Well-Defined Mapping/Examples/Floor of Half Function on Congruence Modulo 6

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.