This preview shows pages 1–3. Sign up to view the full content.
The Real and Complex Number
Systems
Written by
MenGen Tsai
email: [email protected]
1.
2.
3.
4.
5.
6. Fix
b>
1.
(a) If
m, n, p, q
are integers,
n>
0,
q>
0, and
r
=
m/n
=
p/q
, prove
that
(
b
m
)
1
/n
=(
b
p
)
1
/q
.
Hence it makes sense to de±ne
b
r
=(
b
m
)
1
/n
.
(b) Prove that
b
r
+
s
=
b
r
b
s
if
r
and
s
are rational.
(c) If
x
is real, de±ne
B
(
x
) to be the set of all numbers
b
t
, where
t
is
rational and
t
≤
x
. Prove that
b
r
= sup
B
(
r
)
where
r
is rational. Hence it makes sense to de±ne
b
x
= sup
B
(
x
)
for every real
x
.
(d) Prove that
b
x
+
y
=
b
x
b
y
for all real
x
and
y
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Proof:
For (a):
mq
=
np
since
m/n
=
p/q
.
Thus
b
mq
=
b
np
.
By Theorem 1.21 we know that (
b
mq
)
1
/
(
mn
)
=(
b
np
)
1
/
(
mn
)
, that is,
(
b
m
)
1
/n
=(
b
p
)
1
/q
, that is,
b
r
is welldefned.
For (b): Let
r
=
m/n
and
s
=
p/q
where
m, n, p, q
are integers, and
n>
0
, q >
0. Hence (
b
r
+
s
)
nq
=(
b
m/n
+
p/q
)
nq
=(
b
(
mq
+
np
)
/
(
nq
)
)
nq
=
b
mq
+
np
=
b
mq
b
np
=(
b
m/n
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.
 Spring '08
 Ponce
 Integers

Click to edit the document details