Important definitions and formulas - Real Numbers Class 10 Notes | EduRev

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.

Euclid’s Division Lemma

• 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.

Euclid's Division Algorithm

• 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 1: 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.

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.

Rational Number

• The decimal expansion of every rational number is either terminating or non-terminating repeating.
• 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.
• Let x = p/q be a rational number such that prime factorisation of q is of the form 2ⁿ 5^m, where n, m are some non-negative integers. Then x has a terminating decimal expansion.
• 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.

• 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.
  
  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.

  View Solution

• If p is a prime and p divides a², then p divides a, where a is a positive integer.

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

LCM & HCF of Three Numbers 

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