**Question 1 :** Is zero a rational number? Can you write it in the form $\frac{p} {q}$ , where $p$ and $q$ are integers and $q \neq 0$?

**Answer:** Yes, zero is a rational number. It can be written as $\frac{0}{1}$ ,$\frac{0}{2}$ , etc., in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

**Question 2 :** Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ .

**Answer:** To find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ , we may add the given numbers and divide by 2, and repeat the process.

$x_{1}$= $\frac{\frac{3}{5}+\frac{4}{5}}{2}$ = $\frac{7}{5\times2}$ = $\frac{7}{10}$

$\frac{7}{10} + \frac{4}{5} = \frac{7+8}{10} = \frac{15}{10}$

$x_{2}= \frac{15}{10\times2} = \frac{15}{20} = \frac{3}{4}$

$\frac{3}{4} + \frac{4}{5} = \frac{15+16}{20} = \frac{31}{20}$

$x_{3}= \frac{31}{20\times2} = \frac{31}{40} $

$\frac{31}{40} + \frac{4}{5} = \frac{31+32}{40} = \frac{63}{40}$

$x_{4}= \frac{63}{40\times2} = \frac{63}{80}$

$\frac{63}{80} + \frac{4}{5} = \frac{63+64}{80} = \frac{127}{80}$

$x_{5}= \frac{127}{80\times2} = \frac{127}{160}$

$x_{1}$,$x_{2}$,$x_{3}$,$x_{4}$,$x_{5}$ are the 5 rational numbers.

(**Note** : Many answers are possible to this question. There are of course infinitely many rational numbers between *p* and *q*.)

**Question 3** : State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

**Answer **:(i) True. Since the collection of whole numbers contains all the natural

numbers.

(ii) False. Negative integers are not whole numbers.

(iii) False. Numbers such as $\frac{2}{3},\frac{3}{4},\frac{-3}{5}$ etc., are rational numbers but not whole numbers.

**Question 4:** Find a rational number lying between $\sqrt{2}$ and $\sqrt{3}$ .

**Answer :** We know that $\sqrt{2}$ = 1.414215 …… .. (i)

And, $\sqrt{3}$ = 1.732050 …… .. (ii)

From (i) and (ii) we conclude that rational number 1.6 lies between $\sqrt{2}$ and $\sqrt{3}$.

**Question 5:** State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form $\sqrt{m}$ , where m is a natural number.

(iii) Every real number is an irrational number.

Answer :

(i) True. All irrational and rational numbers together make up the collection of real

numbers *R*.

(ii) False. For example, between $\sqrt{2}$ and $\sqrt{3}$ there are infinitely many numbers and these cannot be represented in the form $\sqrt{m}$ where *m* is a natural number.

**Question 6: **To locate $\sqrt{n}$ , where n is a positive integer, on the number line, what do we first locate?**Answer:** We can locate $\sqrt{n}$ , for any positive integer *n* , after $\sqrt{n-1}$ has been located.

**Question 7:** Represent $\sqrt{2}$ on the number line.

**Answer:** On the number line, take OA = 1 unit. Draw BA = 1 unit, perpendicular to OA. Join

OB. By Pythagoras theorem OB = $\sqrt{2}$ units.

With centre O and radius OB, draw an arc, which cuts the number line at C. Then C

corresponds to $\sqrt{2}$ .

**Question 8:** Represent $\sqrt{3}$ on the number line.

**Answer:** First represent $\sqrt{2}$ as explained in the above example.

Construct BD of unit length perpendicular to OB. Then, using Pythagoras theorem, we note

that

$OD=\sqrt{\left(\sqrt{2}\right)^{2}+1^{2}}=\sqrt{3}$

Using a compass, with centre O, and radius OD, draw an arc which cuts the number line at E. The point E corresponds to $\sqrt{3}$

**Question 9:** Write the following in decimal form and say what kind of decimal expansion each has :

(1) $\frac{36}{100}$

(2) $\frac{1}{11}$

(3) $4\frac{1}{8}$

(4) $\frac{3}{13}$

(5) $\frac{2}{11}$

(6) $\frac{329}{400}$

**Answer:**

(1) $0.36, terminating.$

(2) $0.\overline{09} , non-terminating ~recurring.$

(3) $4.125, terminating.$

(4) $0.\overline{230769} , non-terminating ~recurring.$

(5) $0.\overline{18} , non-terminating~recurring.$

(6) $0.8225, terminating.$

**Question 10 : **You know that $\frac{1}{7}$ = 0.142857 . Can you predict what the decimal expansions of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are, without actually doing the long division? If so, how?

