Lectures on Theory of Numbers

Md. Shah Noor

Associate Professor
Department of Mathematics
Shahjalal University of Science and Technology
Sylhet, Bangladesh
email: noorms100@gmail.com
web:http://www.sust.edu
Contents

1 Theory of Divisibility 3
1.1 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Greatest Common Divisor and Least Common Multiple . . . . . . . . . . 5
1.3 Factorization in Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . 8

2 Arithmetic Functions and Diophantine Equations 12

2.1 Arithmetic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Congruences 23
3.1 Congruences and Its Properties . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 Fermat’s Theorem, Euler’s Theorem and Wilson’s Theorem . . . . . . . . 26

4 Solutions of Congruences 31
4.1 Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Quadratic Residuacity 36
5.1 Quadratic Residue and Nonresidues . . . . . . . . . . . . . . . . . . . . . 36
5.2 Legendre Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6 Sets, The Real Number System, and Functions 39

6.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.1.1 Cartesian Product Sets and their visualization . . . . . . . . . . . 40
6.2 The Real Number System . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.2.1 The Field Properties and the Order Properties of R . . . . . . . . 40
6.2.2 The Completeness Properties of R . . . . . . . . . . . . . . . . . . 41
6.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

7 Sequence and Sequence of Functions 44

7.1 Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
7.2 Sequence of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

8 Series and Series of Functions 51

8.1 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
8.1.1 Tests for Absolute Convergence . . . . . . . . . . . . . . . . . . . 52
8.1.2 Tests for Nonabsolute Convergence . . . . . . . . . . . . . . . . . 53
8.2 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1
9 Limit and Continuity of a Function 56
9.1 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
9.2 Continuous Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

10 Differentiation 61

11 Riemann Integration 64

A Reviews 66

2
Chapter 1

Theory of Divisibility

1.1 Divisibility
Definition 1.1.1 (Natural Number). A number used for counting is said to be a nat-
ural number. The set of natural numbers is denoted and defined by N = {1, 2, 3, 4, 5, · · · }.
Definition 1.1.2 (Integer). Any whole number positive, negative, or zero is said to be an
integer. The set of integers is denoted and defined by Z = {· · · , −3, −2, −1, 0, 1, 2, 3, · · · }.
Definition 1.1.3 (Divisibility). A number b is said to be divisible by a number a(a 6= 0)
if there is a number c such that b = ac. We denote this by a | b, read as a divides b. If
no such integer can be found we write a - b, read as a does not divide b.
Example 1.1.1. (1) 3 | 12 because 12 = 3 · 4

(2) 3 - 7 because there is no integer c such that 7 = 3 · c.

If a | b, then a is called a divisor of b and b is called a multiple of a. If a | b and

0 < a < b, then a is called a proper divisor or factor of b. We have b = a · c = (−a) · (−c).

Thus if a | b, then −a | b. For practical purposes we can limit our attention to only

positive divisors of integers. Evidently 1 is a divisor of any integer and 0 is a multiple of

any integer. Any non-zero integer is a divisor or multiple of itself.

Definition 1.1.4 (even number and odd number). A number which is multiple of 2 is
called an even number, otherwise it is called an odd number.
Theorem 1.1.1. For any integer a, b, c the following hold:
(i) a | b ⇒ a | bc

(ii) a | b and b | c ⇒ a | c

(iii) a | b and b | a ⇒ a = ±b

(iv) a | 1 ⇒ a = ±1

3
 (v) a | b and a | c ⇒ a | (bx + cy) for any integers x and y.

The sum, difference, and product of two integers are obviously integers but the quotient

of two integers may or may not be an integer.

Theorem 1.1.2 (Fundamental Theorem of Divisibility ). For any integers a, b(b 6=

0), there exist unique integers q and r such that

a = qb + r, 0 ≤ r < |b|.

Proof. The integer a lies between two consecutive integers of the sequence

· · · , −2|b|, −|b|, 0, |b|, 2|b|, · · ·

So WLOG we may assume that q|b| ≤ a < (q + 1)|b|. Then a − q|b| ≥ 0 and a − q|b| < |b|.

Let a − q|b| = r. Then 0 ≤ r < |b| implies that

a = qb + r when b > 0

= (−q)b + r when b < 0

Hence the existence of q and r is proved.

For uniqueness, let a = q1 b + r1 , 0 ≤ r1 < |b|. Then qb + r = a = q1 b + r1 , 0 ≤

|r1 − r| < |b|.

⇒ (q − q1 )b = r1 − r

⇒ |q − q1 ||b| = |r1 − r|

⇒ |q − q1 ||b| < |b|

⇒ |q − q1 | < 1
Since q and q1 are both integers, so the above relation hold if q − q1 = 0 ⇒ q = q1 .

Consequently, 0 = r1 − r ⇒ r = r1 .

Thus, q and r are unique.

Hence the theorem.

Definition 1.1.5 (Prime number). Any integer p which exceeds unity is called a prime
if it has no integral divisors except ±p and ±1.

For example 2, 3, 5, etc. are primes.

4
Definition 1.1.6 (Composite number). An integer c > 1 which has divisors other than
±c and ±1 is called a composite number.

For example 6, 30, 45, etc. are composite numbers.

Note. 1 is neither prime nor composite.

1.2 Greatest Common Divisor and Least Common

Multiple
Definition 1.2.1 (Greatest Common Divisor(g.c.d. )). The greatest integer g that di-
vides both of two integers a and b is called the g.c.d. of a and b and is denoted by
(a, b) = g.

For example (20, 30) = 10, (−6, 9) = 3. Note that (a, b) ≥ 1, (a, 0) = |a|, (0, 0) = ∞ and,

(a, a) = a for any integer a 6= 0.

Definition 1.2.2 (Relatively prime). If (a, b) = 1 then the integers a and b are called
relatively prime or co-prime.

Theorem 1.2.1 ( Euclidean Algorithm(330-275 B.C.)). Let a and b be two positive inte-
gers where b - a. Let r0 = a, r1 = b and apply the division algorithm repeatedly to obtain
a set of remainders r2 , r3 , · · · , rn , rn+1 defined successively by the relations:

r 0 = q1 r 1 + r 2 , 0 < r 2 < r 1
r 1 = q2 r 2 + r 3 , 0 < r 3 < r 2
r 2 = q2 r 3 + r 4 , 0 < r 4 < r 3
··· ··· ··· ··· ···
rn−2 = qn−1 rn−1 + rn , 0 < rn < rn−1
rn−1 = qn rn + rn+1 , where rn+1 = 0.

Then rn , the last nonzero remainder in this process, is (a, b), the g.c.d. of a and b.

Proof. Since r1 , r2 , · · · , rn+1 are decreasing and nonnegative, so there is a stage at

which rn+1 = 0. The last relation rn−1 = qn rn shows that rn | rn−1 and hence the

penultimate relation implies that rn | rn−2 . By induction we see that rn | ri , ∀i. In

particular, rn | r1 = b and rn | r0 = a, so rn is a common divisor of a and b.

Now let d be any common divisor of a and b. Then d | a = r0 and d | b = r1 . The

definition of r2 shows that d | r2 . The next relation shows that d | r3 . By induction,

d | ri , ∀i, so d | rn . This shows that rn = d.

5
 Hence rn is the required g.c.d. of a and b.

The following theorem states that the g.c.d. of any two integers can be expressed as

an integral linear combination of them.

Theorem 1.2.2. If g = (a, b), then there exists integers x and y such that g = ax + by.

Proof. Consider the linear combinations au + bv, where u, v ∈ Z. This set of integers

includes positive and negative values, and 0 by the choice of u = v = 0. Choose x and y

so that ax + by is the least positive integer l in the set. Thus ax + by = l.

Now we prove that l | a and l | b. Suppose that l - a. Then there exist integers q and

r such that

a = ql + r, 0 < r < l.

⇒ r = a − ql = a − q(ax + by) = a(1 − qx) + b(−qy).

Thus r is in the set {au + bv}. This contradicts the fact that l is the least positive integer

in the set {ax + by}. Thus l | a. Similarly, we can show that l | b.

Now, since g is the g.c.d. of a and b, we may write a = gA, b = gB, and l = ax + by =

g(Ax + By). Thus, g | l and so we conclude that g ≤ l. Now g < l is impossible as g is

the g.c.d. of a and b. Thus, g = l = ax + by.

Corollary 1.2.3. (a, b) = 1 ⇐⇒ ∃ x, y ∈ Z such that ax + by = 1.

Example 1.2.1. 1. Find the g.c.d. of 1769 and 2378 using Euclidean algorithm

2. Hence express it as an integral linear combination of these numbers.

Solution: 1. We have:
2378 = 1 · 1769 + 609
1769 = 2 · 609 + 551
609 = 1 · 551 + 58
551 = 9 · 58 + 29
58 = 2 · 29
Hence by the Euclidean algorithm (1769, 2378) = 29.

6
2. Now
29 = 551 − 9 · 58
= 551 − 9 · (609 − 1 · 551)
= 10 · 551 − 9 · 609
= 10(1769 − 2 · 609) − 9 · 609
= 10 · 1769 − 29 · 609
= 10 · 1769 − 29 · (2378 − 1 · 1769)
= 39 · 1769 + (−29) · 2378
∴ 29 = 39 · 1769 + (−29) · 2378.

Definition 1.2.3 (Least Common Multiple). The least common multiple(l.c.m. ) of two
or more non zero integers is the smallest positive integer that is divisible by all of them.
If l is the l.c.m. of a and b, we denote it by [a, b] = l.

For example [20, 30] = 60. Note that there is no greatest common multiple of two or

more integers. Also observe that g.c.d. and l.c.m. are unique.

Theorem 1.2.4. If ab > 0, then [a, b](a, b) = ab.

Proof. Let [a, b] = l and (a, b) = g. Then a | l and b | l gives ab | la and ab | lb

∴ ab | (la, lb)

⇒ ab | l(a, b)

⇒ ab | lg · · · (1)
ab ab ab
On the other hand, since a | g
and b | g
, so g
is a common multiple of a and b. Therefore,
ab
l| g
⇒ lg | ab · · · (2) From (1) and (2) we conclude that lg = ab ⇒ [a, b](a, b) = ab.

¯
¯
Note. [a, b]¯abm where m is an integer,that is, l.c.m. of any two or more integers divides
any common multiple of them.

Theorem 1.2.5. Let p be a prime and a be any integer. Then either p | a or (a, p) = 1.

Proof. If p | a, then there is nothing to prove. If p - a, then we shall prove that

(a, p) = 1. Let (a, p) = g. Then g | a and g | p. But p is a prime, so g | p gives g = 1 or

g = p. If g = p, then g | a =⇒ p | a, which is not possible.

Hence g = 1 and (a, p) = 1.

Theorem 1.2.6. If p is a prime and p | ab, then p | a or p | b.

7
Proof. If p | a, then there is nothing to prove. If p - a, then (a, p) = 1. Since p | ab and

(a, p) = 1, so p | b.

Theorem 1.2.7 (Euclid’s Lemma). If a | bc and (a, b) = 1, then a | c.

Proof. Since (a, b) = 1, there exist integers x and y such that

ax + by = 1.

Since a | bc ⇒ bc = am, m ∈ Z.

Then c = c · 1 = c(ax + by) = acx + bcy

= acx + amy

= a(cx + my)

Hence a | c

1.3 Factorization in Prime Numbers

Theorem 1.3.1 (Fundamental Theorem of Arithmetic or Unique Factorization Theo-
rem). Every positive integer N > 1 can be expressed product of primes in one and only
one way if we do not distinguish between two same prime factors.

Proof. Let p1 > 1 be the least divisor of N . Then p1 < N . Evidently, p1 is a prime.

We write N = p1 N1 . If N1 is a prime, N has been expressed as a product of two primes

not necessarily distinct.

If N1 is composite, its least divisor p1 > 1 is a prime. We write N1 = p2 N2 , and

proceed with N2 as before. After a finite number of such steps we obtain a factorization

N = p1 p2 p3 · · · pn

of N into primes. Suppose that

N = q1 q2 q3 · · · qr

is a second factorization of N .

8
 Then the prime q1 evidently divides one of the primes pi , say p1 . Hence q1 = p1 and

q2 q3 · · · qr = p2 p3 · · · pn . Similarly q2 is equal to one of the factors on the right, say p2 .

Proceeding in this manner we conclude that r = n and that q1 , q2 , q3 , · · · , qr are identical

with p1 , p2 , p3 , · · · , pn in some order.

