Page 1
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 0
A. NUMBER SYSTEM
(i) Natural Numbers : The counting numbers 1, 2, 3, 4,..... are called Natural Numbers. The set of
natural numbers is denoted by N. Thus N = {1, 2, 3, 4,......}. N is also denoted by ?
+
or Z
+
(ii) Whole Numbers : Natural numbers including zero are called whole numbers. The set of whole
numbers, is denoted by W. Thus W = {0, 1, 2,......}. W is also called as set of non-negative integers.
(iii) Integers : The numbers..... – 3, – 2, –1, 0, 1, 2, 3..... are called integers and the set is
denoted by ? or Z. Thus ? (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........}
(a) Set of positive integers, denoted by ?
+
and consists of {1, 2, 3, ...........}
(b) Set of negative integers, denoted by ?
–
and consists of {..........., –3, –2, –1}
(c) Set of non-negative integers {0, 1, 2, 3,...........}
(d) Set of non-positive integers {...., –3, –2, –1, 0}
(iv) Even Integers : Integers which are divisible by 2 are called even integers. e.g. 0, ± 2, ± 4,.....
(v) Odd Integers : Integers which are not divisible by 2 are called as odd integers. e.g. ± 1, ± 3, .......
(vi) Prime Number : Let ‘p’ be a natural number, ‘p’ is said to be prime if it has exactly two distinct
factors, namely 1 and itself. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,......
Remark : (a) ‘1’ is neither prime nor composite.
(b) ‘2’ is the only even prime number.
(vii) Composite Number : Let ‘a’ be a natural number, ‘a’ is said to be composite if, it has atleast
three distinct factors.
(viii) Co-prime Numbers : Two natural numbers (not necessarily prime) are coprime, if their H.C.F.
(Highest common factor) is one. e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
These numbers are also called as relatively prime numbers.
Remark : (a) Number which are not prime are composite numbers (except 1)
(b) ‘4’ is the smallest composite number.
(c) Two distinct prime numbers are always co-prime but converse need not be true.
(d) Consecutive numbers are always co-prime numbers.
(ix) Twin Prime Numbers : If the difference between two prime numbers is two, then the numbers
are called as twin prime numbers. e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
(x) Rational Numbers : All the numbers those can be represented in the form p/q, where p and q
are integers and q ? 0, are called rational numbers and their set is denoted by Q.
Thus Q = {
q
p
: p, q ? ? ? and q ? 0}. It may be noted that every integer is a rational numbers. If
not integer then either finite or recurring.
(xi) Irrational Numbers : There are real numbers which cannot be expressed in p/q form. These
numbers are called irrational numbers and their set is denoted by Q
c
or Q’. (i.e. complementary
BASIC MATHEMATICS & LOGARITHM
BASIC MATHEMATICS & LOGARITHM
Page 2
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 0
A. NUMBER SYSTEM
(i) Natural Numbers : The counting numbers 1, 2, 3, 4,..... are called Natural Numbers. The set of
natural numbers is denoted by N. Thus N = {1, 2, 3, 4,......}. N is also denoted by ?
+
or Z
+
(ii) Whole Numbers : Natural numbers including zero are called whole numbers. The set of whole
numbers, is denoted by W. Thus W = {0, 1, 2,......}. W is also called as set of non-negative integers.
(iii) Integers : The numbers..... – 3, – 2, –1, 0, 1, 2, 3..... are called integers and the set is
denoted by ? or Z. Thus ? (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........}
(a) Set of positive integers, denoted by ?
+
and consists of {1, 2, 3, ...........}
(b) Set of negative integers, denoted by ?
–
and consists of {..........., –3, –2, –1}
(c) Set of non-negative integers {0, 1, 2, 3,...........}
(d) Set of non-positive integers {...., –3, –2, –1, 0}
(iv) Even Integers : Integers which are divisible by 2 are called even integers. e.g. 0, ± 2, ± 4,.....
(v) Odd Integers : Integers which are not divisible by 2 are called as odd integers. e.g. ± 1, ± 3, .......
(vi) Prime Number : Let ‘p’ be a natural number, ‘p’ is said to be prime if it has exactly two distinct
factors, namely 1 and itself. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,......
Remark : (a) ‘1’ is neither prime nor composite.
(b) ‘2’ is the only even prime number.
(vii) Composite Number : Let ‘a’ be a natural number, ‘a’ is said to be composite if, it has atleast
three distinct factors.
(viii) Co-prime Numbers : Two natural numbers (not necessarily prime) are coprime, if their H.C.F.
(Highest common factor) is one. e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
These numbers are also called as relatively prime numbers.
Remark : (a) Number which are not prime are composite numbers (except 1)
(b) ‘4’ is the smallest composite number.
(c) Two distinct prime numbers are always co-prime but converse need not be true.
(d) Consecutive numbers are always co-prime numbers.
(ix) Twin Prime Numbers : If the difference between two prime numbers is two, then the numbers
are called as twin prime numbers. e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
(x) Rational Numbers : All the numbers those can be represented in the form p/q, where p and q
are integers and q ? 0, are called rational numbers and their set is denoted by Q.
Thus Q = {
q
p
: p, q ? ? ? and q ? 0}. It may be noted that every integer is a rational numbers. If
not integer then either finite or recurring.
