| Latest Maths NCERT Books Solution | ||||||
|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
|---|---|---|
| Example 1 to 3 (Before Exercise 1.1) | Exercise 1.1 | |
Chapter 1 Rational Numbers
Welcome to this comprehensive solutions guide for Chapter 1, "Rational Numbers," meticulously aligned with the latest Class 8 NCERT mathematics textbook for the academic session 2024-25. This chapter revisits the rational number system, building significantly upon the foundational concepts introduced in Class 7. While previous studies focused on defining rational numbers (those expressible as $\frac{p}{q}$, where $p, q$ are integers and $q \neq 0$) and performing basic arithmetic operations, this chapter delves much deeper into the structural properties that govern these numbers under addition and multiplication. Mastering these properties is not just about computation; it's about understanding the underlying logic of the number system, which is crucial for success in algebra and beyond. These solutions provide detailed, step-by-step explanations for all exercises, aiming to solidify understanding and demonstrate practical applications.
A primary focus of this chapter, thoroughly explored in these solutions, is the verification and application of the fundamental properties of rational numbers. Students will find clear illustrations of:
- Closure Property: Demonstrating that the sum, difference, and product of any two rational numbers is always another rational number (though division is closed only if the divisor is non-zero).
- Commutative Property: Showing that the order of numbers does not affect the result for addition ($\frac{a}{b} + \frac{c}{d} = \frac{c}{d} + \frac{a}{b}$) and multiplication ($\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}$).
- Associative Property: Illustrating that the grouping of numbers does not impact the outcome for addition ($(\frac{a}{b} + \frac{c}{d}) + \frac{e}{f} = \frac{a}{b} + (\frac{c}{d} + \frac{e}{f})$) and multiplication ($(\frac{a}{b} \times \frac{c}{d}) \times \frac{e}{f} = \frac{a}{b} \times (\frac{c}{d} \times \frac{e}{f})$).
The crucial Distributive Property of Multiplication over Addition is given significant attention. This property, expressed as $\frac{a}{b} \times (\frac{c}{d} + \frac{e}{f}) = (\frac{a}{b} \times \frac{c}{d}) + (\frac{a}{b} \times \frac{e}{f})$, forms a vital link between multiplication and addition and is frequently used for simplifying complex calculations and in algebraic manipulations. The solutions provide numerous examples where applying this property, often in conjunction with commutativity and associativity (allowing rearrangement and regrouping of terms), leads to more efficient and elegant solutions compared to direct computation following BODMAS rules alone. Logical breakdowns show how identifying common factors or strategically grouping terms simplifies expressions involving multiple operations on rational numbers.
While representing rational numbers on the number line remains a conceptual tool, the detailed construction methods might be less emphasized compared to finding numbers between given rationals. This chapter focuses heavily on the density property of rational numbers – the idea that between any two distinct rational numbers, there lie infinitely many others. The solutions demonstrate systematic methods for finding these intermediate numbers:
- The Averaging Method: Finding the mean ($\frac{x+y}{2}$) of two rational numbers $x$ and $y$ yields a rational number exactly between them. This can be repeated.
- The Common Denominator Method: Converting the given rational numbers (say $\frac{a}{b}$ and $\frac{c}{d}$) into equivalent fractions with a large common denominator (using the $\text{LCM}$ or simply $b \times d$ and then potentially multiplying numerator and denominator by 10 or more) allows easy identification of numerous rational numbers lying between them by choosing intermediate numerators.
Regarding the rationalized syllabus for 2024-25, Chapter 1, "Rational Numbers," in the Class 8 NCERT textbook has sharpened its focus primarily on the properties of rational numbers (Closure, Commutativity, Associativity, Distributivity, Identity, Inverse) and their application in computations and simplification. Methods for finding rational numbers between two given numbers are also retained. The rationalization involved streamlining exercises and potentially reducing the emphasis on detailed geometric constructions for representing numbers on the number line compared to previous editions. By meticulously working through these comprehensive solutions, students can gain a profound understanding of the structure and properties of the rational number system, significantly enhance their computational and simplification skills, and build the essential confidence needed for tackling more advanced algebraic concepts involving these numbers.