Corollary 1.3.2. An integer n > 1 can be uniquely written as n = pα1 1 pα2 2 pα3 3 · · · pαk k
where p1 , p2 , p3 , · · · , pk are distinct primes and α1 , α2 , α3 , · · · , αk are positive integers.

Proof. By Theorem 1.3.1 we have

n = p1 p2 p3 · · · pr

Since the prime factors are not necessarily distinct, by collecting the same primes we can

write the above equation as

n = pα1 1 pα2 2 pα3 3 · · · pαk k .

This representation is called the ::::::::::::

canonical:::::::::::::::::::
representation or the standard representation

or the standard factorization of n.

Motivation:. N = 2 · 3 · 5 = 30
P = N + 1 = 30 + 1 = 31, not divisible by the prime in the composition of P .

Theorem 1.3.3 (Euclid). The number of primes is infinite.

Proof. Let there are only a finite number of primes, say p1 , p2 , p3 , · · · , pk , where pk is

the largest prime. We define

N = p1 p2 p3 · · · pk

⇒N + 1 = p1 p2 p3 · · · pk + 1

⇒P = p1 p2 p3 · · · pk + 1, where P = N + 1.

Now if P is a prime then pk is not the largest prime. If P is not a prime, then by

Fundamental Theorem of Arithmetic, P can be factored into prime factors. So there

exist a prime p such that p | P .

9
 Since there are only prime p1 , p2 , p3 , · · · , pk , so p will be one of them, say p = pi , 1 ≤

i ≤ k. Now p | P ⇒ pi | P , but this impossible by definition of P . Hence it is a

contradiction. Thus the number of primes is infinite.

Example 1.3.1. Show that one of the two consecutive integers is divisible by 2.
Solution: Let the two consecutive integers be n, n + 1. If n be odd, then n + 1 is even
and is divisible by 2. If n is even then it is divisible by 2.

Example 1.3.2. Show that one of the three consecutive integers is divisible by 3.
Solution: Let n, n + 1, n + 2 be any three consecutive integers. Here n must be any
one of the forms 3m, 3m + 1, 3m + 2. When n = 3m, the first integer is divisible by 3,
when n = 3m+1, n+2 = 3m+3 which is divisible by 3, when n = 3m+2, n+1 = 3m+3
which is divisible by 3.

Example 1.3.3. Show that the product of any three consecutive integers is divisible by
6.
Solution: Let n be any integers. Then n(n + 1)(n + 2) is divisible by 6.
a5 −
1 =
Example 1.3.4. If 2p − 1 is a prime, then show that p is itself a prime.
(a−1).
Solution: Suppose that p is not a prime. Then it is composite, say p = ab. Then
(a4 +
2p − 1 = 2ab − 1 = (2a )b − 1 which is divisible by 2a − 1. So, 2ab − 1 or 2p − 1 is a composite 3
a + 1)
number which contradicts our given assumption.
So p must be a prime.

Example 1.3.5. If 2n + 1 is a prime, then n is a power of 2.

Solution: If n is not a power of 2, then suppose that n = 2m × q where q is odd and
m m m
greater than 1. Now 22 ×q + 1 = (22 )q + 1 which is divisible by 22 + 1. This contradicts
that 2n + 1 is a prime. Hence n must be a power of 2.

10
 Problem Plus 1

1. If a = qb + r then prove that (a, b) = (b, r).

2.

11
Chapter 2

Arithmetic Functions and

Diophantine Equations

2.1 Arithmetic Functions

Definition 2.1.1 (Arithmetic Function). A function f : N → C, whose domain is the
set of natural numbers and range is the subset of the set of complex numbers, is called the
arithmetic function, number theoretic function or numerical function.

Definition 2.1.2 (Totally Multiplicative Function). An arithmetic function f is called

completely multiplicative or totally multiplicative if f (ab) = f (a)f (b). Again f is called
multiplicative if f (ab) = f (a)f (b) where (a, b) = 1. If f is multiplicative then f (1) = 1.

Definition 2.1.3. The function

X
τ : N → N defined by τ (n) = 1
d|n
d≥1

denotes the number of all positive divisors of n, where n is a positive integer.

Note. Some authors write d(n) for τ (n).

Definition 2.1.4. The function

X
σ : N → N defined by σ(n) = d
d|n
d≥1

denotes the sum of all positive divisors of n, where n is a positive integer.

Definition 2.1.5. The function

Y
P : N → N defined by P (n) = d
d|n
d≥1

denotes the product of all positive divisors of n, where n is a positive integer.

12
Definition 2.1.6. The function
X
σk : N → N defined by σk (n) = dk
d|n
d≥1

denotes the sum of the k-th powers of the divisors of n, where n is a positive integer.

Example 2.1.1.PLet n=6. The divisors of 6 are d : 1, 2, 3, 6.

σ1 (n) = σ(n) = d|n=6 d = 1 + 2 + 3 + 6 = 12
0
σ0 (n) = 1P + 20 + 30 + 60 = 1 + 1 + 1 + 1 = 4= τ (n)
σ3 (n) = d|n=6 d3 = 13 + 23 + 33 + 63 = 1 + 8 + 27 + 216 = 252

Remark. If n is a prime, then n has only two positive divisors, 1 and n itself.

∴ τ (n) = 2, σ(n) = n + 1, and P (n) = n.1 = n

Definition 2.1.7 (Euler’s φ function). Let n ≥ 1 be a positive integer. Then φ(n) denotes
the number of numbers not exceeding n, which are relatively prime to n. This φ(n) is
called Euler’s φ function.

Example 2.1.2. The functions τ, σ, σk , and φ are all arithmetic functions.

Example 2.1.3. Following table lists τ (n), σ(n) and P (n) for n = 1, 2, 3, · · · , 10.

n 1 2 3 4 5 6 7 8 9 10
τ (n) 1 2 2 3 2 4 2 4 3 4
σ(n) 1 3 4 7 6 12 8 15 13 18
P (n) 1 2 3 8 5 36 7 64 27 100
φ(n) 1 1 2 2 4 2 6 4 6 4

τ (n), σ(n), P (n), and φ(n)

Definition 2.1.8 (Gauss Function(Greatest Integer Function)). Let x be a real number.

Then we denote by [x], the greatest integer not greater than x, and call it a Gauss function.
£1¤ £ −3 ¤
Example 2.1.4.
£ ¤ [5] = 5, [5.25] = 5, 5
= 0, 2
= −2, [π] =
3, [x] = [x]

For each real number x, we can write

x = [x] + a, where 0 ≤ a < 1

Here [x] and a are respectively called the integral and fractional part of x.

Theorem 2.1.1 (Goldbach Conjecture). This states that every even integer bigger than
2 is the sum of two primes.

13
Example 2.1.5.
4=2+2
6=3+3
8=3+5
10 = 3 + 7
50 = 3 + 47
100 = 29 + 71
It is an unsolved problem. Goldbach verified up to 1,00,000 at least.

Now let us provide the general formulae for τ and σ when n is a composite number

as follows:
Theorem 2.1.2. If n = pα1 1 pα2 2 pα3 3 · · · pαk k where p1 , p2 , p3 , · · · , pk are distinct primes and
α1 , α2 , α3 , · · · , αk are positive integers, then

k
Y
(i) τ (n) = (α1 + 1)(α2 + 1) · · · (αk + 1) = (αi + 1)
i=1
k
pα1 +1 − 1 pα2 2 +1 − 1 pαk +1 − 1 Y pαi i +1 − 1
(ii) σ(n) = 1 · ··· k =
p1 − 1 p2 − 1 pk − 1 i=1
pi − 1
Proof. We shall prove the theorem by induction on k. Let k = 1, that is, n has only

one prime factor, say n = pα , where α is a positive integer. Then the positive divisors of

n are 1, p, p2 , · · · , pα . We have
pα+1 − 1
τ (n) = α + 1 and σ(n) = 1 + p + p2 + · · · + pα =
p−1
Next suppose that (i) and (ii) hold when n has k − 1 distinct prime factors(k ≥ 2).
α
Write n = n0 pαk k where n0 = pα1 1 pα2 2 pα3 3 · · · pk−1
k−1
, (n0 , pαk k ) = 1, and all p’s are distinct

primes. Moreover, any divisor of n0 is a divisor of n. Hence any divisor of n is of the form

d0 ptk , 0 ≤ t ≤ αk , d0 ≥ 1, d0 | n0 .
X
Now τ (n) = 1
d|n
d≥1
X X X X
= 1+ 1+ 1 + ··· + 1
¯
d0 pk k ¯n
d0 |n d0 pk |n d0 p2k |n α

= τ (n0 ) + τ (n0 ) + τ (n0 ) + · · · + τ (n0 )

| {z }
αk +1 terms
0
= τ (n )(αk + 1)

14
 By induction hypothesis we have τ (n0 ) = (α1 + 1)(α2 + 1) · · · (αk−1 + 1)
k
Y
∴ (i) τ (n) = (α1 + 1)(α2 + 1) · · · (αk−1 + 1)(αk + 1) = (αi + 1)
i=1

X
Again σ(n) = d
d|n
d≥1
X X X
= d0 + d0 pk + · · · + d0 pαk k
d0 |n0 d0 |n0 d0 |n0
d0 ≥1 d0 ≥1 d0 ≥1
     
X 0 X 0  X 0  αk
= d+ d  pk + · · · +  d  pk
d0 |n0 d0 |n0 d0 |n0
d0 ≥1 d0 ≥1 d0 ≥1
 
X 0
= d  (1 + pk + · · · + pαk k )
d0 |n0
d0 ≥1

pαk k +1 − 1
0
= σ(n )
pk − 1

By induction hypothesis
α +1
pα1 1 +1 − 1 pα2 2 +1 − 1 p k−1 − 1
σ(n0 ) = · · · · k−1
p1 − 1 p2 − 1 pk−1 − 1

α +1 k
pα1 1 +1 − 1 pα2 2 +1 − 1 p k−1 − 1 pαk k +1 − 1 Y pαi i +1 − 1
∴ (ii) σ(n) = · · · · k−1 · =
p1 − 1 p2 − 1 pk−1 − 1 pk − 1 i=1
pi − 1

k k k−1
Theorem
¡ ¢ 2.1.3. If p is a prime and k is a positive integer, then φ(p ) = p − p =
k 1
p 1− p .

Proof. The positive integers less than or equal to pk which are not relatively prime to pk

are multiples of p, i.e., these are p, 2p, 3p, · · · , pk−1 p. Hence deleting these pk−1 numbers

from pk numbers: 1, 2, 3, · · · , pk we obtain

φ(pk ) = pk − pk−1 ,

positive integers not exceeding pk which are relatively prime to pk .

15
Corollary 2.1.4. If p is a prime, then φ(p) = p − 1.

Theorem 2.1.5. ( Prof. Dr. Md. Fazlur Rahman [?]) The number theoretic functions
τ, σ and, φ are multiplicative.

Theorem 2.1.6. If a and b are relatively prime positive integers, then φ(ab) = φ(a)φ(b).

Motivation:. Assume (a, b) = (3, 4) = 1, ab = 12 and consider the following ab numbers

arranged in b rows and a columns

a − columns
1 k=2 3 → φ(a) = 2
b − rows 4 5 6
7 8 9
10 11 12
↓
φ(b) = 2

Motivation for Theorem 2.1.6

Proof. Considering the product ab, we write the first ab integers in b rows and a

columns. Thus we get

1 2 3 4 ··· k ··· a
a+1 a+2 a+3 a+4 ··· a+k ··· 2a
2a + 1 2a + 2 2a + 3 2a + 4 ··· 2a + k ··· 3a
3a + 1 3a + 2 3a + 3 3a + 4 ··· 3a + k ··· 4a
··· ··· ··· ··· ··· ··· ··· ···
(b − 1)a + 1 (b − 1)a + 2 (b − 1)a + 3 (b − 1)a + 4 ··· 3(b − 1)a + k ··· ba

Now considering the k-th column we find that if (k, a) = 1, then every number in k-th

column is prime to a.

Now the first row contains φ(a) integers prime to a and therefore there are φ(a)

columns in each of which every term is prime to a. Let k-th column be such one. Clearly

this column is in A.P.(Arithmetic Progression). So, the terms of this column when divided

by b will leave the remainders 0, 1, 2, · · · , (b − 1). Hence this column contains φ(b) terms

prime to b.

Thus, there are φ(a) columns of numbers prime to a and each column contains φ(b)

numbers prime to b.

16
 Hence in the above arrangements of ab numbers there are φ(a)φ(b) numbers which

are prime to both a and b and so to the product ab. Therefore φ(ab) = φ(a)φ(b). So the

theorem is established.