(xi) Irrational Numbers : There are real numbers which cannot be expressed in p/q form. These
numbers are called irrational numbers and their set is denoted by Q
c
or Q’. (i.e. complementary
BASIC MATHEMATICS & LOGARITHM
BASIC MATHEMATICS & LOGARITHM
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 2
set of Q) e.g.
2
, 1 +
3
, e, ? etc. Irrational numbers can not be expressed as recurring
decimals.
Remark : e ? 2.71 is called Napier’s constant and ? ? 3.14.
Surds : If a is not a perfect nth power, then
n
a
is called a surd of the nth order. .
In an expression of the form
c b
a
?
, the denominator can be rationalized by multiplying numerator
and the denominator by
c b ?
which is called the conjugate of
c b ?
.
If x + y = a +
b
where x, y, a, b are rationals, then x = a and y = b. .
Ex.1 Prove that log
3
5 is irrational.
Sol. Let log
3
5 is rational.
? log
3
5 =
q
p
; where p and q are co-prime numbers
? 3
p/q
= 5 ? 3
p
= 5
q
. which is not possible, hence our assumption is wrong and log
3
5 is irrational.
Ex.2 Simply (make the denominator rational)
2 2 5 3
12
? ?
Sol. The expression =
5 6 6
) 2 2 5 3 ( 12
) 2 2 ( ) 5 3 (
) 2 2 5 3 ( 12
2 2
?
? ?
?
? ?
? ?
=
4
) 2 2 10 2 5 2 2 ( 2
) 1 5 ( ) 1 5 (
) 1 5 )( 2 2 5 3 ( 2 ? ? ?
?
? ? ?
? ? ?
=
2 10 5 1 ? ? ?
Ex.3 Find the factor which will rationalize
3
5 3 ?
Sol. Let x = 3
1/2
and y = 5
1/3
. The L.C.M. of the denominators of the indices 2 and 3 is 6. Hence x
6
and y
6
are
rational. Now x
6
+ y
6
= (x + y) (x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
)
Hence the rationalizing factor required = x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
where x = 3
1/2
and y = 5
1/3
.
Ex.4 Find the square root of 7 + 2 10
Sol. Let
y x 10 2 7 ? ? ?
. Squaring, x + y + 10 2 7 xy 2 ? ?
Hence x + y = 7 and xy = 10. These two relation give x = 5, y = 2. Hence
2 5 10 2 7 ? ? ?
Remark : symbol stands for the positive square root only. .
Ex.5 Prove that
3
2
cannot be represented in the form p + q , where p and q are rational (q > 0 and is not
a perfect square).
Sol. Put
3
2
= p + q . Hence 2 = p
3
+ 3pq + (3p
2
+ q) q ,
Since q is not a perfect square, it must be 3p
2
+ q = 0, which is impossible.
(xii) Real Numbers : The complete set of rational and irrational numbers is the set of real numbers and
is denoted by R. Thus R = Q ? Q
c
. Real numbers can be represented as points of a line. This line is
called as real line or number line
Page 3
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 0
A. NUMBER SYSTEM
(i) Natural Numbers : The counting numbers 1, 2, 3, 4,..... are called Natural Numbers. The set of
natural numbers is denoted by N. Thus N = {1, 2, 3, 4,......}. N is also denoted by ?
+
or Z
+
(ii) Whole Numbers : Natural numbers including zero are called whole numbers. The set of whole
numbers, is denoted by W. Thus W = {0, 1, 2,......}. W is also called as set of non-negative integers.
(iii) Integers : The numbers..... – 3, – 2, –1, 0, 1, 2, 3..... are called integers and the set is
denoted by ? or Z. Thus ? (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........}
(a) Set of positive integers, denoted by ?
+
and consists of {1, 2, 3, ...........}
(b) Set of negative integers, denoted by ?
–
and consists of {..........., –3, –2, –1}
(c) Set of non-negative integers {0, 1, 2, 3,...........}
(d) Set of non-positive integers {...., –3, –2, –1, 0}
(iv) Even Integers : Integers which are divisible by 2 are called even integers. e.g. 0, ± 2, ± 4,.....
(v) Odd Integers : Integers which are not divisible by 2 are called as odd integers. e.g. ± 1, ± 3, .......
(vi) Prime Number : Let ‘p’ be a natural number, ‘p’ is said to be prime if it has exactly two distinct
factors, namely 1 and itself. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,......
Remark : (a) ‘1’ is neither prime nor composite.
(b) ‘2’ is the only even prime number.
(vii) Composite Number : Let ‘a’ be a natural number, ‘a’ is said to be composite if, it has atleast
three distinct factors.
(viii) Co-prime Numbers : Two natural numbers (not necessarily prime) are coprime, if their H.C.F.
(Highest common factor) is one. e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
These numbers are also called as relatively prime numbers.
Remark : (a) Number which are not prime are composite numbers (except 1)
(b) ‘4’ is the smallest composite number.
(c) Two distinct prime numbers are always co-prime but converse need not be true.
(d) Consecutive numbers are always co-prime numbers.
(ix) Twin Prime Numbers : If the difference between two prime numbers is two, then the numbers
are called as twin prime numbers. e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
(x) Rational Numbers : All the numbers those can be represented in the form p/q, where p and q
are integers and q ? 0, are called rational numbers and their set is denoted by Q.
Thus Q = {
q
p
: p, q ? ? ? and q ? 0}. It may be noted that every integer is a rational numbers. If
not integer then either finite or recurring.
