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:

  1. \(a < c\) and \(c < a\)
  2. \(a < c\), and also \(c=a\) and \(d \leq b\).
  3. \(a=c\) and \(b \leq d\), and also \(c < a\)
  4. \(a=c\) 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?