Corollary 2.1.7. If a, b, c, · · · , l are prime to each other, then φ(abc · · · l) = φ(a)φ(b)φ(c) · · · φ(l).

Corollary 2.1.8. If p1 , p2 , p3 , · · · , pk are distinct prime factors of a positive integer n,

then φ(n) = n(1 − p11 )(1 − p12 )(1 − p13 ) · · · (1 − p1k ).

Corollary 2.1.9. If n = pα1 1 pα2 2 pα3 3 · · · pαk k where pi ’s are relatively prime in pairs, then
φ(n) = n(1 − p11 )(1 − p12 )(1 − p13 ) · · · (1 − p1k ).

Example 2.1.6. Find φ(300).

Solution: φ(300) = φ(3.22 .52 ) = φ(3)φ(22 )φ(52 ) = 300 · (1 − 1/3)(1 − 1/2)(1 − 1/5) =
300 · 2/3 · 1/2 · 4/5 = 80

Definition 2.1.9 (Mobius Function). The mobius function µ is defined as

µ : N → {−1, 0, 1}

In other words,


1 if n = 1
µ(n) = 0 if n has a square factor or a2 | n for some a ∈ Z


(−1)k if n = p1 p2 · · · pk where p1 , p2 , · · · , pk are distinct prime factors

Example 2.1.7.

µ(2) = (−1)1 = −1, µ(4) = µ(22 ) = 0, µ(6) = µ(2.3) = (−1)2 = 1

Theorem 2.1.10 (F. Merten’s Lemma). For each integer n ≥ 1

(
X 1 if n = 1
µ(d) =
d|n
0 if n > 1

Proof. If n = 1 then by definition

X X
µ(d) = µ(d) = µ(1) = 1.
d|n d|1

If n > 1 then by Fundamental theorem of arithmetic 1.3.1 let

n = pα1 1 pα2 2 · · · pαr r

17
where pi are distinct primes, αi ≥ 1, i = 1, 2, 3, · · · , r.
X
Therefore µ(d) = µ(1) + [µ(p1 ) + µ(p2 ) + · · · + µ(pr )]
d|n

+ [µ(p1 p2 ) + µ(p2 p3 ) + · · · + µ(pr−1 pr )]

+ · · · + µ(p1 p2 · · · pr )

+ 0 + · · · + 0, (the remaining all terms vanish since they contain

square factors)
Xr r
X
= µ(1) + µ(pi ) + µ(pi pj ) + · · · + µ(p1 p2 · · · pr )
i=1 i<j
i,j=1

= 1 + r(−1) +r C2 (−1)2 + · · · + (−1)r

= 1 +r C1 (−1) +r C2 (−1)2 + · · · + (−1)r

= (1 − 1)r

=0
(
X 1 if n=1
∴ µ(d) =
d|n
0 if n>1

Theorem 2.1.11. If (a, b) = 1, then µ(ab) = µ(a)µ(b).

Proof. If a = b = 1, then the proof is straightforward. If a has a square factor, then ab

has a square factor, so µ(a) = 0, µ(a)µ(b) = 0 and µ(ab) = 0 and hence µ(ab) = µ(a)µ(b).

Similar result holds if b has a square factor.

Now let a = p1 p2 · · · pk and b = q1 q2 · · · ql . Since (a, b) = 1, so all the factors of a

and b are distinct. Therefore µ(a) = (−1)k , µ(b) = (−1)l and µ(ab) = (−1)k+l . Now

µ(ab) = (−1)k+l = (−1)k (−1)l = µ(a)µ(b).

Thus in all cases we have µ(ab) = µ(a)µ(b).

Note. The mobius function µ is not completely multiplicative since µ(4) = µ(22 ) = 0 but
µ(2) · µ(2) = (−1)2 = 1.

18
2.2 Diophantine Equations

An equation which has more than one unknowns is called an indeterminate equation. A

system of equations is called indeterminate if the number of equations is less than that

of the unknowns. Indeterminate equations are called Diophantine equations, after the

name of Greek mathematician Diophantus about 200 A.D., when we seek their solutions

in integers. In this section we shall only consider the linear diophantine equations in two

variables e.g., f (x, y) = ax + by + c = 0.

Definition 2.2.1 (Diophantine Equations). Let a, b, c ∈ Z with ab 6= 0. Then any linear

equation of the form ax + by = c, where the values of x and y are restricted to the set of
integers Z, is called the linear diophantine equation.

The following theorem gives the information when the linear diophantine equation

has solution or not.

Theorem 2.2.1. The linear diophantine equation ax + by = c has a solution if and only
if d | c where d = (a, b).

Proof. First suppose that ax + by = c · · · (1) has a solution (x0 , y0 ). Then ax0 +

by0 = c · · · (2) Since (a, b) = d, so ax0 + by0 is a multiple of d. Therefore ax0 + by0 =

rd, (say) · · · (3) where r ∈ Z. From (2) and (3) we have

c = td ⇒ d | c.

Conversely, suppose that d | c. Then c = td where t ∈ Z · · · (4). Also (a, b) = d gives

d = au + bv, for some u, v ∈ Z ⇒ td = atu + btv · · · (5). From (4) and (5) we have

c = atu + btv ⇒ a(tu) + b(tv) = c. Comparing this with the given equation ax + by = c

we have x = tu, y = tv is a solution.

Theorem 2.2.2. If (x0 , y0 ) is a particular solution of the linear diophantine equation

ax + by = c then any other solution(general solution) of the equation is

b a
x = x0 + t, y = y0 − t,
d d
where t ∈ Z and d := (a, b) | c.

19
Proof. We first show that the expressions x = x0 + db t, y = y0 − ad t represent solutions

of the linear diophantine equation ax + by = c · · · (1). Substituting the values of x

and y in (1), we get

¡ b¢ ¡ a¢
a x0 + t + b y0 − t = c
d d
ax0 + by0 = c
Since (x0 , y0 ) is a particular solution of the linear diophantine equation ax + by = c, the

equation ax + by = c is satisfied by

b a
x = x0 + t, y = y0 − t,
d d

where t ∈ Z.

Now since (x, y) and (x0 , y0 ) are solutions of ax + by = c, so ax + by = c and

ax0 + by0 = c. Therefore

ax + by = ax0 + by0

a(x − x0 ) = b(y0 − y)
a b
(x − x0 ) = (y0 − y)
d d
b a
¡ ¢
So d
| x − x0 and d
| y0 − y as ad , db = 1. It follows that x − x0 = db t, y0 − y = ad t and

hence
b a
x = x0 + t, y = y0 − t,
d d
where t ∈ Z.

Corollary 2.2.3. If (a, b) = 1 and (x0 , y0 ) is a particular solution of the linear diophan-
tine equation ax + by = c, then its general solutions are given by x = x0 + bt, y = y0 − at
where t ∈ Z.
Example 2.2.1. Find the positive integral solutions of the linear diophantine equation
172x + 20y = 1000.
Solution: Here a = 172, b = 20, c = 1000, and d := (a, b) = (172, 20) = 4 obtained
by the following Euclidean algorithm:
172 = 20 · 8 + 12
20 = 12 · 1 + 8
12 = 8 · 1 + 4
8=4·2+0

Since d | c, a solution to the equation 172x + 20y = 1000 exists by Theorem 2.2.1.

20
 Using steps of Euclid’s algorithm from backward direction d = 4 can be expressed as
a linear combination of a = 172 and b = 20.

4 = 12 + (−1) · 8
= 12 + (−1) · {20 + (−1) · 12}
= 2 · 12 + (−1) · 20
= 2 · {172 + (−8) · 20} + (−1) · 20
= 172 · (2) + 20 · (−17)
∴ 172 · (500) + 20 · (−4250) = 1000

It follows that (x0 , y0 ) = (500, −4250) is a particular solution to the equation 172x+20y =
1000. Hence the general solutions given by Theorem 2.2.2 are:
b a
x = x0 + t, y = y0 − t, with t ∈ Z.
d d
That is, x = 500 + 5t, y = −4250 − 43t · · · (1)
Now the positive integral solutions can be obtained by considering the system of in-
equalities
4x = 500 + 5t > 0 and y = −4250 − 43t > 0
500 4250 36
⇒t>− = −100 and t < − = −98
5 43 43
36
⇒ −100 < t < −98
43
Since t is an integer so t = −99. Hence from (1)

x = 500 + 5(−99) = 5 and y = −4250 − 43(−99) = 7

Therefore (x, y) = (5, 7) is the only positive solution of the linear diophantine equation
172x + 20y = 1000.

21
 Problem Plus 2

1. Evaluate σ(τ (φ(7))). [Ans:7]

2. Calculate the value of Euler φ function at 30, 50, 100, 120, 240, and 264.

3. The number theoretic functions τ and σ are multiplicative.

4. Find the positive integral solutions of the linear diophantine equation 18x+5y = 48.

5. Determine the particular solution of the linear diophantine equation 39x + 26y =

105.

6. A man presented a check for Taka 510 to a bank and wished to be paid in 20 and

50 taka notes. In how many ways and means can the cashier meet his request.

7.

22
Chapter 3

Congruences

3.1 Congruences and Its Properties

The concept of congruences was introduced by Carl Friedrich Gauss about 1800, one of

the greatest mathematician of all time. Congruence often arises in every day life. For

instance, if the second of January is Sunday, then 9,16,23 of the same month are all

Sundays, since when they are divided by 7, the remainders are all 2.

A congruence is nothing but a refined statement about divisibility. Congruence can

be treated in the same manner as equations. Let m be a positive integer. Two integers

a and b leave remainders when divided by m. If the remainders are same we say that a

is congruent to be modulo m, and we write

a ≡ b(mod m) or m | a − b.

Alternatively,

Definition 3.1.1 (Congruence). If the difference of two integers a and b is divisible by

m, we shall say that a is congruent to b modulo m, and we write

a ≡ b(mod m) or m | a − b.

Definition 3.1.2 (Incongruence). If the difference of two integers a and b is not divisible
by m, we shall say that a is incongruent to b modulo m, and we write

a 6≡ b(mod m) or m - a − b.

Example 3.1.1. 7 ≡ 3(mod 2), 9 6≡ 4(mod 3), and 13 ≡ 3(mod 5).

Theorem 3.1.1. If a ≡ b(mod m) and c ≡ d(mod m), then

23
 (i) a + c ≡ b + d(mod m)

(ii) a − c ≡ b − d(mod m)

(iii) ac ≡ bd(mod m)

Proof. Here a − b and c − d are both multiples of m, say a − b = t1 m and c − d = t2 m.

(i) (a−b)+(c−d) = (a+c)−(b+d) = (t1 +t2 )m = t3 m, say. So, a+c ≡ b+d(mod m).

(ii) (a−c)−(b−d) = (a−b)−(c−d) = (t1 −t2 )m = t4 m, say. So, a−c ≡ b−d(mod m).

(iii) a − b = t1 m ⇒ (a − b)c = t5 m where t5 = t1 c. So, ac − bc = t5 m · · · (A)

c − d = t2 m ⇒ (c − d)b = t6 m where t6 = t2 b. So, bc − bd = t6 m · · · (B) Then

from (A) and (B), ac − bc + bc − bd = (t5 + t6 )m ⇒ ac − bd = t7 m, say. So,

ac ≡ bd(mod m).

Theorem 3.1.2 (transitive law). If a ≡ b(mod m) and b ≡ c(mod m), then a ≡

c(mod m)

Theorem 3.1.3. a ≡ b(mod m) ⇒ na ≡ nb(mod m), but the converse need not hold.

e.g., 4 · 7 ≡ 4 · 2(mod 10) ⇒ 7 6≡ 2(mod 10) but 7 ≡ 2(mod 5).

Theorem
¡ ¢ 3.1.4 (Cancelation law). If na ≡ nb(mod m) and (n, m) = g, then a ≡
m
b mod g

Proof. Since (n, m) = g, we have n = gN and m = gM where (M, N ) = 1. Given that

m | n(a − b) ⇒ gM | gN (a − b) ⇒ M | N (a − b) ⇒ M | a − b Therefore, a ≡ b(mod M )

¡ ¢
and hence a ≡ b mod mg .

Corollary 3.1.5. If a, b, g are integers such that na ≡ nb(mod m) and (m, n) = 1, then
a ≡ b(mod m).

Definition 3.1.3. Let a, b(b > 0) be any two integers. Then ∃ integers q, r such that
a = bq + r where 0 ≤ r < b. Then r is called the least residue of a modulo m.