(xi) Irrational Numbers : There are real numbers which cannot be expressed in p/q form. These
numbers are called irrational numbers and their set is denoted by Q
c
or Q’. (i.e. complementary
BASIC MATHEMATICS & LOGARITHM
BASIC MATHEMATICS & LOGARITHM
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 2
set of Q) e.g.
2
, 1 +
3
, e, ? etc. Irrational numbers can not be expressed as recurring
decimals.
Remark : e ? 2.71 is called Napier’s constant and ? ? 3.14.
Surds : If a is not a perfect nth power, then
n
a
is called a surd of the nth order. .
In an expression of the form
c b
a
?
, the denominator can be rationalized by multiplying numerator
and the denominator by
c b ?
which is called the conjugate of
c b ?
.
If x + y = a +
b
where x, y, a, b are rationals, then x = a and y = b. .
Ex.1 Prove that log
3
5 is irrational.
Sol. Let log
3
5 is rational.
? log
3
5 =
q
p
; where p and q are co-prime numbers
? 3
p/q
= 5 ? 3
p
= 5
q
. which is not possible, hence our assumption is wrong and log
3
5 is irrational.
Ex.2 Simply (make the denominator rational)
2 2 5 3
12
? ?
Sol. The expression =
5 6 6
) 2 2 5 3 ( 12
) 2 2 ( ) 5 3 (
) 2 2 5 3 ( 12
2 2
?
? ?
?
? ?
? ?
=
4
) 2 2 10 2 5 2 2 ( 2
) 1 5 ( ) 1 5 (
) 1 5 )( 2 2 5 3 ( 2 ? ? ?
?
? ? ?
? ? ?
=
2 10 5 1 ? ? ?
Ex.3 Find the factor which will rationalize
3
5 3 ?
Sol. Let x = 3
1/2
and y = 5
1/3
. The L.C.M. of the denominators of the indices 2 and 3 is 6. Hence x
6
and y
6
are
rational. Now x
6
+ y
6
= (x + y) (x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
)
Hence the rationalizing factor required = x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
where x = 3
1/2
and y = 5
1/3
.
Ex.4 Find the square root of 7 + 2 10
Sol. Let
y x 10 2 7 ? ? ?
. Squaring, x + y + 10 2 7 xy 2 ? ?
Hence x + y = 7 and xy = 10. These two relation give x = 5, y = 2. Hence
2 5 10 2 7 ? ? ?
Remark : symbol stands for the positive square root only. .
Ex.5 Prove that
3
2
cannot be represented in the form p + q , where p and q are rational (q > 0 and is not
a perfect square).
Sol. Put
3
2
= p + q . Hence 2 = p
3
+ 3pq + (3p
2
+ q) q ,
Since q is not a perfect square, it must be 3p
2
+ q = 0, which is impossible.
(xii) Real Numbers : The complete set of rational and irrational numbers is the set of real numbers and
is denoted by R. Thus R = Q ? Q
c
. Real numbers can be represented as points of a line. This line is
called as real line or number line
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 2
All the real numbers follow the order property
i.e. if there are two distinct real numbers a
Negative side Positive side
–3 –2 –1 0 1 2 3 ? ?2
and b then either a < b or a > b.
Remark :
(a) Integers are rational numbers, but converse need not be true.
(b) Negative of an irrational number is an irrational number.
(c) Sum of a rational number and an irrational number is always an irrational number e.g. 2 +
3
(d) The product of a non zero rational number & an irrational number will always be an irrational number.
(e) If a ? Q and b ? Q, then ab = rational number, only if a = 0.
(f) Sum, difference, product and quotient of two irrational numbers need not be an irrational
number (it may be a rational number also).
(xiii) Complex Number : A number of the form a + ib is called complex number, where a, b ? R and
i =
1 ?
. Complex number is usually denoted by C.
Remark : It may be noted that N ? W ? ? ? ? ? Q ? R ? C.
Ex.6 Every number is one of the forms 5n, 5n ± 1, 5n ± 2.
Sol. For if any number is divided by 5, the remainder is one of the numbers 0, 1, 2, 5 – 2, 5 – 1.
Ex.7 Every square number is one of the forms 5n, 5n ± 1.
Sol. The square of every number is one of the forms (5m)
2
, (5m ± 1)
2
, (5m ± 2)
2
. If those are divided by 5,
the remainders are 0, 1, 4; and, since 4 = 5 – 1, the forms are 5n, 5n + 1, and 5n – 1.
Ex.8 Show that the number of primes in N is infinite.
Sol. Suppose the number of primes in N is finite. Let {p
1
, p
2
...., p
n
} be the set of primes in N such that
p
1
< p
2
<.... < p
n
. Consider n = 1 + p
1
p
2
......p
n
. Clearly n is not divisible by any one of p
1
, p
2
,...., p
n
.
Hence n itself is a prime and n has a prime divisor other than p
1
, p
2
....p
n
. This contradicts that the set
of primes is {p
1
, p
2
....,p
n
}. Therefore the number of primes in N is infinite.
Ex.9 If x and y are prime numbers which satisfy x
2
– 2y
2
= 1, solve for x and y.
Sol. x
2
– 2y
2
= 1 gives x
2
= 2y
2
+ 1 and hence x must be an odd number. If x = 2n + 1, then x
2
= (2n + 1)
2
= 4n
2
+ 4n + 1 = 2y
2
+ 1. Therefore y
2
= 2n(n + 1). This means that y
2
is even and hence y is an even
integer. Now, y is also a prime implies that y = 2. This gives x = 3. Thus the only solution is x = 3, y = 2.
