1.1-Introduction to Real Numbers

1.1-Introduction to Real Numbers Important Formulae

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Real numbers are a fundamental concept in mathematics, encompassing a wide range of numbers used in everyday calculations and advanced mathematics. They include all the rational and irrational numbers, providing a complete system for measurement, calculation, and representation of quantities.

1. Definition of Real Numbers

Real numbers can be defined as all the numbers that can be found on the number line. They include:

  • Rational Numbers: Numbers that can be expressed as the quotient of two integers, where the denominator is not zero. Examples include $\frac{1}{2}, -3, 0.75$.
  • Irrational Numbers: Numbers that cannot be expressed as a simple fraction. Their decimal expansions are non-terminating and non-repeating. Examples include $\pi$ and $\sqrt{2}$.
2. Representation of Real Numbers

Real numbers can be represented on a number line, where:

  • Rational numbers can be located at specific points.
  • Irrational numbers are also located on the number line, but cannot be precisely expressed as fractions.

This visualization helps in understanding the density of real numbers, as between any two distinct real numbers, there exists another real number.

3. Classification of Real Numbers

Real numbers can be classified into several subsets:

  • Naturals (N): The set of positive integers starting from 1. $\{1, 2, 3, \ldots\}$
  • Whole Numbers (W): The set of natural numbers including 0. $\{0, 1, 2, 3, \ldots\}$
  • Integers (Z): The set of whole numbers and their negative counterparts. $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$
  • Rational Numbers (Q): Numbers that can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers, and $q \neq 0$. For example, $\frac{3}{4}$.
  • Irrational Numbers: As mentioned earlier, these cannot be expressed as fractions, such as $\sqrt{3}$ or $e$.
4. Properties of Real Numbers

Real numbers possess several important properties:

  • Closure Property: The sum or product of any two real numbers is also a real number.
  • Commutative Property: For addition and multiplication, the order of the numbers does not affect the result. That is, $a + b = b + a$ and $ab = ba$.
  • Associative Property: For addition and multiplication, the grouping of numbers does not affect the result. That is, $(a + b) + c = a + (b + c)$ and $(ab)c = a(bc)$.
  • Distributive Property: This property connects addition and multiplication. That is, $a(b + c) = ab + ac$.
5. Operations with Real Numbers

Real numbers can be added, subtracted, multiplied, and divided (except by zero) following the rules of arithmetic. These operations are essential for solving equations, evaluating expressions, and performing calculations in various fields.

6. Decimal Representation

Real numbers can be represented in decimal form. Rational numbers have either terminating or repeating decimal expansions, while irrational numbers have non-terminating, non-repeating decimal expansions. For example:

  • Terminating: $0.75$
  • Repeating: $0.333\ldots = \frac{1}{3}$
  • Irrational: $\pi \approx 3.14159\ldots$

Understanding real numbers is crucial for further studies in mathematics, as they serve as the foundation for algebra, calculus, and other mathematical concepts.


The real numbers and subsets thereof. The image may be misleading in that it implies there are real numbers that are neither rational or irrational. Could also mislead one about the relative "sizes" of these sets.
Damien Karras, Public domain, via Wikimedia Commons

Express each number as a product of its prime factors:

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Solution:

Express each number as a product of its prime factors:
(i) 140 = $2 \times 2 \times 5 \times 7$
(ii) 156 = $2 \times 2 \times 3 \times 13$
(iii) 3825 = $3 \times 3 \times 5 \times 5 \times 17$
(iv) 5005 = $5 \times 7 \times 11 \times 13$
(v) 7429 = $17 \times 19 \times 23$

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Solution:

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
HCF of 26 and 91: Prime factorization of 26 = $2 \times 13$
Prime factorization of 91 = $7 \times 13$
Common factors = 13
Thus, $HCF = 13$
LCM of 26 and 91: LCM = Product of the highest powers of all prime factors = $2 \times 7 \times 13 = 182$
Verification: $LCM \times HCF = 182 \times 13 = 2366$
Product of the two numbers = $26 \times 91 = 2366$
Thus, $LCM \times HCF = 26 \times 91$

(ii) 510 and 92
HCF of 510 and 92: Prime factorization of 510 = $2 \times 3 \times 5 \times 17$
Prime factorization of 92 = $2^2 \times 23$
Common factor = 2
Thus, $HCF = 2$
LCM of 510 and 92: LCM = Product of the highest powers of all prime factors = $2^2 \times 3 \times 5 \times 17 \times 23 = 2^2 \times 3 \times 5 \times 17 \times 23 = 3460$
Verification: $LCM \times HCF = 3460 \times 2 = 6920$
Product of the two numbers = $510 \times 92 = 6920$
Thus, $LCM \times HCF = 510 \times 92$

