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

Mathematics (Maths) Class 10

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

The document Important definitions and formulas - Real Numbers Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

Real Numbers

  • Real numbers constitute the union of all rational and irrational numbers.
    Important definitions and formulas - Real Numbers Class 10 Notes | EduRev
  • 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.

Try yourself:The largest number that divides 70 and 125, which leaves the remainders 5 and 8, is:
View Solution


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 5m, 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 5m, 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 5m, where n, m are some non-negative integers.

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

Prime Number

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

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