# Relations Problem 2

### Solution

Proof: Let (a,b) and (c,d) be elements of
\(\mathbb{Z}^2\). Suppose that \((a,b) \ll (c,d)\)
and \((c,d) \ll (a,b)\).

By the definition of \(\ll\),
\((a,b) \ll (c,d)\) means that
\(a < c\) or else \(a=c\) and \(b \leq d\).
Similarly
\((c,d) \ll (a,b)\) means that
\(c < a\) or else \(c=a\) and \(d \leq b\).

There are four cases:

- \(a < c\) and \(c < a\)
- \(a < c\), and also \(c=a\) and \(d \leq b\).
- \(a=c\) and \(b \leq d\), and also
\(c < a\)
- \(a=c\) and \(b \leq d\), and also
\(c=a\) and \(d \leq b\).

Of these four cases, the first three are internally
inconsistent. So the only viable possibility
is the fourth case, i.e.
\(a=c\) and \(b \leq d\), and also
\(c=a\) and \(d \leq b\).

We know that \(a=c\).
Since \(b \leq d\) and \(d \leq b\), we also have
that \(b=d\).
So (a,b)=(c,d), which is what we needed to show.

### Self-check

At the start of your proof, did you declare your 2D point
variables and state your assumptions? Did you actually
state the conclusion (a,b)=(c,d) at the end of the proof?

Did you handle all four cases?