This preview shows page 1. Sign up to view the full content.
Section
4.2
(b) We have 260
1
154
0
106
1
48
1
10
3
8
13
2
16
0
1
1
2
5
22
27
and so gcd(154, 260) = 2 = 16(260) + (27)(154).
Multiplying by 2, 4 = 32(260) + (54)(154):
x
= 54,
y
=
32.
79
(c) [BB] Consider the possibility that
154x
+
260y
= 3 for certain integers
x
and
y.
The left side of
this equation is even while the right side is odd, a contradiction.
(d) Suppose we have
196x
+
260y
= 14 for some integers
x
and
y.
Since the left side is divisible by
4, the right side would have to be divisible by 4, which is not the case.
18. (a) [BB] We know that gcd(x,
y)
I
x
and gcd(x,
y)
I
y
and so gcd(x,
y)
I
(mx
+
ny)
(Proposi
tion 4.2.3).
(b) Yes, it is true. Assume gcd(x,
y)
I
d
so that
d
= i(gcd(x,
y))
for some integer
i.
By Theo
rem 4.2.9, there are integers
s
and t such that gcd(x,
y)
=
sx
+
ty.
Hence,
d
=
isx
+
tty,
and
m =
is, n
=
it
will work.
19. (a) Take
x
=
6,
y
= 9, for example.
(b) Suppose7x+5y = 3. Since 7(6)+5(9) = 3aswell,subtractinggives
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details