This document covers Chapter 3 of discrete mathematics focusing on mathematical reasoning, induction, and recursion, with an overview of sequences, summations, and cardinality. It discusses methods for generating terms in sequences, the principles of mathematical induction, and recursive definitions of functions and sets. Key examples illustrate concepts such as proving the sum of odd integers and establishing the countability of certain sets.

1. Discrete
Mathematics
Chapter 3
Mathematical Reasoning, Induction,
and Recursion
大葉大學 資訊工程系 黃鈴玲

2. 3.2 Sequences and Summations
※Sequence (數列)
Def 1. A sequence is a function f from A  Z+
(or A  N) to a set S. We use an to denote f(n),
and call an a term (項) of the sequence.
Example 1. {an} , where an = 1/n , n  Z+
 a1 =1, a2 =1/2 , a3 =1/3, …
Example 2. {bn} , where bn= (-1)n, n  N
 b0 = 1, b1 = -1 , b2 = 1, …
3.2.1

3. Special Integer Sequence
Example 6. What is a rule that can produce the term of
a sequence if the first 10 terms are
1, 2, 2, 3, 3, 3, 4, 4, 4, 4?
Sol :
規則：數字 i 出現 i 次
數字 i 出現之前共有 1+2+3…….+(i -1) = i(i -1)/2 項
ai(i-1)/2 +1= ai(i-1)/2 +2 =… = ai(i+1)/2 = i
3.2.2
A common problem in discrete mathematics is
finding a formula for constructing the term of a sequence.
方法：找出 ai  ai+1的變化；加減某數或者乘除某數？

4. Example 7. How can we produce the terms of a
sequence if the first 10 terms are
5, 11, 17, 23, 29, 35,41, 47, 53, 59?
Sol :
a1 = 5
a2 =11 = 5 + 6
a3 =17 = 11 + 6 = 5 + 6  2
:
:
 an= 5 + 6  (n-1) = 6n-1
3.2.3

5. Example 8. Conjecture a simple formula for an if
the first 10 terms of the sequence {an} are
1, 7, 25, 79, 241, 727, 2185, 6559, 19681,59047?
Sol：
顯然非等差數列
後項除以前項的值接近3
 猜測數列為 3n  …
比較：
{3n} : 3, 9, 27, 81, 243, 729, 2187,…
{an} : 1, 7, 25, 79, 241, 727, 2185,…
 an = 3n - 2 , n  1
3.2.4

6.  Summations
Here, the variable j is call the index of summation,
m is the lower limit, and n is the upper limit.
3.2.5
n
m
m
n
m
j
j a
a
a
a 


 

 
1
Example 10.
Example 13. (Double summation)
55
25
16
9
4
1
5
1
2








j
j
60
)
4
3
2
1
(
6
6
)
3
2
(
4
1
4
1
4
1
3
1








 

 

  i
i
i j
i
i
i
i
ij

7. Example 14.
Table 2. Some useful summation formulae
3.2.6
6
4
2
0
}
4
,
2
,
0
{






S
S
1
,
1
)
1
(
)
1
(
1
0

-
-



 r
r
r
a
ar
n
n
k
k
2
)
1
(
)
2
(
1




n
n
k
n
k
6
)
1
2
)(
1
(
)
3
(
1
2 




n
n
n
k
n
k

8. Cardinality
Def 4. The sets A and B have the same cardinality
(size) if and only if there is a one-to-one
correspondence (1-1,onto 的function) from A to B.
Def 5. A set that is either finite or has the same
cardinality as Z+ (or N) is called countable (可數).
A set that is not countable is called uncountable.
3.2.7

9. 3.2.8
Example 18. Show that the set of odd positive
integers is a countable set.
Pf: (Figure 1)
Z+ : 1 2 3 4 5 6 7 8 …
……
{ 正奇數 } : 1 3 5 7 9 11 13 15 …
f : Z+  {正奇數}
f (n) = 2n – 1 is 1-1 & onto.

10. Example 19. Show that the set of positive rational number (Q+)
is countable.
...
,
,
,
,
,
,
,
,
1
4
2
3
3
2
4
1
3
1
1
3
1
2
2
1
1
1
∴ Z+ : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 …
Q+ :
(注意，因 等於 ，故 不算)
※Note. R is uncountable. (Example 20)
Exercise :
9,13,17,38
3.2.9
Pf: Q+ = { a / b | a, b Z+ }
5
1

2
2
1
1
2
2
(Figure 2)
1
2
1
3
1
4
1
5
1
1

2
1
2
2
2
3
2
4
2
5

3
2
3
1
3
3
3
4
3
5

4
2
4
1



11. 3.3 Mathematical Induction(數學歸納法)
Note : Mathematical induction can be used only to
prove results obtained in some other way. It is
not a tool for discovering formulae or theorems.
(p.239)
P(n) : a propositional function (e.g. n ≦ 2n)
A proof by mathematical induction (MI) that P(n) is
true for every nZ+ consists of two steps :
1. Basis step : The proposition P(1) is shown to be
true.(若 n 從 0 開始則證 P(0)為真 )
2. Inductive step : the implication P(k) → P(k+1) is
shown to be true for every kZ+
3.3.1

