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 Template:Math 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, Template:Math

Problem solved!