# 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?