Math 373 - hw 7 solutions
5.30, 5.36, 6.6, 6.8, 6.10;
Tuesday, 19 Oct 2010
5.30 Show that there exist a rational number a and an irrational number b such that ab is irrational.
By result 5.21, there exist irrational numbers x, y such that a = xy is rational. Let b =
because y is irrational. Moreover, we have
1
1
y.
Then b is irrational
1
ab = (xy ) y = xy y = x1 = x.
Thus, ab = x is irrational.
6
4
2
5.36 Disprove the statement: There is a real number x such that x + x + 1 = 2x .
To disprove this statement we prove that x6 + x4 + 1 6= 2x2 for all x ∈ R. Let x ∈ R. Then there are two cases
to consider x = 0 and x 6= 0.
i. Suppose that x = 0. Then
x6 + x4 + 1 = 1 6= 0 = 2x2 .
ii. Suppose that x =
6 0. It suffices to show that x6 + x4 + 1 − 2x2 > 0. Since x 6= 0, we have x2 > 0 and so
6
2 3
x = (x ) > 0. Hence,
x6 + x4 + 1 − 2x2 = x6 + (x2 − 1)2 ≥ x6 > 0.
Hence, x6 + x4 + 1 − 2x2 > 0.
Therefore, in both cases, x6 + x4 + 1 6= 2x2 .
6.6 Find a formula for 1 + 4 + 7 + · · · + (3n − 2) for positive integers n, and then verify your formula by mathematical
induction.
By result 6.3 we have 1 + 2 + · · · + n =
n(n+1)
.
2
We use this to derive a formula as follows
n
z
}|
{
1 + 4 + 7 + · · · + (3n − 2) = 3(1 + 2 + · · · + n) − 2 1 + 1 + · · · + 1
=3
n(n + 1)
n(3n − 1)
− 2n =
.
2
2
So we use induction to prove
P (n) : 1 + 4 + 7 + · · · + (3n − 2) =
n(3n − 1)
2
for all n ∈ N.
Base step: For n = 1, we have 1 = 1(3−1)
. Hence, P (1) is true.
2
Inductive step: Now suppose that that P (k) holds for some k ∈ N. Then
1 + 4 + 7 + · · · + (3k − 2) =
k(3k − 1)
.
2
We must now verify that P (k + 1) is true: by P (k) we have
k(3k − 1)
+ (3(k + 1) − 2)
2
k(3k − 1)
=
+ 3k + 1
2
1
= (3k 2 + 5k + 2)
2
1
= (k + 1)(3k + 2)
2
(k + 1)(3(k + 1) − 1)
=
.
2
Hence, P (k + 1) is true. Therefore, by the principle of mathematical induction
1 + 4 + 7 + · · · + (3k − 2) + (3(k + 1) − 2) =
1 + 4 + 7 + · · · + (3n − 2) =
for all n ∈ N.
n(3n − 1)
.
2
6.8 (a) We have seen that 12 + 22 + · · · + n2 is the number of squares in an n × n square composed of n2 1 × 1 squares.
What does 13 + 23 + 33 + · · · + n3 represent geometrically?
The sum 13 + 23 + · · · + n3 is the number of cubes (composed of 1 × 1 cubes) in an n × n × n cube composed
of n3 1 × 1 cubes.
(b) Use mathematical induction to prove that 13 + 23 + 33 + · · · + n3 =
n2 (n+1)2
4
for every positive integer n.
For n ∈ N let P (n) denote the above equation.
2
2
Base step: For n = 1, we have 13 = 1 = 1 (1+1)
. Hence, P (1) is true.
4
Inductive step: Now suppose that that P (k) holds for some k ∈ N. Then
k 2 (k + 1)2
.
4
We must now verify that P (k + 1) is true: by P (k) we have
13 + 23 + · · · + k 3 =
k 2 (k + 1)2
+ (k + 1)3
4
(k + 1)2 2
=
(k + 4(k + 1))
4
(k + 1)2 2
(k + 4k + 4)
=
4
(k + 1)2 (k + 2)2
=
.
4
Hence, P (k + 1) is true. Therefore, by the principle of mathematical induction
13 + 23 + · · · + k 3 + (k + 1)3 =
13 + 23 + 33 + · · · + n3 =
n2 (n + 1)2
4
for every positive integer n.
6.10 Let r 6= 1 be a real number. Use induction to prove that a + ar + ar2 + · · · + arn−1 =
[Note that a is also a real number.]
a(1−r n )
1−r .
For n ∈ N let P (n) denote the above equation.
1
)
Base step: For n = 1, we have a = a(1−r
1−r . Hence, P (1) is true.
Inductive step: Now suppose that that P (k) holds for some k ∈ N. Then
a + ar + ar2 + · · · + ark−1 =
a(1 − rk )
.
1−r
We must now verify that P (k + 1) is true: by P (k) we have
a(1 − rk )
+ ar(k+1)−1
1−r
a
=
((1 − rk ) + rk (1 − r))
1−r
a
(1 − rk+1 )
=
1−r
a(1 − rk+1 )
.
=
1−r
Hence, P (k + 1) is true. Therefore, by the principle of mathematical induction
a + ar + ar2 + · · · + ark−1 + ar(k+1)−1 =
a + ar + ar2 + · · · + arn−1 =
for all n ∈ N.
a(1 − rn )
1−r