# Lecture 4

## Review

1. Let $F$ be a field. Let $a,b,c,...,z\in F$ . Using he field axioms, simplify

    $$
    (x-a)(x-b)(x-c)...(x-z)
    $$

    $x\in F$, it must be at least one $0$ in the product...

2. Suppose $A,B\subset\mathbb{R}$. Suppose $A$ and $B$ are nonempty and bounded above,$A\subset B$. WHat can you say about $\sup A$ and $\sup B$? Please justify.

    $$
    \forall x\in A, x\in B.  sup\ A\leq sup\ B
    $$

    Any UB of $B$ is also an UB of $A$.

    $sup\ B$ is an UB of $B$ by def
    $sup\ B$ is an UB of $A$

## Continue

### Archimedean property

(Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.

<details>
<summary>Proof</summary>

Suppose the property is false, then $\exist x,y\in \mathbb{R}$ with $x>0$ such that $\forall v\in \mathbb{N}$, nx\leq y$

Let $A=\{nx:n\in\mathbb{N}\}$. Then $A\neq\phi$ (Since $x\in A$) and $A$ is bounded above by $y$. Since $\mathbb{R}$ has LUBP, $sup\ A$ exists. Let $\alpha=\sup A$.

$x>0\implies \alpha-x<\alpha$, $\alpha-x$ is not an upper bound of $A$. (Since $\alpha$ is the LUB of $A$) $\implies \exist m\in \mathbb{N}$ such that $mx>\alpha-x$ by definition of $A$.

This implies $(m+1)x>\alpha$

Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.

</details>

### $\mathbb{Q}$ is dense in $\mathbb{R}$

$\mathbb{Q}$ is dense in $\mathbb{R}$ if $x,y\in \mathbb{R}$ and $x<y$, then $\exists p\in \mathbb{Q}$ such that $x<p<y$.

Some thoughts:

$$
x<\frac{m}{n}<y\iff nx<m<ny
$$

<details>
<summary>Proof</summary>

Let $x,y\in\mathbb{R}$, with $x<y$. We'll find $n\in \mathbb{N},\mathbb{m}\in \mathbb{Z}$ such that $nx<m<ny$.

By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\exist m_1\in \mathbb{N}$ such that $m_1\cdot 1>nx$, $\exist m_2\in \mathbb{N}$ such that $m_2\cdot 1>-nx$.

So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$

</details>

### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$

Notation $\mathbb{R}_{>0}$= the set of positive numbers.

#### Theorem 1.21

$\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist$ unique $y\in \mathbb{R}_{>0}$ such that $y^n=x$.

(Because of this Theorem we can define $x^{1/x}=y$ and $\sqrt{x}=y$)

Proof:

We cna assume $n\geq 2$ (For $n=1,y=x$)

Step 1 (uniqueness): If $0<y_1<y_2$, then $y_1^n<y_2^n$ (by properties of ordered field)

Step 2 (existence): Let $E=\{t\in \mathbb{R}_{>0}: t^n<x\}$ We want to let $y=sup\ E$, but to do this we need to check 2 things.

1. To show $E\neq \phi$:

    If $x\geq 1$, then $1/2\in E$.

    If $x<1$, then $x\in E$.

2. To show $E$ is bounded above. We need to find an upper bound.

    If $x\geq 1$, then $x\in E$

    If $x<1$, then $1 \in E$.

So we can let $y=sup\ E$

Step 2b ($y^n\geq x$) Suppose for contradiction $y^n<x$.

Thoughts: If we can find $h>0$ such that $(y+h)^n<x$, then $y+h\in E$. This would contradict the facts $y$ is an upper bound of $E$.

$$
(y+h)^n=y^n+ny^{n-1}h+{more\ terms}
$$

We want $ny^{n-1}h+{more\ terms}<x-y^n$

Observe: If $0<a<b$, then

$$
\frac{b^n-a^n}{b-a}=b^{n-1}+b^{n-2}a+...+a^{n-1}\leq b^{n-1}+b^{n-2}b+...+b^{n-1}=nb^{n-1}
$$