B. DIVISIBILITY TEST :
(i) A number will be divisible by 2 iff the digit at the unit place is divisible by 2.
(ii) A number will be divisible by 3 iff the sum of its digits of the number is divisible by 3.
(iii) A number will be divisible by 4 iff last two digits of the number together are divisible by 4.
(iv) A number will be divisible by 5 iff digit at the unit place is either 0 or 5.
(v) A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 & sum
of all digits of the number is divisible by 3.
Page 4
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 0
A. NUMBER SYSTEM
(i) Natural Numbers : The counting numbers 1, 2, 3, 4,..... are called Natural Numbers. The set of
natural numbers is denoted by N. Thus N = {1, 2, 3, 4,......}. N is also denoted by ?
+
or Z
+
(ii) Whole Numbers : Natural numbers including zero are called whole numbers. The set of whole
numbers, is denoted by W. Thus W = {0, 1, 2,......}. W is also called as set of non-negative integers.
(iii) Integers : The numbers..... – 3, – 2, –1, 0, 1, 2, 3..... are called integers and the set is
denoted by ? or Z. Thus ? (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........}
(a) Set of positive integers, denoted by ?
+
and consists of {1, 2, 3, ...........}
(b) Set of negative integers, denoted by ?
–
and consists of {..........., –3, –2, –1}
(c) Set of non-negative integers {0, 1, 2, 3,...........}
(d) Set of non-positive integers {...., –3, –2, –1, 0}
(iv) Even Integers : Integers which are divisible by 2 are called even integers. e.g. 0, ± 2, ± 4,.....
(v) Odd Integers : Integers which are not divisible by 2 are called as odd integers. e.g. ± 1, ± 3, .......
(vi) Prime Number : Let ‘p’ be a natural number, ‘p’ is said to be prime if it has exactly two distinct
factors, namely 1 and itself. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,......
Remark : (a) ‘1’ is neither prime nor composite.
(b) ‘2’ is the only even prime number.
(vii) Composite Number : Let ‘a’ be a natural number, ‘a’ is said to be composite if, it has atleast
three distinct factors.
(viii) Co-prime Numbers : Two natural numbers (not necessarily prime) are coprime, if their H.C.F.
(Highest common factor) is one. e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
These numbers are also called as relatively prime numbers.
Remark : (a) Number which are not prime are composite numbers (except 1)
(b) ‘4’ is the smallest composite number.
(c) Two distinct prime numbers are always co-prime but converse need not be true.
(d) Consecutive numbers are always co-prime numbers.
(ix) Twin Prime Numbers : If the difference between two prime numbers is two, then the numbers
are called as twin prime numbers. e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
(x) Rational Numbers : All the numbers those can be represented in the form p/q, where p and q
are integers and q ? 0, are called rational numbers and their set is denoted by Q.
Thus Q = {
q
p
: p, q ? ? ? and q ? 0}. It may be noted that every integer is a rational numbers. If
not integer then either finite or recurring.
(xi) Irrational Numbers : There are real numbers which cannot be expressed in p/q form. These
numbers are called irrational numbers and their set is denoted by Q
c
or Q’. (i.e. complementary
BASIC MATHEMATICS & LOGARITHM
BASIC MATHEMATICS & LOGARITHM
Basic Mathematics & Logarithm – Nirmaan TYCRP
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi | 011-32044009 2
set of Q) e.g.
2
, 1 +
3
, e, ? etc. Irrational numbers can not be expressed as recurring
decimals.
Remark : e ? 2.71 is called Napier’s constant and ? ? 3.14.
Surds : If a is not a perfect nth power, then
n
a
is called a surd of the nth order. .
In an expression of the form
c b
a
?
, the denominator can be rationalized by multiplying numerator
and the denominator by
c b ?
which is called the conjugate of
c b ?
.
If x + y = a +
b
where x, y, a, b are rationals, then x = a and y = b. .
Ex.1 Prove that log
3
5 is irrational.
Sol. Let log
3
5 is rational.
? log
3
5 =
q
p
; where p and q are co-prime numbers
? 3
p/q
= 5 ? 3
p
= 5
q
. which is not possible, hence our assumption is wrong and log
3
5 is irrational.
Ex.2 Simply (make the denominator rational)
2 2 5 3
12
? ?
Sol. The expression =
5 6 6
) 2 2 5 3 ( 12
) 2 2 ( ) 5 3 (
) 2 2 5 3 ( 12
2 2
?
? ?
?
? ?
? ?
=
4
) 2 2 10 2 5 2 2 ( 2
) 1 5 ( ) 1 5 (
) 1 5 )( 2 2 5 3 ( 2 ? ? ?
?
? ? ?
? ? ?
=
2 10 5 1 ? ? ?
Ex.3 Find the factor which will rationalize
3
5 3 ?
Sol. Let x = 3
1/2
and y = 5
1/3
. The L.C.M. of the denominators of the indices 2 and 3 is 6. Hence x
6
and y
6
are
rational. Now x
6
+ y
6
= (x + y) (x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
)
Hence the rationalizing factor required = x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
where x = 3
1/2
and y = 5
1/3
.
Ex.4 Find the square root of 7 + 2 10
Sol. Let
y x 10 2 7 ? ? ?
. Squaring, x + y + 10 2 7 xy 2 ? ?
