Solution manual an introduction to mathematics thinking

Therefore, by the Principle of Mathematical Induction, Are the following true for all positive integer values of n?. ii As induction hypothesis, suppose that the result holds for n = k,

=104

n − 1

r − 1

.Solution:

n − s

r − s

.Solution:

n

r

rs

By definition,

n + 2n



=(n + 2)!

2!n! .Hence, we must solve n2+ 3n + 2 = 72 or n2+ 3n + 70 = 0 By the quadraticformula, we find n = 7 or n = −10

(ii) Suppose that

Hence the assertion is true for n = k + 1

Therefore, by the principle of mathematical induction, the assertion is truefor all n ∈ P

(k + 1)30(k + 1)2− 2(k + 1)(6k3+ 9k2+ k − 1)Working with these last two terms we have,

(k + 1)(30k3+ 90k2+ 90k + 30 − 12k3− 18k2− 2k + 2)

= (k + 1)(18k3+ 72k2+ 88k + 32)

= (k + 1)(k + 2)(18k2+ 36k + 16)Putting the terms together we get

Exercise 4-14:

Prove by induction the following results for all n ∈ P

12+ 32+ 52+ · · · + (2n − 1)2= n(2n − 1)(2n + 1)

3Solution:

(i) For the base case n = 1, we have 2n − 1 = 1, 12 = 1 and 1(1)(3)3 = 1.Hence the formula is true when n = 1

(ii) As induction hypothesis, suppose that the formula holds when n = k,that is suppose

12+ 32+ 52+ · · · + (2k − 1)2= k(2k − 1)(2k + 1)

3Then

12+ · · · + (2k − 1)2+ (2(k + 1) − 1)2=k(2k − 1)(2k + 1)

2,

= (2k + 1)(k(2k − 1) + 3(2k + 1))

3Adding and subtracting 4k and 6 inside the parenthesis gives

Exercise 4-15:

Prove by induction the following results for all n ∈ P

1 · 2 + 2 · 3 + 3 · 4 + · · · + n(n + 1) = n(n + 1)(n + 2)

3Solution:

(i) For the base case n = 1, we have 1 · 2 = 2 and 1(2)(3)3 = 2 Hence theformula is true when n = 1

(ii) As induction hypothesis, suppose that the formula holds when n = k,that is suppose

1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1) = k(k + 1)(k + 2)

3Then

2n+1 = 2 − n + 2

2n

+12

Let S be the set containing n elements

(i) For the base case n = 0, S has no elements so S = ∅ Therefore, its onlysubset is itself There are 20 subsets

(ii) As induction hypothesis, suppose that a set S with n = k elements has

2k subsets

Then for a set S with n = k + 1 elements,

S = {a1, a2, , ak, ak+1}

A subset of S either contains ak+1 or it does not Notice that S is of the firsttype and that ∅ is of the second So we can partition the subsets of S into thosethat contain ak+1and those that do not contain it If they do not contain ak+1then they are subsets of {a1, a2, , ak} By our induction hypothesis there are

2k of these Then a set S with k + 1 elements has in total

# of subsets = # of subsets with ak+1+ # of subsets without ak+1

= 2k+ 2k

= 2k+1That is, the result is true for n = k + 1 whenever it is true for n = k.Therefore, by the Principle of Mathematical Induction, the result is true forall n ∈ P + {0}

i.e 6|(2n3+ 3n2

+ n) for all n ∈ P

Exercise 4-19:

Find an expression forPn

r=1r(r!) and prove that it is correct

Solution: The first few sums give

We shall prove that this result is true for all n ∈ P by induction on n

(i) For the base case n = 1, 1 = 2! − 1, so the result holds

(ii) As induction hypothesis, suppose that the formula holds for n = k, thatis

So the formula holds for n = k + 1 every time it holds for n = k

Therefore, by the Principle of Mathematical Induction,

Are the following true for all positive integer values of n? If so, prove the result;

if not, give a counterexample

n! ≥ 2nSolution:

It is true for all n ∈ P with n ≥ 3 For n = 1, 2 we have 1! < 2 and 2 < 4,which are two counterexamples

Exercise 4-21:

Are the following true for all positive integer values of n? If so, prove the result;

if not, give a counterexample

3|(22n− 1)

It is true To prove it we use induction

(i) For the base case n = 1, 22− 1 = 3 and 3|3 The result is true for n = 1.(ii) As induction hypothesis, suppose that the result holds for n = k, that is

3|(22k− 1)Then 22(k+1)− 1 = (22k· 4) − 1 By induction hypothesis

22k≡ 1 (mod 3)multiplying by 4 ≡ 1 (mod 3) gives

22(k+1)≡ 1 (mod 3)

