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

Continue

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}

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