12. Example 1. Use MI to prove that the sum of the first n odd
positive integers is n2.
Note. 不用MI就可以得証:
Pf : Let P(n) denote the proposition that
Basis step : P(1) is true , since 1=12
Inductive step : Suppose that P(k) is true for a positive
integer k,
i.e., 1+3+5+…+(2k-1)=k2
Note that 1+3+5+…+(2k-1)+(2k+1) = k2+2k+1= (k+1)2
∴ P(k+1) is true
By induction, P(n) is true for all nZ+
2
1
1
)
1
(
2
)
1
2
( n
n
n
n
n
i
i
n
i
n
i

 


-


-

-
2
1
)
1
2
( n
i
n
i



-
3.3.2

13. Example 2. Use MI to prove the inequality
n<2n for all nZ+
pf : Let P(n) be the proposition “ n < 2n ”.
Basis step : P(1) is true since 1 < 21 .
Inductive step :
Assume that P(k) is true for a positive integer k,
i.e., k < 2k.
Consider P(k+1) :
k + 1 < 2k + 1  2k + 2k =2k + 1
∴ P(k+1) is true.
By MI, P(n) is true for all nZ+.
3.3.3

14. k
Hk
1
...
3
1
2
1
1 




2
1
2
n
H n 

Example 6. The harmonic numbers Hk, k =1,2,3,…, are
defined by . Use MI to show that
Pf : Let P(n) be the proposition that “ ”.
Basis step : P(0) is true, since .
Inductive step : Assume that P(k) is true for some k,
i.e.,
Consider P(k+1) :
3.3.4
whenever n is a nonnegative integer.
2
/
1
2
n
H n 

2
/
0
1
1
1
20 


 H
H
2
/
1
2
k
H k 


15. ∴P(k+1) is true.
By MI, P(n) is true for all nZ+.
3.3.5
1
2
2
1
2
2
1
1
2
1
2
1
3
1
2
1
1
1













k
k
k
k
k
H 

1
2
2
1
2
2
1
1
2
1







 k
k
k
k
H 
1
2
1
2
2
1
1
2
1
)
2
1
( 







 k
k
k
k

k
k
k
k
k
k
k
2
2
1
2
2
1
2
2
1
)
2
1
(








 
k
k
k
k
2
2
2
)
2
1
(




2
1
1



k

16. ※The 2nd principle of mathematical induction:
( 又稱為強數學歸納法 strong induction)
 Basis step 相同
 Inductive step : Assume P(k) is true for all k  n
Show that P(n+1) is also true.
3.3.6

17. Example 14. Show that if nZ and n >1, then n can be written
as the product of primes.
Pf : Let P(n) be the proposition that n can be written as the
product of primes.
Basis : P(2) is true, since 2 is a prime number
Inductive : Assume P(k) is true for all k  n.
Consider P(n+1) :
Case 1 : n+1 is prime  P(n+1) is true
Case 2 : n+1 is composite,
i.e., n+1=ab where 2  a  b ＜ n+1
By the induction hypothesis, both a and b can be
written as the product of primes.
 P(n+1) is true.
By 2nd MI, P(n) is true if nZ and n >1.
Note: 此題無法用 1st MI 證 Exercise : 3,11,17
3.3.7

18. 3.4 Recursive Definitions.
Def. The process of defining an object in terms of itself
is called recursion(遞迴).
e.g. We specify the terms of a sequence using
(1) an explicit formula:
an=2n, n=0,1,2,…
(2) a recursive form:
a0=1,
an+1=2an , n=0,1,2,…
Example 1. Suppose that f is defined recursively by
f(0)=3 , f(n+1)=2f(n)+3
Find f(1), f(2), f(3), f(4).
3.4.1

19. Example 2. Give an inductive (recursive) definition of
the factorial function F(n) = n!.
Sol :
initial value : F(0) = 1
recursive form : F(n+1) = (n+1)! = n!  (n+1)
= F(n)  (n+1)
Example 5. The Fibonacci numbers f0, f1, f2…,are
defined by : f0 = 0 ,
f1 = 1 ,
fn = fn-1 + fn-2 , for n = 2,3,4,…
what is f4 ?
Sol :
f4 = f3 + f2 = (f2 + f1) + (f1 + f0) = f2 + 2
= (f1 + f0) + 2 = 3
3.4.2

20. Example 6. Show that fn > a n-2 , where 3
2
5
1


 n
,
a
Pf: ( By 2nd MI )
Let P(n) be the statement fn >a n-2 .
Basis: f3 = 2 > a
so that P(3) and P(4) are true.
Inductive: Assume that P(k) is true, 3 k  n, n  4.
We must show that P(n+1) is true.
fn+1 = fn + fn-1 > a n-2 + a n-3
= a n-3(a +1)
∵ a +1= a 2
∴ fn+1 > a n-3  a 2 = a n-1
We get that P(n+1) is true.
By 2nd MI , P(n) is true for all n  3
2
5
3
3 2
4



 a
f
3.4.3

21. ※Recursively defined sets.
Example 7. Let S be defined recursively by
3S
x+yS if xS and yS.
Show that S is the of positive integers divisible by 3
(i.e., S = { 3, 6, 9, 12, 15, 18, … }
Pf:
Let A be the set of all positive integers divisible by 3.
We need to prove that A=S.
(i) A  S : (By MI)
Let P(n) be the statement that 3nS
…
(ii) S  A : (利用S的定義)
(1) 3  A ,
(2) if xA,yA, then 3|x and 3|y.
 3|(x+y)  x+yA
∴S  A
S = A
3.4.4

