The map $f:\mathbb{Z}_3 \to \mathbb{Z}_6$ given by $f(x + 3\mathbb{Z}) = x + 6\mathbb{Z}$ is not well-defined

Solution 1:

If a function $f \colon \mathbb{Z}_3 \to \mathbb{Z}_6$ with $f(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ would exist, then $$ [0] = f(\overline{0}) = f(\overline{3}) = [3], $$ where we used for the second equality that $\overline{0} = \overline{3}$. By the definition of $\mathbb{Z}_6$ we have for all $x, y \in \mathbb{Z}$ that $[x] = [y]$ if and only if $x - y$ is divisible by $6$, i.e. if $x - y \in 6\mathbb{Z} = \{6n \mid n \in \mathbb{Z}\}$. So $[0] = [3]$ is equivalent to $3 - 0 = 3$ being divisible by $6$, which doesn’t hold.

This contradiction shows that no such function $f$ exists.

More generally one can show that there exists a function $g \colon \mathbb{Z}_n \to \mathbb{Z}_m$ with $g(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ if and only if $m$ divides $n$.

Solution 2:

It is easy to show that

$\mathbb Z_3=\left\{\begin{array}\{ \{...,-6,-3,0,3,6,...\},\\ \{...,-5,-2,1,4,7,...\},\\ \{...,-4,-1,2,5,8,...\}\end{array}\right\}$

and that

$\mathbb Z_6=\left\{\begin{array}\{ \{...,-12,-6,0,6,12,...\},\\ \{...,-11,-5,1,7,13,...\},\\ \{...,-10,-4,2,8,14,...\},\\ \{...,-9,-3,3,9,15,...\},\\ \{...,-8,-2,4,10,16,...\},\\ \{...,-7,-1,5,11,17,...\}\end{array}\right\}$

, because $\mathbb Z_3$ and $\mathbb Z_3$ are sets of equivalence classes. Let the relation $f$ be represented by a diagram:

Here, the relation lines show that $f(\bar1)=[1]$, $f(\bar2)=[2]$, and $f(\bar3)=[3]$.

But consider that $\bar0\cong_3\bar3$, which means they are the same element in $\mathbb Z_3$ (Remember, $\mathbb Z_3$ is a set of equivalence classes). Then, consider that $f(\bar0)=f(\bar3)=[3]$. Putting this into the diagram yields

, which shows that $f$ is clearly not a function, because a function maps each element in its domain to exactly $1$ element in its co-domain. $Q.E.D.$

Solution 3:

You have a rule that given an element of $\mathbb{Z}_3$, you can write down an element of $\mathbb{Z}_6$. The problem is that your rule is not well-defined. To be well-defined means that even if you have two different ways to describe the same input, you get the same output.

To be more explicit, observe that in $\mathbb{Z}_3$, the elements $\overline{0}$ and $\overline{3}$ are the same object even though they look different (they are defined by different numbers). Because these are the same element, they should go to the same place under the map $f$. Unfortunately, $f(\overline{0})=[0]$ and $f(\overline{3})=[3]$ and $[0]\not=[3]$ in $\mathbb{Z}_6$. Therefore $f$ can't be a function!

The formal definition of well-defined is:

• "If $a=b$, then $f(a)=f(b)$." (Technically, this is not a good definition because writing $f(\cdot)$ implies that $f$ is a function).

If we write functions in terms of relations, well-defined can be expressed as

• "If $(a,b),(a,c)\in f$, then $b=c$.

Solution 4:

$\mathbb Z_3 = \{\overline{0},\overline{1}, \overline{3}\}$

where $\overline{0} = \{....,0,3,6,9....\}$

$\overline{1} = \{.....,1,4,7,10.....\}$

$\overline{2} = \{......,2,5,8,11....\}$

$\overline{3} = \overline{0}; \overline{4} = \overline{1};\overline{5} = \overline{2}$.

So if $f$ is well defined it'd have to be that $f(\overline {0}) = [0]$ and $f(\overline{0}) = f(\overline{3}) = [3]$ so $[0] = [3]$.

But $\mathbb 6 = \{[0],[1],[2],[3],[4],[5]\}$

where $[0] = \{....0,6,12,18\}$

and $[1] = \{.....,1,7,13, 19\}$

etc.

Note $[0] = \{....0,6,12...\} \ne \{.....3,9,15...\} = [3]$. But $\overline{0} = \overline{3}$

So $f(\overline{0}= [0]$ if we write $\overline{0}$ as $\overline{0}$. But $f(\overline{0} = [3]$ if we write $\overline{0}$ as $\overline{3}$.

So $f$ gives different and unequal output for different ways we interpret equal input. So $f$ is poorly defined as it doesn't consistantly give equal output for equal input.