This shows that 3|(22(k+1)− 1)

That is, the result is true for n = k + 1 whenever it is true for n = k.Therefore, by the Principle of Mathematical Induction, the result is true forall n ∈ P

Exercise 4-22:

Are the following true for all positive integer values of n? If so, prove the result;

if not, give a counterexample

7|(5n+ n + 1)

Solution:

It is false For n = 3 53+ 3 + 1 = 129 and 129 = 7(18) + 3 so 7 6 | 129

Exercise 4-23:

Are the following true for all positive integer values of n? If so, prove the result;

if not, give a counterexample

(a + b)|(a2n− b2n)Solution:

It is true, we prove it by induction on n

(i) For the base case n = 1, (a + b)|(a2− b2) The result is true for n = 1.(ii) As induction hypothesis, suppose that the result holds for n = k, that is

(a + b)|(a2k− b2k)Then

a2(k+1)− b2(k+1)= a2k+2− b2k+2

adding and subtracting a2b2k gives

= a2a2k− a2b2k+ a2b2k− b2b2k

= a2(a2k− b2k) + b2k(a2− b2)

By induction hypothesis (a + b)|(a − b ), also (a + b)|(a − b ), it follows that(a + b)|(a2(k+1)− b2(k+1)).

That is, the result is true for n = k + 1 whenever it is true for n = k.Therefore, by the Principle of Mathematical Induction, the result is true forall n ∈ P

Exercise 4-24:

Prove that the sum of the first n terms of the arithmetic progression

a + (a + d) + (a + 2d) + · · · + [a + (n − 1)d]

is n

2[2a + (n − 1)d]; that is, n

2 times the sum of the first and last terms.Solution:

We prove it by induction on the number or terms n

(i) For the base case n = 1, the first term of the arithmetic progression is aand 12[2a + 0] = a The result is true for n = 1

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

(ii) As induction hypothesis assume that the result holds for n = k, that is

pj

.Proof The expression pj
stands for an integer defined by

pj



j!(p − j)!=

(p)(p − 1) · · · (p + 1 − j)(1)(2) · · · (j) .Since j ≥ 1, the factor p appears in the numerator, but since j ≤ p − 1, p doesnot appear as a factor in the denominator Since p is a prime, no factor of p(except 1) appears in the denominator Thus p | pj

(i) Let p be a prime Clearly the result holds for any n = 1

(ii) Suppose kp ≡ k mod p (induction hypothesis) then

(k + 1)p = kp+ p1
kp−1+ p2
kp−2+ · · · + p−1p
k + p

p
1

= kp+ 1 (mod p), (by the lemma above)

= k + 1 (mod p) (by the induction hypothesis)

Therefore np≡ n (mod p) for all integers ≥ 1

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

(i) The result is true for n = 2

(ii) Suppose it is true for n = k A system on k + 1 points may be formedfrom a system on k points by adding an (k + 1)th point and k lines givingk(k − 1)/2 + k = k(k + 1)/2 lines and the result holds for n = k + 1 The resultholds for all n ≥ 2 by the principle of mathematical induction

Exercise 4-30:

Find an expression for

1 − 3 + 5 − 7 + 9 − 11 + · · · + (−1)n−1(2n − 1)and prove that it is correct

We shall prove that this result is true for all n ∈ P by induction on n

(i) We have seen above that the result is true for n = 1

(ii) Suppose that the result is true when n = k; that is suppose

We will prove this result by induction on n

(i) For the base case n = 1, if x26= 1,

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

To prove that this is the correct formula we will apply Mathematical tion

Induc-(i) We have seen that the formula holds if n = 2

(ii) Suppose that the formula holds when n = k for k ≥ 2; that is suppose

  (k + 1)2− 1(k + 1)2



= (k + 1)2k

(k + 2)k(k + 1)2

= [(k + 1) + 1]

2(k + 1)Therefore the formula also holds for n = k + 1

Hence, by the Principle of Mathematical Induction,

r=1 1

r diverges.]Solution:

(i) For n = 1, 1 +12 = 1 +12 so the result holds

(ii) By induction hypothesis assume that the result is true for n = k that is

So the result holds for n = k + 1 whenever it is true for n = k

By the Principle of Mathematical Induction the result holds for all n ∈ P

(2a)5b +6

2

(2a)4b2+6

3

(2a)3b3

+64

(2a)2b4+6

5

(2a)b5+6



a2(−1)3+5

4

a(−1)4+5

5

(−1)5

(4x2)3(−3y3) +4

2

(4x2)2(−3y3)2

+43

(4x2)(−3y3) +4

4

(−3y3)4



a2(b + c) +3

2

a(b + c)2+3

1

(b + c)3

= a3+ 3a2b + 3a2c + 3ab2+ 6abc + 3ac2+ b3+ 3b2c + 3bc2+ c3

4

(.01)4

(0.02)9+ (0.02)10,

= 1 + 0.2 + 45(0.0004) + 120(0.000008)+ 210(0.00000016) + 252(32 · 10−10) +

We only want a precision of three decimal places The sum of the fifth andsixth terms is 0.0000336 + 0.000000806 = 0.000034406 < 0.0001 Because thelast five terms are all smaller we only need to consider the first four terms ofthe binomial expansion

