Show that $c = \max(a, b)$ on $\mathbb{Z}_2$ is not a binary operation Let $*: \mathbb{Z}_2\times\mathbb{Z}_2 \to \mathbb{Z}_2$, be defined as $[a] * [b] = [c]$, where $c = \max\{a, b\}$, for all $[a], [b] \in \mathbb{Z}_2$. Prove that $*$ is not a binary operation on $\mathbb{Z}_2$. Hint: is it well defined?

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I don't want someone to solve the problem for me, but just help clear up my misunderstanding or point me in the right direction.

I don't see how $*$ is not a binary operation, nor how it isn't well defined.

A binary operation is a function on a set $S$ that associates every ordered pair of elements in $S$ to another element of $S$.

Here, $*$ associated ordered pairs of elements $\mathbb{Z}_2$ to another element $\mathbb{Z}_2$, through the max function. It has four possible inputs:

$$
\begin{align}
(0, 0) &amp; \mapsto 0 \\
(0, 1) &amp; \mapsto 1 \\
(1, 0) &amp; \mapsto 1 \\
(1, 1) &amp; \mapsto 1
\end{align}
$$

So it associates every unique input to one output, and meets the definition of a function. Every possible input returns an output that is in $S$ ($\mathbb{Z}_2$). 

How is it not well-defined?
The definition of $[a]\ast[b]$ is defined using representatives of the congruence classes $[a],[b]$. In order for it to be well-defined, it must work the same on all pairs of representatives of $[a]$ and $[b]$. For instance, it must satisfy $[0]\ast[1]=[2]\ast[1]$ since $[0]=[2]$.
Note that $[1]=[3]$. But whe have $[2]*[1]=[\max\{2,1\}]=[2]=[0]$, but $[2]*[3]=[\max\{2,3\}]=[3]=[1]$, so if you operate $[2]$ with $[1]$, and $[2]$ with $[3]$, you get two different elements, so this operation is not well defined.