 # Real Numbers Class 10 Maths Notes

Table of Content
• Real Numbers Important Definition
• Euclid's Division Lemma
• Proof of Euclid's Division Lemma
• HCF (Highest common factor)
• What is Prime Numbers
• What is Composite Numbers
• Fundamental Theorem of Arithmetic
• HCF and LCM by prime factorization method
• Irrational Numbers
• How to prove the irrational numbers or Rational numbers

• ## Real Numbers:Important Definition

### Natural and Whole Number

What is Natural Numbers?
Set of counting numbers is called the Natural Numbers
N = {1,2,3,4,5,...}
What is whole number?
Set of Natural numbers plus Zero is called the Whole Numbers
W= {0,1,2,3,4,5,....}
Note:
So all natural Number are whole number but all whole numbers are not natural numbers

### Integers

What are Integers Numbers
Integers is the set of all the whole number plus the negative of Natural Numbers
Z={..,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,...}
Note
1) So ,integers contains all the whole number plus negative of all the natural numbers
2)the natural numbers without zero are commonly referred to as positive integers
3)The negative of a positive integer is defined as a number that produces 0 when it is added to the corresponding positive integer
4)natural numbers with zero are referred to as non-negative integers
5) The natural numbers form a subset of the integers.

### Rational and Irrational Numbers

#### Rational Number

: A number is called rational if it can be expressed in the form p/q where p and q are integers ( q> 0).
Example : 1/2, 4/3 ,5/7 ,1 etc.
Important Points to Note
• every integers, natural and whole number is a rational number as they can be expressed in terms of p/q
• There are infinite rational number between two rational number
• They either have termination decimal expression or repeating non terminating decimal expression
• The sum, difference and the product of two rational numbers is always a rational number. The quotient of a division of one rational number by a non-zero rational number is a rational number. Rational numbers satisfy the closure property under addition, subtraction, multiplication and division.

#### Irrational Number

: A number is called rational if it cannot be expressed in the form p/q where p and q are integers ( q> 0).
Example : √3,√2,√5,p etc.
Important Points to Note
• Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem we can represent the irrational numbers on the number line.
• They have non-terminating and non-repeating decimal expression
• The sum, difference, multiplication and division of irrational numbers are not always irrational. Irrational numbers do not satisfy the closure property under addition, subtraction, multiplication and division

### Real Numbers:

• All rational and all irrational number makes the collection of real numbers. It is denoted by the letter R
• We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line
• The sum or difference of a rational number and an irrational number is an irrational number.
• The product or division of a rational number with an irrational number is an irrational number.
• This process of visualization of representing a decimal expansion on the number line is known as the process of successive magnification

Real numbers satisfy the commutative, associative and distributive laws. These can be stated as :
$a+b= b+a$
Commutative Law of Multiplication:
$a \times b=b \times a$
$a + (b+c)=(a+b) +c$
Associative Law of Multiplication:
$a \times (b \times c)=(a \times b) \times c$
Distributive Law:
$a \times (b + c)=(a \times b) + (a \times c)$
or
$(a + b) \times c=(a \times c) + (b \times c)$

Real Numbers Examples
1
-2/3
3/4
√2
√2 + 5

### Laws of exponents:

Let a > 0 be a real number and p and q be rational numbers. Then, we have
1. $a^p.a^q=a^{p+q}$
2. $\frac {a^p}{a^q} =a^{p-q}$
3. $(a^p)^q=a^{pq}$
4. $a^p.b^p=(ab)^p$

## Euclid's Division Lemma

For a and b any two positive integer, we can always find unique integer q and r such that
$a=bq + r \; , \; 0 \leq r < b$

It is basic concept and it is restatement of division
• a is called dividend
• b is called divisor
• q is called quotient
• r is called remainder.
If Remainder r =0, then b is divisor of a.
Example
for 11 and 2
$11= 2 \times 5 + 1$
for 10 and 2
$11= 2 \times 5$

## Proof of Euclid's Division Lemma

Here we need to argue that q and r are no unique.So we are going to prove using contradiction method
Let us assume q and r are not unique i.e. let there exists another pair $q_0$ and $r_0$ i.e. $a = bq_0 + r_0$, where $0 \leq r_0 < b$

