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Class - 10 Mathematics Real number notes, Study notes of Mathematics

Mathematics

Class - 10 mathematics notes Chapter - 1 Real number Summarizes whole chapter in easy way to learn

Typology: Study notes

2024/2025

Available from 07/07/2025

ashwini-singh-rajput 🇮🇳

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Mathematics

Chapter 1- Real Numbers

■ Real Numbers:

⮚ Real numbers constitute the union of all rational and irrational numbers.

⮚ Any real number can be plotted on the number line.

⮚ We can say that any number is a real number, except for complex numbers. Examples of real

numbers I include –5, 2/3,

3.75, 7 and so on.

■Euclid's division lemma: Given positive integers a and b there exist whole number q and r

satisfy- satisfy ing a = bq + r, 0  r < b.

Euclid's Division Algorithm: In order to compute the HCF of two positive integers, say a and b,

with a > b by using Euclid’s algorithm we follow the following steps:

STEP I: Apply Euclid's division lemma to a and b and obtain whole numbers q1 and r1 such that a =

b q q1 + r1, 0  r1 < b.

STEP II: If r1 = 0, b is the HCF of a and b.

STEP III: If r1  0, apply Euclid's division lemma to b and r1 and obtain two whole numbers q1 and

r2 s such that b = q1r1 + r2.

STEP IV: If r2 = 0, then r1 is the HCF of a and b.

Class 10th

Short Note

Partial preview of the text

Download Class - 10 Mathematics Real number notes and more Study notes Mathematics in PDF only on Docsity!

Mathematics

Chapter 1- Real Numbers

■ Real Numbers:

⮚ Real numbers constitute the union of all rational and irrational numbers. ⮚ Any real number can be plotted on the number line. ⮚ We can say that any number is a real number, except for complex numbers. Examples of real

numbers I include – 5 , 2 / 3 ,

3. 75 , 7 and so on.

■ Euclid's division lemma: Given positive integers a and b there exist whole number q and r

satisfy- satisfy ing a = bq + r , 0  r < b.

Euclid's Division Algorithm: In order to compute the HCF of two positive integers, say a and b ,

with a > b by using Euclid’s algorithm we follow the following steps:

STEP I: Apply Euclid's division lemma to a and b and obtain whole numbers q 1 and r 1 such that a =

b q q 1 + r 1 , 0  r 1 < b.

STEP II: If r 1 = 0 , b is the HCF of a and b.

STEP III: If r 1  0 , apply Euclid's division lemma to b and r 1 and obtain two whole numbers q 1 and

r 2 s such that b = q 1 r 1 + r 2.

STEP IV: If r 2 = 0, then r 1 is the HCF of a and b.

Class 10th

Short Note

STEP V: If r 2 =8 0 , then apply Euclid's division lemma to r 1 and r 2 continue the above process till the

m remainder rn is zero. The divisor at this stage i.e. rn-1 , or the non-zero remainder at the

previous s stage, is the HCF of a and b.

The Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

⬥ Example 1: 60 = 2 × 2 × 3 × 5 OR, 60 = 2 × 3 × 2 × 5

⬥ Example 2: Consider the numbers 4 n , where n is a natural number. Check whether there

is any u value of n for which 4 n^ ends with the digit zero.

Solution: If the number 4 n , for any n , were to end with the digit zero, then it would be divisible by

5. That is, the prime factorization of 4 n^ would contain the prime 5. This is not possible because 4 n^ =

2 2-n ; so the only prime in the factorization of 4 n^ is 2. So, the uniqueness of the Fundamental Theorem

of Arithmetic guarantees that there are no other primes in the factorization of 4 n. So, there is no

natural number n for which 4 n^ ends with the digit zero.

Prime Factorisation:

⬥ Prime Factorisation is the method of expressing a natural number as a product of prime

numbers.

⬥ Example: 30 = 2 × 3 × 5 is the prime factorization of 30.

Method of Finding LCM:

As we know, the smallest of the common multiples of two or more numbers is called the lowest common multiple (LCM).

Example: To find the Least Common Multiple (L.C.M) of 90 and 210

1. 90 = 2 × 3 × 3 × 5

210 = 2 × 3 × 5 × 7

2. LCM of 90 and 210 = 2 × 3 × 3 × 5 × 7 which is 630

Method of Finding HCF:

We know that the greatest number that divides each of the given numbers without leaving any remainder is the highest common factor (HCF) of two or more given numbers. HCF can be found using two methods – Prime factorisation and Euclid’s division algorithm.

⬥ Example – To find the HCF of 20 and 24

20 = 2 × 2 × 5 and 24 = 2 × 2 × 2 × 3

⬥ The factor common to 20 and 24 is 2 × 2, which is 4 , which in turn is the HCF of 20 and 24.

■ Euclid’s Division Algorithm:

⬥ It is the repeated use of Euclid’s division lemma to find the HCF of two numbers.

⬥ Example: To find the HCF of 18 and 30

■ Rational numbers: -

• A number which can be expressed in the form of p / q , where p and q are integers and q  0.

⬥ Example –: 2/5, -1/4, 0.16, 2.1 3

■ Irrational Numbers:

The number which are not rational is called irrational numbers.

