(c) Let
x
be a rational number for which
x
2
is an integer. Suppose, for the sake of contra
diction, that
x
is not an integer.
(d) Let
p
and
q
be primes for which
p
+
q
is also prime. Suppose, for the sake of contradic
tion, neither
p
nor
q
is equal to 2.
(e) Let
‘
be a line and let
T
be a triangle. Suppose, for the sake of contradiction, that
‘
intersects all three sides of
T
.
(f) Let
C
1
and
C
2
be distinct circles. Suppose, for the sake of contradiction, there are three
(or more) distinct points where
C
1
and
C
2
intersect.
(g) Suppose, for the sake of contradiction, there were only finitely many primes.
20.5 Let
x
and
x
+
1 be consecutive integers. Suppose, for the sake of contradiction
x
and
x
+
1
were both even.
Since
x
is even, there is an integer
a
with
x
=
2
a
. Since
x
+
1 is even, there is an integer
b
with
x
+
1
=
2
b
.
Adding 1 to both sides of
x
=
2
a
we have
x
+
1
=
2
a
+
1, and since
x
+
1
=
2
b
, we have
2
b
=
2
a
+
1, which we can rewrite as 2
(
b

a
) =
1 and so
b

a
=
1
2
.
Since
b
and
a
are integers, so is
1
2
. But
1
2
is not an integer.
⇒⇐
Therefore, consecutive integers
cannot both be even.
20.6 We could write a proof nearly identical to the one from the previous problem (20.5), but here
is another approach.
Suppose
x
and
x
+
1 were consecutive odd integers. Then
x
+
1 and
x
+
2 are consecutive
even
integers (easy to prove), contradicting Exercise 20.5.
⇒⇐
Therefore, consecutive integers
cannot both be odd.
20.7 Let
p
and
q
be primes for which
p
+
q
is also prime. Suppose neither
p
nor
q
is 2. It follows
that
p
and
q
are not divisible by 2, and so are odd. Furthermore,
p
,
q
>
2, so
p
+
q
>
4.
Since
p
and
q
are odd,
p
+
q
is even, and so divisible by 2. Therefore
p
+
q
is not prime.
⇒⇐
Therefore one of
p
or
q
must be 2.
20.8 Suppose, for the sake of contradiction, there is a real number
x
whose square is negative. By
the trichotomy property, one of the following holds:
x
>
0,
x
=
0, or
x
<
0.
(a) If
x
>
0, multiplying both sides by
x
(a positive quantity) gives
x
2
>
x
·
0
=
0, so
x
2
is
positive.
⇒⇐
(b) If
x
=
0, then
x
2
=
0.
⇒⇐
(c) If
x
<
0, then
x
2
= (

x
)(

x
)
and from case (a) we again have
x
2
>
0.
⇒⇐
78
Mathematics: A Discrete Introduction
In all cases we reach a contradiction, and so
x
2
is not negative.
20.9 Let
a
and
b
be real numbers with
ab
=
0. Suppose, for the sake of contradiction, that neither
a
nor
b
is 0. Since
b
6
=
0, it has a reciprocal
b

1
. Thus
ab
=
0
⇒
ab
·
b

1
=
0
·
b

1
⇒
a
=
0
which contradicts
a
6
=
0.
⇒⇐
Therefore
a
=
0 or
b
=
0.
20.10 We must show that 1
<
√
a
<
a
; we prove each part of this inequality separately.
Suppose, for the sake of contradiction, that
√
a
6
>
1. This means that
√
a
≤
1. Since
√
a
≤
1,
we have
(
√
a
)
2
≤
1
2
which gives
a
≤
1, but
a
>
1.
⇒⇐
Therefore
√
a
>
1.
Suppose, for the sake of contradiction, that
√
a
6
<
a
. This means that
√
a
≥
a
. Squaring both
sides gives
a
≥
a
2
and then dividing both sides by
a
gives 1
≥
a
, but
a
>
1.
⇒⇐
Therefore
√
a
<
a
.
20.11 Suppose 4

n
and, for the sake of contradiction, 4

(
n
+
2
)
. Then there are integers
a
and
b
such that
n
=
4
a
and
n
+
2
=
4
b
. Subtracting the former from the latter we have 2
=
4
b

4
a
which leads to
b

a
=
1
2
which is impossible because
a
and
b
are integers (and
1
2
is not).
⇒⇐
Therefore
n
+
2 is not divisible by 4.
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 Fall '13
 RyanKinser