Here r := 0, 1, 2, · · · , b − 1 form a complete set of least residues modulo b.

24
 r, a + r, 2a + r, 3a + r, 4a + r are

1, 4, 2, 0, 3 respectively.

Motivation:. Let r = 6 be any integer, a = 3, b = 5, and (a, b) = 1. Then the least

residues modulo b of

Theorem 3.1.6 (Least Residue Theorem). If a and b(b > 0) are relatively prime and if
r is any integer, then the least residues modulo b of

r, a + r, 2a + r, 3a + r, · · · , (b − 1)a + r · · · (1)

are
0, 1, 2, 3, · · · , (b − 1) · · · (2)
in some rearranged order.

Proof. By definition, the least residue modulo b of any integer is one of the numbers

(2). There are b numbers (1) and b numbers (2). The theorem will be established if we

can show that no two of the numbers (1) have the same least residue (2). Any two of the

numbers of (1) may be denoted by

sa + r, 0 ≤ s < b and ta + r, 0 ≤ t < b with s > t.

If they have the same least residue modulo b, they are congruent. Then

sa + r ≡ ta + r(mod b)

sa ≡ ta(mod b)

s ≡ t(mod b), since (a, b) = 1

But s − t is positive and less than b so that it is not divisible by b. This contradiction

shows that no two distinct numbers (1) have the same least residue modulo b. Hence the

theorem is proved.

Corollary 3.1.7. If (a, b) = 1, then a, 2a, 3a, · · · , (b − 1)a are congruent modulo b to
1, 2, 3, · · · , (b − 1) in some order.

Proof. Since (a, b) = 1, no one of ia is divisible by b; i := 1, 2, 3, · · · , b − 1. So, no

residue is zero. Thus, the corollary is established.

25
3.2 Fermat’s Theorem, Euler’s Theorem and Wil-
son’s Theorem
Theorem 3.2.1 (Fermat’s Theorem). If p is a prime and (a, p) = 1, then ap−1 ≡
1(mod p).

This theorem was conjectured by Fermat and was first proven by Euler in 1736. It is

also called Fermat’s little theorem.

Proof. By the least residue theorem the integers a, 2a, 3a, · · · , (p − 1)a are congruent

modulo p to 1, 2, 3, · · · , p − 1 in some order. The product of the first set is congruent to

that of the second set. Then

ap−1 1 · 2 · 3 · · · · · p − 1 ≡ 1 · 2 · 3 · · · · · p − 1(mod p)

ap−1 (p − 1)! ≡ (p − 1)!(mod p)

ap−1 ≡ 1(mod p), since ((p − 1)!, p) = 1

Hence the theorem is proved.

Corollary 3.2.2. If p is a prime and a is any integer, then ap ≡ a(mod p).

Note. The converse of the Fermat’s theorem is not true, i.e. if ap−1 ≡ 1(mod p), p need
not be prime.

Motivation:. The positive integers not exceeding m = 8 and relatively prime to m = 8

are a := 1, 3, 5, and 7. Now construct the following congruences with a = 3:
3.1 ≡ 3, 3.3 ≡ 1, 3.5 ≡ 7, 3.7 ≡ 5(mod m)
After multiplying we get:

34 .1.3.5.7 ≡ 3.1.7.5(mod m)
⇒34 ≡ 1(mod m), since (1.3.5.7, m) = 1
∴ 3φ(8) ≡ 1(mod m)
⇒aφ(m) ≡ 1(mod m) where (a, m) = 1.

This result is known as the Euler’s theorem.

Theorem 3.2.3 (Euler’s Theorem). If m be any positive integer and a be any integer
prime to m, i.e. (a, m) = 1, then aφ(m) ≡ 1(mod m).

Proof. Let a1 , a2 , a3 , · · · , aφ(m) be the positive integers, in ascending order, less than m

and prime to m. Now consider the products aa1 , aa2 , aa3 , · · · , aaφ(m) . Since (a, m) = 1,

26
it follows that aa1 , aa2 , aa3 , · · · , aaφ(m) are congruent modulo m, not necessarily in order

of appearance to a1 , a2 , a3 , · · · , aφ(m) .

Then
aa1 ≡ a01 (mod m)

aa2 ≡ a02 (mod m)

aa3 ≡ a03 (mod m)

... ... ...

aaφ(m) ≡ a0φ(m) (mod m)

where a01 , a02 , a03 , · · · , a0φ(m) are the integers a1 , a2 , a3 , · · · , aφ(m) in some order.

On taking the product of these φ(m) congruences, we get

aa1 · aa2 · aa3 · · · · aaφ(m) ≡ a01 · a02 · a03 · · · · · a0φ(m) (mod m)

≡ a1 · a2 · a3 · · · · · aφ(m) (mod m)
¡ ¢
⇒ aφ(m) a1 a2 a3 · · · aφ(m) ≡ a1 a2 a3 · · · aφ(m) (mod m)
¡ ¢
Since (ai , m) = 1 for each i, we have a1 a2 a3 · · · aφ(m) , m = 1 and hence by the Cancela-

tion law(Theorem 3.1.4) aφ(m) ≡ 1(mod m).

Corollary 3.2.4. Deduce Farmat’s theorem from Euler’s theorem.

Proof. Since (a, p) = 1, by Euler’s theorem aφ(p) ≡ 1(mod p). But if p is a prime, then

φ(p) = p − 1. Hence ap−1 ≡ 1(mod p).

Motivation:. If p = 11, each of the product of p−3 2

pairs, namely 2 · 6, 3 · 4, 5 · 9, 7 · 8 ≡
1(mod 11). Then 10! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 = (1 · 10)(2 · 6)(3 · 4)(5 · 9)(7 · 8) ≡
10(mod 11) ≡ −1(mod 11) ⇒ (p − 1)! ≡ −1(mod p).
Theorem 3.2.5 (Wilson’s Theorem). If p is a prime, then (p − 1)! ≡ −1(mod p).

Proof. Let a denote any one of the numbers 1, 2, 3, · · · , p − 1. Clearly (a, p) = 1, since

p is a prime. Then by the corollary of the least residue theorem 3.1.7 the integers

a, 2a, 3a, · · · , (p − 1)a · · · (1)

are congruent modulo p to

1, 2, 3, · · · , p − 1 · · · (2)

27
in some order. Hence there exists one and only one number in (1), say ka which will be

congruent to 1 with modulo p, i.e.,

ka ≡ 1(mod p) · · · (3).

If k = a then from (3) we have p | a2 − 1, i.e., p | a − 1 or p | a + 1. Since p is prime and

a < p, it follows that either a + 1 = p or a − 1 = 0 ⇒ a = 1, p − 1

For all other values of a, k differ from a. This means that for every a belonging to

the p − 3 numbers

2, 3, · · · , p − 2 · · · (4)

there exists one and only one number k also belonging to (4) but different from a such

that

ka ≡ ak ≡ 1(mod p).

p−3
Thus, the numbers in (4) can be arranged into 2
pairs such that the product of

the integers of each pair is congruent to 1(mod p). Taking product of all these pairs we

obtain
2 · 3 · · · · p − 2 ≡ 1(mod p)

⇒1 · 2 · 3 · · · (p − 2) · (p − 1) ≡ (p − 1)(mod p)

⇒(p − 1)! ≡ 1(mod p)

Proof. [Alternative] Let p be a prime and (x, p) = 1. Then by Farmat’s theorem we

have xp−1 − 1 ≡ 0(mod p). This equation has p − 1 roots with modulo p and the roots

are

1, 2, 3, · · · , p − 1

∴ xp−1 − 1 ≡ (x − 1)(x − 2)(x − 3) · · · {x − (p − 1)}(mod p). This congruent relation is

satisfied for all values of x. Letting x = 0 we have:

− 1 ≡ (−1)(−2)(−3) · · · {−(p − 1)}(mod p)

(−1)p−1 (p − 1)! ≡ −1(mod p) · · · (1)

28
Since p is a prime, so either p = 2 or p is an odd prime.

Case-I: If p = 2 then (1) implies that

−(2 − 1)! ≡ −1(mod 2)

−1 ≡ −1(mod 2)

−1 + 2 ≡ −1(mod 2)

(2 − 1)! ≡ −1(mod 2)

(p − 1)! ≡ −1(mod p)

Case-I: If p is an odd prime, then (−1)p−1 = 1 and hence (1) yields that (p − 1)! ≡

−1(mod p).

Theorem 3.2.6 (Converse of Wilson’s Theorem). If (p − 1)! ≡ −1(mod p), then p is a

prime.

Proof. Suppose that p is not a prime. Then p is composite and so there exist a divisor

d where 1 < d < p. Furthermore since d is a factor of 1 · 2 · 3 · · · (p − 2) · (p − 1), so we

can write
(p − 1)! ≡ 0(mod d)

(p − 1)! 6≡ −1(mod d)

(p − 1)! 6≡ −1(mod p)
This is a contradiction to the converse statement. It follows that p must be a prime.

29
 Problem Plus chapter 3

1. Congruence is an equivalence relation.

30
Chapter 4

Solutions of Congruences

4.1 Linear Congruences

Definition 4.1.1 (Polynomial congruence). If m - an i.e., (an , m) = 1 then the congru-
ence an xn + an−1 xn−1 + · · · + a1 x + a0 ≡ 0(mod m) is said to be a polynomial congruence
of degree n where a’s are restricted to the integers.

Definition 4.1.2 (Linear congruence). If m - a i.e., (a, m) = 1 then the congruence

ax ≡ b(mod m) is said to be a linear congruence where a, b ∈ Z.

Theorem 4.1.1. The linear congruence ax ≡ b(mod m) has a unique solution if (a, m) =
1.

Proof. Given (a, m) = 1 and ax ≡ b(mod m) · · · (1)

Therefore
aφ(n)−1 ax ≡aφ(n)−1 b(mod m) (multiplying (1) by aφ(n)−1 )

⇒ aφ(n) x ≡baφ(n)−1 (mod m) · · · (2)

But aφ(n) ≡1(mod m) (Euler’s theorem)

⇒ x ≡aφ(n) x(mod m) (multiplying both sides by x)

⇒ x ≡baφ(n)−1 (mod m) (using transitive law 3.1.2 on (2))

which is the required solution of the linear congruence ax ≡ b(mod m).

31
Uniqueness: Let x1 and x2 be two solutions of the linear congruence (1). Then

ax1 ≡ b(mod m) and ax2 ≡ b(mod m)

∴ax1 − ax2 ≡ 0(mod m)

⇒a(x1 − x2 ) ≡ 0(mod m)

⇒x1 − x2 ≡ 0(mod m) since (a, m) = 1

⇒x1 ≡ x2 (mod m)

Thus, the solution is unique.

Theorem 4.1.2. If (a, m) = g, g > 1 then the congruence ax ≡ b(mod m) has

(i) no root if g - b
m m
(ii) exactly g incongruent roots if g | b. The roots are x0 , x0 + g
, x0 +2· g
,··· , x0 +
(g − 1) · mg where x0 is the unique root of ag x ≡ gb (mod mg ).

Proof. Given (a, m) = g, g > 1

∴ a = gA, m = gM, where (A, M ) = 1 for some A, M ∈ Z

(i) Now ax ≡ b(mod m) ⇒ m | (ax − b) ⇒ gM | (gAx − b.) This will be possible

only if b has a factor g i.e., b = gB for some B ∈ Z ⇒ g | b. Thus, if g - b then

ax ≡ b(mod m) has no root.

(ii) If if g | b ⇒ b = gB for some B ∈ Z. Again

ax ≡ b(mod m)

⇒gAx ≡ gB(mod gM )

⇒Ax ≡ B(mod M ) · · · (1) which has a unique solution, say x0 , by (i) since (A, M ) = 1

⇒M | (Ax − B)

⇒gM | g(Ax0 − B), since x = x0 is a solution to (1)

⇒m | (ax0 − b)

⇒ax0 ≡ b(mod m) · · · (2)

∴ x0 is a solution of the given congruence.

32
 Suppose x0 is any other solution of ax ≡ b(mod m). Then ax0 ≡ b(mod m) · · · (3)

Applying symmetric relation1 on (2) we obtain b ≡ ax0 (mod m) · · · (4).

