Set theory: Problem 2

Solution

Proof: Let (p,q) be an element of $$\mathbb{Z}^2$$ and suppose that (p,q) is an element of $$A \cap B$$.

Then (p,q) is an element of A, so $$3pq + 15p -5q -25 \ge 0$$. But (p,q) is also an element of B, so $$q \ge 0$$.

Notice that $$3pq + 15p -5q -25 = (q+5)(3p-5)$$. So we know that $$(q+5)(3p-5) \ge 0$$. So q+5 and 3p-5 are either both non-negative or both non-positive. Since $$q \ge 0$$, $$q +5 \ge 0$$. So we must have $$3p-5 \ge 0$$.

Now, if $$3p-5 \ge 0$$, then $$3p \ge 5$$. So $$p \ge \frac{5}{3}$$, and therefore $$p \ge 0$$.

Since (p,q) be an element of $$\mathbb{Z}^2$$ and $$p \ge 0$$, (p,q) must be an element of C, which is what we needed to show.

You don't have to repeat the definitions of A, B, and C at the start of the proof. That's considered background information.

The pair you're manipulating in the proof doesn't have to be named (p,q). It could be named something like (s,t) or (a,b). You could also introduce it as x, and then later reveal its internal structure e.g. "x must have the form (p,q), where p and q are integers."

Self-check

Do your main outline and key steps look similar to the above? In particular, look at the middle part of your proof where you're doing the algebra. Does yours start with $$3pq + 15p -5q -25 \ge 0$$ and $$q \ge 0$$? Does it conclude with $$p \ge 0$$?

Did you declare your variable (p,q) (or whatever you called this variable)? Although the definition of the set A uses a variable named (p,q), the scope of that definition is just the material inside the set brackets. (p,q) is not defined globally. So, in particular, it's not defined as your proof begins.

Did you mention that (p,q) is an ordered pair of integers? This isn't a major part of this proof, because the type is constant across all three sets and the claim would also hold if these were pairs of real numbers. However, good style and complete correctness require that you declare the type of (p,q). You could lose points for not doing so, depending on the mood of the person grading your proof.

Buggy proof

Proof: $$(p,q)\in \mathbb{Z}^2$$,   $$(p,q)\in A \cap B$$.

$$(p,q)\in A \cap B \rightarrow 3pq + 15p -5q -25 \ge 0$$.     $$(p,q)\in B \rightarrow q \ge 0$$.

$$3pq + 15p -5q -25 = (q+5)(3p-5)$$.

$$(q+5)(3p-5) \ge 0$$.

$$q +5 \ge 0$$

$$3p-5 \ge 0 \rightarrow 3p \ge 5 \rightarrow p \ge \frac{5}{3} \ge 0$$

$$(p,q)\in C$$

Could your roommate follow this argument? I thought not. The course staff can do this only because we already know what the correct proof looks like. If this was some novel theorem proved in your thesis, how many people would read and appreciate it?

Use connector words so that an intelligent reader can easily understand how each statement follows from the previous ones.