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Math 104  Homework 1 Solutions
Lectures 2 and 4, Fall 2011
1.
Ross, 3.5. (a) Show that

b
 ≤
a
if and only if

a
≤
b
≤
a
.
(b) Prove that

a
  
b
 ≤ 
a

b

for all
a,b
∈
R
.
Proof.
(a) Suppose that

b
 ≤
a
. Note that this implicitly includes the assumption that
a
≥
0.
There are two cases to consider. First, if
b
≥
0, then

b

=
b
so our assumptions is precisely that
b
≤
a
. Since

a
≤
0
≤
b
, we have

a
≤
b
≤
a
as required. Instead, if
b <
0, then

b

=

b
, so our
assumptions becomes

b
≤
a,
or
b
≥ 
a.
Since
b <
0
≤
a
, we again have

a
≤
b
≤
a
. Thus in either case

a
≤
b
≤
a
.
Conversely suppose that

a
≤
b
≤
a
. If
b
≥
0, then

b

=
b
and
b
≤
a
gives us

b
 ≤
a
. If
b <
0,
then

b

=

b
and

a
≤
b
gives
a
≥ 
b
=

b

. Thus again in either case we have

b
 ≤
a
.
(b) By part (a), to establish the required inequality it is enough to show that

a

b
 ≤ 
a
  
b
 ≤ 
a

b

.
(In other words, apply part (a) with

a
  
b

playing the role of “
b
” and

a

b

playing the role of
“
a
”.) From the triangle inequality, we know that

a

=

(
a

b
) +
b
 ≤ 
a

b

+

b

,
so

a
  
b
 ≤ 
a

b

. Switching the roles of
a
and
b
shows that also

b
  
a
 ≤ 
b

a

; since

b

a

=

a

b

, this gives

a
  
b
 ≥ 
a

b

.
Thus

a

b
 ≤ 
a
  
b
 ≤ 
a

b

as required.
2.
Ross, 3.6. (a) Prove that

a
+
b
+
c
 ≤ 
a

+

b

+

c

for all
a,b,c
∈
R
.
(b) Use induction to prove

a
1
+
a
2
+
···
+
a
n
 ≤ 
a
1

+

a
2

+
···
+

a
n

for
n
numbers
a
1
,a
2
,...,a
n
.
Proof.
(a) By the triangle inequality, we have

a
+
b
+
c

=

a
+ (
b
+
c
)
 ≤ 
a

+

b
+
c

.
Again by the triangle inequality,

b
+
c
 ≤ 
b

+

c

, so

a
+
b
+
c
 ≤ 
a

+

b
+
c
 ≤ 
a

+

b

+

c

as claimed.
(b) The base case
n
= 2 is the statement of the triangle inequality. By way of induction, suppose
that the statement is true for some
k
: i.e. suppose that

a
1
+
···
+
a
k
 ≤ 
a
1

+
···
+

a
k

.
Then
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 Fall '08
 RIEMAN
 Math

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