krit.club logo

Real Numbers - Introduction

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Fundamental Theorem of Arithmetic states that every composite number can be uniquely expressed (factorized) as a product of primes, except for the order in which the prime factors occur. For example, 210=2×3×5×7210 = 2 \times 3 \times 5 \times 7.

•

HCF (Highest Common Factor) is the product of the smallest power of each common prime factor in the numbers.

•

LCM (Least Common Multiple) is the product of the greatest power of each prime factor involved in the numbers.

•

For any two positive integers aa and bb, the relationship between their HCF and LCM is given by HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b.

•

A number is Irrational if it cannot be expressed in the form pq\frac{p}{q} where p,qp, q are integers and q≠0q \neq 0. If pp is a prime number, then p\sqrt{p} is always irrational.

•

The decimal expansion of a rational number pq\frac{p}{q} is terminating if the prime factorization of qq is of the form 2n5m2^n 5^m, where nn and mm are non-negative integers. Otherwise, it is non-terminating repeating.

📐Formulae

HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b

q=2n×5mq = 2^n \times 5^m

If p divides a2, then p divides a (where p is prime).\text{If } p \text{ divides } a^2, \text{ then } p \text{ divides } a \text{ (where } p \text{ is prime).}

💡Examples

Problem 1:

Find the HCFHCF and LCMLCM of 9696 and 404404 using prime factorization.

Solution:

96=25×3196 = 2^5 \times 3^1 404=22×1011404 = 2^2 \times 101^1 HCF(96,404)=22=4HCF(96, 404) = 2^2 = 4 LCM(96,404)=25×31×1011=32×3×101=9696LCM(96, 404) = 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 9696

Explanation:

We first find the prime factors. For HCF, we take the lowest power of common factors (222^2). For LCM, we take the highest power of all factors present (25,31,10112^5, 3^1, 101^1).

Problem 2:

Check if the product of HCFHCF and LCMLCM for numbers 2626 and 9191 equals the product of the numbers.

Solution:

26=2×1326 = 2 \times 13 91=7×1391 = 7 \times 13 HCF=13,LCM=2×7×13=182HCF = 13, \quad LCM = 2 \times 7 \times 13 = 182 HCF×LCM=13×182=2366HCF \times LCM = 13 \times 182 = 2366 a×b=26×91=2366a \times b = 26 \times 91 = 2366 Verification: 2366−23660\begin{array}{r} 2366 \\ -2366 \\ \hline 0 \end{array}

Explanation:

We calculate HCF and LCM first, then multiply them. We then multiply the original numbers. Since both results are 23662366, the formula HCF×LCM=a×bHCF \times LCM = a \times b is verified.

Problem 3:

Without actual division, state whether 133125\frac{13}{3125} has a terminating or non-terminating repeating decimal expansion.

Solution:

The denominator is q=3125q = 3125. Prime factorization of 31253125: 3125=55=20×553125 = 5^5 = 2^0 \times 5^5 Since the denominator is in the form 2n×5m2^n \times 5^m, the decimal expansion is terminating.

Explanation:

By the theorem on rational numbers, if the denominator's prime factors only consist of 22s and/or 55s, the decimal will terminate.