Real Numbers
Class9Mathematics1Chapter2Maharashtra Board
Notes
Topics to be learn : Part 1

Number System :
(i) Natural numbers : The counting numbers 1, 2, 3, 4, ... are called natural numbers.
 These numbers are written in the set form as N = {1, 2, 3, 4, ... }
 The set of natural numbers is adequate for addition and multiplication but not for subtraction and division.
 Examples : 2 + 5 = 7, 2 × 5 = 10. Here, 7 and 10 are natural numbers but 2 −5 ∉ N; 2 ÷ 5 ∉
(ii) Whole numbers : The union of set of natural numbers and zero is a set of whole numbers. It is denoted by W.
 These numbers are written in the set form as W = {0, 1, 2, 3, ... }
(iii) Integers : The set of natural numbers, zero and opposite of all natural numbers is called set of integers and denoted by I.
 These numbers are written in the set form as I = { ... , −3, −2, − 1, 0, 1, 2, 3, ... }
 The order relation on I on the number line is defined as follows : ... , −3 < − 2 < − 1 < 0 < 1 < 2 < 3, ...
 The set of integers is adequate for addition, subtraction and multiplication but not for division. e.g. 3 ÷ 9 ∉
(iv) Rational Numbers : If p and q are integers (q ≠ 0), then the number \(\frac{p}{q}\) is called a rational number
 The set of rational numbers is denoted by Q.
 Q = { \(\frac{p}{q}\)  p, q ∉ I and q ≠ 0} ……(Rule method)
 The set of real numbers is denoted by R.
Know This :
(i) In the deﬁnition of rational numbers, the condition that q ≠ 0 is necessary, since division by zero is not deﬁned. (ii) Integers are also rational numbers because every integer p = \(\frac{p}{1}\). Hence, I ⊆ Q. (iii) N ⊆ W ⊆ I ⊆ Q ⊆ R. 
Properties of rational numbers :
(A) Properties of addition of rational numbers :
If a, b, c are any rational numbers, then
(i) Commutative property : a + b = b + a
(ii) Associative property : (a + b) + c = a + (b + c)
(iii) Additive identity : a + 0 = 0 + a = a
 0 is called the identity element for addition]
(iv) Additive inverse : a + ( — a) = 0
 For two rational numbers to be additive inverses of each other: their sum should be 0.
(B) Properties of multiplication of rational numbers :
(i) Commutative property : a × b = b × a ()
(ii) Associative property : (a × b) × c = a × (b × c)
(iii) Multiplicative identity : a × 1 = 1 × a = a
 1 is called the identity element for multiplication.
(iv) Multiplicative inverse : a × \(\frac{1}{1}\) = \(\frac{1}{a}\) × a = 1 (a ≠ 0)
 For two rational numbers to be multiplicative inverses of each other; their product should be 1.
Know This :
Decimal Representation of Rational Numbers :
The decimal representation of rational numbers is either (i) terminating or (ii) non−terminating recurring. Example :

To express the recurring decimal in \(\frac{p}{q}\) form
Ex. (1) Express the recurring decimal 0.777.... in \(\frac{p}{q}\) form.
Solution : Let x = 0.777... = \(0.\dot 7\)
Here, digit 7 is the only recurring digit after the decimal point. Hence to convert \(0.\dot 7\) into \(\frac{p}{q}\) form multiply \(0.\dot 7\) by 10.
∴ 10 x = 7.777... = \(7.\dot 7\)
∴ 10x − x = \(7.\dot 7\) − \(0.\dot 7\)
∴ 9x = 7
∴ x = \(\frac{7}{9}\)
∴ 0.777... = \(\frac{7}{9}\)
Answer : \(\frac{7}{9}\) is the \(\frac{p}{q}\) form of the recurring decimal 0.777....
Ex. (2) Express the recurring decimal 7.529529529.... in \(\frac{p}{q}\) form.
Let x = 7.529529529... = \(7.\overline{529}\)
Here, three digit 5, 2 and 9 recurring after the decimal point. Hence to convert \(7.\overline{529}\) into \(\frac{p}{q}\) form multiply \(7.\overline{529}\) by 1000.
∴ 1000 x = \(7529.\overline{529}\)
∴ 1000x − x = 7529. − \(7.\overline{529}\)
∴ 999x = 7522.0
∴ x = \(\frac{7522}{999}\)
∴ \(7.\overline{529}\) = \(\frac{7522}{999}\)
Answer : \(\frac{7522}{999}\) is the \(\frac{p}{q}\) form of the recurring decimal 7.529529529....
Remember :
(1) Note the number of recurring digits after decimal point in the given rational number. Accordingly multiply it by 10, 100, 1000 e.g.