Hence x + y = 7 and xy = 10. These two relation give x = 5, y = 2. Hence
2 5 10 2 7 ? ? ?
Remark : symbol stands for the positive square root only. .
Ex.5 Prove that
3
2
cannot be represented in the form p + q , where p and q are rational (q > 0 and is not
a perfect square).
Sol. Put
3
2
= p + q . Hence 2 = p
3
+ 3pq + (3p
2
+ q) q ,
Since q is not a perfect square, it must be 3p
2
+ q = 0, which is impossible.
(xii) Real Numbers : The complete set of rational and irrational numbers is the set of real numbers and
is denoted by R. Thus R = Q ? Q
c
. Real numbers can be represented as points of a line. This line is
called as real line or number line
Basic Mathematics & Logarithm – Nirmaan TYCRP
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All the real numbers follow the order property
i.e. if there are two distinct real numbers a
Negative side Positive side
–3 –2 –1 0 1 2 3 ? ?2
and b then either a < b or a > b.
Remark :
(a) Integers are rational numbers, but converse need not be true.
(b) Negative of an irrational number is an irrational number.
(c) Sum of a rational number and an irrational number is always an irrational number e.g. 2 +
3
(d) The product of a non zero rational number & an irrational number will always be an irrational number.
(e) If a ? Q and b ? Q, then ab = rational number, only if a = 0.
(f) Sum, difference, product and quotient of two irrational numbers need not be an irrational
number (it may be a rational number also).
(xiii) Complex Number : A number of the form a + ib is called complex number, where a, b ? R and
i =
1 ?
. Complex number is usually denoted by C.
Remark : It may be noted that N ? W ? ? ? ? ? Q ? R ? C.
Ex.6 Every number is one of the forms 5n, 5n ± 1, 5n ± 2.
Sol. For if any number is divided by 5, the remainder is one of the numbers 0, 1, 2, 5 – 2, 5 – 1.
Ex.7 Every square number is one of the forms 5n, 5n ± 1.
Sol. The square of every number is one of the forms (5m)
2
, (5m ± 1)
2
, (5m ± 2)
2
. If those are divided by 5,
the remainders are 0, 1, 4; and, since 4 = 5 – 1, the forms are 5n, 5n + 1, and 5n – 1.
Ex.8 Show that the number of primes in N is infinite.
Sol. Suppose the number of primes in N is finite. Let {p
1
, p
2
...., p
n
} be the set of primes in N such that
p
1
< p
2
<.... < p
n
. Consider n = 1 + p
1
p
2
......p
n
. Clearly n is not divisible by any one of p
1
, p
2
,...., p
n
.
Hence n itself is a prime and n has a prime divisor other than p
1
, p
2
....p
n
. This contradicts that the set
of primes is {p
1
, p
2
....,p
n
}. Therefore the number of primes in N is infinite.
Ex.9 If x and y are prime numbers which satisfy x
2
– 2y
2
= 1, solve for x and y.
Sol. x
2
– 2y
2
= 1 gives x
2
= 2y
2
+ 1 and hence x must be an odd number. If x = 2n + 1, then x
2
= (2n + 1)
2
= 4n
2
+ 4n + 1 = 2y
2
+ 1. Therefore y
2
= 2n(n + 1). This means that y
2
is even and hence y is an even
integer. Now, y is also a prime implies that y = 2. This gives x = 3. Thus the only solution is x = 3, y = 2.
B. DIVISIBILITY TEST :
(i) A number will be divisible by 2 iff the digit at the unit place is divisible by 2.
(ii) A number will be divisible by 3 iff the sum of its digits of the number is divisible by 3.
(iii) A number will be divisible by 4 iff last two digits of the number together are divisible by 4.
(iv) A number will be divisible by 5 iff digit at the unit place is either 0 or 5.
(v) A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 & sum
of all digits of the number is divisible by 3.
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(vi) A number will be divisible by 8 iff the last 3 digits, all together, is divisible by 8.
(vii) A number will be divisible by 9 iff sum of all it’s digits is divisible by 9.
(viii) A number will be divisible by 10 iff it’s last digit is 0.
(ix) A number will be divisible by 11 iff the difference between the sum of the digits at even places
and sum of the digits at odd places is a multiple of 11.
e.g. 1298, 1221, 123321, 12344321, 1234554321, 123456654321, 795432
Ex.10 Prove that :
(a) the sum ab +ba is multiple of 11;
(b) a three-digit number written by one and the same digit is entirely divisible by 37.
Sol. (a)
ba ab ?
= (10a + b) + (10b + a) = 11(a + b);
(b)
aaa
= 100a + 10a + a = 111a = 37.3a.
Ex.11 Prove that the difference 10
25
– 7 is divisible by 3.
Sol. Write the given difference in the form 10
25
– 7 = (10
25
– 1) – 6. The number 10
25
– 1 =
? ? ?
digits 25
9 .. 99
is
divisible by 3(and 9). Since the numbers (10
25
– 1) and 6 are divisible by 3, the number 10
25
– 7, being
their difference, is also divisible by 3 without a remainder.
Ex.12 If the number A 3 6 4 0 5 4 8 9 8 1 2 7 0 6 4 4 B is divisible by 99 then the ordered pair of digits
(A, B) is
Sol. S
O
= A + 37 ; S
E
= B + 34 ? ? A ? B + 3 = 0 or 11 and A + B + 71is a multiple of 9
? A ? B = ?
