chapter 1 er system real number and numb · we have been ideally made familiar with the number...

MATHEMATICS for CLASS 10 (NEW pattern for CBSE board) By-DEEPAK SIR 9811 29 16 04

1 | P a g e S H R I S A I M A S T E R S T U I T I O N C E N T R E , A R J U N N A G , S J E , N D

INTRODUCTION

We have been ideally made familiar with the Number System in detail in class IX. We perhaps

know now how to apply the basic arithmetic operations viz. addition, subtraction, multiplication

and division on the natural numbers, integers, rational as well as irrational numbers. We have

been taught to locate irrational numbers on the number line as well, we also are aware of Laws of

Exponents of Real Numbers. I’m not absolutely right though I hate to admit it. (Pardon me, if

you are quite an expert in all these!)

In our current discussion, we shall do a recall of divisibility on integers and shall state

some important properties such as, Euclid’s Division Lemma, Euclid’s Division

Algorithm and the Fundamental Theorem of Arithmetic. All these shall be used in the

remaining part of this chapter to explore and gain more knowledge about integers and real

numbers. We shall use Euclid’s Lemma to find HCF (Highest Common Factor)of integers;

Fundamental Theorem of Arithmetic shall be used to find the HCF as well as LCM (Lowest

Common Multiple) of integers. We shall learn to check irrationality of an irrational

number by using contradiction .At the last we shall learn how to decipher without actual

division, whether a rational number p/q, q=/= 0 (say) has a terminating or non-terminating

decimal expansion.

Before we start, we shall have a quick recap of the Number System for your benefit.

01. Natural numbers: The numbers used in ordinary counting i.e. 1, 2, 3… are called

natural numbers (and positive integers as well). The collection (set) of natural numbers is

denoted by N. Also if we include 0 to the set of natural numbers, we get set of the whole

numbers which is denoted by the symbol W.

02. Integers :The numbers ...-3,- 2, -1,0,1, 2,3,... are called integers. The set of integers is

denoted by the symbol I or Z. Though now we use Z to symbolize the set of integers.

➢ Also from the above discussion, it is evident that integers are of three types viz.: a) Positive integers i.e. Z + =1,2,3,... b) Negative integers i.e. Z- = - 1,- 2,- 3,... c) Zero integer i.e. non-positive and non-negative integer. 03. Rational numbers: A number of the form p/q, where p and q are integers and q≠ 0 , is

called a rational number. The set of rational numbers is denoted by Q.

➢ Zero being an integer, is also a rational number. 04. Irrational numbers: An irrational number has a non-terminating and non-repeating

decimal representation i.e.it can not be expressed in the form of p/q . The set of irrational

numbers is denoted by the letter I. Few examples of irrational numbers are √2, 5√ 7, 8+√ 3,3- √7,

π,... etc.

➢ Note that πis irrational while 22/7 is rational. 05. Real numbers :The set of all numbers either rationalism called real number. Set of all the

real numbers is denoted by R.

06. Laws of exponents or Indices:

REAL NUMBER AND NUMBER SYSTEM CHAPTER 1

By-DEEPAK SIR 9811291604



✓ The above discussion has been done for the benefit of the reader. I understand you have

been taught all these many a times, yet you tend to forget them. They can be memorized, just

start doing this– regular revision for a few days and use them as often you can. And believe me,

after some time they will be onyour tips! And obviously, those who are familiar with all these,

they may skip this discussion.

1. write how every positive integer is written when divided by 2.










9. Show that every positive even integer is of the form 2 q and that every positive odd integer is of the form 2q + 1 for some integer q.

SOLUTION Let a be a given positive integer. On dividing a by 2, let q be the quotient and r be the remainder. Then, by Euclid's algorithm, we have

a = 2q + r, where 0 < r < 2

=> a = 2q + r, where r = 0 or r =1

=> a =2q or a = 2q + l.

When a = 2q for some integer q, then clearly a is even. Also, an integer can be either even or odd. Hence,

an odd integer is of the form 2q +1 for some integer q. Thus, every positive even integer is of the form 2q

and every positive odd integer is of the form 2q + l for some integer q.

10. Show that every positive odd integer is of the, form (4 q +1) or(4q + 3) for some integer q.

Let a be a given positive odd integer.

