Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 18 July 2014

Lecture 24

A field is a set $𝔽$ with operations $\begin{matrix} + : & 𝔽 \times 𝔽 & ⟶ & 𝔽 \\ (a, b) & ⟼ & a + b \end{matrix} \begin{matrix} 𝔽 \times 𝔽 & ⟶ & 𝔽 \\ (a, b) & ⟼ & a b \end{matrix}$ such that

(a)	if $a, b, c \in 𝔽$ then $(a + b) + c = a + (b + c),$
(b)	if $a, b \in 𝔽$ then $a + b = b + a,$
(c)	there exists $0 \in 𝔽$ such that if $a \in 𝔽$ then $0 + a = a$ and $a + 0 = a,$
(d)	if $a \in 𝔽$ then there exists $- a \in 𝔽$ such that $a + (- a) = 0$ and $(- a) + a = 0,$
(e)	if $a, b, c \in 𝔽$ then $(a b) c = a (b c),$
(f)	if $a, b, c \in 𝔽$ then $(a + b) c = a c + b c and c (a + b) = c a + c b,$
(g)	there exists $1 \in 𝔽$ such that if $a \in 𝔽$ then $1 \cdot a = a$ and $a \cdot 1 = a,$
(h)	if $a \in 𝔽$ and $a \neq 0$ then there exists $a^{- 1} \in 𝔽$ such that $a a^{- 1} = 1$ and $a^{- 1} a = 1,$
(i)	if $a, b \in 𝔽$ then $a b = b a .$

Let $𝔽$ be a field and let $a \in 𝔽 .$ Show that $a \cdot 0 = 0 .$

Proof.

$\begin{matrix} a \cdot 0 & = & a \cdot (0 + 0), by (c) \\ = & a \cdot 0 + a \cdot 0, by (f). \end{matrix}$ Add $- a \cdot 0$ to each side and use (d) to get $0 = a \cdot 0 .$

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Let $𝔽$ be a field and let $a \in 𝔽 .$ Show that $- (- a) = a .$

Proof.

By (d), $- (- a) + (- a) = 0 = a + (- a) .$ Add $a$ to each side and use (d) to get $- (- a) = a .$

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Let $𝔽$ be a field and let $a \in 𝔽$ with $a \neq 0 .$ Show that ${(a^{- 1})}^{- 1} = a .$

Proof.

By (h), ${(a^{- 1})}^{- 1} \cdot a^{- 1} = 1 = a \cdot a^{- 1} .$ Multiply each side by $a$ and use (h) and (g) to get ${(a^{- 1})}^{- 1} = a .$

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Let $𝔽$ be a field and let $a \in 𝔽 .$ Show that $a (- 1) = - a .$

Proof.

By (f), $a (- 1) + a \cdot 1 = a (- 1 + 1) = a \cdot 0 = 0,$ where the last equality follows from Example 1. So, by (g), $a (- 1) + a = 0 .$ Add $- a$ to each side and use (d) and (c) to get $a (- 1) = - a .$

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Let $𝔽$ be a field and let $a, b \in 𝔽 .$ Show that $(- a) b = - a b .$

Proof.

(-a)b+ab = (-a+a)b,by (f) = 0·b,by (d) = 0,by Example 1. Add $- a b$ to each side and use (d) and (c) to get $(- a) b = - a b .$

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Let $𝔽$ be a field and let $a, b \in 𝔽 .$ Show that $(- a) (- b) = a b .$

Proof.

$\begin{matrix} (- a) (- b) & = & - (a (- b)), by Example 5, \\ = & - (- a b), by Example 5, \\ = & a b, by Example 2. \end{matrix}$

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Let $S$ be a set.

A total order on $S$ is a relation $\leq$ on $S$ such that

(a)	if $x, y, z \in S$ and $x \leq y$ and $y \leq z$ then $x \leq z,$
(b)	if $x, y \in S$ and $x \leq y$ and $y \leq x$ then $x = y,$
(b)	if $x, y \in S$ and $x \leq y$ or $y \leq x$ then $x = y$ or $y \leq x .$

An ordered field is a field $𝔽$ with a total order $\leq$ such that

(a)	if $a, b, c \in 𝔽$ and $a \leq b$ then $a + c \leq b + c,$
(b)	if $a, b \in 𝔽$ and $a \geq 0$ and $b \geq 0$ then $a b \geq 0 .$

Let $𝔽$ be an ordered field and let $a \in 𝔽$ with $a > 0 .$ Show that $- a < 0 .$

Proof.