3 or 8 and A + B = 1 or 10 Ans. : (9, 1)
Ex.13 Consider a number N = 2 1 P 5 3 Q 4. Find the number of ordered pairs (P, Q) so that the number ‘N is
divisible by 44, is
Sol. S
O
= P + 9, S
E
= Q + 6 ? S
O
– S
E
= P – Q + 3
‘N’ is divisible is 11 if P – Q +3 = 0, 11
P – Q = –3 .......(i) or P – Q = 8 ....(ii)
N is divisible by 4 if Q = 0, 2, 4, 6, 8
From Equation (i)
Q = 0 P = –3 (not possible) Q = 2 P = –1 (not possible)
Q = 4 P = 1 Q = 6 P = 3 Q = 8 P = 5
? number of ordered pairs is 3
From equation (ii)
Q = 0 P = 8 Q = 2 P = 10 (not possible) similarly Q ? 4, 6, 8
? No. of ordered pairs is 1
? total number of ordered pairs, so that number ‘N’ is divisible by 44, is 4
Ex.14 Prove that the square of any prime number p ? 5, when divided by 12, gives 1 as remainder.
Sol. When divided by 6, a natural number can give as a remainder only the numbers 0, 1, 2, 3, 4 and 5.
Therefore, any natural number has one of the following forms :
6k, 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.
Page 5
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A. NUMBER SYSTEM
(i) Natural Numbers : The counting numbers 1, 2, 3, 4,..... are called Natural Numbers. The set of
natural numbers is denoted by N. Thus N = {1, 2, 3, 4,......}. N is also denoted by ?
+
or Z
+
(ii) Whole Numbers : Natural numbers including zero are called whole numbers. The set of whole
numbers, is denoted by W. Thus W = {0, 1, 2,......}. W is also called as set of non-negative integers.
(iii) Integers : The numbers..... – 3, – 2, –1, 0, 1, 2, 3..... are called integers and the set is
denoted by ? or Z. Thus ? (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........}
(a) Set of positive integers, denoted by ?
+
and consists of {1, 2, 3, ...........}
(b) Set of negative integers, denoted by ?
–
and consists of {..........., –3, –2, –1}
(c) Set of non-negative integers {0, 1, 2, 3,...........}
(d) Set of non-positive integers {...., –3, –2, –1, 0}
(iv) Even Integers : Integers which are divisible by 2 are called even integers. e.g. 0, ± 2, ± 4,.....
(v) Odd Integers : Integers which are not divisible by 2 are called as odd integers. e.g. ± 1, ± 3, .......
(vi) Prime Number : Let ‘p’ be a natural number, ‘p’ is said to be prime if it has exactly two distinct
factors, namely 1 and itself. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,......
Remark : (a) ‘1’ is neither prime nor composite.
(b) ‘2’ is the only even prime number.
(vii) Composite Number : Let ‘a’ be a natural number, ‘a’ is said to be composite if, it has atleast
three distinct factors.
(viii) Co-prime Numbers : Two natural numbers (not necessarily prime) are coprime, if their H.C.F.
(Highest common factor) is one. e.g. (1, 2), (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
These numbers are also called as relatively prime numbers.
Remark : (a) Number which are not prime are composite numbers (except 1)
(b) ‘4’ is the smallest composite number.
(c) Two distinct prime numbers are always co-prime but converse need not be true.
(d) Consecutive numbers are always co-prime numbers.
(ix) Twin Prime Numbers : If the difference between two prime numbers is two, then the numbers
are called as twin prime numbers. e.g. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
(x) Rational Numbers : All the numbers those can be represented in the form p/q, where p and q
are integers and q ? 0, are called rational numbers and their set is denoted by Q.
Thus Q = {
q
p
: p, q ? ? ? and q ? 0}. It may be noted that every integer is a rational numbers. If
not integer then either finite or recurring.
(xi) Irrational Numbers : There are real numbers which cannot be expressed in p/q form. These
numbers are called irrational numbers and their set is denoted by Q
c
or Q’. (i.e. complementary
BASIC MATHEMATICS & LOGARITHM
BASIC MATHEMATICS & LOGARITHM
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set of Q) e.g.
2
, 1 +
3
, e, ? etc. Irrational numbers can not be expressed as recurring
decimals.
Remark : e ? 2.71 is called Napier’s constant and ? ? 3.14.
Surds : If a is not a perfect nth power, then
n
a
is called a surd of the nth order. .
In an expression of the form
c b
a
?
, the denominator can be rationalized by multiplying numerator
and the denominator by
c b ?
which is called the conjugate of
c b ?
.
If x + y = a +
b
where x, y, a, b are rationals, then x = a and y = b. .
Ex.1 Prove that log
3
5 is irrational.
Sol. Let log
3
5 is rational.
? log
3
5 =
q
p
; where p and q are co-prime numbers
? 3
p/q
= 5 ? 3
p
= 5
q
. which is not possible, hence our assumption is wrong and log
3
5 is irrational.
Ex.2 Simply (make the denominator rational)
2 2 5 3
12
? ?
Sol. The expression =
5 6 6
) 2 2 5 3 ( 12
) 2 2 ( ) 5 3 (
) 2 2 5 3 ( 12
2 2
?
? ?
?
? ?
? ?
=
4
) 2 2 10 2 5 2 2 ( 2
) 1 5 ( ) 1 5 (
) 1 5 )( 2 2 5 3 ( 2 ? ? ?
?
? ? ?
? ? ?