On dividing a by 4, let q be the quotient and r be the remainder. Then, by Euclid's algorithm, we have

a = 4q + r, where 0 < r < 4 => a = 4q + r, where r = 0,1,2,3 => a =4q or a =4q + l or a =4q + 2 or a

=4q+ 3. But, a = 4q and a = 4q + 2 = 2(2q +1) are clearly even. Thus, when a is odd, it is of the form a

= (4q +1) or (4q + 3) for some integer

11. .Show that every positive odd integer is of the form (6 q +1 )or(6q+ 3) or(6q + 5) for some integer q.

sol. Let a be a given positive odd integer.

On dividing aby 6, let q be the quotient and r be the remainder. Then, by Euclid's algorithm, we have

a = 6q + r, where 0 a -6q + r, where r =0,1,2,3,4,5 => a = 6q or a = 6q +1 or a

= 6q + 2 or a = 6q + 3 or a = 6^ + 4 or a = 6q + 5.

But, a = 6q, a = 6q + 2, a = 6q + 4 give even values of a. Thus, when a is odd, it is of the form

6q +1 .or 6q + 3 or 6q + 5 for some integer90 = 45 x 2 + 0

12. Show that every positive odd integer is of the form (8q +1 )or(8q+ 3) or(8q + 5) for some integer q.



13. Ex3.Use Eculid's division demma, to show that the cube of any positive integer is of the form 9m,

9m + 1 or 9m + 8.

Solution: Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4) Hence, x3 is either of form 9 m or 9m +1 or 9m + 8. 14. Show that one and only one out of n, n + 2, n + 4 are divisible by 3, where n is any positive

integer.

Solution : On dividing by 3, let be the quotient and be the remainder. Then, n = 3q + r, where 0

n = 3q + r, where = 0, 1, 2 => n = 3q or n = 3^ + 1 or n = 3q+ 2.

Case 1. If n = 3q, then n is divisible by 3.

Case 2. If n = 3q +1, then (n + 2) = 3q + 3 = 3(q +1), which is divisible by 3.

So in this case, (n + 2) is divisible by 3. Case 3. when n = 3q+2, then (n + 4) = 3q + 6 = 3(q + 2),

which is divisible by 3.So, in this case, (n + 4) is divisible by 3.Hence, one and only one out of n, n + 2, n

+ 4 is divisible by 3.

15. Use Euclid's Division Lemma to show that the square of any positive integer is either of the form 3 m or 3 m + I for some integer m. [CBSE 2008]

SOLUTION Let a be an arbitrary positive integer.

Then, by Euclid's Division Algorithm, corresponding to the positive integers a and 3, there exist non-negative integers q and r such that

a = 3q + r, where 0 < r< 3 => a2 =9q2 +r2 +6qr ... (i), where0



01. The meaning of Lemma and Algorithm: A lemma is an already proven statement

which helps in proving another statement. Also algorithm basically means the steps. It is a

series of some rules which are given step wise to solve similar kind of problems.

02. Euclid’s Division Lemma: Given two positive integers a and b, there exist unique

integers q and r satisfying , a=b q + r, where 0 < r < b.. Here a, b and r are called dividend,

divisor, quotient and the remainder respectively.

Dividend = (divisor x quotient) + remainder.

Suppose we divide 117 by 14. Then, we get 8 as quotient and 5 as remainder.

Here dividend =117, divisor = 14, quotient = 8 and remainder = 5.

Clearly, 117 =(14x 8)+5.



03. Finding HCF of two positive integers a and b (such that a > b) using Euclid’s division

algorithm:

STEP1-By applying Euclid’s division algorithm, find q and rwhere , a=b q + r, where 0 < r < b

STEP2-If r = 0 , then the HCF of the numbers a and b is “ b ”. If r≠ 0 , then apply the division

algorithm to b and r taking bas the new dividend and r as the new divisor.

STEP3-Continue this process till the remainder comes zero. When the remainder comes zero, the

divisor at that stage is the required HCF of the numbers a and b.



