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80
Solutions to Exercises
26. [BB] As suggested by the hint, consider the set
S
of all positive linear combinations of
a
and
b.
Since
S
contains
a
=
1a
+
Ob
if
a
>
0 and
a
otherwise,
S
is not empty so, by the WellOrdering Principle,
S
contains a smallest element
g.
Since
9
E
S,
we know that
9
=
ma
+
nb
for integers
a
and
b;
hence, we have only to prove that
9
is
th~
greatest common divisor of
a
and
b
..
First we prove that
9
I
a.
Write
a
=
qg
+
r
with 0 :5
r
<
g
..
Note that
r
isa linear combination of
a
and
b
since
r
=
a

qg
=
a

q(ma
+
nb)=
(1 
qa)m
+
(qn)b.
Since
9
is the smallest positive linear
combination of
a
and
b,
we have
r
= 0, so
9
I
a
as'desired.Similarly,
9
I
b.
Finally, if c
I
a
and c
I
b,
then c
I
(ma
+
nb),
so c
I
g,
so c:5
g.
27. [BB] Since gcd(63, 273) = 21, lcm(63, 273) =
63~t3)
= 819 by formula (2) of this section.
Since gcd(56, 200) = 8, lcm(56, 200) =
.56(~OO)
= 1400 by formula (2) of this section .
.
.
:~
. ,.
28.
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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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