Proving the Equation: 1/p c(p,n) = (-1)^{n-1}/n (mod p)

In summary, the conversation discusses proving the equation $\frac{1}{p}c(p,n) = (-1)^{n-1}/n (mod p)$ and different ways of approaching it, such as using Wilson's Theorem and expanding the combination. The conversation also touches on the meaning of 1/n and -1/n mod p and the notation for congruences. Ultimately, it is suggested to write out the LHS as a fraction with negative representatives in the numerator, which should nicely cancel to give the desired result.
  • #1
ehrenfest
2,020
1

Homework Statement


Prove that

[tex] \frac{1}{p} c(p,n) = (-1)^{n-1}/n (mod p) [/tex]

I expanded that combination in every way I could think and I tried to use Wilson's Theorem and I couldn't get :(

Homework Equations


The Attempt at a Solution

 
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  • #2
That's p choose n, right?

Try writing the LHS out as a fraction with the stuff in the numerator as negative representatives. It should nicely cancel to give the result.
 
  • #3
What do you mean "negative representatives"?
 
  • #4
-(p-1)/2, -(p-2)/2,..., -1 for odd p
 
Last edited:
  • #5
What is 1/n, or -1/n mod p supposed to mean?
 
  • #6
matt grime said:
What is 1/n, or -1/n mod p supposed to mean?

Usually those would be the multiplicative inverses of n and -n respectively.
 
  • #7
Usually? I beg to differ. Writing 1/n would indicate that the OP hasn't grasped what's going on. As would the fact there is an equals sign. I can't think of anyone who writes 1/2 mod 3 and not -1 0 it is incredibly bad notation. There is a difference from what I infer and what the OP ought to have written.
 
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  • #8
What does OP stand for? Is that me?

I just realized that my book my book defines congruence as

[tex] x \equiv y \mod p [/tex]

when x-y is a rational number whose numerator, in reduced form, is divisible by p.

So, it is like a generalized congruence or something...

Are there different rules for these generalized congruences?

I am not sure why what Gokul43201 wrote cancels nicely?
 

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