Well-Defined Mapping/Examples/Floor of Half Function on Congruence Modulo 6
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Example of Mapping which is not Well-Defined
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 $\phi$ on $\N / {C_6}$ as follows:
- $\map \phi {\eqclass x {C_6} } = \eqclass {\floor {x / 2} } {C_6}$
where $\floor {\, \cdot \,}$ denotes the floor function.
Then $\phi$ is not a well-defined mapping.
Proof
Consider $x = 4$ and $x' = 10$.
We have that:
- $4 = 6 \times 0 + 4$
and:
- $10 = 6 \times 1 + 4$
and so:
- $4 \mathrel {C_6} {10}$
That is:
- $\eqclass 4 {C_6} = \eqclass {10} {C_6}$
However, we have that:
- $\map \phi {\eqclass 4 {C_6} } = \eqclass 2 {C_6}$
while:
- $\map \phi {\eqclass {10} {C_6} } = \eqclass 5 {C_6}$
But $\eqclass 2 {C_6} \ne \eqclass 5 {C_6}$
Hence we have $x, x' \in \N$ for which it is not the case that:
- $\map \phi {\eqclass x {C_6} } = \map \phi {\eqclass {x'} {C_6} }$
That is, $\phi$ is not well-defined.
$\blacksquare$
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