Lecture 4

Review

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…
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 $A$

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Archimedean property

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

Proof

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 $A$ .

$\mathbb{Q}$ 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

Proof

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 $ny>1+nx\geq 1+(m-1)=m$

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

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.

To show $E\neq \phi$ :

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

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

Lecture 4

Review

Continue

Archimedean property

Proof

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

Proof

2∈R\sqrt{2}\in \mathbb{R}2​∈R, (xn∈R)(\sqrt[n]{x}\in\mathbb{R})(nx​∈R)

Theorem 1.21

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

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