Modern Analysis 1
Homework 02
1. Let
F
be a complete ordered field and
Q
its prime subfield. Fix
a >
1 in
F
and for
x
∈
X
define
A
(
x
) =
{
a
q
:
Q
3
q < x
}
.
(i) Show that
A
(
x
) is nonempty and bounded above, whence it is possible to
define
a
x
= sup
A
(
x
)
.
(ii) Show that if also
y
∈
F
then
a
x
+
y
=
a
x
a
y
.
Solution
For convenience, we shall simply write
X
=
A
(
x
) and
Y
=
A
(
y
).
(i) The fact that
F
is nonempty and bounded above only needs
F
to be
Archimedean. Indeed, we may choose a positive integer
m
such that
m >

x
(in which case
m
∈
Q
and

m < x
so
a

m
∈
X
) and a positive integer
n
such that
x < n
(in which case if
Q
3
q < x
then certainly
q < n
so that
a
q
< a
n
and
a
n
is an upper bound for
X
). As
F
is complete, sup
X
exists
and we may call it
a
x
; we address below the fact that if
x
happens to lie in
Q
then this supremum coincides with
a
x
as defined ‘algebraically’.
(ii) Let us introduce
Z
=
XY
=
{
a
q
a
r
:
Q
3
q < x,
Q
3
r < y
}
.
As
X
and
Y
contain only positive elements, we know (HW01) that
sup
Z
= sup
X
sup
Y
=
a
x
a
y
.
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 Fall '09
 Robinson
 ax, Prime number, ax ay

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