# Integers mod p

1. Jul 20, 2005

### gazzo

Hey, umm... I can't find an answer for this anywhere.

if we have a group $\mathbb{Z}/p\mathbb{Z}$ (for sufficient p) under multiplication modulo p, is divsion defined

$$\frac{a}{b} = ab^{-1}$$

ie in $\mathbb{Z}/5\mathbb{Z} = \{1,2,3,4\}$; would $\frac{3}{2}$ be $(3)(2^{-1}) \equiv (3)(3) \equiv 4$

Maybe i've completely understood modulo arithmetic

2. Jul 20, 2005

### Timbuqtu

Seems alright to me.

By the way, the multiplicative group of integers modulo p is usually denoted by $$(\mathbb{Z}/p\mathbb{Z})^*$$ and is defined as the set of elements of $$\mathbb{Z}/p\mathbb{Z}$$ which have an inverse under multiplication. $$\mathbb{Z}/p\mathbb{Z}$$ is a group only under addition. So:

$$\mathbb{Z}/5\mathbb{Z} = \{0,1,2,3,4\}$$ and $$(\mathbb{Z}/5\mathbb{Z})^*=\{1,2,3,4\}$$

$$\mathbb{Z}/9\mathbb{Z} = \{0,1,2,3,4,5,6,7,8\}$$ and $$(\mathbb{Z}/9\mathbb{Z})^*=\{1,2,4,5,7,8\}$$

3. Jul 20, 2005

### matt grime

inverses of all non-zero elements are defined for ALL p when p is a prime. and only for elements coprime to p when p is not a prime (actually this implies they all are invertible if p is a prime).

i would never say anything like "i completely understand SUBJECT" since there is always someone cleverer than you who understands more about it.

4. Jul 20, 2005

### gazzo

hehe that's true :) thanks.

5. Jul 20, 2005

### matt grime

if you'd like to put your knowledge to the test then how abhout this:

let p be a prime and work mod p.

show that x^2=1 mod p has exactly two solutions.

hence show that (p-1)!=-1 mod p

hint: every element x has a unique y such that xy=1, paur them up. what can you not pair with a distinct inverse? see previous question).

show that pCr (p choose r) is 0 mod p unless r=1 or p (when it is 1)

if you know group theory explain why a^{p-1}=1 mod p. if you don't know group theory, use the previous exercise to show it by considering (1+1+..+1)^p {a 1's added together} to show why.