=
2 10 5 1 ? ? ?
Ex.3 Find the factor which will rationalize
3
5 3 ?
Sol. Let x = 3
1/2
and y = 5
1/3
. The L.C.M. of the denominators of the indices 2 and 3 is 6. Hence x
6
and y
6
are
rational. Now x
6
+ y
6
= (x + y) (x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
)
Hence the rationalizing factor required = x
5
– x
4
y + x
3
y
2
– x
2
y
3
+ xy
4
– y
5
where x = 3
1/2
and y = 5
1/3
.
Ex.4 Find the square root of 7 + 2 10
Sol. Let
y x 10 2 7 ? ? ?
. Squaring, x + y + 10 2 7 xy 2 ? ?
Hence x + y = 7 and xy = 10. These two relation give x = 5, y = 2. Hence
2 5 10 2 7 ? ? ?
Remark : symbol stands for the positive square root only. .
Ex.5 Prove that
3
2
cannot be represented in the form p + q , where p and q are rational (q > 0 and is not
a perfect square).
Sol. Put
3
2
= p + q . Hence 2 = p
3
+ 3pq + (3p
2
+ q) q ,
Since q is not a perfect square, it must be 3p
2
+ q = 0, which is impossible.
(xii) Real Numbers : The complete set of rational and irrational numbers is the set of real numbers and
is denoted by R. Thus R = Q ? Q
c
. Real numbers can be represented as points of a line. This line is
called as real line or number line
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All the real numbers follow the order property
i.e. if there are two distinct real numbers a
Negative side Positive side
–3 –2 –1 0 1 2 3 ? ?2
and b then either a < b or a > b.
Remark :
(a) Integers are rational numbers, but converse need not be true.
(b) Negative of an irrational number is an irrational number.
(c) Sum of a rational number and an irrational number is always an irrational number e.g. 2 +
3
(d) The product of a non zero rational number & an irrational number will always be an irrational number.
(e) If a ? Q and b ? Q, then ab = rational number, only if a = 0.
(f) Sum, difference, product and quotient of two irrational numbers need not be an irrational
number (it may be a rational number also).
(xiii) Complex Number : A number of the form a + ib is called complex number, where a, b ? R and
i =
1 ?
. Complex number is usually denoted by C.
Remark : It may be noted that N ? W ? ? ? ? ? Q ? R ? C.
Ex.6 Every number is one of the forms 5n, 5n ± 1, 5n ± 2.
Sol. For if any number is divided by 5, the remainder is one of the numbers 0, 1, 2, 5 – 2, 5 – 1.
Ex.7 Every square number is one of the forms 5n, 5n ± 1.
Sol. The square of every number is one of the forms (5m)
2
, (5m ± 1)
2
, (5m ± 2)
2
. If those are divided by 5,
the remainders are 0, 1, 4; and, since 4 = 5 – 1, the forms are 5n, 5n + 1, and 5n – 1.
Ex.8 Show that the number of primes in N is infinite.
Sol. Suppose the number of primes in N is finite. Let {p
1
, p
2
...., p
n
} be the set of primes in N such that
p
1
< p
2
<.... < p
n
. Consider n = 1 + p
1
p
2
......p
n
. Clearly n is not divisible by any one of p
1
, p
2
,...., p
n
.
Hence n itself is a prime and n has a prime divisor other than p
1
, p
2
....p
n
. This contradicts that the set
of primes is {p
1
, p
2
....,p
n
}. Therefore the number of primes in N is infinite.
Ex.9 If x and y are prime numbers which satisfy x
2
– 2y
2
= 1, solve for x and y.
Sol. x
2
– 2y
2
= 1 gives x
2
= 2y
2
+ 1 and hence x must be an odd number. If x = 2n + 1, then x
2
= (2n + 1)
2
= 4n
2
+ 4n + 1 = 2y
2
+ 1. Therefore y
2
= 2n(n + 1). This means that y
2
is even and hence y is an even
integer. Now, y is also a prime implies that y = 2. This gives x = 3. Thus the only solution is x = 3, y = 2.
B. DIVISIBILITY TEST :
(i) A number will be divisible by 2 iff the digit at the unit place is divisible by 2.
(ii) A number will be divisible by 3 iff the sum of its digits of the number is divisible by 3.
(iii) A number will be divisible by 4 iff last two digits of the number together are divisible by 4.
(iv) A number will be divisible by 5 iff digit at the unit place is either 0 or 5.
(v) A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 & sum
of all digits of the number is divisible by 3.
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(vi) A number will be divisible by 8 iff the last 3 digits, all together, is divisible by 8.
(vii) A number will be divisible by 9 iff sum of all it’s digits is divisible by 9.
(viii) A number will be divisible by 10 iff it’s last digit is 0.
(ix) A number will be divisible by 11 iff the difference between the sum of the digits at even places
and sum of the digits at odd places is a multiple of 11.
e.g. 1298, 1221, 123321, 12344321, 1234554321, 123456654321, 795432
Ex.10 Prove that :
(a) the sum ab +ba is multiple of 11;
(b) a three-digit number written by one and the same digit is entirely divisible by 37.
Sol. (a)
ba ab ?
= (10a + b) + (10b + a) = 11(a + b);
(b)
aaa
= 100a + 10a + a = 111a = 37.3a.
Ex.11 Prove that the difference 10
25
– 7 is divisible by 3.
