Important Definitions & Formulas: Real Numbers | Mathematics (Maths) Class 10 PDF Download

The chapter on "Real Numbers" is crucial for understanding the properties and relationships of numbers, forming the basis for solving a wide range of mathematical problems in various fields.

This doc provides Class 10 Maths Formulas for Real Numbers that can help you to succeed in board exams and in future competitive exams.

Important Definitions

1. Real Numbers

• Real numbers constitute the union of all rational and irrational numbers.
  
• In general, all the arithmetic operations can be performed on these numbers and they can be represented on the number line, also.

2. Fundamental Theorem of Arithmetic

• Every composite number can be factorised as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
• Example: 54=2×3×3×3
• Therefore, 54 is represented as a product of prime factors (One 2 and three 3s) ignoring the arrangement of the factors.
• Also, the number 120 can be factorized as 2 × 2 × 3 × 5. The prime factors are 2, 2, 3, and 5. The order of the factors doesn't matter, as the Fundamental Theorem of Arithmetic guarantees a unique factorization regardless of the arrangement of the prime factors.

Example 1: What is the HCF of 36 and 48 using prime factorisation? 
Solution: To find the HCF using prime factorisation, follow these steps:
1. Find the prime factors of each number.
Prime factors of 36: 2 × 2 × 3 × 3 (2^2 × 3^2)
Prime factors of 48: 2 × 2 × 2 × 2 × 3 (2^4 × 3)
2. Find the common prime factors.
Common prime factors: 2 × 2 × 3 (2^2 × 3)
3. Multiply the common prime factors to get the HCF.
HCF = 2 × 2 × 3 = 12
So, the HCF of 36 and 48 using prime factorisation is 12. 

3. Rational Numbers

The decimal expansion of every rational number is either terminating or non-terminating repeating.

• Terminating Decimal Representation: Let x be a rational number with a terminating decimal representation. Then we can express x as p/q where p and q are coprime, and the prime factorisation of q is of the form 2ⁿ 5^m, where n, m are some non-negative integers.

• Non-terminating Decimal Representation:The rational number p/q will have a non-terminating repeating (recurring) decimal representation if, in standard form, the prime factorisation of q is not of the form 2ⁿ 5^m, where n, m are some non-negative integers.

4. Irrational Number

• A number is irrational if and only if, its decimal representation is non-terminating and non-repeating (non-recurring).
  OR

• A number that cannot be expressed in the form of p/q, q ≠ 0 and p, q ∈ I, will be an irrational number. The set of irrational numbers is generally denoted by S.
  
  Question for Important Definitions & Formulas: Real Numbers

  Try yourself:Find out the decimal expansion of 13/125.

  • A.

    Terminating decimal expansion
  • B.

    Non-terminating repeating decimal expansion
  • C.

    Non-terminating non-repeating decimal expansion
  • D.

    None of these

  Explanation

  Since the denominator, 125 = 5³ is of the form 2^m5ⁿ where m = 0 and n = 3.
  So 13/125 has a terminating decimal expansion.
  Therefore, the decimal expansion of 13/125​ is 0.1040.Thus, the correct answer is:  Terminating decimal expansion.

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5. Prime Number

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and the number itself. 

In other words, a prime number cannot be formed by multiplying two smaller natural numbers. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, and so on. 

Important Formulas with Examples

1. LCM & HCF of Two Numbers

For any two positive integers p and q, we have:
HCF (p, q) x LCM [p, q] = p x q

Example 2: There are two positive numbers such that one exceeds the other by 20 and their LCM and HCF are 504 and 4 respectively. Find the numbers.
Solution: Let the numbers be 'x' and 'x + 20'.
We know that:
HCF (p, q) x LCM (p, q) = p x q
x(x + 20) = 4 x 504
x² + 20x = 2016
x² + (56 - 36)x - 2016 = 0
(x + 56)(x - 36) = 0
x = -56 & x = 36
x cannot be negative.
So two numbers are: x = 36 & (x + 20) = 56

Example 3: Two positive numbers have a difference of 15. If their LCM (Least Common Multiple) is 360 and their HCF (Highest Common Factor) is 5, what are the values of these numbers?

Let's assume the smaller number is 'x'. Since one number exceeds the other by 15, the larger number would be 'x + 15'.

Using the relationship between LCM, HCF, and the product of two numbers, we have:

HCF (p, q) x LCM (p, q) = p x q

Substituting the given values, we get:

5 x 360 = x(x + 15)

1800 = x²+ 15x

Rearranging the equation, we have:

x² + 15x - 1800 = 0

To factorize this quadratic equation, we find two numbers whose sum is 15 and product is -1800. One such pair is 45 and -40.

(x + 45) (x - 40) = 0

Setting each factor to zero, we get:

x + 45 = 0 or x - 40 = 0

x = -45 or x = 40

Since the numbers are specified to be positive, we discard the negative value. Thus, the smaller number 'x' is 40.

The larger number would be x + 15:

40 + 15 = 55

Therefore, the two numbers are 40 and 55, respectively.

2. LCM & HCF of Three Numbers 

For any three positive integers a, b and c, we have:

3. Euclid’s Division Lemma (Deleted from NCERT)

• Given positive integers a and b, there exist unique integers q and r satisfying:
  a = bq + r, 0 ≤ r < b.
• This lemma is essentially equivalent to:
  dividend = (divisor x quotient) + remainder 
  where q is called the quotient and r is called the remainder.
• Lemma is a proven statement used for proving another statement.

Example 4:

Let's apply Euclid's Division Lemma to find the quotient and remainder when 87 is divided by 9.

According to the lemma, for any positive integers a and b, there exist unique integers q and r satisfying:

a = bq + r, 0 ≤ r < b.

In our case, a = 87 and b = 9. We need to find the values of q and r.

By applying Euclid's Division Lemma, we have:

87 = 9q + r

To find the quotient and remainder, we can divide 87 by 9:

87 ÷ 9 = 9 remainder 6

So, in this case, the quotient (q) is 9, and the remainder (r) is 6.

Hence, when 87 is divided by 9, the quotient is 9, and the remainder is 6, as per Euclid's Division Lemma.

4. Euclid's Division Algorithm

(Deleted from NCERT)

• This is based on Euclid's Division Lemma. It is a method used to find the H.C.F. of two numbers.
• Let there be two numbers a and b where a > b, the HCF is obtained by the following method:
  Step I: We apply Euclid’s Division Lemma to find two integers q and r, such that a = bq + r, 0 ≤ r < b.
  Step II: If r = 0, then b is the required HCF.
  Step III: If r ≠ 0, then again obtain two integers using Euclid’s Division Lemma and continue till the remainder becomes zero. The divisor when the remainder becomes zero is the required HCF.

Note: It can be extended for all integers, except zero i.e., b ≠ 0.

Example 5: Find HCF of 56 and 72.
Solution:
Apply lemma to 56 and 72.
Take a bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16
Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8
Again, 8 ≠ 0, consider 16 as the new dividend and 8 as the new divisor. 16 = 8 × 2 + 0.
Since the remainder is zero, divisor 8 is HCF.