Download Math 3000 Section 003 Intro to Abstract Math Homework 6

Math 3000 Section 003 Intro to Abstract Math Homework 6
Department of Mathematical and Statistical Sciences
University of Colorado Denver, Spring 2012
Solutions (April 2, 2012)
Please note that these solutions are only suggestions; different answers or proofs are always possible.
• Section 6.1: The Principle of Mathematical Induction
1. Exercise 6.10: Let r 6= 1 be a real number. Use induction to prove that
a + ar + ar2 + · · · + arn−1 =
a(1 − rn )
1−r
for every positive integer n.
Solution: We will prove by (weak) mathematical induction.
1. Base Step: For n = 1, the formula
a(1−r1 )
1−r
= a is satisfied.
2. Induction Hypothesis: Now assume that a+ar +· · ·+arn−1 =
a(1−rn )
1−r
for n = 1, 2, . . . , k.
3. Inductive Step: Then consider a+ar+· · ·+ark−1 +ark where a+ar+· · ·+ark−1 =
by our induction hypothesis. Using some basic algebra, we find that
a(1−rk )
1−r
a(1 − rk )
+ ark
1−r
a(1 − rk ) ark (1 − r)
a(1 − rk + rk − rk+1 )
a(1 − rk+1 )
=
+
=
=
.
1−r
1−r
1−r
1−r
a + ar + ar2 + · · · + ark−1 + ark =
• Section 6.2: A More General Principle of Mathematical Induction
2. Exercise 6.22: Prove that if A1 , A2 , . . . , An are any n ≥ 2 sets, then
A1 ∩ A2 ∩ · · · ∩ An = A1 ∪ A2 ∪ · · · ∪ An .
Solution: We will proof by (strong) mathematical induction.
1. Base Step: For n = 2, the identity A1 ∩ A2 = A1 ∪ A2 is true by De Morgan’s law.
2. Induction Hypothesis: Now assume that A1 ∩ · · · ∩ An = A1 ∪· · ·∪An for n = 2, 3, . . . , k.
3. Inductive Step: Then consider A1 ∩ · · · ∩ Ak ∩ Ak+1 and let B = A1 ∩ · · · ∩ Ak so that
A1 ∩ · · · ∩ Ak ∩ Ak+1 = B ∩ Ak+1 = B ∪ Ak+1
(*)
by our base step or induction hypothesis for n = 2, and
B = A1 ∩ · · · ∩ Ak = A1 ∪ · · · ∪ Ak
(**)
by our induction hypothesis for n = k. Combining equations (*) and (**), we obtain
A1 ∩ · · · ∩ Ak ∩ Ak+1 = B ∪ Ak+1 = A1 ∩ · · · ∩ Ak = A1 ∪ · · · ∪ Ak .
Remark: Note how the inductive step from k to k + 1 in the proof of Exercise 6.10 only
depends on the induction hypothesis for n = k (weak induction), whereas the proof of Exercise
6.22 also uses the induction hypothesis for n = 2 which forms the base step (strong induction).
Math 3000-003 Intro to Abstract Math Homework 6, UC Denver, Spring 2012 (Solutions)
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• Section 6.3: Proof by Minimum Counterexample
3. Exercise 6.30: Use proof by minimum counterexample to prove that
3 | (2n + 2n+1 )
for every nonnegative integer n.
Solution: Note that 2n + 2n+1 = 2n (1 + 2) = 3 · 2n is clearly divisible by 3. The simplicity
of this statement makes it a good example to practice a proof by minimum counterexample.
Proof by Minimum Counterexample. By contradiction, let us assume that there exists a nonnegative integer n such that 3 - (2n + 2n+1 ), and let m be the smallest such integer (so we
assume that m is the minimum counterexample). Because 20 + 21 = 3 is divisible by 3, it follows that m ≥ 1 and 3 | (2k +2k+1 ) for all k = 0, 1, . . . , m−1. In particular, then 2m−1 +2m ≡ 0
(mod 3) and by the laws of modular arithmetic we find that 2(2m−1 + 2m ) = 2m + 2m+1 ≡ 0
(mod 3) in contradiction to our initial assumption that 3 - (2m +2m+1 ). Hence our assumption
must have been wrong, and 3 | (2n + 2n+1 ) for every nonnegative integer n.
• Section 6.4: The Strong Principle of Mathematical Induction
4. Exercise 6.34: A sequence {an } is defined recursively by a1 = 1, a2 = 4, a3 = 9, and
an = an−1 − an−2 + an−3 + 2(2n − 3)
for n ≥ 4. Conjecture a formula for an and prove that your conjecture is correct.
Solution: We conjecture that an = n2 for every positive integer n (clearly true for n = 1, 2, 3).
Proof. Because our conjecture is true for n = 1, 2, 3, it suffices to verify the recursive relationship defined by the identity an−1 − an−2 + an−3 + 2(2n − 3) = an by algebraic substitution:
an−1 − an−2 + an−3 + 2(2n − 3) = (n − 1)2 − (n − 2)2 + (n − 3)2 + 2(2n − 3)
= (n2 − 2n + 1) − (n2 − 4n + 4) + (n2 − 6n + 9) + 4n − 6 = n2 = an .
5. Exercise 6.36∗ : Consider the following sequence of equalities:
1
=0+1
2+3+4
=1+8
5+6+7+8+9
= 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64
...
(a) What is the next equality in this sequence?
(b) Now develop a general conjecture and prove that your conjecture is correct by induction.
Solution: For (a), it is not difficult to see that the next equality in this sequence is
17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 = 64 + 125
both of which equal 189. For (b), we conjecture that a general identity is given by
(n+1)2
(n2 + 1) + (n2 + 2) + · · · + (n + 2n + 1) =
X
k = n3 + (n + 1)3
k=n2 +1
for every nonnegative integer n (which by the above is true for n = 0, 1, 2, 3, 4).
Math 3000-003 Intro to Abstract Math Homework 6, UC Denver, Spring 2012 (Solutions)
3
P
Proof. We can use the known relationship that 1 + 2 + 3 + · · · + n = nk=1 k = n(n+1)
to write
2

