Solutions to Homework 2
February 9, 2007
General Comments:
This problem set will probably end up being the
hardest in the course for many of you. Your score will, in all likelihood, be
significantly lower than on the previous homeworks.
If you are frustrated
with your performance, don’t give up.
Many people are new to this kind
of methodology and it will take some time for it to sink in. There are lots
of induction problems in the practice exercises for you to try. See me or a
TA or grader if you’re having trouble with them. Also,
read these solutions
carefully
. They are an example of good mathematical exposition and may
help you to write better, clearer, and more concise proofs next time around.
2.2: 24
Let
P
(
n
) be the statement that
n
3

n
is divisible by 3.
(a) We prove
P
(
n
) for
n
= 0
,
1
,
2
,
3
, . . .
by induction.
Base case:
P
(0) says that 0
3

0 = 0 is divisible by 3, which is clearly
true.
Inductive step:
Assume
P
(
n
). We want to show
P
(
n
+1): (
n
+1)
3

(
n
+ 1) is divisible by 3. Note that
(
n
+ 1)
3

(
n
+ 1)
=
n
3
+ 3
n
2
+ 3
n
+ 1

(
n
+ 1)
=
(
n
3

n
) + 3
n
2
+ 3
n
By the induction assumption,
n
3

n
is divisible by 3. Clearly 3
n
2
+ 3
n
is divisible by 3. Hence, their sum, which is (
n
+1)
3

(
n
+1) is divisible
by 3.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2.2: 26
Let
P
(
n
) be the statement: 13
n
+ 6 is divisible by 7 if
n
is even.
We prove
P
(
n
) by strong induction for
n
≥
0.
Basis:
P
(0) is 13
0
+ 6 is divisible by 7. Since 13
0
+ 6 = 7, this is clearly
true.
Inductive step: Suppose
P
(0)
, . . . , P
(
n
) are true.
We prove
P
(
n
+ 1),
which is the statement that 13
n
+1
+ 6 is divisible by 7 if
n
+ 1 is even.
If
n
+1 is odd, there is nothing to prove. If
n
+1 is even, then so is
n

1.
Moreover,
n

1
≥
0 (since if
n
+ 1 is even,
n
+ 1 is at least 2). By
P
(
n

1),
13
n

1
+ 6 is divisible by 7. Note that
13
n
+1
+ 6 = 13
2
(13
n

1
+ 6)

13
2
(6) + 6 = 13
2
(13
n

1
+ 6)

1008
.
Since 13
n

1
+ 6 is divisible by 7 (by
P
(
n

1)), so is 13
2
(13
n

1
+ 6). It is
easy to check that 1008 is also divisible by 7. Thus, 13
2
(13
n

1
+ 6)

1008 is
divisible by 7, which is what we want to show.
2.3:
12
The obvious generalization is
A
1
∪
. . .
∪
A
n
=
A
1
∩
. . .
∩
A
n
for
n
≥
2.
Let’s prove this by strong induction. Let
P
(
n
) be the statement
A
1
∪
. . .
∪
A
n
=
A
1
∩
. . .
∩
A
n
. We prove
P
(
n
) for
n
≥
2. You were told you could assume the
base case. (You should be able to prove it though.) Assume
P
(2)
, . . . , P
(
n
).
Let’s prove
P
(
n
+ 1):
A
1
∪
. . .
∪
A
n
∪
A
n
+1
=
(
A
1
∪
. . .
∪
A
n
)
∪
A
n
+1
[note the grouping here]
=
A
1
∪
. . .
∪
A
n
∩
A
n
+1
[by
P
(2)]
=
A
1
∩
. . .
∩
A
n
∩
A
n
+1
[by
P
(
n
)]
2.3: 19
(a) You also have to show that
P
(1),
P
(2),
P
(3), and
P
(4) are true. (These
become the base cases.) If you can show that
P
(0),
P
(1),
P
(2), and
P
(3) are true, then you can’t use the induction hypothesis as stated,
because it says that
P
(
n
+1) is true
for
n >
3 (which means that
n
+1
will be at least 5!) if
P
(
n
),
P
(
n

1),
P
(
n
= 2), and
P
(
n

3) are true.
You lost 1 if you did this.
If you prove that
P
(4),
P
(5),
P
(6), and
P
(7) are true, then you can use the induction hypothesis as stated, but
you’ll end up proving that
P
(
n
) holds for all
n
≥
4, not for all positive
n
, which is what the problem asks you to do. You lost 1 if you did this.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 
 Mathematical Induction, Inductive Reasoning, Natural number, base case, inductive hypothesis, inductive step

Click to edit the document details