The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n. The proof consists of two steps: The base case (or initial case): prove that the statement holds for 0, or 1. The induction step (or inductive step, or step case): prove that for every n, if the statement holds for n, then it holds for n + 1. In other words, assume that the statement holds for some arbitrary natural number n, and prove that the statement holds for n + 1

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