Mathematical Proof/Appendix/Answer Key/Mathematical Proof/Methods of Proof/Constructive Proof

Problem 1
First, we wish to show that $$A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C)$$. Let $$x \in A\cup(B\cap C)$$. Then $$x \in A$$ or $$x \in B\cap C$$.

Case 1: $$x \in A$$

$$x \in A \subset A\cup B$$ so that $$x \in A\cup B$$ $$x \in A \subset A\cup C$$ so that $$x \in A\cup C$$ $$x \in A\cup B$$ and $$x \in A\cup C$$ so that $$x \in (A\cup B)\cap (A\cup C)$$

Case 2: $$x \in B\cap C$$

$$x \in B$$ and $$x \in C$$ $$x \in B \subset A\cup B$$so that $$x \in A\cup B$$ $$x \in C \subset A\cup C$$ so that $$x \in A\cup C$$ $$x \in A\cup B$$ and $$x \in A\cup C$$ so that $$x \in (A\cup B)\cap (A\cup C)$$

Since in both cases, $$x \in (A\cup B)\cap (A\cup C)$$, we know that $$A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C)$$ Now we wish to show that $$(A\cup B)\cap(A\cup C) \subset A\cup(B\cap C)$$. Let $$x \in (A\cup B)\cap (A\cup C)$$. Then $$x \in A\cup B$$ and $$x \in A\cup C$$.

Case 1a: $$x \in A$$

$$x \in A \subset A\cup(B\cap C)$$, so $$x \in A\cup(B\cap C)$$

Case 1b: $$x \in B$$

We can't actually conclude anything we want with just this, so we have to also to consider the case $$x \in A\cup C$$.

Case 2a: $$x \in A$$ : [see Case 1a]

Case 2b: $$x \in C$$

We now have $$x \in B$$ and $$x \in C$$ so that $$x \in B\cap C$$ Of course, since $$x \in B\cap C \subset A\cup(B\cap C)$$, it follows that $$x \in A\cup(B\cap C)$$. Since both cases 2a and 2b yield $$x \in A\cup(B\cap C)$$, we know that it follows from 1b.

Since in both cases 1a and 1b, $$x \in A\cup(B\cap C)$$, we know that $$(A\cup B)\cap(A\cup C) \subset A\cup(B\cap C)$$.

Since both $$A\cup(B\cap C) \subset (A\cup B)\cap(A\cup C)$$ and $$(A\cup B)\cap(A\cup C) \subset A\cup(B\cap C)$$, it follows (finally) that $$A\cup(B\cap C) = (A\cup B)\cap(A\cup C)$$.

--will continue later, feel free to refine it if you feel it can be--

Problem 3
  Because the question asks about the square of a number, you can substitute the definition of an odd number $2n + 1$ into the number to be squared. So, say x is that number, then $$x^2 = (2n+1)^2 = (2n+1)(2n+1)$$   Multiply both factors together $$(2n+1)(2n+1) = 4n^2 + 4n + 1$$   Factor out a two for the first two terms $$4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1$$   The factor $$2n^2 + 2n$$ will always be a natural number. As such, it fits the definition of an odd number, $2n + 1$ 
 * Problem solved!