**Answer: **

$\frac{2}{7} = 2\times\frac{1}{7} = 0.\overline{285714}$

$\frac{3}{7} = 2\times\frac{1}{7} = 0.\overline{428571}$

$\frac{4}{7} = 4\times\frac{1}{7} = 0.\overline{571428}$

$\frac{5}{7} = 5\times\frac{1}{7} = 0.\overline{714285}$

$\frac{6}{7} = 2\times\frac{6}{7} = 0.\overline{857142}$

**Question 11 :** Express 0.99999 … in the form $\frac{p}{q}$ .

**Answer:** $ x = 0.99999 … = 0.\overline{9}$

One digit is repeating, we multiply by 10.

$10 x = 9.\overline{9} \Rightarrow 10 x = 9 + x \Rightarrow 9 x = 9 \Rightarrow x = 1$

The answer makes sense as $0.\overline{9}$ is infinitely close to 1, i.e., we can make the difference between 1 and 0.99 …… as small as we wish by taking enough 9’s.

**Question 12:** Look at several examples of rational numbers in the form $\frac{p}{q} (q \neq 0 )$, where *p* and *q* are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer:

$\frac{2}{5} = 0.4 , \frac{3}{2} =1.5 ,\frac{7}{8} = 0.875 ,\frac{7}{10} = 0.7 $

All the denominators are either 2 (or its power), 5 (or its power) or a combination of both.

**Question 13:** Find the value of 1.999 …. in the form $\frac{p}{q} $, where *p* and *q* are integers and *q* ≠ 0 .

**Answer:** Let $x = 1.999$ … (i)

$\Rightarrow 10 x = 19.999 $ … (ii)

Subtracting (i) from (ii), we get $9 x = 18$

$\Rightarrow x = 18/9 = 2 $

**Question 14:** Express $0.12\overline{3}$ in the form of $\frac{p}{q} $, where $p,q \in Z$ and $q \neq 0$ .

**Answer: ** Let $x = 0.12333 …$ … (i)

$\Rightarrow 100 x = 12.333 … $ … (ii)

$\Rightarrow 1000 x = 123.333 …$ … (iii)

Subtracting (ii) from (iii), we get $900 x = 111$

$\Rightarrow x = \frac{111}{900} = \frac{37}{300} $

**Question 15 : **Visualise 7.8324 on the number line using successive magnification.

**Answer:**

Step 1. The given number 7.8324 lies between 7 and 8 on the number line.

Step 2. Magnify the interval between 7 and 8 and divide it into 10 equal parts.

Step 3. The given number lies between 7.8 and 7.9.

Step 4. Magnify the interval between 7.8 and 7.9 and divide it into 10 equal parts.

Step 5. The given number lies between 7.83 and 7.84.

Step 6. Magnify the interval between 7.83 and 7.84 and divide it into 10 equal parts.

Step 7. The given number lies between 7.832 and 7.833.

Step 8. Magnify the interval between 7.832 and 7.833 and divide it into 10 equal parts.

Step 9. The given number is the fourth division of the given interval.

**Question 16: **Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = $\frac{c}{d} $ . This seems to contradict d the fact that π is irrational. How will you resolve this contradiction?

**Answer : **

With a scale or tape we get only an approximate rational number as the result of our

measurement. That is why π can be approximately represented as a quotient of two

rational numbers. As a matter of mathematical truth, it is irrational.

**Question 17: ** Represent $\sqrt{9.3}$ on the number line.

Answer: To represent $\sqrt{9.3}$ To represent 9.3 , draw a segment of 9.3 units on the number line. Let A represent 9.3. Extend it by 1 cm. Show point $\frac{10.3}{2} = 5.15$ by O on the number line. With O as centre and radius 5.15 units, draw a semi-circle. Draw AB perpendicular to OA to cut the semi-circle at B.

The length AB is $\sqrt{9.3}$ units.