(2) If the prime factors of the denominator of a rational number are 2 or 5 only then its decimal expansion is terminating. If the prime factors are other than 2 or 5 also then its decimal expansion is non terminating and recurring. 
Irrational and real numbers
Irrational number :
 The number which is not rational is called an irrational number.
 The set of irrational numbers is denoted by 'Q’.
 The decimal representation of an irrational number is non−terminating, non−recurring.
Real Numbers :
 The numbers which are represented by points on a number line are real numbers.
 Every point on the number line is associated with a unique real number and every real number is associated with a unique point on the number line.
 Every rational number is a real number. But every real number may not be a rational number.
 Numbers \(\sqrt{2},\sqrt{3},\sqrt{5}\) .. etc. can be shown on a number line but these are not the rational numbers.
Proof : \(\sqrt{2}\) is irrational number.
This can be proved using indirect proof.
Let us assume that \(\sqrt{2}\) is rational. So \(\sqrt{2}\) can be expressed in \(\frac{p}{q}\) form.
\(\frac{p}{q}\) is the reduced form of rational number hence p and q have no common factor other than 1.
\(\sqrt{2}\) = \(\frac{p}{q}\), ∴ 2 = \(\frac{p^2}{q^2}\) …..(Squaring both the sides)
∴ 2q^{2} = p^{2}
∴ p^{2} is even.
∴ p is also even means 2 is a factor of p. ....(I)
∴ p = 2t ∴ p^{2} = 4t^{2} …t ∈ I
∴ 2q^{2} = 4t^{2} ( ∴ p^{2} = 2q^{2}) ∴ q^{2} = 2t^{2} ∴ q2 is even. ∴ q is even.
∴ 2 is a factor of q. .... (II)
From the statement (I) and (II), 2 is a common factor of p and q both.
This is contradictory because in \(\frac{p}{q}\) ; we have assumed that p and q have no common factor except 1.
∴ Our assumption that \(\sqrt{2}\) is rational is wrong.
∴ \(\sqrt{2}\) is irrational number.
Similarly, one can prove that \(\sqrt{3}\), \(\sqrt{5}\) are irrational numbers.
Numbers \(\sqrt{2},\sqrt{3},\sqrt{5}\) can be shown on a number line.
We know that every rational number is a real number. But \(\sqrt{2},\sqrt{3},\sqrt{5}\) π, 3 +\(\sqrt{5}\) etc. are not rational numbers. It means that Every real number may not be a rational number.
Decimal form of irrational numbers :
Let us ﬁnd the square root of 2 using division method :
∴ \(\sqrt{2}\) = 1.41421…
In the above division, numbers after decimal point are unending, means it is non−terminating. Even no group of numbers or a single number is repeating in its quotient. So decimal expansion of such numbers is non terminating, non−recurring.
 A number whose decimal expansion is non−terminating, non−recurring is irrational.
Properties of irrational numbers
 Addition or subtraction of a rational number with irrational number is an irrational number.
 Multiplication or division of non zero rational number with irrational number is also an irrational number.
 Addition, subtraction, multiplication and division of two irrational numbers can be either a rational or irrational number.
Properties of order relation on Real numbers :
 If a and b are two real numbers then only one of the relations holds good.
 e. a = b or a < b or a > b
 If a < b and b < c then a < c
 If a < b then a + c < b + c
 If a < b and c > 0 then ac < bc and If c < 0 then ac > bc
Square root of a Negative number :
We know that, if \(\sqrt{a}\) = b then b^{2 }= a.
Hence if \(\sqrt{5}\) = x then x^{2} = 5.
Similarly we know that square of any real number is always non−negative. It means that square of any real number is never negative.
But ( \(\sqrt{5}\) )^{2} = − 5, ∴ \(\sqrt{5}\) is not a real number.
Hence square root of a negative real number is not a real number.
Root of positive rational number :
If n is a positive integer and x^{n} = a, then x is the n^{th} root of a . x = \(\sqrt [n] {a}\) . This root may be rational or irrational.
For example, 2^{5} = 32, ∴ 2 is the 5th root of 32, but if x^{5} = 2 then x = \(\sqrt [5] {2}\) , which is an irrational number.
Surds :
If n is an integer greater than 1 and if n^{th} root of a is x, and x is a positive real number, then it is written as x^{n} = a or \(\sqrt [n] {a}\) = x.
If a is a positive rational number and n^{th} root of a is x and if x is an irrational number, then x is called a surd.
In the surd \(\sqrt [n] {a}\) the symbol √¯ is called radical sign, n is called the order of the surd and a is called the radicand.
Examples :
(1) If a = 7, n = 3, then \(\sqrt [3] {7}\) is a surd because \(\sqrt [3] {7}\) is an irrational number.
