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

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SUMMARY

The theorem states that for a prime number p and an integer i, where 0 < i < p, the equation p!/[(p− i)! * i] = 1/p holds true. The discussion emphasizes the need for clarity in proving this theorem and suggests using mathematical induction as a method for verification. It also highlights the importance of correctly framing the theorem and following the forum's template for mathematical discussions. The proof involves demonstrating that p divides the binomial coefficient p choose i.

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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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