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

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Homework Help Overview

The discussion revolves around proving the equation 1/p c(p,n) = (-1)^{n-1}/n (mod p), which involves combinatorial concepts and modular arithmetic. The subject area includes combinatorics and number theory.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the left-hand side (LHS) and suggest rewriting it with negative representatives. Questions arise about the meaning of terms like "negative representatives" and the notation used for modular inverses. There is also a query regarding the definition of congruence in the context of the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations and clarifying terminology. Some guidance has been offered regarding rewriting expressions, but there is no consensus on the notation or the underlying concepts yet.

Contextual Notes

Participants are questioning the notation used in the problem statement and discussing the implications of generalized congruences as defined in a textbook. There is a concern about the clarity of the original poster's understanding of modular arithmetic.

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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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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.
 
What do you mean "negative representatives"?
 
-(p-1)/2, -(p-2)/2,..., -1 for odd p
 
Last edited:
What is 1/n, or -1/n mod p supposed to mean?
 
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.
 
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.
 
Last edited:
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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