22. Example 8. The set of strings over an alphabet 
is denoted by *. The empty string is denoted
by l, and wx* whenever w* and x.
eg.  = { a, b, c }
* = { l, a , b , c , aa , ab , ac , ba , bb , bc, …
abcabccba, …}
Example 9. Give a recursive definition of l(w),
the length of the string w*
Sol :
initial value : l(l)=0
recursive def : l(wx)=l(w)+1 if w*, x.
la lb lc
3.4.5

23. Exercise 3,13, 25, 49
Exercise 39. When does a string belong to the
set A of bit strings defined recursively by
lA
0x1A if xA.
Sol :
A={l, 01 , 0011, 000111, …}
∴當bit string a = 000…011…1 時
aA n個 n個
0l1
3.4.6

24.  Ackermann’s function
A(m, n) = 2n if m = 0
0 if m  1 and n = 0
2 if m  1 and n = 1
A(m-1, A(m, n-1)) if m  1 and n  2
Exercise 49 Show that A(m,2)=4 whenever m  1
Pf :
A(m,2) = A(m-1, A(m,1)) = A(m-1,2) whenever m  1.
A(m,2) = A(m-1,2) = A(m-2,2) = … = A(0,2) = 4.
3.4.7

25. 3.5 Recursive algorithms.
※ Sometimes we can reduce the solution to a
problem with a particular set of input to the
solution of the same problem with smaller
input values.
eg. gcd(a,b) = gcd(b mod a, a) (when a < b)
Def 1. An algorithm is called recursive if it
solves a problem by reducing it to an instance
of the same problem with smaller input.
3.5.1

26. Example 1. Give a recursive algorithm for
computing an, where aR {0}, nN.
Sol :
recursive definition of an :
initial value : a0=1
recursive def : an = a  an-1.
Algorithm 1.
Procedure power( a : nonzero real number,
n : nonnegative integer )
if n = 0 then power(a, n):=1
else power(a, n):= a * power(a, n-1).
∴
3.5.2

27. Example 4. Find gcd(a,b) with 0a<b
Sol :
Algorithm 3.
procedure gcd(a,b : nonnegative integers with a<b)
if a=0 then gcd(a,b) := b
else gcd(a,b) := gcd(b mod a, a).
Example 5. Search x in a1, a2,…,an by Linear Search
Sol : Alg. 4
procedure search (i, j, x)
if ai = x then location := i
else if i = j then location := 0
else search(i+1, j, x)
從ai,ai+1,…aj 中找 x
call
search(1, n, x)
3.5.3

28. Example 6. Search x from a1,a2,…,an by binary
search.
Sol : Alg. 5
procedure binary_search (x , i , j)
m := (i+j) / 2
if x = am then location := m
else if (x < am and i < m) then
binary_search(x, i, m-1)
else if (x > am and j > m) then
binary_search(x, m+1, j)
else location := 0
call binary_search(x, 1, n)
search x from ai, ai+1, …, aj
3.5.4
表示左半邊
ai, ai+1, …, am-1
至少還有一個元素

29. Example 7. Give the value of n!, nZ+
Sol :
Note : n! = n  (n-1)!
Alg. 6 (Recursive Procedure)
procedure factorial (n: positive integer)
if n = 1 then factorial (n) := 1
else factorial (n) := n  factorial (n-1)
Alg. 7 (Iterative Procedure)
procedure iterative_factorial (n : positive integer)
x := 1
for i := 1 to n
x := i  x
{ x = n! }
3.5.5

30. ※ iterative alg. 的計算次數通常比 recursive alg.少
※ Find Fibonacci numbers
(Note : f0=0, f1=1, fn=fn-1+fn-2 for n2)
Alg. 8 (Recursive Fibonacci)
procedure Fibonacci (n : nonnegative integer)
if n = 0 then Fibonacci (0) := 0
else if n = 1 then Fibonacci (1) := 1
else Fibonacci (n) := Fibonacci (n-1)+Fibonacci (n-2)
3.5.6

31. Alg.9 (Iterative Fibonacci)
procedure iterative_fibonacci (n: nonnegative integer)
if n = 0 then y := 0 // y = f0
else begin
x := 0
y := 1 // y = f1
for i := 1 to n-1
begin
z := x + y
x := y
y := z
end
end
{y is fn }
Exercise : 5 , 27
i = 1 i = 2 i = 3
z f2 f3 f4
x f1 f2 f3
y f2 f3 f4
3.5.7