Modulo Arithmetic: Division Defined?

[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

 

[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\}

 

[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.

 

[gazzo]

hehe that's true :) thanks.

 

[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.