⬥ Example – 2 ,  , e

⬥ Theorem: Let p be a prime number. If p divides a^2 , then p divides a , where a is a positive integer. ■ D

e ■ Determining the nature of the decimal expansion of Rational Number:

(1) Let x be a rational number whose decimal expansion terminates. Then x can be expressed in

the form p / q , where p & q are co-prime, and the prime factorization of q is of the form 2 m^ × 5 n , where

m , n are non-negative integers.

(2) Let x = p / q be a rational number, such that the prime factorization of q is of the form 2 m^ × 5 n ,

where m , n are non-negative integers. Then, x has a terminating decimal expansion which

terminates after k places of decimals, where k is the larger of m & n.

(3) Let x = p / q be a rational number, such that the prime factorization of q is not the form of 2 m

× 5 n , where m , n are non-negative integers. Then, x has a non-terminating repeating decimal

expansion.

Class - 10 Mathematics Real number notes, Study notes of Mathematics

Related documents

Partial preview of the text

Download Class - 10 Mathematics Real number notes and more Study notes Mathematics in PDF only on Docsity!

Mathematics

Chapter 1- Real Numbers

■ Real Numbers:

numbers I include – 5 , 2 / 3 ,

3. 75 , 7 and so on.

■ Euclid's division lemma: Given positive integers a and b there exist whole number q and r

satisfy- satisfy ing a = bq + r , 0  r < b.

Euclid's Division Algorithm: In order to compute the HCF of two positive integers, say a and b ,

with a > b by using Euclid’s algorithm we follow the following steps:

STEP I: Apply Euclid's division lemma to a and b and obtain whole numbers q 1 and r 1 such that a =

b q q 1 + r 1 , 0  r 1 < b.

STEP II: If r 1 = 0 , b is the HCF of a and b.

STEP III: If r 1  0 , apply Euclid's division lemma to b and r 1 and obtain two whole numbers q 1 and

r 2 s such that b = q 1 r 1 + r 2.

STEP IV: If r 2 = 0, then r 1 is the HCF of a and b.

Class 10th

STEP V: If r 2 =8 0 , then apply Euclid's division lemma to r 1 and r 2 continue the above process till the

m remainder rn is zero. The divisor at this stage i.e. rn-1 , or the non-zero remainder at the

previous s stage, is the HCF of a and b.

The Fundamental Theorem of Arithmetic:

⬥ Example 1: 60 = 2 × 2 × 3 × 5 OR, 60 = 2 × 3 × 2 × 5

⬥ Example 2: Consider the numbers 4 n , where n is a natural number. Check whether there

is any u value of n for which 4 n^ ends with the digit zero.

Solution: If the number 4 n , for any n , were to end with the digit zero, then it would be divisible by

5. That is, the prime factorization of 4 n^ would contain the prime 5. This is not possible because 4 n^ =

2 2-n ; so the only prime in the factorization of 4 n^ is 2. So, the uniqueness of the Fundamental Theorem

of Arithmetic guarantees that there are no other primes in the factorization of 4 n. So, there is no

natural number n for which 4 n^ ends with the digit zero.

Prime Factorisation:

⬥ Prime Factorisation is the method of expressing a natural number as a product of prime

⬥ Example: 30 = 2 × 3 × 5 is the prime factorization of 30.

Method of Finding LCM:

Example: To find the Least Common Multiple (L.C.M) of 90 and 210

1. 90 = 2 × 3 × 3 × 5

210 = 2 × 3 × 5 × 7

2. LCM of 90 and 210 = 2 × 3 × 3 × 5 × 7 which is 630

Method of Finding HCF:

⬥ Example – To find the HCF of 20 and 24

20 = 2 × 2 × 5 and 24 = 2 × 2 × 2 × 3

⬥ The factor common to 20 and 24 is 2 × 2, which is 4 , which in turn is the HCF of 20 and 24.

■ Euclid’s Division Algorithm:

⬥ It is the repeated use of Euclid’s division lemma to find the HCF of two numbers.

⬥ Example: To find the HCF of 18 and 30

■ Rational numbers: -

• A number which can be expressed in the form of p / q , where p and q are integers and q  0.

⬥ Example –: 2/5, -1/4, 0.16, 2.1 3

■ Irrational Numbers:

⬥ Example – 2 ,  , e

⬥ Theorem: Let p be a prime number. If p divides a^2 , then p divides a , where a is a positive integer. ■ D

e ■ Determining the nature of the decimal expansion of Rational Number:

(1) Let x be a rational number whose decimal expansion terminates. Then x can be expressed in

the form p / q , where p & q are co-prime, and the prime factorization of q is of the form 2 m^ × 5 n , where

m , n are non-negative integers.

(2) Let x = p / q be a rational number, such that the prime factorization of q is of the form 2 m^ × 5 n ,

where m , n are non-negative integers. Then, x has a terminating decimal expansion which

terminates after k places of decimals, where k is the larger of m & n.

(3) Let x = p / q be a rational number, such that the prime factorization of q is not the form of 2 m

× 5 n , where m , n are non-negative integers. Then, x has a non-terminating repeating decimal