= 1 +10

1

0.02 +10

2

0.0004 +10

3

0.000008 + · · ·

x5

11

.Solution:

Fifth term is the one for k = 4, i.e

11

4

(2x6)7

6

.Solution:

The rthterm in the expansion of (x2+x3)6is

6r

(x2)6−r 3

x

r

=6r



3rx12−3r

Thus, for the term containing x5, we must have 12 − 3r = 5, or 3r = 7, so r isnot an integer There is no term containing x5 in this expansion

Exercise 4-43:

If p is a prime, prove that

(a + b)p≡ ap+ bp (mod p) for all a, b ∈ Z

Solution:

p

r
= p(p−1)···(p−r+1)

r! Thus p0
= p

p
= 1 p

1
= p For other values of r (i.e r = 2, 3, , p − 1) nofactor in the denominator r! divides p Hence r! divides (p − 1) · · · (p − r + 1).Thus pr
is a multiple of p (true for r = 1 also)

Therefore (a + b)p = Pp

r=0 p

r
arbb−r ≡ ap+ bp (mod p) since terms for

1 ≤ r < p are all ≡ 0 (mod p)

r is the percentage increase)

Exercise 4-45:

Prove that

n0

+n1

+n2

+ · · · +n



−n1

+n2

+ · · · +n

+n2

We will prove this result by induction on n

(i) For the base case n = 1,

2

(1 − x)2 = x

So the result is true for n = 1

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

That is, the result is true for n = k + 1 whenever it is true for n = k.Therefore, by the Principle of Mathematical Induction, the result is true forall n ∈ P.

n

= 1 + n · 1

n+

n2

Let us look at the first few cases separating them in n = 2m and n = 2m + 1,

to see if there is a pattern to the sums

We know prove these expressions by induction on m For n = 2m

(i) We have seen that the formula holds for the base case m = 1 n = 2.(ii) Suppose that the formula holds when m = k, that is for n = 2k,

[12− 32+ + (4k − 3)2− (4k − 1)2] = −8(k2)

Hence, by the principle of Mathematical induction the expression for n = 2mholds for all m ∈ P.

Both expressions for 12− 32+ 52− 72+ · · · + (−1)n−1(2n − 1)2 for n oddand even are correct

Problem 4-49:

Prove that a convex n-gon contains 12n(n − 3) diagonals (A convex n-gon is apolygon with n sides such that all the line segments, joining two nonadjacentvertices, lie inside the polygon.)

Solution 1:

The smallest convex n-gon contains three vertices So n ≥ 3 We will provethis result by induction on n, where n is the number of vertices

(i) For the base case n = 3, a triangle has 0 diagonals and 12(3)(3 − 3) = 0

So the result is true for n = 1

(ii) As induction hypothesis, suppose that the result holds for n = k with

k ≥ 3, that is the number of diagonals of a convex k-gon is 12k(k − 3)

Let A be a convex (k + 1)-gon with consecutive vertices a1, a2, , ak, ak+1.Consider the diagonal a1ak that divides A into a triangle B with vertices

a1, ak, ak+1, and a k-gon C with vertices a1, a2, , ak

of this form Therefore

That is, the result is true for n = k + 1 whenever it is true for n = k

Therefore, by the Principle Mathematical Induction, the result is true for allintegers n, with n ≥ 3

Solution 2:

As an alternative solution, we can solve this question without using induction.Let A be the convex n-gon with vertices a1, a2, · · · , an All lines joining twodifferent vertices are a side of the n-gon or a diagonal And by convexity, allthe lines joining two nonadjacent vertices (no sides) lie inside A There are n2
lines joining two different vertices, and A has n sides, so

So the result is true for n = 1.

