# Set theory: Problem 2

### Hint 1

Make sure you know how to read the set-builder notation defining A, B, and C. In particular, these sets contain ordered pairs of integers. Are you tempted to write something like the following?

Let m and n be integers. Suppose m and n are in A. ....

m is an integer but A contains pairs of integers (aka 2D points). So m cannot be an element of A.

If you are having trouble with this part, go to office hours and have someone walk you through more examples.

### Hint 2

To prove a subset inclusion like this, your proof should start by choosing an element z from the smaller set. The proof should end on a conclusion that z is in the larger set. The smaller set here is $$A \cap B$$ and the larger set is C. So the start of your proof should assume that ...? And your conclusion at the end should be ...? Write those down before you start trying to fill in the middle of the proof.

### Hint 3

Not sure how the variables p and q relate to the variables s, t, x, y? Notice that p and q are just temporary local variables, used to describe the structure of an element of A. Similarly for s, t, x, y. So, for example, you could paraphrase the definition of C as: the set of all 2D points whose first coordinates are non-negative.

In the main body of your proof, you'll need to set up the name for the pair you're manipulating. Perhaps you might call it (m,n). When you apply the definition of A, you'll match (m,n) with the local variable (p,q) in A's definition. That is, set m=p, n=q. So if (m,n) is an element of A, we substitute m=p and n=q into A's defining equation, to get that $$3mn + 15m -5n -25 \ge 0$$.

### Hint 4

Have you tried factoring that big long polynomial?