MthT 430 Spivak Problem 1.21
MthT 430 Spivak Problem 1.21
21.
Assume that if
|x
-
x
0
|
<
min
æ
è
e
2(|y
0
| + 1)
,1
ö
ø
,
|y
-
y
0
|
<
e
2(|x
0
| + 1)
.
then
|x y
-
x
0
y
0
|
<
e
.
The
trick
is to write x y
-
x
0
y
0
in terms of x
-
x
0
and y
-
y
0
. There are several ways to do this, but the one which works best is
x y
-
x
0
y
0
= x(y
-
y
0
+ y
0
)
-
x
0
y
0
= x (y
-
y
0
) + (x
-
x
0
) y
0
=
I
+
II
.
Then
|
I
|
£
|x||y
-
y
0
|
£
(|x
0
| + |x
-
x
0
|)|y
-
y
0
|
£
(|x
0
| + 1)|y
-
y
0
|
<
e
/2,
|
II
|
£
|x
-
x
0
||y
0
|
<
e
/2.
Thus
|x y
-
x
0
y
0
|
£
|
I
| + |
II
|
<
e
/2 +
e
/2
=
e
.
File translated from T
E
X by
T
T
H
, version 3.77.
On 04 Aug 2007, 15:37.