(2) If a = 27, n = 3 then \(\sqrt [3] {27}\) is not a surd because \(\sqrt [3] {27}\) = 3 is not an irrational number.
(3) \(\sqrt [3] {8}\) is a surd or not ?
Let = p, ∴ p^{3} = 8. We know 2 is cube−root of 8.
2 is not an irrational number. .
∴ \(\sqrt [3] {8}\) is not a surd.
 This year we are going to study surds of order 2 only, means
 The surds of order 2 is called Quadratic surd.
Simplest form of a Surd :
A surd of order 'n' is said to be in its simplest form if it has the following properties:
(1) The radicand has no factor which is the nth power of a rational number.
(2) The radicand is not a fraction.
(3) 'n' is the least order of the surd.
e.g. (i) 27 = \(\sqrt {9×3}\) = \(\sqrt {9}×\sqrt {9}\) = \(3\sqrt {3}\)
(ii) 50 = \(\sqrt {25×2}\) = \(\sqrt {25}×\sqrt {2}\) = \(5\sqrt {2}\)
\(\sqrt {2}\),\(\sqrt {3}\), \(\sqrt {5}\) .... these type of surds are in the simplest form which cannot be simplified further.
Similar or like surds :
If p and q are rational numbers then p\(\sqrt {a}\), q\(\sqrt {a}\) are called like surds. Two surds are said to be like surds if their order is equal and radicands are equal.
e.g. 3\(\sqrt {2}\), −4\(\sqrt {2}\), \(\frac{5}{3}\sqrt {2}\) are some like surds.
[Note : 3\(\sqrt {2}\), −4\(\sqrt {5}\), \(\frac{5}{3}\sqrt {3}\) are not some like surds.]
Question : \(\sqrt {45}\) and \(\sqrt {80}\) are the surds of order 2, so their order is equal but radicands are not same, Are these like surds?
Let us simplify \(\sqrt {45}\) and \(\sqrt {80}\)
\(\sqrt {45}\) = \(\sqrt {9×5}\) = \(\sqrt {9}×\sqrt {5}\) = \(3\sqrt {5}\) …..(i)
\(\sqrt {80}\) = \(\sqrt {16×5}\) = \(\sqrt {16}×\sqrt {5}\) = \(4\sqrt {5}\) ….(ii)
\(3\sqrt {5}\) and \(4\sqrt {5}\) are now similar or like surds means \(\sqrt {45}\) and \(\sqrt {80}\) are similar surds.
Remember :
In the simplest form of the surds if order of the surds and redicand are equal then the surds are similar or like surds. 
Comparison of surds :
Two surds of the same order can be compared by comparing the radicands.
Let a and b be two positive real numbers.
If a < b, then a × a < b × a.
∴ a^{2} < ab ….(1)
If a < b, then a × b < b × b.
∴ ab < b^{2 }….(2)
From (1) and (2), a^{2} < ab < b^{2}.
∴ a^{2} < b^{2}
If a > b, then a^{2} > b^{2} and if
a = b, then a^{2} = b^{2}.
Question : Compare the surds \(2\sqrt {7}\) and \(5\sqrt {3}\) .
\(2\sqrt {7}\) = \(\sqrt {4}×\sqrt{7}\) = \(\sqrt {4×7}\) = \(\sqrt {28}\)
\(5\sqrt {3}\) = \(\sqrt {25}×\sqrt{3}\) = \(\sqrt {25×3}\) = \(\sqrt {75}\
∴ \(\sqrt {28}\) < \(\sqrt {75}\
∴ \(2\sqrt {7}\) < \(5\sqrt {3}\)
Operations on like surds :
Mathematical operations like addition, subtraction, multiplication and division can be done on like surds.
Examples :
(i) 7\(\sqrt {3}\) + 29\(\sqrt {3}\) = (7 + 29)\(\sqrt {3}\) = 36\(\sqrt {3}\)
(ii) 7\(\sqrt {5}\) − 29\(\sqrt {5}\) = (7 − 29)\(\sqrt {5}\) = −22\(\sqrt {5}\)
(iii) 7\(\sqrt {5}\) × 2\(\sqrt {7}\) = (7 × 2)\(\sqrt {5×7}\) = 14\(\sqrt {35}\)
(iv) 21\(\sqrt {125}\) ÷ 3\(\sqrt {5}\) = \(\frac{21\sqrt {125}}{3\sqrt{5}}\)
= \(7\sqrt{\frac{125}{5}}\) = \(7\sqrt {25}\) = 7 × 5 = 35 ..(rational number)
Note that product and quotient of two surds may be rational numbers.