Sol. Write the given difference in the form 10
25
– 7 = (10
25
– 1) – 6. The number 10
25
– 1 =
? ? ?
digits 25
9 .. 99
is
divisible by 3(and 9). Since the numbers (10
25
– 1) and 6 are divisible by 3, the number 10
25
– 7, being
their difference, is also divisible by 3 without a remainder.
Ex.12 If the number A 3 6 4 0 5 4 8 9 8 1 2 7 0 6 4 4 B is divisible by 99 then the ordered pair of digits
(A, B) is
Sol. S
O
= A + 37 ; S
E
= B + 34 ? ? A ? B + 3 = 0 or 11 and A + B + 71is a multiple of 9
? A ? B = ?
3 or 8 and A + B = 1 or 10 Ans. : (9, 1)
Ex.13 Consider a number N = 2 1 P 5 3 Q 4. Find the number of ordered pairs (P, Q) so that the number ‘N is
divisible by 44, is
Sol. S
O
= P + 9, S
E
= Q + 6 ? S
O
– S
E
= P – Q + 3
‘N’ is divisible is 11 if P – Q +3 = 0, 11
P – Q = –3 .......(i) or P – Q = 8 ....(ii)
N is divisible by 4 if Q = 0, 2, 4, 6, 8
From Equation (i)
Q = 0 P = –3 (not possible) Q = 2 P = –1 (not possible)
Q = 4 P = 1 Q = 6 P = 3 Q = 8 P = 5
? number of ordered pairs is 3
From equation (ii)
Q = 0 P = 8 Q = 2 P = 10 (not possible) similarly Q ? 4, 6, 8
? No. of ordered pairs is 1
? total number of ordered pairs, so that number ‘N’ is divisible by 44, is 4
Ex.14 Prove that the square of any prime number p ? 5, when divided by 12, gives 1 as remainder.
Sol. When divided by 6, a natural number can give as a remainder only the numbers 0, 1, 2, 3, 4 and 5.
Therefore, any natural number has one of the following forms :
6k, 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.
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it is obvious that the numbers 6k, 6k + 2, 6k + 3, and 6k + 4 are composite. Therefore, the prime
number p ? 5 has the form 6k + 1 or 6k + 5.
If p = 6k + 1, then p
2
= (6k + 1)
2
= 36k
2
+ 12k + 1.
If p = 6k + 5, then p
2
= (6k + 5)
2
= 36k
2
+ 60k + 25 = 12(3k
2
+ 5k + 2) + 1.
Thus, in both cases, when dividing p
2
by 12, the remainder is equal to 1.
Ex.15 Prove that for every positive integer n, 1
n
+ 8
n
– 3
n
– 6
n
is divisible by 10.
Sol. Since 10 is the product of two primes 2 and 5, it will suffice to show that the given expression is
divisible both by 2 and 5. To do so, we shall use the simple fact that if a and b be any positive integers,
then a
n
– b
n
is always divisible by a –b.
Writing A ? 1
n
+ 8
n
– 3
n
– 6
n
, = (8
n
– 3
n
) – (6
n
– 1
n
),
we find that 8
n
– 3
n
and 6
n
– 1
n
are both divisible by 5, and consequently A is divisible by 5 (= 8 – 3 =
6 – 1). Again, writing A = (8
n
– 6
n
) – (3
n
– 1
n
), we find that A is divisible by 2(= 8 – 6 = 3 – 1). Hence
A is divisible by 10.
C. (1) LCM AND HCF
(i) HCF is highest common factor between any two or more numbers (or algebraic expression)
when only take numbers Its called highest common divisor.
(ii) LCM is least common multiple between any two or more numbers (or algebraic expression)
(iii) Multiplication of LCM and HCF of two numbers is equal to multiplication of two numbers.
(iv) LCM of ?
?
?
?
?
?
?
?
m
,
q
p
,
b
a ?
=
LCM of (a, p, )
HCF of (b, q, m)
?
(v) HCF of ?
?
?
?
?
?
?
?
m
,
q
p
,
b
a ?
=
HCF of (a, p, )
LCM of (b, q, m)
?
(vi) LCM of rational and irrational number is not defined.
(2) Remainder Theorem : Let P(x) be any polynomial of degree greater than or equal to one and
‘a’ be any real number. If P(x)is divided (x – a), then the remainder is equal to P(a).
(3) Factor Theorem : Let P(x) be polynomial of degree greater than of equal to 1 and ‘a’ be a real
number such that P(a) = 0, then (x – a) is a factor of P(x). Conversely, if (x – a) is a factor of
P(x), then P(a) = 0.
(4) Some Important Identities :
(i) (a + b)
2
= a
2
+ 2ab + b
2
= (a – b)
2
+ 4ab
(ii) (a – b)
2
= a
2
– 2ab + b
2
= (a + b)
2
– 4ab
(iii) a
2
– b
2
= (a + b) (a – b)
(iv) (a + b)
3
= a
3
+ b
3
+ 3ab (a + b)
(v) (a – b)
3
= a
3
– b
3
– 3ab (a – b)
(vi) a
3
+ b
3
= (a + b)
3
– 3ab (a + b) = (a + b) (a
2
+ b
2
– ab)
(vii) a
3
– b
3
= (a – b)
3
+ 3ab (a – b) = (a – b) (a
2
+ b
2
+ ab)
(viii) (a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca = a
2
+ b
2
+ c
2
+ 2abc
?
?
?
?
?
?
? ?
c
1
b
1
a
1
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