Lecture 5

Review

In each case, determine (with justification) whether the claim or its negation is the true statement.

(a) For all real numbers satisfying $a<b$, there exists an $n\in \mathbb{N}$ such that $a+1/n<b$.

negation: $\exists a<b$, $\forall n\in \mathbb{N}$ such that $a+1/n\geq b$.

By Archimedean property, the statement is true.

(b) There exists a real number $x>0$ such that $x<1/n$ for all $n\in \mathbb{N}$.

The statement is ambiguous because we can arrange the statement in two ways.

$\exists x\in \mathbb{R}_{>0}$ such that $\forall n\in \mathbb{N},x\leq \frac{1}{n}$

negation: $\forall x\in \mathbb{R}_{>0}$, $\exists n\in \mathbb{N}$, such that $x\leq \frac{1}{n}$.

The statement is true, let $x=\frac{1}{n+1}$.

New Materials

Continue on the theorem

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 $E$.

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

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

The fact tells us $(y+h)^n-y^n\leq n(y+h)^{n-1}h$.

And we want $n(y+h)^{n-1} h+y^n<x$.

So want $h<\frac{x-y^h}{n(y+h)^{n-1}}$.

To achieve this, choosey any $h>0$ satisfying $h<1$ and $h<\frac{x-y^h}{n(y+h)^{n-1}}$

[For actual proof, see the text.]

Step 2c showing ($y^n\leq x$)

Suppose for contradiction $y^n>x$

Thoughts: Find $k>0$ such that $(y-k)^n>x$.

Then $y-k$ is an upper bound for $E$, which contradicts the fact that $y$ is the least upper bound of $E$.

$y^n-(y-k)^n\leq ny^{n-1}k$.

We want $y^n-ny^{n-1}k\geq x$.

So want $k\leq \frac{y^n-x}{ny^{n-1}}$

[For actual proof, see the text.]

Complex numbers

1. $=\{a+bi:a,b\in \mathbb{R}\}$.

Conjugate: $z=a+bi,\bar{z}=a-bi$.

Theorem 1.31 (see text)

Pure computational proof: boring...

$z\bar{z}=a^2-(bi)^2=a^2+b^2$

You can also use vector sum for representing operation in complex numbers.

Theorem 1.33 (see text)

More computation and still, boring...

some fun theorems:

• $|Re\ z|\leq |z|$ (equal when no imaginary parts)
• $|z+w|\leq |z|+|w|$ (equal when both $z,w$ have no imaginary parts) (Triangle inequality)

Proof for $|z+w|\leq |z|+|w|$:

Since

Theorem 1.35 Cauchy-Schwarz inequality

If $\vec{a},\vec{b}\in \mathbb{C}^n$, then

Remark: The proof is very tricky.

To help us motivate the proof in text, let's consider the special case of real numbers.

Proof for real numbers:

Let $A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j$, want to show $C^2\leq AB$

Note: if $B=0$, then $b_1=b_2=...=0$, so $C=0$ and we are done, so we may assume $B\neq 0$ so $B>0$.

Clever step: For any $t\in \mathbb{R}$,

let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$

to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.

Euclidean spaces

Nothing much to say. lol.

Normal dot product as inner product.

read text... Theorem 1.37