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

(2i + 1) + [2(i + 1) + 1] + · · · + [2(i + k − 1) + 1] = k3 where i = k(k − 1)

Then for i0= k(k+1)2 ,

(2i0+ 1) + [2(i0+ 1) + 1] + · · · + [2(i0+ k − 1) + 1] + [2(i0+ k) + 1]

If we add and subtract 2k2

= (2i0+ 1) + · · · + [2(i0+ k − 1) + 1] + [2(i0+ k) + 1] − 2k2+ 2k2

= (2i0− 2k + 1) + · · · + [2(i0+ k − 1) − 2k + 1] + [2(i0+ k) + 1] + 2k2noticing that i0− i = k, we have

(2i + 1) + [2(i + 1) + 1] + · · · [2(i + k − 1) + 1] + [2(i0+ k) + 1] + 2k2

+ 2k2

= k3+ k(k + 1) + 2k + 1 + 2k2

= k3+ 3k2+ 3k + 1

= (k + 1)3.That is, the result is true for n = k + 1 whenever it is true for n = k.Therefore, by the Principle of Mathematical Induction, the result is true forall n ∈ P

Note that this result is always true if n = 0, for any r, and also if r = 1, for any

n We shall take as induction hypothesis

r! | (n + 1)(n + 2) · · · (n + r) whenever n + r = k

and use induction on k

(i) The base case is k = 1 This means that n = 0 and r = 1, so the result

is true in this case

(ii) Assume that r! | (n + 1)(n + 2) · · · (n + r) whenever n + r = k Let

By the Principle of Mathematical Induction, the result is true whenever

and prove by induction that your expression is correct

Solution: Note that r is fixed in the above expression, and also r ≥ 1 for

i + r − 1 to come after i Every product in the sum can be expressed as aquotient of factorials

i(i + 1)(i + 2) · · · (i + r − 1) =(i − 1 + r)!

(i − 1)! .

Multiplying and dividing by r! and then using the definition of the binomialcoefficients, we obtain

Substituting this into the sum gives

+ · · · +r + n − 1

r



Try to find a formula for the sumPr+n−1

s=r s

r
in brackets on the right side.For n = 1; rr

= 1

For n = 2; Pr+1

s=r s r

= Pr+2 s=r s

r
+ r+3 r

We shall fix r ≥ 1, and use induction on n to prove this

(i) The base case is n = 1 In this case there is only one term on the left side,1(2)(3) · · · (r) = r! = r! r+1r+1
, and the result is true

(ii) Assume inductively that the result is true for n = k so

Now let n = k + 1, so, using the induction hypothesis,



= r!r + k + 1

r + 1

,

using Proposition 4.32 Therefore the result holds for n = k + 1 whenever it istrue for n = k By the Principle of Mathematical Induction, the result is truefor all positive integers n and r

Note that the result can also be written as

We will prove the result by induction on n

(i) For the base case n = 1 If x ≡ 1 (mod 2) then x = 1 + 2m for someinteger m By squaring x we get

x2= 1 + 4m + 4m2= 1 + 4m(m + 1)Because m and m+1 are consecutive integers one of them is even so m(m+1) =2j for some integer j, therefore 4m(m + 1) = 8m It is clear that x ≡ 1 (mod 8)

So the result is true for n = 1

(ii) As induction hypothesis, suppose that the result holds for n = k, that is

if x ≡ 1 (mod 2) then

x2k ≡ 1 (mod 2k+2)Then x2 k

= 1 + 2k+2` for some integer ` Squaring x2 k

we get(x2k)2 = 1 + 2k+3` + (2k+2)2`2

= 1 + 2k+3(` + 2k+1`2)

because ` + 2k+1`2

∈ Z then x2 k+1

≡ 1 (mod 2k+3)

So the result is true for n = k + 1 whenever it is true for n = k

By the Principle of Mathematical Induction, the result is true for all n ∈ P

Problem 4-54:

Is 2n− 1 the minimum number of moves required to solve the Tower of Hanoipuzzle in Example 4.17 Give reasons for your answer

Solution:

Indeed, 2n− 1 is the minimum number of moves required to solve the Tower

of Hanoi puzzle We can prove this by induction on the number of discs Fromthe example we already know that the puzzle can be solved with 2n− 1 number

of moves, if we show that at least 2n− 1 moves are needed then we would bedone

(i) For n = 1 one disc has to be transferred to another peg in at least 21− 1 = 1move

(ii) As induction hypothesis suppose that to transfer k discs to another peg, atleast 2k− 1 moves are needed

Now suppose that there are k + 1 discs on one peg When the k + 1 discsare all transferred, the largest disc will be in another peg So there is a movewhere the largest disc is transferred to that peg When this happens the other

k discs have to be in the remaining peg (see figure of example) By inductionhypothesis, to reach this stage at least 2k− 1 moves are needed to transfer thefirst k discs onto that peg Now, by the base case, transferring the largest disccan be accomplished in at least 1 move Finally, from this new stage to the end

of the puzzle, the k discs on the remaining peg move back onto the largest disc.Again, this can be done in at least 2n− 1 more moves