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

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

[x]_R = \{y \in S | (x, y) \in R\}

([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\}

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)


we then associate each with a negative integer

so we can also define + and * of negative integer

We check if this + and * is well defined

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