Now using transitive relation on (3) and (4) we get

ax0 ≡ ax0 (mod m) · · · (5)

⇒m | (ax0 − ax0 )

⇒gM | gA(x0 − x0 )

⇒M | A(x0 − x0 )

⇒M | (x0 − x0 ), since (A, M ) = 1

⇒x0 − x0 = M k, for some k ∈ Z

But division algorithm ⇒k = gq + r, 0 ≤ r < g, q ∈ Z

So x0 = x0 + M (gq + r) = x0 + gM q + M r
m
⇒x0 = x0 + mq + r
g
m
⇒x0 ≡ x0 + r(mod m)
g
Thus, if ax ≡ b(mod m) admits a solution x0 , then it has g incongruent solutions,

namely:
m m m
x0 , x 0 + , x0 + 2 · , · · · , x0 + (g − 1) · .
g g g

Example 4.1.1. Solve 35x ≡ 4(mod 9)

Solution: Here (35, 11) = 1. Therefore the given linear congruence has a unique solution.
35x ≡ 4(mod 9)
⇒36x − x ≡ 4(mod 9)
⇒ − x ≡ 4(mod 9), by Theorem 3.1.1
⇒x ≡ −4(mod 9)
⇒x ≡ −4(mod 9), by Theorem 3.1.3
⇒x ≡ −9 + 5(mod 9)
⇒x ≡ 5(mod 9)
∴ x = 5 is a solution of the given linear congruence

1
If a ≡ b(mod m) then b ≡ a(mod m)

33
Example 4.1.2. Solve 18x ≡ 30(mod 42)
Solution: Here a = 18, b = 30, m = 42, g = (18, 42) = 6 and g | b. Therefore the given
congruence has exactly 6 solution. Now

18x ≡ 30(mod 42)

18 30 42
⇒ x ≡ (mod )
6 6 6
⇒ 3x ≡ 5(mod 7) · · · (1)
⇒ 15x ≡ 25(mod 7)
⇒ x ≡ 4(mod 7)
∴ x = x0 = 4 is a particular solution to the linear congruence (1)
m
Hence the 6 roots of the given congruence 18x ≡ 30(mod 42) are: x = x0 + i · g
, i=
1, 2, 3, 4, 5 i.e., 4,11,18,25,32, and 39.

34
 Problem Plus chapter 4

1. Congruence is an equivalence relation.

35
Chapter 5

Quadratic Residuacity

5.1 Quadratic Residue and Nonresidues

Definition 5.1.1 (Quadratic residue and Quadratic nonresidue). Let p > 2 and p - a,
then a is called quadratic residue modulo p, often denoted by aRp, if x2 ≡ a(mod p) is
solvable. In other words, a is called quadratic nonresidue modulo p and is often denoted
by aN p,.
Example 5.1.1. Quadratic residue modulo 11 are 1, 3, 4, 5, and 9 because 1 = 12 , 3 =
52 , 4 = 22 , 5 = 42 , 9 = 32
Quadratic nonresidue modulo 11 are: 2, 6, 7, 8, 10 because none of the quadratic congru-
ences 2 ≡ x2 (mod 11), 6 ≡ x2 (mod 11), 7 ≡ x2 (mod 11), 8 ≡ x2 (mod 11), 10 ≡
x2 (mod 11)

5.2 Legendre Symbol

Definition 5.2.1 (Legendre Symbol). If p > 2 and (a, p) = 1, then the Legendre symbol
is defined by
à ! (
a 1 if a is a quadratic residue modulo p
=
p −1 if a is a quadratic nonresidue modulo p

Example 5.2.1. Solution:

Theorem 5.2.1. Let p be an odd prime and let a and b denote integers relatively prime
to p. Then
à !
p−1
a
(i) p
≡a 2 (mod p)

à !à ! à !
a b ab
(ii) p p
= p

à ! à !
a b
(iii) a ≡ b(mod p) =⇒ p
= p

36
 Ã ! Ã ! Ã ! Ã ! Ã !
p−1
a2 a2 b b 1 −1
(iv) p
= 1, p
= p
, p
= 1, p
= (−1) 2

Proof.

37
 Problem Plus chapter 5

1. Congruence is an equivalence relation.

38
Chapter 6

Sets, The Real Number System, and

Functions

6.1 Sets
Definition 6.1.1 (Set). A set is a well-defined list, collection, or class of objects.

Examples include the following:

Example 6.1.1. (a) A = {1, 5, 8, 10}
(b) B = {2n : n ∈ N }

Does EVERY collection of objects make up a set?

The set operations are binary operations and are some useful tools to obtain new

sets from the old ones through combining(union), intersecting(intersection), and taking

differences(difference or complement).
Definition 6.1.2 (Union of two sets). The union of two sets A and B is mathematically
defined as
A ∪ B = {x : x ∈ A or x ∈ B or x ∈ both}.
Definition 6.1.3 (Intersection of two sets). The intersection of two sets A and B is
mathematically defined as
A ∩ B = {x : x ∈ A, x ∈ B}.
Definition 6.1.4 (Difference of two sets). The difference of two sets A and B is
mathematically defined as
A ∼ B = {x ∈ A : x ∈
/ B}.

The difference of A and B is often termed as the complement of B relative to A

Definition 6.1.5 (Venn diagram). Venn diagram is a convenient way to depict the
relations among the sets.

39
6.1.1 Cartesian Product Sets and their visualization
Definition 6.1.6 (Cartesian Product set). The cartesian product of two nonvoid sets
A and B is mathematically defined as

A × B = {(a, b) : a ∈ A, b ∈ B}.

Problem 6.1.1. Visualize A × B where A = {x : 1 ≤ x ≤ 2} and B = {x : 0 ≤ x ≤

1 or 2 ≤ x ≤ 3}.

6.2 The Real Number System

6.2.1 The Field Properties and the Order Properties of R

The real number system R can be described as a complete ordered field.

• The field properties of R, often called the algebraic properties, are based on the two

operations of addition and multiplication.

• The order properties of R are based on inequalities of numbers.

• The completeness properties of R are based on some completeness axioms which

will be discussed later.

The historical development of real number system R was from the positive inte-

gers, called the natural numbers N to the rational numbers Q to the full real numbers

R := Q ∪ Q0 , where the irrational numbers Q0 were obtained either as cuts of rationals

(Dedekind), or as Cauchy sequences or as suprema of sets of rationals (Supremum prop-

erty of R). The existence of irrationals is guaranteed by any of the three theorems called

the Completeness Axiom of R.

Definition 6.2.1 (Field Properties of R). On the set of real numbers there are two
binary operations, denoted by + and · and called addition and multiplication, respectively.
These operations satisfy the following properties:

(A1) closure property of addition

(A2) commutative property of addition

(A3) associative property of addition

(A4) existence of an additive identity, called zero element, in R

40
(A5) existence of an additive inverse, called negative element, in R

(M1) closure property of multiplication

(M2) commutative property of multiplication

(M3) associative property of multiplication

(M4) existence of a multiplicative identity, called unit element, in R

(M5) existence of a multiplicative inverse, called reciprocals, in R

(D ) distributive property of multiplication over addition

The real number system R is a FIELD because it obeys the field axioms.

Definition 6.2.2 (Order Properties of R). On the field of real numbers there is a
binary relation denoted by <. For all x, y, z ∈ R this relation satisfies the following
properties:

(O1) Law of Trichotomy: either x < y or x = y or y < x

(O2) Transitive Law: x < y and y < z implies x < z

(O3) compatibility of < and + : x < y implies x + z < y + z

(O4) compatibility of < and · : 0 < z and x < y implies x.z < y.z

The real number system R is an ORDERED FIELD because it obeys the order

axioms.

6.2.2 The Completeness Properties of R

√
Theorem 6.2.1. 2 is not a rational number.

√ √ m
Proof. If 2 be a rational number, assume 2 = n
where (m, n) = 1 and m, n ∈

Z, n 6= 0 =⇒ m2 = 2n2 =⇒ m is an even integer, so m = 2k, say. So (2k)2 = 2n2 =⇒

√
n2 = 2k 2 =⇒ n is an even integer, a contradiction to the assumption. Hence 2 is not

a rational number.

Definition 6.2.3 (Intervals). Open interval: (a, b) = {x ∈ R : a < x < b} and closed
interval: [a, b] = {x ∈ R : a ≤ x ≤ b}.

41
Definition 6.2.4 (Bounded Set). Let S ⊂ R. Then u ∈ R is said to be an upper bound
of S if s ≤ u for all s ∈ S and l ∈ R is said to be a lower bound of S if l ≤ s for all
s ∈ S. Set S is bounded above if it has an upper bound and bounded below if it has a
lower bound. It is said to be bounded if it has both upper bound and lower bound. It is
unbounded if it lacks either an upper bound or a lower bound.
Definition 6.2.5 (Suprema and Infima). If S is bounded above, then an upper bound
u ∈ R is said to be a least upper bound (lub) or supremum of S if no number smaller than
u is an upper bound of S. Similarly greatest lower bound(glb) or infimum is defined.
Theorem 6.2.2. An upper bound u of a nonempty set S ⊆ R is the supremum of S if
and only if for each ² > 0 there exists an s² ∈ S such that u − ² < s² .
Example 6.2.1. • If S = (0, 1), the sup S = 1, and inf S = 0.

• If S = [0, 2] the sup S = 2, greatest element of S and inf S = 0, the least element
os S.

• If S = { n1 : n ∈ N}, the sup S = 1, greatest element of S and inf S = 0. But the set
has no least element.
Example 6.2.2. Since every real number is an upper bound for the empty set, so the
empty set has no supremum. Similarly it has no infimum.
Theorem 6.2.3 (Completeness Axiom of R/The Supremum Property of R).
Every nonempty set of real numbers that has an upper bound has a supremum in R.
Theorem 6.2.4 (Completeness Axiom of R/The Infimum Property of R). Every
nonempty set of real numbers that has a lower bound has an infimum in R.
Theorem 6.2.5. There exists a positive real number x such that x2 = 2.

Proof. Let S = {s ∈ R : 0 ≤ s, s2 < 2}. Since 1 ∈ S, the set is not empty. Also,

S is bounded above by 2, because if t > 2, then t2 > 4 so that t ∈

/ S. Therefore the

Supremum Property implies that the set S has a supremum in R and let x = sup S(here
√
sup S = 2).

Thus after some mathematical reasoning we can prove that x is a real number.

Definition 6.2.6. A cut or Dedekind cut of the real line is a pair of nonempty subsets
L and R of R such that L ∪ R = R and for every x ∈ L and y ∈ R, x < y. In terms of
geometry, L is the left-hand set and R is the right-hand set.
Example 6.2.3. If (−∞, 1] ∪ (1, ∞) = R, then L = (−∞, 1] and R = (1, ∞)
Theorem 6.2.6 (Completeness Axiom of R). Let L and R define the cut of the real
line. Then there is one and only one real number z such that for every x ∈ L and y ∈ R,
x ≤ z ≤ y.

42
 √
Example 6.2.4.√ We have
√ earlier proved that √2 is not√a rational
√ number and observe
here that (−∞, 2)
√ ∪ ( 2, ∞) 6= R, but (−∞, 2) ∪ { 2} ∪ ( 2, ∞) = R. So by the
Dedekind theorem 2 is a real number.

Thus, from the above discussion we may conclude that R is complete BUT Q is not.

6.3 Functions
Definition 6.3.1 (Function). A function f from a set A into a set B, denoted by
f : A → B, is a rule that assigns to each element x ∈ A a unique element y ∈ B, and we
write y = f (x).

Or more precisely, A function f from a set A into a set B, denoted by f : A → B,

is a subset f of the cartesian product A × B such that for each x ∈ A there exists a

unique y ∈ B, and we write y = f (x). Set A is called the domain of f and the set

{y ∈ B : y = f (x)} is called the range of f.

Example 6.3.1. The equation y = 2x + 1 defines a function but x2 + y 2 = 1 does not.
Definition 6.3.2 (Monotone function). A function f is said to be increasing if f (x1 ) ≤
f (x2 ) for x1 ≤ x2 and decreasing if f (x1 ) ≥ f (x2 ) for x1 ≤ x2 . A function is said to be
monotone if it is either increasing or decreasing.
Definition 6.3.3 (Injective/one-one function). A function f : A → B is said to be
one-one if x1 6= x2 , then f (x1 ) 6= f (x2 ).
Definition 6.3.4 (Surjective/onto function). A function f is said to be onto if f (A) =
B.
Definition 6.3.5 (Bijective function/1-1 correspondence). A function f is said to
be bijective if it is both injective and surjective.
Definition 6.3.6 (Inverse function). A function, denoted by f −1 , is said to be the
inverse of f if f is one-one. Thus f −1 is related to f as follows:

x = f −1 (y) if and only if y = f (x).

Problem Plus 6

1.

2.