**Question 18: **If *a *and *b* are natural numbers, then $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$ is rational. Is it true?

**Answer: ** $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = (\sqrt{a})^{2}-(\sqrt{b})^{2}= a-b$ ,which is a rational number.

**Question 19: **Simplify : $\frac{4+\sqrt{6}}{4-\sqrt{6}}+\frac{4-\sqrt{6}}{4+\sqrt{6}}$

**Answer:** $\frac{4+\sqrt{6}}{4-\sqrt{6}}+\frac{4-\sqrt{6}}{4+\sqrt{6}}$

$=\frac{\left(4+\sqrt{6}\right)^{2}+\left(4-\sqrt{6}\right)^{2}}{\left(4-\sqrt{6}\right)\left(4+\sqrt{6}\right)}$

$=\frac{16+6+8\sqrt{6}+16+6-8\sqrt{6}}{16-6}$

$=\frac{32+12}{10}$

$=\frac{44}{10}$

$=4.4$

**Question 20:** If $x=2-\sqrt{3}$ ,then find the value of $\left(x+\frac{1}{x}\right)^{3}$.

**Answer: **$x=2-\sqrt{3}$

$\Rightarrow\frac{1}{x}=\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$

So, $\left(x+\frac{1}{x}\right)^{3}=\left(2-\sqrt{3}+2+\sqrt{3}\right)^{3}=4^{3}=64$

**Question 21: ** If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$ , find the values of a and b.

**Answer: **

$\frac{\sqrt{3}-1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=a+b\sqrt{3}$

$\Rightarrow\frac{3+1-2\sqrt{3}}{3-1}=a+b\sqrt{3}$

**Question 22:** Simplify the following by rationalising the denominator : $\frac{1}{\sqrt{5}-\sqrt{2}-\sqrt{7}}$

**Answer :** We have , $\frac{1}{\sqrt{5}-\sqrt{2}-\sqrt{7}}$

$={\frac{1}{\left(\sqrt{5}-\sqrt{2}\right)-\sqrt{7}}=\frac{\left[\left(\sqrt{5}-\sqrt{2}\right)+\sqrt{7}\right]}{\left[\left(\sqrt{5}-\sqrt{2}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{2}+\sqrt{7}\right)\right]}}$

[Rationalising the denominator]

$=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\sqrt{7}}{\left(\sqrt{5}-\sqrt{2}\right)^{2}-\left(\sqrt{7}\right)^{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\sqrt{7}}{5+2-2\sqrt{5}\sqrt{2-7}}$

$=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\sqrt{7}}{-2\sqrt{10}}=\frac{-\left[\left(\sqrt{5}-\sqrt{2}\right)+\sqrt{7}\right]\sqrt{10}}{2\sqrt{10}\sqrt{10}}$

[Multiplying numerator and denominator by $\sqrt{10}$ ]

$\frac{-\sqrt{50}+\sqrt{20}-\sqrt{70}}{20}=\frac{\sqrt{20}-\sqrt{70}-\sqrt{50}}{20}=\frac{2\sqrt{5}-\sqrt{70}-5\sqrt{2}}{20}$

**Question 23:** Find the value of $x^{a-b}\times x^{b-c}\times x^{c-a}$

**Answer :**

$x^{a-b}\times x^{b-c}\times x^{c-a}=x^{a-b+b-c+c-a}=x^{0}=1$

**Question 24: **Show that $\frac{x^{a\left(bc\right)}}{x^{b\left(ac\right)}}\div\left(\frac{x^{b}}{x^{a}}\right)^{c}=1$

**Answer:**

$\frac{x^{a\left(bc\right)}}{x^{b\left(ac\right)}}\div\left(\frac{x^{b}}{x^{a}}\right)^{c}=\frac{x^{ab-ac}}{x^{ba-bc}}\div x^{\left(b-a\right)c}$

$=x^{ab-ac-ba+bc}\times\frac{1}{x^{bc-ac}}$

$=x^{-ac+bc}\times x^{ac-bc}=x^{-ac+bc+ac-bc}=x^{0}=1$

**Question 25:** Prove that $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$

**Answer:**

$\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}=\frac{x^{b}+x^{a}}{x^{b}+x^{a}}=1$