NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

Q2 Complete the following table:

Commutative for

Answer: In rational numbers, a $÷$ b $\ne$ b $÷$ a

also a-b $\ne$ b-a

Q3 Complete the following table:

Associative for

Answer: For associative in multiplication:- a $\times$ (b $\times$ c) = (a $\times$ b) $\times$ c

(i) $\{\frac{7}{5}\times \left(\frac{-3}{12}\right)\}+\{\frac{7}{5}\times \frac{5}{12}\}$ (ii) $\{\frac{9}{16}\times \frac{4}{12}\}+\{\frac{9}{16}\times \frac{-3}{9}\}$

Answer: (i) Using distributivity, a(b+c) = ab + ac

$$\begin{gathered} \{\frac{7}{5}\times \left(\frac{-3}{12}\right)\}+\{\frac{7}{5}\times \frac{5}{12}\}=\frac{7}{5}(\frac{-3}{12}+\frac{5}{12}) \\ =\frac{7}{5}\times \frac{1}{6}=\frac{7}{30} \end{gathered}$$

(ii) Using distributivity of multiplication over addition and subtraction,

$$\begin{gathered} \{\frac{9}{16}\times \frac{4}{12}\}+\{\frac{9}{16}\times \frac{-3}{9}\}=\frac{9}{16}(\frac{4}{12}-\frac{3}{9}) \\ =\frac{9}{16}\times 0=0 \end{gathered}$$

Q5 Write the rational number for each point labeled with a letter:

Answer: (i) In this, we can see that 1 is divided into 5 parts each, so when we are moving from zero to the right-hand side, it is easy to observe that

All the numbers should contain 5 in their denominator. Thus, A is equal to $\frac{1}{5}$ , B is equal to $\frac{4}{5}$ , C is equal to $\frac{5}{5}=1$ , D is equal to $\frac{8}{5}$ , E is equal to $\frac{9}{5}$

(ii) Here we see that 1 is divided in 6 parts each. So when we move from zero towards left we observe that

All the numbers should contain 6 in their denominator. Thus, F is equal to $\frac{-2}{6}$ , G is equal to $\frac{-5}{6}$ , H is equal to $\frac{-7}{6}$ , I is equal to $\frac{-8}{6}$ , J is equal to $\frac{-11}{6}$

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers - Exercise: 1.1

Q1 (i) Using appropriate properties find. $(i)\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$

Answer: By using the commutativity property of numbers, we get,

$-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}=-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}$

(Now we will use distributivity of numbers)

= $$\begin{gathered} =\frac{3}{5}(\frac{-2}{3}+\frac{-1}{6})+\frac{5}{2} \\ =\frac{3}{5}\times \frac{-5}{6}+\frac{5}{2} \\ =\frac{-1}{2}+\frac{5}{2} \\ =\frac{-1+5}{2}=\frac{4}{2} \\ =2 \end{gathered}$$

Q1 (ii) Using appropriate properties find. $(ii)\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

Answer: By using commutativity, we get

$\frac{2}{5}\times \left(\frac{-3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}=\frac{2}{5}\times \left(\frac{-3}{7}\right)+\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$

Now by distributivity,

$$\begin{gathered} =\frac{2}{5}(\frac{-3}{7}+\frac{1}{14})-\frac{1}{4} \\ =\frac{2}{5}\times \frac{-5}{14}-\frac{1}{4}=\frac{-1}{7}-\frac{1}{4}=\frac{-4-7}{28}=\frac{-11}{28} \end{gathered}$$

Q2 Write the additive inverse of each of the following:

(i) $\frac{2}{8}$ (ii) $\frac{-5}{9}$ (iii) $\frac{-6}{-5}$ (iv) $\frac{2}{-9}$ (v) $\frac{19}{-6}$

Answer: (i) The additive inverse of $\frac{2}{8}$ is $\frac{-2}{8}$ because $\frac{2}{8}+\frac{-2}{8}=\frac{2-2}{8}=0$

(ii) The additive inverse of $\frac{-5}{9}$ is $\frac{5}{9}$ because $\frac{-5}{9}+\frac{5}{9}=\frac{-5+5}{9}=0$

(iii) The additive inverse of $\frac{-6}{-5}$ is $\frac{6}{-5}$ because $\frac{-6}{-5}+\frac{6}{-5}=\frac{-6+6}{-5}=0$

(iv) The additive inverse of $\frac{2}{-9}$ is $\frac{-2}{-9}$ because $\frac{2}{-9}+\frac{-2}{-9}=\frac{2-2}{-9}=0$

(v) The additive inverse of $\frac{19}{-6}$ is $\frac{-19}{-6}$ because $\frac{19}{-6}+\frac{-19}{-6}=\frac{19-19}{-6}=0$