43
Chapter 7

Sequence and Sequence of Functions

7.1 Sequence
Definition 7.1.1 (Sequence). A sequence in R is a function from the set N of natural
numbers into the set R of real numbers. Thus, if X : N → R is a sequence, then the value
of X at n ∈ N is denoted by xn rather than X(n). We denote sequence by the notations:

X, (xn ), < xn > .

What is the difference between the SET and SEQUENCE?

Definition 7.1.2 (Convergence of Sequence). A real number x is said to be a limit

of a sequence xn , written as xn → x or limn→∞ xn = x if for every ² > 0 there exists
a natural number K := K(²) such that |xn − x| < ² for all n ≥ K. If x is the limit
of a sequence, we say that xn converges to x or xn has the limit x. If a sequence has a
limit we say that the sequence is convergent; if it has no limit, we say that the sequence
is divergent.
1
Example 7.1.1. (i) xn = n
is a convergent sequence.
2n
(ii) xn = n+2
is a convergent sequence.

(ii) X =< 2, 4, 6, 8, .......... > is a divergent sequence.

(iv) xn = (−1)n is a bounded but divergent sequence.

Theorem 7.1.1. Every convergent sequence is bounded.

2n
Example 7.1.2. xn = n+2

Theorem 7.1.2 (Archimedean Property). If x ∈ R, then there exists nx ∈ N such

that x < nx .

This suggests that N, the set of natural numbers is unbounded.

44
Example 7.1.3. Show that lim(1/n) = 0.
Solution: For given ² > 0, 1/² > 0. Hence by Archimedean Property there exists a
natural number K := K(²) such that 1/² < K, then for any n ∈ N such that n ≥ K =⇒
1/² < K ≤ n so that 1/n < ². Therefore, if n ≥ K, then | n1 − 0| = n1 < ².

Theorem 7.1.3. The sum, difference, scalar multiple, product, and division (not always)
of two convergent sequences are convergent.

Theorem 7.1.4 (Squeeze Theorem). If xn ≤ yn ≤ zn for n ≥ K and both xn , zn → x,

then yn → x

Example 7.1.4. Use the above theorem to show that lim( sinn n ) = 0

Theorem 7.1.5 (Absolute Convergence). If |xn | → 0, then xn → 0.

n
Example 7.1.5. Use the above theorem to show that lim( (−1)
n
)=0

Theorem 7.1.6 (Ratio Test). Let xn be a sequence of positive real numbers such that
L = lim(xn+1 /xn ) exists. If L < 1, then xn converges and lim(xn ) = 0.

Example 7.1.6. Use the above theorem to show that lim(n!/nn ) = 0

Theorem 7.1.7 (Monotone Convergence Theorem). A monotone sequence of real

numbers is convergent iff it is bounded. Further, if < xn > is a bounded increasing
sequence, then xn → sup{xn } and if < xn > is a bounded decreasing sequence, then
xn → inf{xn }

Example 7.1.7. Consider the sequence xn+1 = (xn + a/xn )/2 where a > 0, x1 > 0.
(This calculates the square roots of a). Here xn − xn+1 ≥ 0 =⇒ xn+1 ≤ xn for all
n ≥ 2. It follows from Monotone
√ Convergence Theorem that x = lim(xn ) exists. So
x = (x + a/x)/2 =⇒ x = a.

Theorem 7.1.8 (The Bolzano-Weierstrass Theorem). A bounded sequence of real

numbers has a convergent subsequence

Example 7.1.8. Find a convergent subsequence of xn = (−1)n

Definition 7.1.3 (Cauchy Sequence). A sequence of real numbers xn is said to be a

Cauchy sequence if for every ² > 0 there exists a natural number K := K(²) such that
|xm − xn | < ² for all m, n ≥ K.

Theorem 7.1.9 (Cauchy Convergence Criterion). A sequence of real numbers is

convergent if and only if it is Cauchy sequence.
1
Example 7.1.9. xn = n
is a Cauchy sequence.

45
7.2 Sequence of Functions

What we have studied so far is the sequence of constants i.e., sequence of constant

functions. In this section we shall discuss about the sequence of arbitrary functions.

Therefore, the sequence of function is a sequence whose terms are functions rather

than real numbers. Sequences of functions arise naturally in real analysis and are espe-

cially useful in obtaining approximations to a given function and defining new functions

from known ones.

Let’s begin with an example:

Example 7.2.1. Consider the sequence of functions fn : A = [0, 1] → R defined by

fn (x) := xn . We have fn (0) = 0 for al n and fn (X) → 0 if x < 1, but fn (1) = 1 for all
n. Thus, fn (x) converges to a function f (x) where
(
0 for 0 ≤ x < 1
f (x) :=
1 if x = 0,

Note that the limit of the sequence of functions depend on x that is the sequence of

functions converges pointwise. Also note that the functions in the sequence are continuous

but the limit function is not.

Now let’s go through the following example:

sin x
Example 7.2.2. Consider the sequence of functions fn : R → R defined by fn (x) := n
.
We have f (x) := limn→∞ sinn x = (sin x) limn→∞ n1 = (sin x) · 0 = 0 for all x ∈ R.

Here note that the limit of the sequence of functions does not depend on x that is the

sequence of functions converges uniformly. Also note that the functions in the sequence

are continuous and so does the limit function.

Therefore according to these examples the following two definitions are in order:

Definition 7.2.1 (Pointwise Convergence). Let < fn > be a sequence of functions

on B ⊆ R to R, let A ⊆ B, and f : A ⊆ B → R. Then the sequence < fn > converges
pointwise on A to f if, for EACH x ∈ A, limn→∞ fn (x) = f (x).

In other words, fn → f (pointwise) on A if for every ² > 0 and for each x ∈ A, there
exists a natural number K := K(², x) such that if n ≥ K

|fn (x) − f (x)| < ².

46
Definition 7.2.2 (Uniform Convergence). Let < fn > be a sequence of functions on
B ⊆ R to R, let A ⊆ B, and f : A ⊆ B → R. Then the sequence < fn > converges
uniformly on A to f if, for ALL x ∈ A, limn→∞ fn (x) = f (x).

In other words, fn → f (uniformly) on A if for every ² > 0 and for all x ∈ A, there
exists a natural number K := K(²) (depending on ² but not on x) such that if n ≥ K

|fn (x) − f (x)| < ².

Notations:

• fn → f (pointwise) on A ⇒ fn → f on A or fn (x) → f (x) for x ∈ A

• fn → f (uniformly) on A ⇒ fn ⇒ f on A or fn (x) ⇒ f (x) for x ∈ A

Example 7.2.3. Discuss the uniform convergence of the sequence < fn > where f : A =
[0, 1] → R is defined by fn (x) = nx , n ∈ N.
Solution: First find the pointwise convergence and then the uniform convergence. Clearly,
fn (x) → f = 0 on A. We claim that this convergence is uniform as well. To this end, let
² > 0, then 1/² > 0, by Archimedean theorem, ∃K ∈ N such that 1/² < K ⇒ 1/K < ².
Now, if n ≥ K, then ¯ ¯ ¯x ¯
¯ ¯ ¯ ¯
¯fn (x) − f (x)¯ = ¯ − 0¯
n
x
=
n
1
≤ for all x ∈ A
n
1
<
K
<²
Therefore, fn ⇒ f on A
Example 7.2.4. Discuss the uniform convergence of the sequence < fn > where f : A =
[0, 1] → R is defined by fn (x) = 1 − nx , n ∈ N.
Solution: First find the pointwise convergence and then the uniform convergence. Clearly,
fn (x) → f = 1 on A. We claim that this convergence is uniform as well. To this end, let
² > 0, then 1/² > 0, by Archimedean theorem, ∃K ∈ N such that 1/² < K ⇒ 1/K < ².
Now, if n ≥ K, then ¯ ¯ ¯ ¯
¯ ¯ ¯ x ¯
¯fn (x) − f (x)¯ = ¯1 − − 1¯
n
x
=
n
1
≤ for all x ∈ A
n
1
< <²
K
Therefore, fn ⇒ f on A

47
Example 7.2.5. Discuss the uniform convergence of the sequence < fn > where f : A =
x
[0, 1] → ∞ is defined by fn (x) = 1+nx , n ∈ N.
Solution: First find the pointwise convergence and then the uniform convergence. Clearly,
fn (x) → f = 0 on A. We claim that this convergence is uniform as well. To this end, let
² > 0, then 1/² > 0, by Archimedean theorem, ∃K ∈ N such that 1/² < K ⇒ 1/K < ².
Now, if n ≥ K, then
¯ ¯ ¯ x ¯ ¯x¯
¯ ¯ ¯ ¯ ¯ ¯
¯fn (x) − f (x)¯ = ¯ − 0¯ < ¯ ¯ = 1/n for all x ∈ A
1 + nx nx
< 1/K < ²
Therefore, fn ⇒ f on A
Lemma 7.2.1. A sequence of functions fn on B ⊆ R does not converge uniformly on
A ⊆ B to a function f : A → R iff for some ²0 > 0 there is a subsequence fnk of fn and
a sequence xk in A such that
¯ ¯
¯ ¯
¯fnk (xk ) − f (xk )¯ ≥ ²0 for all k ∈ N.

Example 7.2.6. Discuss the uniform convergence of the sequence < fn > where fn :
nx
A = [0, ∞) → R is defined by fn (x) = 1+n 2 x2 , n ∈ N.

Solution: First find the pointwise convergence and then the uniform convergence. Clearly,
fn (x) → f = 0 on A. We claim that this convergence is not uniform on A. To this end,
let ² > 0, and let nk = n, xk = 1/n, then
¯ ¯ ¯ ¯
¯ ¯ ¯ ¯
¯fnk (xk ) − f (xk )¯ = ¯fn (1/n) − f (1/n)¯ = 1/2 ≥ ² for all n ∈ N.

Therefore, fn is not uniformly convergent to f on A

Theorem 7.2.2 (Interchange of Limit and Continuity). Let < fn > be a sequence
of functions on a set A ⊆ R and suppose that fn converges uniformly on A to a function
f : A → R. Then f is continuous on A.
Theorem 7.2.3 (Interchange of Limit and Derivative). (J. R. Marsden [?]) Let
< fn > be a sequence of differentiable functions on a set A = (a, b) ⊆ R converging
pointwise to f on A. Suppose that the derivatives < fn0 > are continuous and converge
uniformly to a function g. Then f is differentiable and f 0 = g, i.e.,
¡ ¢0
lim fn0 = lim fn
n→∞ n→∞

Theorem 7.2.4 (Interchange of Limit and Integral). Let < fn > be a sequence of
functions that are (Riemann) integrable on A = [a, b] and suppose that < fn > converges
Rb
uniformly on A to a function f . Then f is (Riemann) integrable on A and a f (x)dx =
Rb
limn→∞ a fn (x)dx i.e.,
Z b Z b
lim fn (x)dx = lim fn (x)dx
n→∞ a a n→∞
.

48
Theorem 7.2.5 (Bounded Convergence Theorem). Let < fn > be a sequence of
(Riemann) integrable functions on A = [a, b] and suppose that < fn > converges on A
to a (Riemann) integrable function f . Suppose also that there exits M > 0 such that
|fn (x)| ≤ M for all x ∈ A. Then
Z b Z b
f (x)dx = lim fn (x)dx.
a n→∞ a

49
 Problem Plus 7

1.

2.

50
Chapter 8

Series and Series of Functions

8.1 Series
Definition 8.1.1
P (Series).
PSum of the terms of an infinite sequence is called a series.
Given a series ∞ x
n=1 n = xn = x1 + x2 + x3 + · · · , let sn denote its n-th partial sum:
n
X
sn = xk = x1 + x2 + x3 + · · · + xn .
k=1

If the sequence
P sn is convergent, i.e. if x is a real number such that lim(sn ) = x, then the
series xn is called convergent and we write
n
X ∞
X
lim xk = xk = x1 + x2 + x3 + · · · = x.
n→∞
k=1 k=1

The number x is called the sum of the series. Otherwise, the series is divergent.

P∞ Pn
Note that k=1 xk = limn→∞ k=1 xk .

Theorem 8.1.1. The geometric series

∞
X
arn−1 = a + ar + ar2 + · · ·
n=1

is convergent if |r| < 1 and its sum is

∞
X a
arn−1 = |r| < 1
n=1
1−r

If |r| ≥ 1, the series is divergent.

P∞ 2n 1−n
Example 8.1.1. Test
P∞ 2n 1−n whetherP∞ the series n=1 2 3 is convergent or divergent.
n−1
Solution: n=1 2 3 = n=1 4(4/3) is a geometric series with a = 4 and r =
4/3 > 1. So, the series is divergent.