Example 1 to 3 (Before Exercise 1.1)
Example 1: Find $\frac{3}{7}$ + $\left( \frac{-6}{11} \right)$ + $\left( \frac{-8}{21} \right)$ + $\left( \frac{5}{22} \right)$
Answer:
We need to find the sum of the given rational numbers:
$\frac{3}{7} + \left( \frac{-6}{11} \right) + \left( \frac{-8}{21} \right) + \left( \frac{5}{22} \right)$
Simplify the signs: $\frac{3}{7} - \frac{6}{11} - \frac{8}{21} + \frac{5}{22}$
To add and subtract these fractions, we need a common denominator, which is the LCM of 7, 11, 21, and 22.
Let's find the LCM:
$\begin{array}{c|cccc} 2 & 7 \;, & 11 \;, & 21 \;, & 22 \\ \hline 3 & 7 \; , & 11 \; , & 21 \; , & 11 \\ \hline 7 & 7 \; , & 11 \; , & 7 \; , & 11 \\ \hline 11 & 1 \; , & 11 \; , & 1 \; , & 11 \\ \hline & 1 \; , & 1 \; , & 1 \; , & 1 \end{array}$LCM$(7, 11, 21, 22) = 2 \times 3 \times 7 \times 11 = 462$.
Now, convert each fraction to have the denominator 462:
$\frac{3}{7} = \frac{3 \times 66}{7 \times 66} = \frac{198}{462}$
$\frac{6}{11} = \frac{6 \times 42}{11 \times 42} = \frac{252}{462}$
$\frac{8}{21} = \frac{8 \times 22}{21 \times 22} = \frac{176}{462}$
$\frac{5}{22} = \frac{5 \times 21}{22 \times 21} = \frac{105}{462}$
Combine the fractions:
$\frac{198}{462} - \frac{252}{462} - \frac{176}{462} + \frac{105}{462} = \frac{198 - 252 - 176 + 105}{462}$
Group positive and negative terms in the numerator:
Numerator $= (198 + 105) - (252 + 176) = 303 - 428 = -125$.
The result is $\frac{-125}{462}$.
The fraction cannot be simplified further as 125 ($=5^3$) and 462 ($=2 \times 3 \times 7 \times 11$) have no common factors.
Thus, the final answer is $\mathbf{\frac{-125}{462}}$.
Example 2: Find $\frac{-4}{5}$ × $\frac{3}{7}$ × $\frac{15}{16}$ × $\left( \frac{-14}{9} \right)$
Answer:
We need to find the product: $\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \left( \frac{-14}{9} \right)$
Multiply the numerators and denominators, cancelling common factors where possible. Note that the product of two negative numbers is positive.
Rewrite the expression to show potential cancellations:
$\frac{(-4) \times 3 \times 15 \times (-14)}{5 \times 7 \times 16 \times 9}$
Cancel common factors:
$\frac{\cancel{-4}^{\text{ }1}}{\cancel{5}_{\text{ }1}} \times \frac{\cancel{3}^{\text{ }1}}{\cancel{7}_{\text{ }1}} \times \frac{\cancel{15}^{\text{ }3}}{\cancel{16}_{\text{ }4}} \times \frac{\cancel{-14}^{\text{ }2}}{\cancel{9}_{\text{ }3}}$
(We used $\cancel{-4}$ and $\cancel{-14}$ to indicate the negative signs will cancel out to positive overall)
The expression becomes:
$\frac{1}{1} \times \frac{1}{1} \times \frac{3}{4} \times \frac{2}{3}$
Cancel further common factors:
$\frac{1}{1} \times \frac{1}{1} \times \frac{\cancel{3}^{\text{ }1}}{\cancel{4}_{\text{ }2}} \times \frac{\cancel{2}^{\text{ }1}}{\cancel{3}_{\text{ }1}}$
The expression simplifies to:
$\frac{1}{1} \times \frac{1}{1} \times \frac{1}{2} \times \frac{1}{1} = \frac{1 \times 1 \times 1 \times 1}{1 \times 1 \times 2 \times 1} = \frac{1}{2}$
The final answer is $\mathbf{\frac{1}{2}}$.
Example 3: Find $\frac{2}{5}$ × $ \frac{-3}{7}$ - $\frac{1}{14}$ - $\frac{3}{7}$ × $\frac{3}{5}$
Answer:
We need to evaluate the expression: $\frac{2}{5} \times \frac{-3}{7}$ - $\frac{1}{14}$ - $\frac{3}{7}$ × $\frac{3}{5}$
According to the order of operations, perform multiplications first.
