelpful. Clarifications are requested regarding the exact nature of the theorem being proved, as some participants believe the statement is incomplete. Suggestions include using induction to demonstrate that p divides the binomial coefficient "p choose i." The conversation emphasizes the need for clear definitions and adherence to forum templates for effective communication.
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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