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122
Solutions to Exercises
23. Let
g
=
gcd(al,
...
,an),
as defined in Exercise 22. We first prove by induction on
n
that
g
divides
each
ai.
If
n
=
3, we know
g
=
gcd(al, gcd(a2'
a3))
divides both
al
and gcd(a2'
a3)'
Since gcd(a2'
a3)
divides
a2
and
a3,
by definition (see Definition 4.2.4), we conclude that
g
divides each of
al, a2
and
a3.
Now assume that
k
~
3 and that the result is true for
n
=
k.
Given
k
+
1 nonzero integers
al,"" ak+l,
we know that
g
=
gcd(al, gcd(a2,""
ak+l)
divides both
al
and gcd(a2,""
ak+l).
By the induction hypothesis, gcd(a2'
...
,ak+l)
divides each of
a2, .
..
,ak+l,
hence,
g
divides each
of
al, .
.. , ak+l.
By the Principle of Mathematical Induction, the result is proved.
Now let c be any common divisor of al,
...
,an'
Since gcd(a2'
...
,an)
is an integral linear combina
tion of
a2,
...
,an
by Exercise 22, c is a divisor of gcd(a2'
...
,an)'
Since c also divides
al
and
g
is
an integral linear combination of
al
and gcd(a2'
...
,an),
c is a divisor of
g.
Certainly then c ::;
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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