Section 4.2
n
8. [BB] We have
a
=
qln
+
r and
b
=
q2n
+
r for some integers
ql, q2
and r. Subtracting,
a

b
=
(qln
+ r) 
(q2n
+ r)
=
{ql

q2)n.
Thus,
n
1
(a

b)
as required.
9.
[BB]
() If
lOa
+
b
=
7k
for some integer
k,
then
5{lOa
+
b)
=
35k,
so
49a
+
a
+
5b
=
35k
and
a
+
5b
=
35k

49a
is divisible by 7.
(f)
Conversely, if
a
+
5b
=
7k,
then
lO{a
+
5b)
=
70k,
so
lOa
+
50b
=
lOa
+
b
+
49b
=
70k
and
lOa
+
b
=
70k

49b
is divisible by 7.
10. (a) [BB] True. Since
alb, b
=
ax
for some integer
x.
Since
b
1
(c),
e
=
by
for some integer
y.
Thus, e
=
by
=
axy
=
a{
xy).
Since
xy
is an integer,
ale.
(b) False. Let
a
=
3,
b
=
15, e
=
3. Then 3115, so
alb
and e 1
b,
but
ae
=
9
~
15.
(c) True. Since
alb, b
=
ax
for some integer
x.Thus, be
=
axe
=
a{xe).
Since
xc
is an integer,
a
1
be.
(d) True. Since
alb, b
=
ax
for some integer
x.
Since e 1
d, d
=
cy
for some integer
y.
Thus,
bd
=
axcy
=
ae{ xy).
Since
xy
is an integer,
ae
1
bd.
(e) True. Since
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 Summer '10
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 Graph Theory, Addition, Integers, 7 K, 35K, 3 5k, 7 0k

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