Well-Defined Mapping/Examples/Square Function on Congruence Modulo 6

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Example of Well-Defined Mapping

Let $x \mathrel {C_6} y$ be the equivalence relation defined on the natural numbers as congruence modulo $6$:

$x \mathrel {C_6} y \iff x \equiv y \pmod 6$

defined as:

$\forall x, y \in \N: x \equiv y \pmod 6 \iff \exists k, l, m \in \N, m < 6: 6 k + m = x \text { and } 6 l + m = y$

Let $\eqclass x {C_6}$ denote the equivalence class of $x$ under $C_6$.

Let $\N / {C_6}$ denote the quotient set of $\N$ by $C_6$.


Let us define the mapping $\mathrm {sq}$ on $\N / {C_6}$ as follows:

$\map {\mathrm {sq} } {\eqclass x {C_6} } = m^2$

where $x = 6 k + m$ for some $k, m \in \N$ such that $m < 6$.


Then $\mathrm {sq}$ is a well-defined mapping.


Proof

Let:

$x \mathrel {C_6} x'$

for arbitrary $x, x' \in \N$.

We need to demonstrate that:

$\map {\mathrm {sq} } {\eqclass x {C_6} } = \map {\mathrm {sq} } {\eqclass {x'} {C_6} }$


Let $m, n \in \N$ such that:

$\map {\mathrm {sq} } {\eqclass x {C_6} } = m^2$
$\map {\mathrm {sq} } {\eqclass {x'} {C_6} } = n^2$


We have by definition of $C_6$ that:

$x \mathrel {C_6} x' \iff \exists k, l, m \in \N, m < 6: 6 k + m = x \text { and } 6 l + m = x'$

Hence:

$\map {\mathrm {sq} } {\eqclass {x'} {C_6} } = m^2$

But for $m^2 \ne n^2$ where $0 \le m < 6$ and $0 \le n < 6$:

$m = n$

and the result follows.

$\blacksquare$


Sources

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