Equivalence of Definitions of Modulo Multiplication

$(1)$ implies $(2)$
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $1$.

Then $\times_m$ is defined as:
 * $\eqclass a m \times_m \eqclass b m = \eqclass {a b} m$

By the Division Theorem:

Hence $p \in \eqclass a m$

Thus:

Thus:
 * $p q \in \eqclass {a b} m$

By the Division Theorem:
 * $p q = s + k m$

for some $k \in \Z$ and $0 \le s < m$

Then:
 * $s \in \eqclass {p q} m$

while $s$ equals the remainder after $p q$ has been divided by $m$.

That is:
 * $p \times_m q$ equals the remainder after $p q$ has been divided by $m$.

Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $2$.

$(2)$ implies $(1)$
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $2$.

Then by definition:
 * $\forall a, b \in \Z_m: a \times_m b$ equals the remainder after $a b$ has been divided by $m$

where:
 * $\Z_m = \set {0, 1, \ldots, m - 1}$

Thus by definition of residue class:

Let $s$ equal the remainder after $a b$ has been divided by $m$:
 * $a b = s + k m$

where $k \in \Z$.

Hence:
 * $s \in \eqclass {a b} m$

So as Modulo Multiplication is Well-Defined:
 * $\eqclass a m \times_m \eqclass b m = \eqclass {a b} m$

Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $1$.

$(2)$ implies $(3)$
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $2$.

Then by definition:
 * $\forall a, b \in \Z_m: a \times_m b$ equals the remainder after $a b$ has been divided by $m$

where:
 * $\Z_m = \set {0, 1, \ldots, m - 1}$

Let $s$ equal the remainder after $a b$ has been divided by $m$.

That is, there exists $k \in \Z$ such that:
 * $a b = s + k m$

where $0 \le s < m$.

It follows that:
 * $k m \le a b$

but:
 * $\paren {k + 1} m > a b$

Thus we have:

where $k$ is the largest integer such that $k m \le a b$.

Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $3$.

$(3)$ implies $(2)$
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $3$.

Then by definition:
 * $a \times_m b := a b - k m$

where $k$ is the largest integer such that $k m \le a b$.

Let $s$ be defined as:
 * $s = a b - k m$

By definition:
 * $0 \le s < m$

That is, $s$ equals the remainder after $a b$ has been divided by $m$.

Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $2$.