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