This preview shows page 1. Sign up to view the full content.
128
Solutions to Exercises
3) Let
k
::::: 2 and assume
P(k)
is true. We must prove that
P(k
1)
is true. So we suppose that
aI,
... ,
akl
are
k
1 nonnegative numbers. The idea is now to apply
P( k)
to the
k
numbers
aI,
a2,
... , akl, k.yal··· akl
since the
kth
root of the product of these numbers is
Thus,
and so
k/
_1
k/
H_1
y
a
l···ak_l(al···ak_d
k

1
=
y(al
... akl)
k1
~(al
...
ak_I)6
=
(al··
.ak_d
l
/
k

l
=
k.yal·· ·akl·
k.yal··· akl
=
~(al·.·
akl) k .yal
...
akl
al
+
a2
+ .
.. +
akl
+
k.yal··· akl
::;
k
'
usingP(k)
(
1)
al+···+akl
k.yal
...
akl
1 
k::;
k
(
k1)
al+···+akl
k.yal
...
akl
k::; k
al
+ .
.. +
akl
k1
which is
P(k

1),
as desired.
41. (With thanks to M. Oleson) (a) Assume to the contrary that
mn
 m 
n
=
am
+
bn
where
a
and
b
are nonnegative integers. Then
m(n

1 
a)
=
n(l
+
b),
so
n
I
m(n

1 
a).
Since m and
n
are
relatively prime,
n
I
(n
 1 
a)
by Corollary 4.2.10, which is impossible since 0
<
n
 1 
a
<
n.
(b) There exist integers
This is the end of the preview. Sign up
to
access the rest of the document.
 Summer '10
 any
 Graph Theory, Negative Numbers

Click to edit the document details