1. Show that any number of the form 4n ,n ϵ N can never end with the digit 0.

SOLUTION If 4n ends with 0, then it must have 5 as a factor. But, we know that the only prime factor of 4"

is 2. Also, we know from the fundamental theorem of arithmetic, That the prime factorization of each

number is unique.4n can never end with 0.

2. Show that any number of the form 6n, n ϵ N can never end with digit 0.

SOLUTION If 6n ends with 0, then it must have 5 as a factor.But, we know that the only prime factors of

6" are 2 and 3. Also, we know from the fundamental theorem of arithmetic that the prime factorization of

each number is unique. 6n can never end with 0.



Q 16

Funda:

Formula:

Application:

Q17

Funda:

Formula:

Application:

Q18



Q19

Q20

Q21

Q22



HOME WORK EDL-1



Understanding the meaning of HCF:

The common divisor of 12 and 24 is 1,2,3,4,6,12. The largest among all these is 12 . this is called GREATEST Common divisor ( GCD) or heigest

common Factor (HCF)



❖ Here in our syllabus, Euclid’s division lemma is stated for only positive integers. However it can be extended for all the integers except zero (as b≠ 0).

04. Fundamental Theorem of Arithmetic:

Every composite number can be expressed (i.e. factorised) as a product of primes, and this

factorization is unique, apart from the order in which the prime factors occur.

05. Finding the HCF and LCM of two integers using Fundamental Theorem of Arithmetic:

❖ Do you notice that to find the HCF (Highest Common Factor), we take the smallest

exponents of the Common prime factors. Where as in the case of LCM (Lowest Common

Multiple), we take those factors which are with the largest exponents.

Remark :To find the LCM (or HCF) of two integers a and b, we can use the relation given here

if we know already their HCF (or LCM):

a x b= (HCF) x (LCM).

For any three number a , b , c then we have

𝐻𝐶𝐹(𝑎, 𝑏, 𝑐) =𝑎 𝑥 𝑏 𝑥 𝑐 𝑥 𝐿𝐶𝑀(𝑎, 𝑏, 𝑐)

𝐿𝐶𝑀(𝑎, 𝑏) 𝑥 𝐿𝐶𝑀(𝑎, 𝑐) 𝑥 𝐿𝐶𝑀(𝑏, 𝑐)

𝐿𝐶𝑀(𝑎, 𝑏, 𝑐) =𝑎 𝑥 𝑏 𝑥 𝑐 𝑥 𝐻𝐶𝐹(𝑎, 𝑏, 𝑐)

𝐻𝐶𝐹(𝑎, 𝑏) 𝑥 𝐻𝐶𝐹(𝑎, 𝑐) 𝑥 𝐻𝐶𝐹(𝑏, 𝑐)

You must remember that a prime number is divisible by only itself and one. Also note

that one (i.e. 1) is not considered as a prime number.

STEP1-Factorize each of the given integers and then express them as a product of powers of the primes.

STEP2- To find the HCF, identify the common prime factorsand then find the smallest exponent (power)

of these common factors. Now raise these common prime factors to their smallest exponents and

multiply each of these to obtain the HCF.

STEP3- To find the LCM, list all the prime factors occurring in the prime factorization of the given

integers. For each of these factors, find the greatest exponentand raise each prime factor to the

greatest exponent and multiply each of these to obtain the LCM.



06. Theorem: If a prime number p divides a2, then p also divides a where a is a positive integer.

That is, if a2= p q, then a= p λ where q and λ are positive integers.



Home work EDL-2



HOME WORK EDL-3

07. Condition for a Rational number to have Terminating Decimal Expansion: A

rational number

having terminating decimal expansion can always be expressed in the form of p/q where p and q

are co primes and the prime factorization of denominator i.e., q is of the form 2n 5m

where m and n are non-negative integers.

❖ If a rational number expressed in the form of p/q is such that q is not of the form 2 5 m n,

then decimal expansion of p/q is non- terminating i.e. it has repeating decimal expansion.

Remark If for a rational number p/q , the denominator q is of the form 2m 5n then, it

terminates after k

places of decimals where k is the largest of m and n.



Home work EDL-4



EXTRA QUESTION



TEST QUESTION



2 MARKS BORD QUESTION



3 MARKS

chapter 1 er system real number and numb · we have been ideally made familiar with the number...

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