SOLUTIONS TO HOMEWORK # 2
3.5 (a)
Prove

b
 ≤
a
⇔ 
a
≤
b
≤
a
. Since

b
 ≥
0 we have
a
≥
0. Consider two
cases:
Case 1:
b
≥
0; then

b

=
b
and we immediately have
b
≥ 
a
. Also

b
 ≤
a
⇔
b
≤
a
so in this case our statement holds.
Case 2:
b <
0; then

b

=

b,b
≤
a
. Also

b
 ≤
a
⇔ 
b
≤
a
⇔
b
≥ 
a
so in this
case our statement holds too.
We are done.
(b) Show
ﬂ
ﬂ

a
  
b

ﬂ
ﬂ
≤
ﬂ
ﬂ
a

b
ﬂ
ﬂ
.
From (a), it will suﬃce to show

ﬂ
ﬂ
a

b
ﬂ
ﬂ
≤ 
a
  
b
 ≤
ﬂ
ﬂ
a

b
ﬂ
ﬂ
. Note

a

=

a
+
b

b
 ≤ 
a

b

+

b

using the triangle inequality
(1)

b

=

b
+
a

a
 ≤ 
b

a

+

a

same reason
(2)
Thus

a
  
b
 ≤ 
a

b

,
from the top inequality (1), and

b
  
a
 ≤ 
b

a

=

a

b

,
so

a

b
 ≤ 
a
  
b

, from (2)
.
Thus

a

b
 ≤ 
a
  
b
 ≤ 
a

b

, and this proves the claim.
3.7 (a)
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 Spring '08
 OLSAVSKYGREGORM
 Supremum, Order theory, sup

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