First multiplication: $\frac{2}{5} \times \frac{-3}{7} = \frac{2 \times (-3)}{5 \times 7} = \frac{-6}{35}$
Second multiplication: $\frac{3}{7} \times \frac{3}{5} = \frac{3 \times 3}{7 \times 5} = \frac{9}{35}$
Substitute these results back into the expression:
$\frac{-6}{35} - \frac{1}{14} - \frac{9}{35}$
Group the terms with the same denominator (35):
$\left( \frac{-6}{35} - \frac{9}{35} \right) - \frac{1}{14}$
Combine the terms with denominator 35:
$\frac{-6 - 9}{35} - \frac{1}{14} = \frac{-15}{35} - \frac{1}{14}$
Simplify the fraction $\frac{-15}{35}$:
$\frac{-15 \div 5}{35 \div 5} = \frac{-3}{7}$
The expression is now: $\frac{-3}{7} - \frac{1}{14}$
Find a common denominator for 7 and 14, which is 14. Convert $\frac{-3}{7}$:
$\frac{-3}{7} = \frac{-3 \times 2}{7 \times 2} = \frac{-6}{14}$
Perform the final subtraction:
$\frac{-6}{14} - \frac{1}{14} = \frac{-6 - 1}{14} = \frac{-7}{14}$
Simplify the result $\frac{-7}{14}$:
$\frac{-7 \div 7}{14 \div 7} = \frac{-1}{2}$
The final answer is $\mathbf{\frac{-1}{2}}$.
Exercise 1.1
Question 1. Name the property under multiplication used in each of the following.
(i) $\frac{-4}{5}$ × 1 = 1 × $\frac{-4}{5}$ = $-\frac{4}{5}$
(ii) $-\frac{13}{17}$ × $\frac{-2}{7}$ = $\frac{-2}{7}$ × $\frac{-13}{17}$
(iii) $\frac{-19}{29}$ × $\frac{29}{-19}$ = 1
Answer:
We need to identify the mathematical property of multiplication demonstrated in each given expression.
(i) $\frac{-4}{5}$ × 1 = 1 × $\frac{-4}{5}$ = $-\frac{4}{5}$
This shows that multiplying any number by 1 results in the number itself. The number 1 is the multiplicative identity.
Property used: Multiplicative Identity
(ii) $-\frac{13}{17}$ × $\frac{-2}{7}$ = $\frac{-2}{7}$ × $\frac{-13}{17}$
This shows that the order of the factors does not affect the product.
Property used: Commutativity of Multiplication (or Commutative Property)
(iii) $\frac{-19}{29}$ × $\frac{29}{-19}$ = 1
This shows that the product of a number and its reciprocal (multiplicative inverse) is 1.
Property used: Multiplicative Inverse (or Reciprocal Property)
Question 2. Tell what property allows you to compute $\frac{1}{3}$ × $\left( 6 \times \frac{4}{3} \right)$ as $\left( \frac{1}{3} \times 6 \right)$ × $\frac{4}{3}$ .
Answer:
The given computation involves changing the grouping of factors in a multiplication expression without changing the order of the factors.
The expression is changed from $a \times (b \times c)$ to $(a \times b) \times c$, where $a = \frac{1}{3}$, $b = 6$, and $c = \frac{4}{3}$.
This property states that the way in which factors are grouped in a multiplication does not change the product.
This property is called the Associative Property of Multiplication.
Question 3. The product of two rational numbers is always a _______.
Answer:
We need to complete the given statement about the product of two rational numbers.
A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not zero.
Let's consider two rational numbers, say $\frac{a}{b}$ and $\frac{c}{d}$, where $a, b, c, d$ are integers and $b \neq 0, d \neq 0$.
The product of these two rational numbers is:
$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$
Since $a$ and $c$ are integers, their product $a \times c$ is also an integer.
Since $b$ and $d$ are non-zero integers, their product $b \times d$ is also a non-zero integer.
Therefore, the product $\frac{a \times c}{b \times d}$ is an integer divided by a non-zero integer, which fits the definition of a rational number.
The statement is: The product of two rational numbers is always a rational number.