$\mathbb{R}$ - The set of real numbers. Think of it as all decimals that terminate to the left of the decimal point.

This is "real" analysis because of the real numbers. What are the real numbers?

Real numbers

$0$

$0.5$

$2.7$

$0.16666\ldots$

$-\frac{5}{4} = -1.25$

$\sqrt{2} = 1.414\ldots$

Not real numbers

$i=\sqrt{-1}$

$\ldots 2222.1$

$\infty$

$\mathbb{N} = \{1,2,3,4,\ldots\}$ - The set of natural numbers.

$\mathbb{Z} = \{\ldots,-3,-2,-1,0,1,2,3\ldots\}$ - The set of integers.

$\mathbb{Q} = \{\frac{a}{b} : a\in\Z,\ b\in\N\}$ - The set of rational numbers.

$\mathbb{R}\setminus\mathbb{Q}$ - The set of irrational numbers.

Some important sets of real numbers

The Positive Numbers

Definition 1. The set $\mathbb{P}$ , called the positive real numbers, is the nonempty subset of $\mathbb{R}$ with the following three properties:

If $a$ and $b$ belong to $\mathbb{P}$ , then $a+b$ belongs to $\mathbb{P}$ .
If $a$ and $b$ belong to $\mathbb{P}$ , then $ab$ belongs to $\mathbb{P}$ .
If $a$ belongs to $\mathbb{R}$ , then exactly one of the following holds:

$a\in\mathbb{P},\quad a=0,\quad -a\in\mathbb{P}$ .

Property 1. says that $\mathbb{P}$ is closed under addition

Property 2. says that $\mathbb{P}$ is closed under multiplication.

Property 3. is called the Trichotomy Property.

What can we say about $\mathbb{P}$ ?

Is $1\in\mathbb{P}$ ? What about $2$ ? Are squares in $\mathbb{P}$ ?

Proposition 1. $1\in\mathbb{P}.$

Proof. Assume toward a contradiction that $1\notin\mathbb{P}$ .

Proof. Assume toward a contradiction that $1\notin\mathbb{P}$ . By the Trichotomoy Property, either $-1\in\mathbb{P}$ , or $1=0$ . Obiously $1\neq 0$ , hence, it must be the case that $-1\in\mathbb{P}$ .

Proof. Assume toward a contradiction that $1\notin\mathbb{P}$ . By the Trichotomoy Property, either $-1\in\mathbb{P}$ , or $1=0$ . Obiously $1\neq 0$ , hence, it must be the case that $-1\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under multiplication, we have

$1 = (-1)(-1)\in\mathbb{P}.$

This is a contradiction by the Trichotomy Property. We conclude that $1\in\mathbb{P}$ . $\Box$

Proof. Assume toward a contradiction that $1\notin\mathbb{P}$ . By the Trichotomoy Property, either $-1\in\mathbb{P}$ , or $1=0$ . Obiously $1\neq 0$ , hence, it must be the case that $-1\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under multiplication, we have

$1 = (-1)(-1)\in\mathbb{P}.$

Proof. From Proposition 1 we see that $1\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under addtion, we have
$2 = 1 + 1\in\mathbb{P}.\ \Box$

Proposition 2. $2\in\mathbb{P}.$

Prove the following:

Is $3\in\mathbb{P}$ ?

What about $4$ ?

How many examples would you need to check before you would KNOW that all natural numbers are positive?

What can we say about $\mathbb{P}$ ?

Proposition 3. $\N\subset\mathbb{P}$

Equivalent statements to $\N\subset \mathbb{P}$ :

If $x\in \N$ , then $x\in \mathbb{P}$ .
$\forall\,x\in \N,\ x\in \mathbb{P}$ .
$x\in \mathbb{P}$ for all $x\in \N$ .
If $x\notin \mathbb{P}$ , then $x\notin \N$ .

This is the perfect type of statement to prove using mathematical induction.

What can we say about $\mathbb{P}$ ?

Proof. We will use induction to prove $n\in\mathbb{P}$ for all $n\in\N$ . We have already proved $1\in\mathbb{P}$ . This finishes the base case.

Now, assume the claim is true for some $k\in\N$ , that is $k\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under addition, and $1\in\mathbb{P}$ , we conclude $k+1\in\mathbb{P}.$ This completes the inductive step, and shows that $n\in\mathbb{P}$ for all $n\in\N$ . $\Box$

Proof. We will use induction to prove $n\in\mathbb{P}$ for all $n\in\N$ .

Proof. We will use induction to prove $n\in\mathbb{P}$ for all $n\in\N$ . We have already proved $1\in\mathbb{P}$ . This finishes the base case.

Now, assume the claim is true for some $k\in\N$ , that is $k\in\mathbb{P}$ .

Proof. We will use induction to prove $n\in\mathbb{P}$ for all $n\in\N$ . We have already proved $1\in\mathbb{P}$ . This finishes the base case.

Now, assume the claim is true for some $k\in\N$ , that is $k\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under addition, and $1\in\mathbb{P}$ , we conclude $k+1\in\mathbb{P}.$

Proof. We will use induction to prove $n\in\mathbb{P}$ for all $n\in\N$ . We have already proved $1\in\mathbb{P}$ . This finishes the base case.

Now, assume the claim is true for some $k\in\N$ , that is $k\in\mathbb{P}$ . Since $\mathbb{P}$ is closed under addition, and $1\in\mathbb{P}$ , we conclude $k+1\in\mathbb{P}.$ This completes the inductive step, and shows that $n\in\mathbb{P}$ for all $n\in\N$ . $\Box$

Proposition 3. $\N\subset\mathbb{P}$