=>$bq + r = bq_0 + r_0$
=> $b(q - q_0) = r - r_0$ ................ (I)

Since $0 \leq r < b$ and $0 \leq r_0 < b$, thus $0 \leq r - r_0 < b$ ......... (II)

The above equation (I) tells that b divides (r - r0) and Equation (II) tells us (r - r0) is an integer less than b. Both of these condition can be satisfied only when (r - r0) is 0.
=> $r - r_0 = 0$
=> $r = r_0$

Equation (I) will be, $b(q - q_0) = 0$
Since b > 0, => $(q - q_0) = 0$
=> $q = q_0$

Since $r = r_0$ and $q = q_0$, Therefore q and r are unique.

## HCF (Highest common factor)

Before starting on this topic, we need to prove an important theorem which will be used in finding HCF between two numbers

### Theorem

If a and b are positive integers such that a = bq + r, then every common divisor of a and b is a common divisor of b and r, and vice-versa.

Proof : Let m be a common divisor of a and b. Then,
m| a => a = mx for some integer x
m| b => b = mq2 for some integer y
Now, a = bq + r
=> r = a - bq
=> r = mx - my q
=> r = m( x - yq)
=> m| r
=> m| r and m | b
=> m is a common divisor of b and r.
Hence, a common divisor of a and b is a common divisor of b and r.

### How to find HCF (Highest common factor)

Now HCF of two positive integers can be find using the Euclid's Division Lemma algorithm and above stated theorem

We know that for any two integers a,b. we can write following expression
$a=bq + r \; , \; 0 \leq r < b$
If r=0 ,then
HCF( a,b) =b
If r >0 , then
HCF (a,b) = HCF ( b,r) as already proved from above theorem

Again expressing the integer b,r in Euclid's Division Lemma, we get
b=pr + r1
HCF ( b,r)=HCF ( r,r1)
Similarly successive Euclid 's division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

Examples

Use Euclid's algorithm to find the 65 and 117.

Solution :
Step:1 Since 117 > 65 we apply the division lemma to 117 and 65 to get ,
$117 = 65 \times 1 + 52$
Step:2 Since 52 > 0 , we apply the division lemma to 65 and 52 to get
$65 = 52 \times 1 + 13$
Step:3 Since 13 > 0 , we apply the division lemma to 52 and 13 to get
$52 = 13 \times 4 + 0$
The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 13, the HCF of 117 and 52 is 13.

Some other points to remember
1.The HCF of three numbers can be calculated by first calculating the HCF of first two numbers ,then the calculating the HCF of the HCF of previous two numbers and third number.
2. If HCF ( a,b) =1 ,the a and b are co primes.

## What is Prime Numbers

A Prime Number is a number that cannot be evenly divided by any other number (except 1 or itself).

Examples 2,3,5,7,11,13,17,19,23,29.......
Prime numbers are interesting blocks in Mathematics.They are highly used in cryptography field

Prime Number help
PRIME NUMBERS to 100 ={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}
Rules to find Prime Number
Rule Number 1:
Since even numbers are always divisible by 2, no even number (other than 2) can be a prime number because they'll always have more than two factors. For example, the factor pairs for 6 are: 1,2,3,6 So 6 has four factors and therefore is not a prime number.

Rule Number 2:
Numbers that end in 5 or 0 are always divisible by 5. So no numbers divisible by 5 (other than 5) can be prime numbers because they'll always have more than two factors. They'll have 1 and themselves, and they'll also have 5 and some other factor.

Rule Number 3:
0 and 1 are not prime numbers. It may seem at first glance that they are, but remember that a prime number has only 1 and itself as factors. But 0 has an infinite number of factors because 0 * 1 = 0, and 0 * 2 = 0, and 0 * 3 = 0, and so on. So 0 is far from prime.

In the case of 1, it doesn't have two distinct factors because it's only factor is itself: 1 * 1 = 1. If you change one of those factors to any other integer, you no longer get a product of 1. Since it has only one factor, it's not prime.

Rule Number 4:
Add all of the digits composing the number. If the sum of the digits is divisible by 3, then so is the original number. Example: If the number is 111, add 1 + 1 + 1 = 3. The sum of the digits (3) is divisible by 3, so 111 is also divisible by 3.

