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
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $11 \ \text {(b)}$