Real Numbers, Class 10, Maths Detailed Chapter Notes PDF Download

INTRODUCTION
"God gave us the natural number, all else is the work of man". It was exclaimed by L. Kronecker (1823-1891), the reputed German Mathematician. This statement reveals in a nut shell the significant role of the universe of numbers played in the evolution of human thought.
N : The set of natural numbers,
W : The set of whole numbers,
Z : The set of Integers, 
Q : The set of rationals,
R : The set of Real Numbers

HISTORICAL FACTS

Dedekind was the first modern mathematician to publish in 1872 the mathematically rigorous definition of irrational numbers. He gave explanation of their place in the real Numbers System. He was able to demonstrate the completeness of the real number line. He filled in the 'holes' in the system of Rational numbers with irrational Numbers. This innovation has made Richard Dedekind an immortal figure in the history of Mathematics.

Srinivasa Ramanujan (1887-1920) was one of the most outstanding mathematician that India has produced. He worked on history of Numbers and discovered wonderul properties of numbers. He stated intuitively many complicated result in mathematics. Once a great mathematician Prof. Hardy come to India to see Ramanujan. Prof. Hardy remarked that he has travelled in a taxi with a rather dull number viz. 1729. Ramanujan jumped up and said, Oh! No. 1729 is very interesting number. It is the smallest number which can be expressed as the sum of two cubes in two different
ways.
viz 1729 = 1³ + 12³,
1729 = 9³ + 10³,
1729 = 1³ + 12³ = 9³ + 10³

RECALL

In our day to day life, we deal with different types of numbers which can be broadly classified as follows.

Natural numbers (N) : N = {1, 2, 3, 4...∞}

Remark :

(i) The set N is infinite i.e. it has unlimited members.
(ii) N has the smallest element namely '1'.
(iii) N has no largest element. i.e., give me any natural number, we can find the bigger number from
the given number.
(iv) N does not contain '0' as a member. i.e. '0' is not a member of the set N.

(ii) Whole numbers (W) : W = {0, 1, 2, 3, 4...∞}

Remark :

(i) The set of whole number is infinite (unlimited elements).
(ii) This set has the smallest members as '0'. i.e. '0' the smallest whole number. i.e., set W contain
'0' as a member.
(iii) The set of whole numbers has no largest member.
(iv) Every natural number is a whole number.
(v) Non-zero smallest whole number is '1'.

(iii) Integers (I or Z) : I or Z = {– ∞... –3, –2, –1, 0, +1, +2, +3 ...+ ∞}

Positive integers : {1, 2, 3...},

Negative integers : {.... –4, –3, –2, –1}

Remark : (i) This set Z is infinite.
(ii) It has neither the greatest nor the least element.
(iii) Every natural number is an integer.
(iv) Every whole number is an integer.
(iv) The set of non-negative integer = {0, 1, 2, 3, 4,....}
(v) The set of non-positive integer = {......–4, – 3, – 2, –1, 0}

Rational numbers :– These are real numbers which can be expressed in the form of p/q , where p and q are integers and q = 0. 

Remark :

(i) Every integer is a rational number.

(ii) Every terminating decimal is a rational number.

(iii) Every recurring decimal is a rational number.

(iv) A non- terminating repeating decimal is called a recurring decimal.

(v) Between any two rational numbers there are an infinite number of rational property is known as the density of rational numbers. numbers. This

(vi) If a and b are two rational numbers then lies between a and b.

n rational numbers between two different rational numbers a and b are :

(vii) Every rational number can be reppresented either as a terminating decimal or as a non-terminating

(viii) Types of rational numbers :–

(a) Terminating decimal numbers and

(b) Non-terminating repeating (recurring) decimal numbers

(v) Irrational numbers :– A number is called irrational number, if it can not be written in the form p/q , where p & q are integers and q= 0. All Non-terminating & Non-repeating decimal numbers are Irrational numbers.

Real numbers :– The totality of rational numbers and irrational numbers is called the set of real numbers i.e. rational numbers and irrational numbers taken together are called real numbers. Every real number is either a rational number or an irrational number.

NATURE OF THE DECIMAL EXPANSION OF RATIONAL NUMBERS

Theorem-1 : Let x be a rational number whose decimal expansion terminates. Then we can express x in the form p/q , where p and q are co-primes, and the prime factorisation of q is of the form 2^m × 5ⁿ, where m,n are non-negative integers.

Theorem-2 : Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2^m × 5ⁿ, where m,n are non-negative integers . Then, x has a decimal expansion which terminates.

Theorem-3 : Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2^m × 5ⁿ, Where m,n are non-negative integers . Then, x has a decimal expansion which is non-terminating repeating.

we observe that the prime factorisation of the denominators of these rational numbers are of the form 2m × 5n, where m,n are non-negative integers. Hence, 189/125 has terminating decimal expansion.

we observe that the prime factorisation of the denominator of these rational numbers are not of the form 2^m × 5ⁿ, where m,n are non-negative integers. Hence 17/6 has non-terminating and repeating decimal expansion.