Assume $a \in 𝔽$ and $a > 0 .$
Then $a + (- a) > 0 + (- a),$ by (OFb).
So $0 > - a,$ by (Fd) and (Fc).

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Let $𝔽$ be an ordered field and let $a \in 𝔽$ with $a \neq 0 .$ Show that $a^{2} > 0 .$

Proof.

Case 1: $a > 0 .$ Then $a \cdot a > a \cdot 0,$ by (OFb).
So $a^{2} > 0,$ by F Example 1.

Case 2: $a < 0 .$ Then $- a > 0,$ by Example 1.
Then ${(- a)}^{2} > 0,$ by Case 1.
So $a^{2} > 0,$ by F Example 6.

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Let $𝔽$ be an ordered field. Show that $1 \geq 0 .$

Proof.

By (Fg), $1 = 1^{2} \geq 0, by Example 2.$

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Let $𝔽$ be an ordered field. Let $a \in 𝔽$ with $a > 0 .$ Show that $a^{- 1} > 0 .$

Proof.

Assume $a \in 𝔽$ and $a > 0 .$
By Example 2, $a^{- 2} = {(a^{- 1})}^{2} > 0 .$
So $a {(a^{- 1})}^{2} > a \cdot 0,$ by (OFb).
So $a^{- 1} > 0,$ by (Fh) and F Example 1.

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Let $𝔽$ be an ordered field and let $a, b \in 𝔽 .$ Show that if $a \geq 0$ and $b \geq 0$ then $a + b \geq 0 .$

Proof.

$\begin{matrix} a + b & \geq & 0 + b, by (OFa) \\ \geq & 0 + 0, by (OFa) \\ = & 0, by (Fa). \end{matrix}$

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Let $𝔽$ be an ordered field and let $a, b \in 𝔽 .$ Show that if $0 < x < y$ then $y^{- 1} < x^{- 1} .$

Proof.

Assume $0 < x < y .$
So $x > 0$ and $y > 0 .$
Then, by Example 4, $x^{- 1} > 0$ and $y^{- 1} > 0 .$
Thus, by (OFb) $x^{- 1} y^{- 1} > 0 .$
Since $x < y,$ then $y - x > 0,$ by (OFa).
So, by (OFb), $x^{- 1} y^{- 1} (y - x) > 0 .$
So, by (Fh), $x^{- 1} - y^{- 1} > 0 .$
So, by (OFa), $x^{- 1} > y^{- 1} .$

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Let $𝔽$ be an ordered field. Let $x, y \in 𝔽$ with $x \geq 0$ and $y \geq 0 .$ Then $x \leq y if and only if x^{2} \leq y^{2} .$

Proof.

Assume $x, y \in 𝔽$ and $x \geq 0$ and $y \geq 0 .$
To show:

(a)	If $x \leq y$ then $x^{2} \leq y^{2} .$
(b)	If $x^{2} \leq y^{2}$ then $x \leq y .$

(a) Assume $y \geq x .$
Then $y - x \geq 0$ and, since $x \geq 0$ and $y \geq 0$ then $x + y \geq 0 .$
So $(y - x) (x + y) \geq 0 \cdot (x + y) .$
So $y^{2} - x^{2} \geq 0 .$
So $y^{2} \geq x^{2} .$

(b) Assume $x^{2} \leq y^{2} .$
Then $y^{2} + (- x^{2}) \geq x^{2} + (- x^{2}) = 0 .$
So $y^{2} - x^{2} \geq 0 .$
So $(y - x) (y + x) \geq 0 .$
Since $x \geq 0$ and $y \geq 0$ then $x + y \geq 0 .$

Case 1: $x + y \neq 0 .$
Then $x + y > 0$ and ${(x + y)}^{- 1} > 0 .$
So $(y - x) (y + x) {(y + x)}^{- 1} \geq 0 {(x + y)}^{- 1} = 0 .$
So $y - x \geq 0 .$
So $y \geq x .$

Case 2: $x + y = 0 .$
Then $x = 0$ and $y = 0$ (since $x \geq 0$ and $y \geq 0).$
So $y \geq x .$

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Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100505Lect24.pdf and was given on 5 May 2010.

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