6
Solutions to Exercises
17.
n
2

n
+
5
=
n(n

1)
+
5.
Now either
n

1 or
n
is even, since these integers are consecutive. So
n(
n

1) is even. Since the sum
of
an even integer and the odd integer 5 is odd, the result follows.
18.
[BB]
2X2

4x
+
3
=
2(x
2

2x)
+
3
=
2[(x
_1)2

1]
+
3
=
2(x

1)2
+
1 is the sum
of
1 and a
nonnegative number. So
it
is at least 1 and hence positive.
19. For
a
2

b
2
to be odd, it is necessary and sufficient for one
of
a
or
b
to
be
even while the other is odd.
Here's why.
Case
i:
a,
b
even.
In this case,
a
=
2n
and
b
=
2m
for some integers
m
and
n,
so
a
2

b
2
=
4n
2

4m
2
=
4(n
2

m
2
)
is even.
Case
ii:
a,
b
odd.
In this case,
a
=
2n
+
1 and
b
=
2m
+
1 for some integers
m
and
n,
so
a
2

b
2
=
(4n
2
+
4n
+
1) 
(4m
2
+
4m
+
1)
=
4(n
2
+
n

m
2

m)
is even.
Case
iii:
a
even,
b
odd.
In this case,
a
=
2n
and
b
=
2m+1
for some integers
n
and
m,
so
a
2
_b
2
=
4n
2

(4m2+4m+
1)
=
4(n
2

m
2

m)

1 is odd.
Case
iv:
a
odd,
b
even.
This is similar to Case iii, and the result follows.
20.
[BB] (
+)
To
prove this direction, we establish the contrapositive, that is, we prove that
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Graph Theory, Addition, Integers, Negative and nonnegative numbers

Click to edit the document details