MthT 430 Projects Chapter 1a Inequalities
MthT 430 Projects Chapter 1a Inequalities






Chapter 1, Problems 20, 21 and 22 in Spivak
20.
Prove that if
|x - x0| < e/2 and |y - y0| < e/2,
then
|(x + y) - (x0 + y0)| < e.
Proof: The idea is to express or estimate the expression in the conclusion ,
|(x + y) - (x0 + y0)|,
in terms of expressions in the hypothesis ,
|x - x0| and |y - y0| .

|(x + y) - (x0 + y0)|
= |(x - x0) + (y - y0)|
£ |(x - x0)| + |(y - y0)|
< e/2 + e/2 = e
21.
Prove that if

|x - x0|
< min
æ
è
e

2 (|y0| + 1)
, 1 ö
ø
,
|y - y0|
< min
æ
è
e

2 (|x0| +1)
, 1 ö
ø
,
then
|x y - x0 y0| < e.
Proof: The idea again is to express or estimate the expression in the conclusion ,
|x y - x0 y0|,
in terms of expressions in the hypothesis ,
|x - x0| and |y - y0| .

|x y - x0 y0|
= |x (y - y0) + (x - x0) y0|
£ |x||y - y0| + |y0||x - x0|
= I + II.

I
£ |( x - x0) + x0| |y - y0|
£ (1 + |x0|) e

2 (|x0| + 1)
< e/2.
II
£ |y0| e

2 (|y0| +1)
< e/2.
22.
Show that if x0 ¹ 0, and
|x - x0| < min
æ
è
|x0|2 e

2
, |x0|

2
ö
ø
,
then
ê
ê
1

x
- 1

x0
ê
ê
< e.
Proof: Note that
|x|
= |x0 + (x - x0)|
³ |x0| - |x0|

2
= |x0|

2
.

ê
ê
1

x
- 1

x0
ê
ê
= ê
ê
x - x0

x x0
ê
ê
= 1

|x|
1

|x0|
|x - x0|
£ 2

|x0|
1

|x0|
|x - x0|
< 2

|x0|
1

|x0|
|x0|2 e

2
= e.



File translated from TEX by TTH, version 3.74.
On 07 Sep 2006, 16:13.