51
 P
Theorem 8.1.2. The p-series ∞ 1
n=1 np is convergent if p > 1 and divergent if p ≤ 1.
When p = 1 the series is called the harmonic series.
P
Theorem 8.1.3. If the series ∞ n=1 xn is convergent , then lim(xn ) = 0. But the converse
is not true in general, e.g., harmonic series.

Theorem
P∞ 8.1.4 (The Test for Divergence). If xn → ∞ or xn 9 0 then the series
n=1 xn is divergent.
P n2
Example 8.1.2. Test whether the series ∞ n=1 5n2 +4 is convergent or divergent.
2
Solution: Here xn = 5nn2 +4 → 1/5 6= 0 so The Test for Divergence implies that the series
is divergent.

Theorem 8.1.5. The sum, difference, and scalar multiple of two convergent series are
convergent.
P
Example 8.1.3. Find the sum of the series ∞ 3 1
n=1 ( n(n+1) + 2n ).
Solution:
P∞ 1 Here second series is a geometric
P∞ series with a = Pn1/2 and r = 1/2, so
a 3 1
n=1 2n = 1−r
= 1. The first series is n=1 n(n+1) = 3 limn→∞ i=1 i(i+1) = limn→∞ (1−
1
n+1
) = 3 · 1 = 3. Therefore the sum of the given series is 3 + 1 = 4.
P
Theorem 8.1.6 (Cauchy Criterion for Series). A series xn in R is convergent iff
for each ² > 0 there is a natural number K := K(²) such that if m > n ≥ K, then

|sm − sn | = |xn+1 + xn+2 + · · · + xm | < ².

P
Definition
P 8.1.2. We say that a series xn is absolutely convergent if the series
|xn | is convergent in R. A series is said to be conditionally convergent if it is
convergent but not absolutely convergent.

Theorem 8.1.7. If a series is absolutely convergent, then it is convergent.

8.1.1 Tests for Absolute Convergence

Theorem 8.1.8 (Comparison Test). Let xn and yn be real sequences such that for
some natural number K,
0 ≤ xn ≤ yn for n ≥ K.
P P
Then
P the convergence of y n Pimplies the convergence of xn and the divergence of
xn implies the divergence of yn .
P
Example 8.1.4. Test the convergence of the series ∞ 5
n=1 2n2 +4n+3 .
5
P
Solution: Note that 2n2 +4n+3 < 2n5 2 by the p-series 1
n2
converges and hence by the
Comparison Test the given series is convergent.

Theorem 8.1.9 (Limit Comparison Test). Let xn and yn be positive real sequences
and L = lim(xn /ynP ) P
(a) If L 6= 0, thenP xn is convergent iff Pyn is convergent.
(b) If L = 0 and yn is convergent, then xn is convergent.

52
 P 2
Example 8.1.5. Test the convergence of the series ∞ 2n
√ +3n
n=1 5+n5 .
Solution: Note that the dominant
√ part of the numerator is 2n2 and the dominant
2 2n2
part of the denominator is n . This suggests taking xn = 2n
5 √ +3n ,
5+n 5 y n = √
n5
=
2 2 1/2 P P
n1/2
, limn→∞ xynn = limn→∞ 2n√ +3n · n
5+n5 2
= 1 Since yn = 2 1/n1/2 is divergent
(p-series with p = 1/2 < 1), the given series diverges by the Limit Comparison Test.
Theorem 8.1.10 (Root Test). Let xn be a sequence in R and
r := lim(|xn |1/n ).
P
Then xn is absolutely convergent if r < 1 and divergent if r > 1.
P ¡ 2n+3 ¢n
Example 8.1.6. Test the series ∞ n=1
¡ 2n+3 ¢n 3n+2 p 2n+3
Solution: Root Test with xn = 3n+2 gives: n |xn | = 3n+2 → 2/3 < 1 Thus, the given
series converges by the Root Test.
Theorem 8.1.11 (Ratio Test). Let xn be a sequence in R and
r := lim(|xn+1 |/|xn |).
P
Then xn is absolutely convergent if r < 1 and divergent if r > 1.
P n n3
Example 8.1.7. Test the series ∞ n=1 (−1) 3n for absolutely convergence.
3
Solution: Ratio Test with xn = (−1)n 3nn gives: | an+1
an
| = | 31 ( n+1
n
)3 | = | 13 (1 + 1/n)3 | →
1
3
< 1 Thus, by the Ratio Test, the given series is absolutely convergent and therefore
convergent.
Theorem 8.1.12 (Raabe’s Test). Let xn be a sequence of nonzero real numbers and
³ ³ |xn+1 | ´
r := lim n 1 − )
|xn |
P
Then xn is absolutely convergent if r > 1 and is not absolutely convergent if r < 1.
Theorem 8.1.13 (Integral Test). RLet f be a continuous, positive decreasing function P
∞
on [1, ∞) and xn = f (n). Then if 1 f (x)dx is convergent (divergent), then xn is
convergent (divergent).
P
Example 8.1.8. Test the series ∞ lnn
n=1 n for convergence.
Solution: The function f (x) = (lnx)/x is positive and continuous for x > 1 because the
logarithm function is continuous. But it is not clear whether or not f is decreasing, so
f 0 (x) = 1−lnx
x2 R
. Thus, f 0 (x) < 0 when 1 − lnx < 0, i.e., x > e. So f is decreasing when
∞ R t lnx
x > e. Here 1 lnx x
dx = lim t→∞ 1 x
dx = ∞ Therefore, by the Integral Test the given
series is divergent.

8.1.2 Tests for Nonabsolute Convergence

Definition 8.1.3 (Alternating Series). An alternating series is a series whose terms
are alternately positive and negative. The n-th term of an alternating series is defined by
xn = (−1)n−1 yn , xn = (−1)n yn , xn = (−1)n+1 yn where n ∈ N and yn > 0
Theorem 8.1.14 (Alternating Series Test). Let xn be a decreasing
P sequence of positive
n+1
real numbers with lim(xn ) = 0. Then the alternating series (−1) xn is convergent.
P (−1)n+1 P∞ (−1)n+1
Example 8.1.9. Test the convergence of the series ∞ n=1 n
, n=1
√
n
.

53
8.2 Series of Functions

We have studied the series of constants(i.e. constant functions). He we shall study the

series of functions rather than constants.

Definition 8.2.1 (Pointwise and Uniform Convergence of Series). Let P the function
f and the sequence of functions < fn P > be defined on B ⊆ R and sn = nk=1 fk be the
sequence of partial sums of the series ∞ n=1 fn in B ⊆ R. P∞
Then if sn → f (pointwise)
P∞ on B (i.e., s n → f on B), then n=1 fn (x) → f (x) for
x ∈ B and we write n=1 fn (x) = f (x)(pointwise) P∞
Then if sn → f (uniformly)
P on B (i.e., s n ⇒ f on B), then n=1 fn (x) ⇒ f (x) for
x ∈ B and we write ∞ f
n=1 n (x) = f (x)(unif ormly)

Theorem 8.2.1 (Interchange of LimitPand Sum). Let < fn > be a sequence of

functions on a set A ⊆ R and suppose that fn converges uniformly on A to a function
f : A → R. Then f is continuous on A and
∞
X ∞
X
lim fn (x) = lim fn (x).
x→x0 x→x0
n=1 n=1

Theorem 8.2.2 (Interchange of Sum and Derivative). (J. R. Marsden [?]) IfP < fn >
0 ∞
P∞on a0 set A = (a, b) ⊆ R, the < fn > are continuous, n=1 fn
are differentiable functions
converge pointwise, and n=1 fn converges uniformly, then
³X
∞ ´0 ∞
X
fn = fn0 .
n=1 n=1

Theorem 8.2.3 (Interchange of Integral and Sum). Let < fn > be a sequence of
functions that are (Riemann) integrable on A = [a, b] and suppose that < fn > converges
uniformly on AThen we may interchange the order of integration and summation:
Z b³X
∞ ´ ∞ ³Z b
X ´
fn (x) dx = fn (x)dx .
a n=1 n=1 a

54
 Problem Plus 8

1.

2.

55
Chapter 9

Limit and Continuity of a Function

9.1 Limit of a Function

Definition 9.1.1 (Cluster Point or Accumulation Point). Let A ⊆ ReE. A point
c ∈ R is called a cluster point of A if every δ−neighborhood Vδ (c) := (c − δ, c + δ) of c
contains at least one point of A other than c, i.e.,

Vδ (c) − {c} ∩ A 6= φ

Definition 9.1.2 (Limit Point). Let A ⊆ R. A point c ∈ R is called a limit point of A

if every δ−neighborhood Vδ (c) := (c − δ, c + δ) of c contains at least one point of A, i.e.,

Vδ (c) ∩ A 6= φ

Example 9.1.1.

(a) Let A = (0, 1) ∪ {3}. Then the set of cluster points of A is [0, 1] and the set of limit
points of A is [0, 1] ∪ {3}.

(b) A finite set has no cluster points.

(c) N has no cluster points.

(d) B = { n1 : n ∈ N} has only one cluster point 0.

(e) Let S = {x ∈ R : x ∈ [0, 1] and x is rational}. Then all points of [0, 1] are accumu-
lation points of S.
Definition 9.1.3 (² − δ : Definition of Limit of a Function). Let A ⊆ R, and let
c be a cluster point of A. We say that a real number L is a limit of f at c, written
limx→c f (x) = L or f (x) → L as x → c if, given any ²−neighborhood V² (L) of L, there
exists a δ−neighborhood Vδ (c) of c such that if x 6= c is any point of Vδ (c) ∩ A, then f (x)
belongs to V² (L).

In other words, L is a limit of f at c, written limx→c f (x) = L or f (x) →

L as x → c, if given ² > 0 there exists a δ := δ(²) > 0 such that if 0 < |x − c| < δ,
then |f (x) − L| < ²

56
Example 9.1.2. Apply ² − δ definition of limit to illustrate that

lim (2x − 2) = 6
x→4

Solution: Here f (x) = 2x − 2, c = 4, and L = 6.

Let ² > 0. WANT to find a δ := δ(²) > 0 such that
if 0 < |x − c| < δ, then |f (x) − L| < ² i.e.,
if 0 < |x − 4| < δ, then |f (x) − L| < ²
Now, |f (x) − L| = |(2x − 2) − 6| = 2|x − 4| < 2δ
If we choose δ = 2² =: δ(²), then
if 0 < |x − c| < δ, then |f (x) − L| < ²
Therefore, limx→c f (x) = L ⇒ limx→4 (2x − 2) = 6

Theorem 9.1.1 (Sequential Criterion). Let f : A → R and c be a cluster point of A;

then:
(i) limx→c f (x) = L if and only if
(ii) for every sequence xn in A that converges to c such that x 6= c for all n ∈ N, the
sequence f (xn ) converges to L.

Theorem 9.1.2 (Divergence Criterion). Let f : A → R and c be a cluster point of

A; then:
(i) The function f does not have a limit at c
if and only if
(ii) ∃ a sequence xn 6= c for all n ∈ N such that the sequence xn converges to c but the
sequence f (xn ) does not converges in R

Example 9.1.3. Show that limx→0 x1 does not exist in R.

Solution: Here let f (x) = x1 for x > 0. Take the sequence xn := 1/n for n ∈ N, then
f (xn ) = n does not converge in R. Hence by the Divergence Criterion the given sequence
does not exist in R.

9.2 Continuous Function

Definition 9.2.1 (Continuous Function). Let f : A ⊆ R → R. We say that f is
continuous at a point c ∈ A, written limx→c f (x) = f (c) if, given any ²−neighborhood
V² (f (c)) of f (c), there exists a δ−neighborhood Vδ (c) of c such that if x ∈ Vδ (c) ∩ A, then
f (x) ∈ V² (f (c)).

In other words, f is continuous at a point c ∈ A, written limx→c f (x) = f (c) if, given
any ² > 0 there exists a δ := δ(², x) > 0 such that if |x − c| < δ, then |f (x) − f (c)| < ²

Theorem 9.2.1 (Discontinuity Criterion). Let f : A → R and c ∈ A; then:

(i) The function f is discontinuous at c
if and only if
(ii) ∃ a sequence xn in A such that the sequence xn converges to c, but the sequence f (xn )
does not converges to f (c).

57
 Example 9.2.1.

(a) f (x) = x is continuous on R.

(b) f (x) = x2 is continuous on R.

(c) f (x) = 1/x is continuous on A = (0, ∞).

