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

## Contents

### Problem 1Edit

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

Case 1: ${\displaystyle x\in A}$

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

Case 2: ${\displaystyle x\in B\cap C}$

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

Since in both cases, ${\displaystyle x\in (A\cup B)\cap (A\cup C)}$, we know that ${\displaystyle A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)}$

Now we wish to show that ${\displaystyle (A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)}$. Let ${\displaystyle x\in (A\cup B)\cap (A\cup C)}$. Then ${\displaystyle x\in A\cup B}$ and ${\displaystyle x\in A\cup C}$.

Case 1a: ${\displaystyle x\in A}$

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

Case 1b: ${\displaystyle x\in B}$

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

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

Case 2b: ${\displaystyle x\in C}$

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

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

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

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

### Problem 3Edit

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

${\displaystyle x^{2}=(2n+1)^{2}=(2n+1)(2n+1)}$

2. Multiply both factors together

${\displaystyle (2n+1)(2n+1)=4n^{2}+4n+1}$

3. Factor out a two for the first two terms

${\displaystyle 4n^{2}+4n+1=2(2n^{2}+2n)+1}$

4. The factor ${\displaystyle 2n^{2}+2n}$ will always be a natural number. As such, it fits the definition of an odd number, 2n + 1
Problem solved!