Equivalence of Congruence Definitions

Theorem

The definitions of congruence (in the context of Number Theory):

$(1) \qquad x \equiv y \pmod z \iff \left({x, y}\right) \in \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$

$(2) \qquad x \equiv y \pmod z \iff x \mod z = y \mod z$

$(3) \qquad x \equiv y \pmod z \iff \exists k \in \Z: x - y = k z$

... are equivalent.

Integer definition

Also, if $x, y, z$ are all integers, the definition:

$x \equiv y \pmod z \iff z \backslash \left({x - y}\right)$

is likewise equivalent to the other given definitions.

Proof

Let $x_1, x_2, z \in \R$.

Let $x_1 \equiv x_2 \pmod z$ in the sense of Definition by Equivalence Relation.

That is, let $\mathcal R_z$ be the relation on the set of all $x, y \in \R$:

$\mathcal R_z = \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$

Let $\left({x_1, x_2}\right) \in \mathcal R_z$.

Then by definition, $\exists k \in \Z: x_1 = x_2 + k z$.

So, by definition of the modulo operation, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x_1 \mod z\)	\(=\)	\(\displaystyle \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2 + kz} z}\right \rfloor\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2} z + k}\right \rfloor\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_2 + kz}\right) - z \left \lfloor {\frac {x_2} z}\right \rfloor + k z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_2 - z \left \lfloor {\frac {x_2} z}\right \rfloor\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x_2 \mod z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So:

$x_1 \equiv x_2 \pmod z$

in the sense of Definition by Modulo Operation.

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Now let $x_1 \equiv x_2 \pmod z$ in the sense of Definition by Modulo Operation.

That is, :$x_1 \equiv x_2 (\bmod\, z) \iff x_1 \mod z = x_2 \mod z$.

Let $z = 0$.

Then by definition, $x_1 \mod 0 = x_1$ and $x_2 \mod 0 = x_2$.

So as $x_1 \mod 0 = x_2 \mod 0$ we have that $x_1 = x_2$.

So $x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \pmod z$ in the sense of Definition by Integral Multiple.

Now suppose $z \ne 0$.

Then from definition of the modulo operation:

• $x_1 \mod z = x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor$
• $x_2 \mod z = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$

Thus:

$x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$

and so:

$x_1 - x_2 = z \left({\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor}\right)$

From the definition of the floor function, we see that both $\left \lfloor {\dfrac {x_1} z}\right \rfloor$ and $\left \lfloor {\dfrac {x_2} z}\right \rfloor$ are integers.

Therefore, so is $\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor$ an integer.

So $\exists k \in \Z: x_1 - x_2 = k z$.

Thus $x_1 - x_2 = k z$ and:

$x_1 \equiv x_2 \pmod z$

in the sense of Definition by Integral Multiple.

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Now let $x_1 \equiv x_2 \pmod z$ in the sense of Definition by Integral Multiple.

That is, $\exists k \in \Z: x_1 - x_2 = k z$.

Then $x_1 = x_2 + k z$ and so $\left({x_1, x_2}\right) \in \mathcal R_z$ where:

$\mathcal R_z = \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$

and so

$x_1 \equiv x_2 \pmod z$

in the sense of Definition by Equivalence Relation.

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So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$.

$\blacksquare$

Proof for Integer Definition

By definition of divisor, we have that:

$z \backslash x - y \iff x - y = k z$

which is equivalent to the third definition.

$\blacksquare$