 

(n+1)2
(n+1)2
n2
X
X
X
k=
k − 
(n2 + 1) + (n2 + 2) + · · · + (n + 2n + 1) =
k
(*)
k=n2 +1
k=1
(n + 1)4 + (n + 1)
(n + 1)2 ((n + 1)2 + 1) n2 (n2 + 1)
−
=
=
2
2
2
n4 + 4n3 + 6n2 + 4n + 1 + n2 + 2n + 1 − n4 − n2
=
2
= 2n3 + 3n2 + 3n + 1 = n3 + (n + 1)3 .
k=1
2
− n4 + n2
(**)
Remark: Note that the above proof is a direct proof and does not use the principle of
mathematical induction. The only induction proof that I found myself is more complicated
2
2
P
and also uses the formula for the sum of cubes: 13 + 23 + 33 + . . . + n3 = nk=1 k 3 = n (n+1)
.
4
Proof by Induction. The base step is established by the initial observations, so let us now
assume that the identity is correct for all nonnegative integers up to n − 1 and show that it is
also true for n. Like in the direct proof, we start with (*) but now split up the second sum:
 


 

 

(n−1)2
32
n2
22
12
n2
X
X
X
X
X
X
k + 
k + · · · + 
k + 
k .
k + 
k=
k=12 +1
k=1
k=1
k=22 +1
k=(n−2)2 +1
k=(n−1)2 +1
Because each sum on the right-hand side satisfies our induction hypothesis, it follows that
2
n
X
k = 03 + 13 + 13 + 23 + 23 + 33 + · · · + (n − 2)3 + (n − 1)3 + (n − 1)3 + n3
k=1
=0+2
n−1
X
!
k
3
3
+n =2
k=1
(clearly
Pn2
k=1 k
=
n2 (n2 +1)
2
=
n4 +n2
2
(n − 1)2 n2
4
+ n3 =
(n − 1)2 n2 + 2n3
n4 + n2
=
2
2
is much faster) so that the proof continues as in (**).
• Additional Exercises for Chapter 6
6. Exercise 6.38∗ : In class (or by Result 6.5 in the book), we have shown that
12 + 22 + 32 + · · · + n2 =
n(n + 1)(2n + 1)
6
for every positive integer n.
(a) Use this formula to determine a new formula for
22 + 42 + 62 + · · · + (2n)2
for every positive integer n.
(b) Use that formula and (a) to determine a new formula for
12 + 32 + 52 + · · · + (2n − 1)2
for every positive integer n.
Math 3000-003 Intro to Abstract Math Homework 6, UC Denver, Spring 2012 (Solutions)
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(c) Use (a) and (b) to determine a new formula for
12 − 22 + 32 − 42 + · · · + (−1)n+1 n2
for every positive integer n.
(d) Use mathematical induction to verify the formulas in (b) and (c).
P
P
Solution: For (a), we can write nk=1 (2k)2 = 4 nk=1 k 2 = 2n(n+1)(2n+1)
. For (b), then we
3
Pn
P2n 2 Pn
2n(n+1)(2n+1)
2n(2n+1)(4n+1)
2
2
−
= n(2n−1)(2n+1)
find k=1 (2k − 1) = k=1 k − k=1 (2k) =
6
3
3
after some algebra. For (c), we initially distinguish the two cases in which n is even or odd:
P
Pm
m
2
2−

= m(2m−1)(2m+1)
− 2m(m+1)(2m+1)
(2k
−
1)

k=1 (2k)
k=1
3
3


n

X
=
−m(2m
+
1)
if
n
=
2m for m ∈ N;
(−1)k+1 k 2 = Pm+1
Pm
(m+1)(2m+1)(2m+3)
2m(m+1)(2m+1)
2
2