So, the denominator 8 of 17/8 is of the form 2^m × 5ⁿ, where m,n are non-negative integers. Hence 17/8 has terminating decimal expansion.

Clearly, 455 is not of the form 2^m × 5ⁿ. So, the decimal expansion of 64/455 is non-terminating repeating.

Ex.1 Prove that √2 is not a rational number or there is no rational whose square is 2. (CBSE (outside Delhi ) 2008).

Sol. Let us find the square root of 2 by long division method as shown below.

Clearly, the decimal representation of (√2) is neither terminating nor repeating. We shall prove this by the method of contradiction. If possible, let us assume that (√2) is a rational number.

Then (√2) = a,b where a, b are integers having no common factor other than 1.

(squaring both sides)

⇒ 2 .divides a²
⇒  2 divides a
Therefore let a = 2c for some integer c.
⇒ a² = 4c²
⇒ 2b² = 4c²
⇒ b² = 2c²
⇒ 2 divides b²
⇒ 2 divides b

Thus, 2 is a common factor of a and b.

But, it contradicts our assumption that a and b have no common factor other than 1.

So, our assumption that (√2) is a rational, is wrong.

Hence, √2 is irrational.

Ex.2 Prove that  is irrational.
Sol. Let be rational = p/q , where p and q ∈  Z and p, q have no common factor except 1 also q > 1.

Cubing both sides

  Clearly L.H.S is rational since p, q nave no corrrnon factor.

∴  p³, q also have no common factor while R.H.S. is an integer.
∴ L.H.S  R.H.S which contradicts our assumption that is Irrational

Ex.3 Prove that is irrational. [Sample paper (CBSE) 2008]
Sol. Let be a rational number equals to r
= r
3 = r – 2
Here L.H.S is an irrational number while R.H.S. r – 2 is rational. L.H.S R.H.S
Hence it contradicts our assumption that is rational.
∴ is irrational.

Ex.4 Prove that  is irrational.
Sol. Let be rational number say 'x'  x =

As x, 5 and2 are rational   is a rational number.

  is a rational number

Which is contradication of the fact that √6 is a irrational number.
Hence our supposition is wrong ⇒ is an irrational number.

Ex.5 Given that H.C.F. (306, 657) = 9. Find L.C.M. (306, 657)

Sol. H.C.F. (306, 657) = 9 means H.C.F. of 306 and 657 = 9

Required L.C.M. (306, 657) means required L.C.M. of 306 and 657.

For any two positive integers;

Ex.6 Given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100).

Sol. L.C.M. (150, 100) = 300
⇒ L.C.M. of 150 and 100 = 300

Since, the product of number 150 and 100 = 150 × 100

And , we knoe : H.C.F (150, 100) 

Ex.7 Explain why 7 × 11 × 13 + 13 are composite numbers:

Sol. (i) Let 7 × 11 × 13 + 13 = (7 × 11 + 1) × 13
= (77 + 1) × 13 = 78 × 13  7 × 11 × 13 + 13 = 2 × 3 × 13 × 13
= 2 × 3 × 132 is a composite number as powers of prime occur.

Ex.8 Use Euclid's algorithm to find the HCF of 4052 and 12576.
Sol. Using a = bq + r, where 0  r < b.
Clearly, 12576 > 4052 [a = 12576 , b = 4052]
⇒ 12576 = 4052 × 3 + 420
⇒ 4052 = 420 × 9 + 272
⇒ 420 = 272 × 1 + 148
⇒272 = 148 × 1 + 124
⇒ 148 = 124 × 1 + 24
⇒124 = 24 × 5 + 4
⇒ 24 = 4 × 6 + 0
 

Ex.9 The remainder at this stage is 0. So, the divisor at this stage, i.e., 4 is the HCF of 12576 and 4052.

Sol. Two numbers 1848 and 3058, where 3058 > 1848
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638 [Using Euclid's division algorithm to the given number 1848 and 3058]
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
Therefore HCF of 1848 and 3058 is 22.
HCF (1848 and 3058) = 22
Let us find the HCF of the numbers 1331 and 22.
1331 = 22 × 60 + 11
22 = 11 × 2 + 10
∴HCF of 1331 and 22 is 11
⇒ HCF (22, 1331) = 11
Hence the HCF of the three given numbers 1848, 3058 and 1331 is 11.
HCF (1848, 3058, 1331) = 11

Ex.10 Using Euclid's division, find the HCF of 56, 96 and 404 [Sample paper

Sol. Using Euclid's division algorithm, to 56 and 96.
96 = 56 × 1 + 40
56 = 40 × 1 + 16
40 = 16 × 2 + 8
16 = 8 × 2 + 0
Now to find HCF of 8 and 404
We apply Euclid's division algorithm to 404 and 8
404 = 8 × 50 + 4
8 = 4 × 2 + 0
Hence 4 is the HCF of the given numbers 56, 96 and 404.