Rule Number 5:
Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule repeatedly as necessary. Example: 826. Twice 6 is 12. So, take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also

Rule Number 6:
If you are checking if a number is a prime number,then you don't need to check all the prime number below it. you need to check run for small primes(2,3,5,7) and you're not done factoring, then keep trying bigger and bigger primes (11, 13, 17, 19, 23, etc.) until you find something that works or until you reach primes whose squares are bigger than what you're dividing into. if you don't find then it is a prime Number Reason: If your prime doesn't divide in, then the only potential divisors are bigger primes. Since the square of your prime is bigger than the number, then a bigger prime must have as its remainder a smaller number than your prime. The only smaller number left, since all the smaller primes have been eliminated, is 1. So the number left must be prime, and you're done.
Lets take it with an Example
The number given is 1017
We can observer that 1031 lies between square of 30 and 33. So we can just keep trying till 33
a. 1031 is not divisible by 2
b. 1031 is not divisible by 3
c. 1031 is not divisible by 5
d. 1031 is not divisible by 7
e. 1031 is not divisible by 11
f. 1031 is not divisible by 13
g. 1031 is not divisible by 17
h. 1031 is not divisible by 19
i. 1031 is not divisible by 23
h. 1031 is not divisible by 29
j. 1031 is not divisible by 31
The next prime is 37. So we are done factoring. We can say 1031 is a Prime Number
List of Prime Number till 1000
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101
103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197
199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311
313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431
433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557
569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673
677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811
821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941
947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039

## What is Composite Numbers

A whole number that can be divided evenly by numbers other than 1 or itself

Examples
4,6,8,9,10.....

Since 4 is divided by 2,6 is divided by 2 and 3

## Fundamental Theorem of Arithmetic

Every composite number can be written as the product of power of primes and this factorization is unique
Composite number = Product of primes

Example

$16=2 \times 2 \times 2 \times 2$
This is unique, We cannot express it in any other prime numbers

Few things to take note
• The prime Number can be repeated in the factorization
• The order does not matter
• The prime factorisation of a natural number is unique, except for the order of its factors.

### How to Factorize the Composite Numbers(Prime Factorisation)

1. Few things we need to remember,
a. If the number is even,then it will be divisible by 2
b. If the sums of its digits is divisible by 3,then it is divisible by 3
c. if the end of the number is 0 or 5,then it is divisible by 5
2. We have to start with the small prime number with the rules given in step 1. Once we find the quotient, repeat the same process for the quotient. The last quotient will be a prime number itself

Example

Suppose the composite Number is 168
1. Now 168 is even number.So, we know it will get divided by 2
168 / 2 = 84
2. Again 84 is even number.So, we know it will get divided by 2
84 / 2 = 42
3. Again 42 is even number.So, we know it will get divided by 2
42 / 2 = 21
4. Now sum of digits of 21 is 3.So, we know we can divide it by 3
21 / 3 = 7
5. 7 is a prime Number

So prime factors = 2 X 2 X 2 X 3 X 7
Prime Factorisation Tree Form Here are the prime factors of the composite numbers between 1 and 30.
$4 = 2 \times 2$
$6 = 3 \times 2$
$8 = 2 \times 2 \times 2$
$9 = 3 \times 3$
$10 = 5 \times 24$
$12 = 3 \times 2 \times 2$
$14 = 7 \times 2$
$15 = 5 \times 3$
$16 = 2 \times 2 \times 2 \times 2$
$18 = 3 \times 3 \times 2$
$20 = 5 \times 2 \times 2$
$21= 3 \times 7$
$22=2 \times 11$
$24= 2 \times 2 \times 2 \times 3$
$25= 5 \times 5$
$26= 2 \times 13$
$27= 3 \times 3 \times 3$
$28= 2 \times 2 \times 7$
$30= 2 \times 3 \times 5$

## HCF and LCM by prime factorization method

We can find HCF and LCM between two number by the prime factorization method also
HCF (Highest Common factor) = product of the smallest power of each common factor in the numbers

