MATH 74: HOMEWORK 10 COMMENTS
Book Problem # 14, p. 227
Let
a
and
b
be positive integers. Let
A
=
{
am
+
bn

m,n
∈
Z
and
am
+
bn >
0
}
.
Note that
A
6
=
∅
because
a,b
∈
A
. By the Wellordering principle
A
has a smallest element.
Call this element
c
.
We claim that
c

a
and
c

b
. Suppose for a contradiction that it doesn’t. Then either
c

a
or
c

b
. First we’ll treat the case where
c

a
. In this case the division theorem tells us there
exist unique
q
and
r
such that
a
=
cq
+
r
0
< r < c,
(note
r
cannot equal zero by our assumption that
c

a
. Since
c
∈
A
there exist integers
m
0
and
n
0
such that
c
=
am
0
+
bn
0
. Hence
r
=
a

cq
=
a

(
am
0
+
bn
0
)
q
=
a
(1

m
0
q
) +
b
(

n
0
q
)
.
Since in addition
r >
0 we see that
r
∈
A
. But
r < c
, and
c
is the smallest element of
A
,
and this is a contradiction. Hence
c

a
.
A similar argument shows that
c

b
.
We claim that
c
is in fact the greatest common divisor of
a
and
b
. We must verify two
things:
(i)
c

a
and
c

b
and
(ii) if
c
0

a
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 COURTNEY
 Division, Remainder, Integers, Natural number

Click to edit the document details