(iii) 336 and 54
HCF of 336 and 54: Prime factorization of 336 = $2^4 \times 3 \times 7$
Prime factorization of 54 = $2 \times 3^3$
Common factors = $2 \times 3$
Thus, $HCF = 6$
LCM of 336 and 54: LCM = Product of the highest powers of all prime factors = $2^4 \times 3^3 \times 7 = 3024$
Verification: $LCM \times HCF = 3024 \times 6 = 18144$
Product of the two numbers = $336 \times 54 = 18144$
Thus, $LCM \times HCF = 336 \times 54$

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25

Solution:

Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21

Prime factorisation of 12 = $2^2 \times 3$

Prime factorisation of 15 = $3 \times 5$

Prime factorisation of 21 = $3 \times 7$

HCF = Common prime factors with lowest powers = $3$

LCM = Product of highest powers of all prime factors = $2^2 \times 3 \times 5 \times 7 = 420$

(ii) 17, 23 and 29

Prime factorisation of 17 = $17$ (prime number)

Prime factorisation of 23 = $23$ (prime number)

Prime factorisation of 29 = $29$ (prime number)

HCF = No common prime factors = $1$

LCM = Product of all prime factors = $17 \times 23 \times 29 = 11251$

(iii) 8, 9 and 25

Prime factorisation of 8 = $2^3$

Prime factorisation of 9 = $3^2$

Prime factorisation of 25 = $5^2$

HCF = No common prime factors = $1$

LCM = Product of highest powers of all prime factors = $2^3 \times 3^2 \times 5^2 = 900$

Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

Given that HCF (306, 657) = 9, find LCM (306, 657).

We know that,

HCF × LCM = Product of the two numbers

So, we have:

HCF (306, 657) = 9

Product of 306 and 657 = 306 × 657

Now, using the formula:

9 × LCM (306, 657) = 306 × 657

LCM (306, 657) = (306 × 657) / 9

First, calculate the product of 306 and 657:

306 × 657 = 201,762

Now, divide the product by 9:

LCM (306, 657) = 201,762 / 9 = 22,419

The LCM of 306 and 657 is 22,419.

Check whether $6^n$ can end with the digit 0 for any natural number n.

Solution:

Question: Check whether $6^n$ can end with the digit 0 for any natural number n.

We are tasked with determining whether $6^n$ can end with the digit 0 for any natural number $n$.

To solve this, we need to analyze the units digit of $6^n$ for various values of $n$.

Let's calculate the first few powers of 6:

When $n = 1$, $6^1 = 6$ (units digit is 6).

When $n = 2$, $6^2 = 36$ (units digit is 6).

When $n = 3$, $6^3 = 216$ (units digit is 6).

When $n = 4$, $6^4 = 1296$ (units digit is 6).

It can be observed that for all natural numbers $n$, the units digit of $6^n$ is always 6.

Since 6 does not end in 0, $6^n$ can never end with the digit 0 for any natural number $n$.

Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.

Solution:

Explanation of Why 7×11×13+13 and 7×6×5×4×3×2×1+5 are Composite Numbers

To prove that 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers, we will check if they can be factored into smaller integers.

1. 7×11×13 + 13

We can factor out 13 from the expression:

$7×11×13 + 13 = 13(7×11 + 1)$

Now, simplify the expression inside the parentheses:

$7×11 + 1 = 77 + 1 = 78$

So, the expression becomes:

$13 × 78$

Since this number is the product of 13 and 78, both of which are greater than 1, it is a composite number.

2. 7×6×5×4×3×2×1 + 5

We first calculate the value of $7×6×5×4×3×2×1$:

$7×6×5×4×3×2×1 = 5040$

Now, add 5 to this value:

$5040 + 5 = 5045$

We need to check if 5045 is composite. Let's check if it can be factored:

5045 is divisible by 5 (since it ends in 5), so we divide:

$5045 ÷ 5 = 1009$

Thus, 5045 = 5 × 1009. Since 5045 is the product of 5 and 1009, both of which are greater than 1, it is a composite number.

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

Question:

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

We are required to find the time after which Sonia and Ravi will meet again at the starting point. This is the time after which they will both have completed whole rounds of the field and will be back at the starting point.

Let the time taken for them to meet again be the least common multiple (LCM) of 18 minutes (Sonia's time) and 12 minutes (Ravi's time).

We first find the LCM of 18 and 12.

Factorize 18 and 12:

$18 = 2 \times 3^2$

$12 = 2^2 \times 3$

The LCM is found by taking the highest powers of each prime factor:

$\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$

Thus, Sonia and Ravi will meet again at the starting point after 36 minutes.