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 +: 𝔽×𝔽 𝔽 (a,b) a+b 𝔽×𝔽 𝔽 (a,b) ab such that

(a) if a,b,c𝔽 then (a+b)+c=a+(b+c),
(b) if a,b𝔽 then a+b=b+a,
(c) there exists 0𝔽 such that if a𝔽 then 0+a=a and a+0=a,
(d) if a𝔽 then there exists -a𝔽 such that a+(-a)=0 and (-a)+a=0,
(e) if a,b,c𝔽 then (ab)c=a(bc),
(f) if a,b,c𝔽 then (a+b)c=ac+bc andc(a+b) =ca+cb,
(g) there exists 1𝔽 such that if a𝔽 then 1·a=a and a·1=a,
(h) if a𝔽 and a0 then there exists a-1𝔽 such that aa-1=1 and a-1a=1,
(i) if a,b𝔽 then ab=ba.

Let 𝔽 be a field and let a𝔽. Show that a·0=0.

Proof.

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

Let 𝔽 be a field and let a𝔽. Show that -(-a)=a.

Proof.

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

Let 𝔽 be a field and let a𝔽 with a0. Show that (a-1)-1=a.

Proof.

By (h), (a-1)-1· a-1=1=a·a-1. Multiply each side by a and use (h) and (g) to get (a-1)-1=a.

Let 𝔽 be a field and let a𝔽. Show that a(-1)=-a.

Proof.

By (f), a(-1)+a·1=a (-1+1)=a·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.

Let 𝔽 be a field and let a,b𝔽. Show that (-a)b=-ab.

Proof.

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

Let 𝔽 be a field and let a,b𝔽. Show that (-a)(-b)=ab.

Proof.

(-a)(-b) = -(a(-b)), by Example 5, = -(-ab),by Example 5, = ab,by Example 2.

Let S be a set.

A total order on S is a relation on S such that

(a) if x,y,zS and xy and yz then xz,
(b) if x,yS and xy and yx then x=y,
(b) if x,yS and xy or yx then x=y or yx.

An ordered field is a field 𝔽 with a total order such that

(a) if a,b,c𝔽 and ab then a+cb+c,
(b) if a,b𝔽 and a0 and b0 then ab0.

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

Proof.

Assume a𝔽 and a>0.
Then a+(-a)>0+(-a), by (OFb).
So 0>-a, by (Fd) and (Fc).

Let 𝔽 be an ordered field and let a𝔽 with a0. Show that a2>0.

Proof.

Case 1: a>0. Then a·a>a·0, by (OFb).
So a2>0, by F Example 1.

Case 2: a<0. Then -a>0, by Example 1.
Then (-a)2>0, by Case 1.
So a2>0, by F Example 6.

Let 𝔽 be an ordered field. Show that 10.

Proof.

By (Fg), 1=120,by Example 2.

Let 𝔽 be an ordered field. Let a𝔽 with a>0. Show that a-1>0.

Proof.

Assume a𝔽 and a>0.
By Example 2, a-2=(a-1)2>0.
So a(a-1)2>a·0, by (OFb).
So a-1>0, by (Fh) and F Example 1.

Let 𝔽 be an ordered field and let a,b𝔽. Show that if a0 and b0 then a+b0.

Proof.

a+b 0+b,by (OFa) 0+0,by (OFa) = 0,by (Fa).

Let 𝔽 be an ordered field and let a,b𝔽. 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-1y-1>0.
Since x<y, then y-x>0, by (OFa).
So, by (OFb), x-1y-1(y-x)>0.
So, by (Fh), x-1-y-1>0.
So, by (OFa), x-1>y-1.

Let 𝔽 be an ordered field. Let x,y𝔽 with x0 and y0. Then xyif and only ifx2 y2.

Proof.

Assume x,y𝔽 and x0 and y0.
To show:

(a) If xy then x2y2.
(b) If x2y2 then xy.

(a) Assume yx.
Then y-x0 and, since x0 and y0 then x+y0.
So (y-x)(x+y)0·(x+y).
So y2-x20.
So y2x2.

(b) Assume x2y2.
Then y2+(-x2)x2+(-x2)=0.
So y2-x20.
So (y-x)(y+x)0.
Since x0 and y0 then x+y0.

Case 1: x+y0.
Then x+y>0 and (x+y)-1>0.
So (y-x)(y+x)(y+x)-10(x+y)-1=0.
So y-x0.
So yx.

Case 2: x+y=0.
Then x=0 and y=0 (since x0 and y0).
So yx.

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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