# 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

1. 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

$x^{2}=(2n+1)^{2}=(2n+1)(2n+1)$ 2. Multiply both factors together

$(2n+1)(2n+1)=4n^{2}+4n+1$ 3. Factor out a two for the first two terms

$4n^{2}+4n+1=2(2n^{2}+2n)+1$ 4. 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!