Rationalization of surds :
If the product of two surds is a rational number, then each of the two surds is called a rationalizing factor of the other.
e.g. the product of the surds \(\sqrt {2}\) and \(\sqrt {8}\) is \(\sqrt {16}\) = 4, which is a rational number.
\(\sqrt {2}\) is the rationalizing factor of \(\sqrt {8}\) and \(\sqrt {8}\) is the rationalizing factor of \(\sqrt {2}\).
Question : Find the rationalizing factor of \(\sqrt {27}\)
\(\sqrt {27}\) = \(\sqrt {9× 3}\) = 3\(\sqrt {3}\)
∴ 3(\sqrt {3}\) × (\sqrt {3}\) = 3 × 3 = 9 is a rational number.
∴ (\sqrt {3}\) is the rationalizing factor of 3\(\sqrt {3}\).
Note that, \(\sqrt {27}\) = 3(\sqrt {3}\) means 3(\sqrt {3}\) × 3(\sqrt {3}\) = 9 × 3 = 27.
Hence 3(\sqrt {3}\) is also a rationalizing factor of \(\sqrt {27}\).
Question : Rationalize the denominator of \(\frac{1}{\sqrt {5}}\)
\(\frac{1}{\sqrt {5}}\) = \(\frac{1}{\sqrt {5}}×\frac{\sqrt{5}}{\sqrt{5}}\) = \(\frac{\sqrt{5}}{5}\) ......(multiply numerator and denominator by \(\sqrt{5}\).)
Remember :
The rationalizing factor of a given surd is not unique. If a particular surd is a rationalizing factor of a given surd, then the product of this factor and a rational number is also the rationalizing factor of the given surd. It is convenient to use the simplest of all rationalizing factors of a given surd. It is easy to use the numbers with rational denominators. That is why we rationalize it. 
Binomial quadratic surd :
The sum or difference of two numbers one of which is a quadratic surd, the other either a non−zero rational number or a quadratic surd is called binomial quadratic surd.
Examples :
(1) (i) a + x\(\sqrt {b}\) ; (ii) a − x\(\sqrt {b}\) ; (iii) x\(\sqrt {a}\) + y\(\sqrt {b}\) (iv) x\(\sqrt {a}\) − y\(\sqrt {b}\) are the binomial expressions of quadratic surds,
where \(\sqrt {a}\), \(\sqrt {b}\) are quadratic surds. a, b, x and y are non−zero rational numbers.
(2) (i) 5 + 3\(\sqrt{2}\) (ii) 3 − 2\(\sqrt {3}\) (iii) 2\(\sqrt {3}\) + 3\(\sqrt {5}\) (iv) 3\(\sqrt {2}\) — 5\(\sqrt {3}\) are also the examples of binomial quadratic surds.
The product of two binomial surds \((\sqrt {3}\sqrt {2})\) and \((\sqrt {3}+\sqrt {2})\) is \((\sqrt {3})^2(\sqrt {2})^2\) = 3 – 2 = 1.
1 is a rational number. Hence \((\sqrt {3}\sqrt {2})\) and \((\sqrt {3}+\sqrt {2})\) are the conjugate pairs of each other.
The product of two conjugate binomial surds is a rational number.
Rationalization of the denominator :
The product of two conjugate binomial surds is always a rational number. Using this property, the rationalization of the denominator in the form of binomial surd can be done.
Example :
(1) Rationalize the denominator : \(\frac{5}{3\sqrt{2}+\sqrt{5}}\)
The conjugate pair of \(3\sqrt{2}+\sqrt{5}\) is \(3\sqrt{2}\sqrt{5}\)
\(\frac{5}{3\sqrt{2}+\sqrt{5}}\) = \(\frac{5}{3\sqrt{2}+\sqrt{5}}\) × \(\frac{3\sqrt{2}\sqrt{5}}{3\sqrt{2}\sqrt{5}}\)
= \(\frac{5(3\sqrt{2}\sqrt{5})}{(3\sqrt{2})^2(\sqrt{5})^2}\)
= \(\frac{15\sqrt{2}5\sqrt{5})}{9×25}\)
= \(\frac{15\sqrt{2}5\sqrt{5})}{13}\)
Absolute value :
If x is a real number then absolute value of x is its distance from zero on the number line which is written as  x , and  x  is read as Absolute Value of x or modulus of x.
 If x > 0 then  x  = x If x is positive then absolute value of x is
 If x = 0 then  x = 0 If x is zero then absolute value of x is zero.
 If x < 0 then  x  = −x If x is negative then its absolute value is opposite of
Examples: (1) 3 = 3, −3 = −(−3) = 3, 0 = 0
The absolute value of any real number is never negative.
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