Q3 Verify that – (– x) = x for (i) x = $\frac{11}{15}$ (ii) x = $\frac{-13}{17}$

Answer: (i) We have x = $\frac{11}{15}$

The additive inverse of x = $\frac{11}{15}$ is -x = $\frac{-11}{15}$

The same equality $\frac{11}{15}+\left(\frac{-11}{15}\right)=0$

which implies $-\left(\frac{-11}{15}\right)=\frac{11}{15}$ shows that -(-x) = x

(ii) Additive inverse of x = $\frac{-13}{17}$ is -x = $\frac{13}{17}$ (since $\frac{-13}{17}+\frac{13}{17}=0$ )

The same quality shows that the additive inverse of $\frac{13}{17}$ is $\frac{-13}{17}$

i.e., -(-x) = x

Answer:

(i) The multiplicative inverse of -13 is $\frac{-1}{13}$ because $-13\times \frac{-1}{13}=1$

(ii) The multiplicative inverse of $\frac{-13}{19}$ is $\frac{-19}{13}$ because of i

(iii) The multiplicative inverse of $\frac{1}{5}$ is 5 because $\frac{1}{5}\times 5=1$

(iv) The multiplicative inverse of $\frac{-5}{8}\times \frac{-3}{7}$ is $\frac{56}{15}$ because $\frac{15}{56}\times \frac{56}{15}=1$

(v) The multiplicative inverse of $-1\times \frac{-2}{5}$ is $\frac{5}{2}$ because $\frac{2}{5}\times \frac{5}{2}=1$

(vi) The multiplicative inverse of -1 is -1 because $-1\times -1=1$

Q5 Name the property under multiplication used in each of the following.

(i) $\frac{-4}{5}\times 1=1\times \frac{-4}{5}=\frac{-4}{5}$ (ii) $-\frac{13}{17}\times \frac{-2}{7}=\frac{-2}{7}\times \frac{-13}{17}$ (iii) $\frac{-19}{29}\times \frac{29}{-19}=1$

Answer: (i) Multiplying any number with 1 we get the same number back.

i.e., a $\times$ 1 = 1 $\times$ a = a

Hence 1 is the multiplicative identity for rational numbers.

(ii) Commutativity property states that a $\times$ b = b $\times$ a

(iii) It is the multipicative inverse identity, i.e., $a\times \frac{1}{-a}=1$

Q8 Is $\frac{8}{9}$ the multiplicative inverse of $-1\frac{1}{8}$ ? Why or why not?

Answer:

We know that A is the multiplicative inverse of B if B $\times$ A = 1

Applying this in given question, we get,

$\frac{-9}{8}\times \frac{8}{9}=-1$ Thus $\frac{8}{9}$ is not the multiplicative inverse of $\frac{-9}{8}$ .

Q10 Write.

(i) The rational number that does not have a reciprocal.

(ii) The rational numbers that are equal to their reciprocals.

(iii) The rational number that is equal to its negative.

Answer:

(i) Zero(0). We know that reciprocal of A is $\frac{1}{A}$ . So for 0, its reciprocal is not defined.

(ii) 1 and -1 . (Since $\frac{1}{1}=1$ and $\frac{1}{-1}=-1$ )

(iii) Zero,0. (as -0=0)

Q11 Fill in the blanks.

(i) Zero has ________ reciprocal.

(ii) The numbers ________ and ________ are their own reciprocals.

(iii) The reciprocal of – 5 is ________.

(iv) Reciprocal of $\frac{1}{x}$ , where x ≠ 0 is ________.

(v) The product of two rational numbers is always a _______.

(vi) The reciprocal of a positive rational number is ________.

Answer: (i) Zero has no reciprocal as it's reciprocal is not defined.

(ii) The numbers 1 and -1 are their own reciprocals as $\frac{1}{1}=1$ and $\frac{1}{-1}=-1$

(iii) $\frac{1}{-5}$ . We know that reciprocal of A is $\frac{1}{A}$ .

(iv) Since $\frac{1}{\frac{1}{x}}=x$ .

(v) rational number. We know that if p and q are 2 rational numbers then pq is also a rational number.

(vi) Positive. Since reciprocal of A is $\frac{1}{A}$ , now if A is positive then reciprocal is also positive.

Class 8 maths chapter 1 question answer - Exercise: 1.2

Q1 Represent these numbers on the number line. (i) $\frac{7}{4}$ (ii) $\frac{-5}{6}$

Answer:(i) To represent $\frac{7}{4}$ on a number line, firstly we will divide 1 in 4 parts and draw it on a line such as 1/4, 2/4, 3/4, ........, 9/4. Then will mark the required number.

(ii) To represent $\frac{-5}{6}$ on the number line, firstly we will divide 1 in 6 parts and draw it on the left side of zero on number line such as -1/6, -2/6, .......,-9/4. Then mark the required number on the number line.