Lecture 015

Let $m\in \mathbb{N}$, define a relation on $\mathbb{Z}$ by $(\forall a, b \in \mathbb{Z})$: $$ a \equiv b (\mod m) \iff m|a-b$$

read as "a is congruent to b modulo m" (a and b have the same meaning under modulo m)

Claim: this is an equivalence relation Proof: show reflexive, symmetric, transitive

• R: a-a=0=m*0

• S: yes

• T: yes

Equivalence Class

Definition: let R be an equivalence relation on a set S. For every $x \in S$, we define "equivalence class of x under R", denoted $[x]_R$ by

• all elements in two equivalence are related iff the equivalence classes are equal (need to prove)

($[x]_R$ is a collection of all elements that are equivalent to x under the specific notion of equivalence being discussed.)

Congruent Modulo of All Equivalence Classes

Definition: $S/R = \{[x]_R | x \in S\}$

• $\mathbb{Z}/m\mathbb{Z}$ is for congruent modulo m

• There are only m distinct equivalence classes in $\mathbb{Z} /m \mathbb{Z} = \{[0]_m, [1]_m, ... [m-1]_m\}$

• each of equivalence classes is an infinite set

• if we union them together, $=\mathbb{Z}$

• if we intersect them together, $=\emptyset$

• so $\mathbb{Z}/m\mathbb{Z}$ is a partition on $\mathbb{Z}$ (like $\mathbb{Z}/m \mathbb{Z} = \{[0]_3, [1]_3, [2]_3\}$)

Consider: define ~ on $\mathbb{N}^2$. by $(a, b) ~ (c, d) \iff a+d = b+c$ (same thing as a-b = c-d, but we can't use minus because we haven't define negative number)

So:

• $[(0, 0)]_\sim = {(0, 0), (1, 1), (2, 2) ...}$

• $[(0, 1)]_\sim = {(0, 1), (1, 2), (2, 3) ...}$

• $[(1, 0)]_\sim = {(1, 0), (2, 1), (3, 2) ...}$

we then associate each with a negative integer

• -2 = $[(0, 2)]_\sim = [(1, 3)]_\sim$

• -1 = $[(0, 1)]_\sim$

• 0 = $[(0, 0)]_\sim$

• 1 = $[(1, 0)]_\sim$

• 2 = $[(2, 0)]_\sim$

so we can also define + and * of negative integer

• $[(a, b)]_\sim + [(c, d)]_\sim = [(a+c, b+d)]_\sim$

• $[(a, b)]_\sim \times [(c, d)]_\sim = [(ac+bd, bc+ad)]_\sim$

We check if this + and * is well defined