=
−

k=1 (2k)
k=1 (2k − 1) −
3
3
k=1



= (m + 1)(2m + 1) if n = 2m + 1 for m ∈ N.
If we now substitute m = n/2 and 2m + 1 = n + 1 for the former, and m + 1 = (n + 1)/2 and
P
2m+1 = n for the latter, we obtain that nk=1 (−1)k+1 k 2 = (−1)n+1 n(n+1)
in both cases. For
2
(d), we first (successfully) verify the above formulas for small values of n and then compute
!
n+1
n
X
X
2n(n + 1)(2n + 1) 12(n + 1)2
(2k)2 =
(2k)2 + 4(n + 1)2 =
+
3
3
k=1
k=1
2(n + 1) (2n2 + n) + (6n + 6)
2(n + 1)(2n2 + 7n + 6)
=
=
3
3 =
n+1
X
2
2(n + 1) (n + 1) + 1
2(n + 1)(n + 2)(2n + 3)
=
3
3
=
n+1
X
k+1 2
(−1)
k=1
k =
!
n(2n − 1)(2n + 1) 3(2n + 1)2
+
3
3
k=1
(2n + 1) (2n2 − n) + (6n + 3)
(2n + 1)(2n2 + 5n + 3)
=
=
3
3 (2k − 1) =
k=1
n
X
2(n + 1) + 1
n
X
2
(2k − 1)
+ (2n + 1)2 =
(n + 1) 2(n + 1) − 1
(2n + 1)(n + 1)(2n + 3)
=
3
3
!
k+1 2
(−1)
k=1
= (−1)n+2
= (−1)n+2
k
+ (−1)n+2 (n + 1)2 = (−1)n+1
2(n + 1) + 1
n(n + 1)
2(n + 1)2
+ (−1)n+2
2
2
2(n2 + 2n + 1) − (n2 + n)
n2 + 3n + 2
= (−1)n+2
2
2
(n + 1) (n + 1) + 1
(n + 1)(n + 2)
= (−1)(n+1)+1
2
2
7. Exercise 6.40: Prove that 4n > n3 for every positive integer n.
Solution: There are many possible ways to prove this result; here is a short induction proof.
Math 3000-003 Intro to Abstract Math Homework 6, UC Denver, Spring 2012 (Solutions)
5
1. Base Step: The inequality 4n > n3 holds for n = 1, 2, 3 because 4 > 1, 16 > 8, 64 > 27.
2. Induction Hypothesis: Now assume that 4n > n3 for n = 1, 2, . . . , k, where k ≥ 3.
3. Inductive Step: We need to show that 4k+1 > (k + 1)3 and use the observation that we
can equivalently write 4k+1 = 4·4k = 4k +4k +4k +4k > (k +1)3 = k 3 +3k 2 +3k +1 (now
we should have an immediate idea on what to do). Namely, by our induction hypothesis
and because k ≥ 3, we know that 4k > k 3 = k(k 2 ) ≥ 3k 2 = (3k)k > 3k > 1 and thus
4k+1 = 4 · 4k = 4k + 4k + 4k > k 3 + 3k 3 + 3k + 1 = (k + 1)3 .
8. Exercise 6.50: Evaluate the proposed proof for the following result.
Result. For every positive integer n,
1 + 3 + 5 + . . . + (2n − 1) = n2 .
Proof. We proceed by induction. Since 2 · 1 − 1 = 12 , the formula holds for n = 1.
Assume that 1 + 3 + 5 + · · · + (2k − 1) = k 2 for a positive integer k. We prove that
1 + 3 + 5 + · · · + (2k + 1) = (k + 1)2 . Observe that
1 + 3 + 5 + · · · + (2k + 1) = (k + 1)2
1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = (k + 1)2
k 2 + (2k + 1) = (k + 1)2
(k + 1)2 = (k + 1)2 .
Solution: The first four sentences of this proof are perfect: the reader is informed about the
type of proof (induction), the induction is properly anchored at n = 1, the (weak) induction
hypothesis is stated correctly, and the reader is alerted about the subsequent inductive step.
However, although the (correct) idea of the following computation is apparent, its presentation is not acceptable. In particular, the fifth sentence is logically false: from the stated
assumptions, it cannot be observed that 1 + 3 + 5 + · · · + (2k + 1) = (k + 1)2 ; otherwise the
proof would be trivial. To correct the proof, the fifth sentence including the computation
should be changed as follows: Observe that
1 + 3 + 5 + · · · + (2k + 1)
= 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1)
= k 2 + (2k + 1)
= (k + 1)2 .
Please let me know if you have any questions, comments, corrections, or remarks.
Important Reminder Regarding Spring 2012 Drop Dates
• For non-CLAS students, the last day to drop or withdraw from this course without a
petition and special approval from the academic dean is April 2, 2012 at 5 PM. After this
date, a dean’s signature is required.
• For CLAS students, the last day to drop or withdraw from this course with signatures from
the faculty and dean but without a full petition is April 16, 2012 at 5 PM. After this date,
all schedule changes require a full petition.
Please talk to me in due time if you are not sure whether you should drop the course.