(d) f (x) = 1/x is not continuous at 0.

Theorem 9.2.2 (Combination of Continuous Functions). The sum, difference,

product, scalar multiple, and division (if the denominator is not zero) of two contin-
uous functions are continuous.

Theorem 9.2.3 (Polynomial functions). are always continuous.

Definition 9.2.2 (Bounded Function). A function f : A ⊆ R → R is said to be

bounded on A if there exists a constant M > 0 such that |f (x)| ≤ M for all x ∈ A.

Example 9.2.2.

(a) f (x) = x is bounded on A = [−7, 2] but unbounded on R.

(b) f (x) = 1/x is continuous on A = (0, ∞) but not bounded on A. f (x) is not even
bounded when restricted to the set B = (0, 1).

Theorem 9.2.4 (Boundedness Theorem). Let I := [a, b] be a closed bounded interval

and let f : I → R be continuous on I. Then f is bounded on I.

Definition 9.2.3 (Absolute Extremum of a Function). Let f : A ⊆ R → R be a

function. We say that f has an absolute maximum on A if there is a point x? ∈ A such
that
f (x? ) ≥ f (x) for all x ∈ A.
We say that f has an absolute minimum on A if there is a point x? ∈ A such that

f (x? ) ≤ f (x) for all x ∈ A.

We say that x? is an absolute maximum point for f on A, and that x? is an absolute

minimum point for f on A, if they exist.

Example 9.2.3. 1. Continuous function on a set A does not necessarily have an ab-
solute maximum or an absolute minimum on the set.

2. f (x) = 1/x has neither an absolute maximum nor an absolute minimum on the set
A = (0, ∞).

3. The same function has neither an absolute maximum nor an absolute minimum
when restricted to the open set B = (0, 1).

4. While the same function has BOTH an absolute maximum and an absolute mini-
mum when restricted to the closed set C = [0, 1].

58
 5. The function f (x) = x2 on A = [−1, +1] has two points x = ±1 giving the absolute
maximum and a single point x = 0 yielding the absolute minimum on A.

6. The constant function f (x) = c is such that for x ∈ R is such that EVERY point
of R is BOTH an absolute maximum and an absolute minimum point for f .

Theorem 9.2.5 (Maximum-Minimum(Maximin) Theorem). Let I := [a, b] be a

closed bounded interval and let f : I → R be continuous on I. Then f has an absolute
maximum and an absolute minimum on I.

Theorem 9.2.6 ((Bolzano’s) Intermediate Value Theorem). Let I be any interval

and let f : I → R be continuous on I. If a, b ∈ I and if k ∈ R satisfies f (a) < k < f (b),
then there exists a point c ∈ I between a and b such that f (c) = k.

Corollary 9.2.7. Let I = [a, b] be a closed, bounded interval and let f : I → R be

continuous on I. If k ∈ R is any number satisfying

inf f (I) ≤ k ≤ sup f (I),

then there exists a number c ∈ I such that f (c) = k.

The following corollary provides the Location of Roots:

Corollary 9.2.8. Let I be any interval and let f : I → R be continuous on I. If a < b

are numbers in I such that f (a) < 0 < f (b)(or such that f (a) > 0 > f (b)), then there
exists a point c ∈ (a, b) such that f (c) = 0.

59
 Problem Plus 9

1.

2.

60
Chapter 10

Differentiation

Definition 10.0.4 (The Derivative). Let f : I → R where c ∈ I, an interval. We say

that a real number L is the derivative of f at c if for any given number ² > 0 there
exists a number δ(²) > 0, then
¯ f (x) − f (c) ¯
¯ ¯
¯ − L ¯ < ².
x−c
In this case we say that f is differentiable at c, and we write f 0 (c) for L.
In other words, the derivative of f at c is given by the limit
f (x) − f (c)
f 0 (c) = lim
x→c x−c
Theorem 10.0.9. If f : I → R has a derivative at c ∈ I, then f is continuous at c.

Is the converse true? WHY?

Theorem 10.0.10. The sum, difference, scalar multiple, product, and division(not al-
ways) of two differentiable functions are differentiable.
Theorem 10.0.11 (Rolle’s Theorem). Let f : I = [a, b] → R be continuous on a closed
interval [a, b] and differentiable on the open interval (a, b), and that f (a) = f (b) = 0. Then
there exists at least one point c in (a, b) such that f 0 (c) = 0.

The Mean Value Theorem relates the values of a function to values of its derivative

which is stated as follows.

Theorem 10.0.12 (Mean Value Theorem). Let f : I = [a, b] → R be continuous on
a closed interval [a, b] and differentiable on the open interval (a, b). Then there exists at
least one point c in (a, b) such that f 0 (c) = f (b)−f
b−a
(a)
.

One can easily deduce Roll’s Theorem from Mean Value Theorem. Now we shall

present the Taylor’s Theorem which is simply the generalization of the Mean Value

Theorem.

61
Theorem 10.0.13 (Taylor’s Theorem). Let f : I = [a, b] → R be a function such that
f and its derivatives up to order n are continuous on a closed interval [a, b] and f (n+1)
exists on the open interval (a, b). If x0 ∈ I, then for any x in I there exists a point c
between x and x0 such that
f 00 (x0 ) f (n) (x0 ) f (n+1) (c)
f (x) = f (x0 )+f 0 (x0 )(x−x0 )+ (x−x0 )2 +· · ·+ (x−x0 )n + (x−x0 )(n+1)
2! n! (n + 1)!
where the last term
f (n+1) (c)
Rn (x) = (x − x0 )(n+1)
(n + 1)!
is called the remainder.
Theorem 10.0.14 (Taylor’s Inequality or The Remainder Estimation Theo-
rem). If |f (n+1) (x)| ≤ M for all x ∈ I, and a number x0 ∈ I, then
M
|Rn (x)| ≤ |x − x0 |n+1 .
(n + 1)!
Theorem 10.0.15. If n → ∞ then Rn (x) → 0.

Hence the Taylors Theorem gives us the Taylor’s series:

Theorem 10.0.16 (Taylor’s Series).
f 00 (x0 ) f (n) (x0 )
f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) + (x − x0 )2 + · · · + (x − x0 )n + · · ·
2! n!
Taylor’s Series evaluated at x0 = 0 is called the Maclaurin’s series which is given
as follows:
f 00 (0) 2 f (n) (0) n
f (x) = f (0) + f 0 (0)x + x + ··· + x + ···
2! n!
Example 10.0.4.
• Expand f (x) = ex .
• The function f (x) = sin x about the point x0 = π/2.
Theorem 10.0.17 (L’Hospital’s Rule). Suppose that f and g are differentiable and
g 0 (x) 6= 0 near x0 (except possibly at x0 ). Suppose that limx→x0 f (x) = 0 and limx→x0 g(x) =
0 or that limx→x0 f (x) = ±∞ and limx→x0 g(x) = ±∞ (In other words, we have an inde-
terminate form of type 00 or ∞ ∞
.)
Then
f (x) f 0 (x)
lim = lim 0 .
x→x0 g(x) x→x0 g (x)

If the limit on the right side exists(or is ∞ or −∞).

ln x
Example 10.0.5. • Find limx→1 x−1
tan x−x
• Find limx→0 x3

x2 −4
• Find limx→2 x−2

62
 Problem Plus 10

1.

2.

63
Chapter 11

Riemann Integration

Riemann Integration

64
 Problem Plus 11

1.

2.

65
Appendix A

Reviews

Reviews

(A.1) a + b + c
+d+e+f
+g+h+i
j+k+l

∂u ∂2u ∂3u ∂4u

∂x
+ ∂x ∂y
+ ∂x3
+ ∂y 4
= 0 http://www.sust.edu Shahjalal University
complex number
z }| {
x +
z = |{z} iy
|{z}
real imaginary

Everything will be printed in W H A T E V E R you like.

stack-out

1. Shah Noor
2 Happy

First Name: Mohammad

Second Name: Shah
Third Name: Noor
Spouse: Happy
Son: Rashad

‘quote’ “quote” margin

reverse margin note 1 note
1
This chapter summarizes the important commands

66
 1
2 % 1.simple newton
3
4 function [root] = simple newton(x0,tol);
5 % Finds the root of floating sphere cubic using Newton's method
6 % Input:
7 % x0 − initial guess for root
8 % tol − desired absolute error
9 % Output:
10 % root − approximate root
11 format long e;
12 error = 1e10;
13 x0=0.5; tol = 1e−10;
14 xk=x0;
15 k=0;
16 while error > tol
17 k=k+1;
18 xk = xk − (xkˆ2 − (xkˆ3+1)/3)/(2*xk − xkˆ2);
19 error = abs(xkˆ2−(xkˆ3+1)/3);
20 disp(['Iteration' sprintf('%2i',k) ' approx root ' ...
21 sprintf('%15.13f',xk) ' abs error ' sprintf('%3.2e',error)] )
22 end

1 function ycal = rk4 msn(f,x0,xn,y0,n)

2 % This function solves the
3 % ode y'=f(x,y),y(x0)=y0 using runge−kutta 4th order method
4
5 n = 50 ; x0=0 ; xn=2 ; x1=x0 ; xnp1=xn ; np1= n + 1 ;h
6 =(xnp1−x1)/np1 ; x = linspace(x1, xnp1, np1); ycal(1) = 1/10 ;
7

8 for k = 1 : n
9 f1 = exp(−2*x(k)) − 2*ycal(k) ;
10 f2 = exp(−2*(x(k)+h/2)) − 2*(ycal(k) +h*f1/2) ;
11 f3 = exp(−2*(x(k)+h/2)) − 2*(ycal(k) +h*f2/2) ;
12 f4 = exp(−2*(x(k)+h)) − 2*(ycal(k) +h*f3) ;
13 sf = f1 + 2 * f2 + 2 * f3 + f4 ;
14 ycal(k+1) = ycal(k) + h*sf/6 ;
15 end yexact = (0.1 + x).*exp(−2*x) ; a = [x' ycal' yexact'] ;
16 b=a(:,1)
17 %plot(x,ycal,'r', x,yexact,'b'); grid on;
18 plot(a(:,1),a(:,2),'r', a(:,1),a(:,3),'b'); grid on;
19 xlabel('x');ylabel('ycal and yexact') return

67
 Problem Plus A

1.

2.

68
Index

cancelation law, 24 number

canonical representation, 9 composite, 5
co-prime, 5 even, 3
congruence, 23 integer, 3
linear, 31 natural, 3
polynomial, 31 odd, 3
prime, 4
Diophantine Equations, 19 relatively prime, 5
Euclid, 9 Quadratic
Euclid’s Lemma, 8 nonresidue, 36
Euclidean Algorithm, 5 residue, 36
field standard factorization, 9
ordered standard representation, 9
complete, 40
function Theorem
arithmetic, 12 Euler’s, 26
Euler’s φ, 13 Fermat’s, 26
Gauss, 13 least residue, 25
Greatest Integer [x], 13 unique factorization , 8
mobius, 17 Wilson’s, 27, 29
multiplicative, 12 transitive law, 24
number of divisors τ , 12
product of divisors P, 12
sum of divisors σ, 12
totally multiplicative, 12
Fundamental Theorem of Arithmetic,
8
Fundamental Theorem of Divisibility,
4

Goldbach Conjecture, 13
greatest common divisor(g.c.d.), 5

incongruence, 23

least common multiple(l.c.m.), 7

least residue, 24
Legendre Symbol, 36

69
Bibliography

[1] Fatema Chowdhury, Munibur Rahman Chowdhury, “Theory of Num-

bers”, First Edition, Bangla Academy, Dhaka, Bangladesh, (June 1993)

[2] Prof. Dr. Md. Fazlur Rahman, “Theory of Numbers”, Third Edition,
Titas Publications, Bangladesh, (2009)

[3] T. M. Apostol, “Introduction to Analytic Number Theory”, 512.73

API

[4] D. M. Burton, “Elementary Number Theory” 512.72 BUE

[5] R. G. Bartle, D. R. Sherbert, “Introduction to Real Analysis”, Second

Edition, John Wiley & Sons, Inc,, Singapore, (1994)

[6] W.H. Cornish, Characterization of distributive and modular

semilattices, Math. Japonica, 22 (1977), 159–174.

[7] J. R. Marsden, “Classical Analysis”, Springer-Verlag, New York, (1990)

Md. Shah Noor, Associate Professor, Department of Mathematics, Shahjalal Uni-

versity of Science and Technology, Sylhet, Bangladesh, email:noorms100@gmail.com,
web:http://www.sust.edu

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