Example: Suppose the number are 14,24
1. Lets do prime factorization method for both the numbers
$14=2 \times 7$
$24= 2 \times 2 \times 2 \times 3= 2^3 \times 3$
2. Now as per method 2 is the only factor and lowest power is 1
HCF =2

LCM( Lowest Common Multiple) = Product of the greatest power of each prime factor involved in the number
Example: Suppose the number are 14,24
1. Lets do prime factorization method for both the numbers
$14=2 \times 7$
$24= 2 \times 2 \times 2 \times 3= 2^3 \times 3$
2. Now as per method
$LCM =2^3 \times 7 \times 3=168$
Also important Formula to remember
If a and b are two number,then
$HCF(a,b) \times LCM (a,b) =a \times b$
So if HCF and both the numbers are given, we can calculate the LCM using the formula
$LCM = \frac {a \times b}{HCF}$
if LCM and both the numbers are given, we can calculate the HCF using the formula
$HCF = \frac {a \times b}{LCM}$
These Formula are Valid for two numbers only. This is not true for 3 or more Numbers

We can also LCM using Division algorithm .In this method divide the given numbers by common prime number until the remainder is a prime number or one. LCM will be the product obtained by multiplying all divisors and remaining prime numbers.
Steps are
1. We place number in the line
2. We start dividing the number by least prime number which is common among all of them or group of them
3. Keep dividing by least until we have 1's in the remainder
4. LCM is the product of the divisors

Example
1. Find the LCM of 14 and 20 $LCM = 2 \times 2 \times 7 \times 5=140$
2. Find the LCM of 12 ,18 and 44 $LCM = 2 \times 2 \times 3 \times 3 \times 11=396$

## Irrational Numbers

A number r is called irrational number if it cannot be expressed in the form p/q where p and q are integer and q≠ 0

### Important Theorem

Let p be a prime number ,if p divides a2 ,the p divide a where a is positive number

## Rational numbers

A number r is called rational number if it can be expressed in the form p/q where p and q are integer and q≠ 0
Decimal expressions of rational number are of two types
a. Terminating decimal expression
b. Non-terminating repeating decimal expression

Terminating decimal expression can be written in the form
$\frac {p}{2^n5^m}$
Non terminating repeating decimal expression can not be expressed in this form
$\frac {p}{2^n5^m}$

## How to prove the irrational numbers or Rational numbers

1. We know the definition of each of them,so first try to express the number in the form p/q.If it is not possible then proceed to next step
2. If the proof is irrational number which will be the case always,assume the number to rational in the form p/q and then try to prove it wrong

## Solved Examples

1.Prove that $\sqrt {2}$ is irrational number
Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {2}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {2}=p$
squaring both sides
$2q^2=p^2$

So 2 divides p2,from theorem we know that,
2 will divide p also. p=2c

$2q^2=4c^2$
or
$q^2=2c^2$
So q divided by 2 also

So both p and q are divided by 2 which is contradiction from we assumed
So $\sqrt {2}$ is irrational number
2. Prove that $\sqrt {3}$ is irrational number
Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {3}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {3}=p$
squaring both sides
$3q^2=p^2$

So 3 divides p2,from theorem we know that,
2 will divide p also. p=3c

$3q^2=9c^2$
or
$q^2=3c^2$
So q divided by 3 also

So both p and q are divided by 3 which is contradiction from we assumed
So $\sqrt {3}$ is irrational number

### Quiz Time

Question 1 What is the HCF of 12576, 4052 ?
A. 4
B. 32
C. 196
D. 146
Question 2 HCF (p, q) x LCM (p, q) ?
A. p+q
B. pq
C. p-q
D. p/q
Question 3 HCF (a, b, c) x LCM (a, b, c).
A. a+b+c
B. abc
C. a
D. None of these
Question 4 if p a prime number, divides b2,then
A. p also divides b
B. p may divide b
C. p does not divide b
D. none of these
Question 5 Which of these is a rational number?
A. √5 + √2
B. (√5 + √2)(√5 - √2)
C.√5 - √2
D. (√5 + √2)2
Question 6 Which of these is not prime number?
A. 877
B. 101
C.97
D. 873

• Notes
• NCERT Solutions & Assignments

Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

You can use above books for extra knowledge and practicing different questions.

### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

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