# NCERT solution for class 9 maths number systems ( Chapter 1)

#### Solution for Exercise 1.1

1. Is zero a rational number?Can you write it in the form p/q,where p and q are integers and $$q \ne 0$$.

Yes, we may rewrite 0 as 0/1 , 0/2 , 0/3 (where 0 and 1,2,3 are integers and q=1,2,3 which is not equal to zero).
Because we know that a number is called as rational number if it can be expressed in p/q form.

2. Find six rational numbers between 3 and 4.

There can be infinitely many rational numbers between 3 and 4, one way to take them is 3=21/(6+1) , 4=28/(6+1).
The rational numbers between 3 and 4 therefore, will become 22/7 ,23/7 ,24/7 ,25/7 ,26/7 ,27/7.
Or Another way is i.e. (21/7 + 28/7)/2 i.e. (49/7)/2 i.e. 7/2. As, 21/7 < 7/2 < 28/7.
Rational no. between 21/7 and 7/2 will become (21/7 + 7/2)/2 i.e. (91/14)/2 i.e. 91/28.
As, 21/7 < 91/28 < 7/2 < 28/7.
Similarly, the other group of rational numbers between 3 and 4 will become 175/56 , 91/28 , 7/2 , 203/56, 105/28 , 217/56 .

3. Find five rational numbers between 3/5 and 4/5.

As per the example above, we know that rational number between 3/5 and 4/5 is (3/5 + 4/5)/2 i.e. (7/5)/2 i.e. 7/10. As, 3/5 < 7/10 < 4/5.
Now,a rational number between 3/5 and 7/10 is (3/5 + 7/10)/2 i.e. (13/10)/2 i.e. 13/20. As, 3/5 < 13/20 <7/10 < 4/5. Similarly, 25/40 , 27/40 , 15/20 are rational numbers between 3/5 and 4/5.
Therefore, five rationals between numbers3/5 and 4/5 are 25/40, 13/20, 27/40, 7/10, 15/20.

4.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.

i)True, because natural numbers are 1, 2, 3, 4,.............. and whole numbers are 0, 1, 2, 3, 4, 5,....... i.e. the collections of whole numbers contain all the natural numbers.

ii)False, because negative integers are not whole numbers.

iii)False, because for e.g. 16/23 or 7/2 are rational numbers who are not whole numbers.

#### Solution for Exercise 1.2

1. 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.

i) True, since (Real number = Rational number + Irrational number).

ii)False, because no negative number can be the square root of any natural number.

iii)False, as every irrational number is a real number but opposite is not true.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

No, the square roots of all positive integers are not irrational.
For Example, $$\sqrt{9}$$ = 3 Here, '3' is a rational number. While some numbers have irrational square roots, too.

3. Show how $$\sqrt{5}$$ can be represented on the number line.

We know that, $$\sqrt{5}$$ = $$\sqrt{4 + 1}$$ = $$\sqrt{2^2 + 1^2}$$.
Draw a right angled triangle, OQP, such that OQ = 2 Units and PQ = 1 Unit And $$\angle{OQP} = 90°$$
Now, by using Pythagoras theorem, we have $$OQ^2 = OP^2 + PQ^2 = 2^2 + 1^2$$.

Therefore, OP = $$\sqrt{5}$$.
Now, take O as centre OP = $$\sqrt{5}$$ as radius, draw an arc, which intersects the line at point R.
Hence, the point R represents $$\sqrt{5}$$ .

4. Classroom activity (constructing the 'square root spiral').

Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion.
Start with a point O and draw a line segment $$OP_1$$ of unit length. Draw a line segment $$P_1P_2$$ perpendicular to $$OP_1$$ of unit length.
Now, draw a line segment $$P_2P_3$$ perpendicular to $$OP_2$$ . Then draw a line segment $$P_3P_4$$ perpendicular to $$OP_3$$.
Continuing in this manner, you can get the line segment $$P_(n-1)Pn$$ by drawing a line segment of unit length perpendicular to $$OP_(n-1)$$ .
In This manner, you will have created the points $$P_2$$, $$P_3$$, $$P_4$$, $$P_5$$,........ and joined them to create a beautiful spiral depicting $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{4}$$,......

#### Solution for Exercise 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has:
i) 36/100
ii) 1/11
iii $$4\frac{1}{8}$$
iv)3/13
v)2/11
vi)329/400

i) Clearly, 36/100 can be written as 0.36.
Therefore, it is a terminating decimal.

ii) Dividing 1 by 11, we get$$\require{enclose} \begin{array}{rll} 0.0909 && \hbox{(Explanations)} \\[-3pt] 11 \enclose{longdiv}{100}\kern-.2ex \\[-3pt] \underline{99\phantom{00}} && \hbox{(9 \times 11 = 99)} \\[-3pt] \phantom{0}100 && \hbox{(100 - 99 = 1)} \\[-3pt] \underline{\phantom{0}99} && \hbox{(9 \times 11 = 99)} \\[-3pt] \phantom{00}1 \end{array}$$Hence, 1/11 can be written as 0.090909...
Therefore, it is a non-terminating decimal.

iii) We have, $$4\frac{1}{8}$$ = $$\frac{4×8+1}{8}$$ = $$\frac{33}{8}$$
Dividing 33 by 8, we get, $$\require{enclose} \begin{array}{rll} 4.125 && \hbox{(Explanations)} \\[-3pt] 8 \enclose{longdiv}{33}\kern-.2ex \\[-3pt] \underline{32\phantom{00}} && \hbox{(8 \times 4 = 32)} \\[-3pt] \phantom{0}10 && \hbox{(33 - 32 = 1)} \\[-3pt] \underline{8\phantom{00}} && \hbox{(8 \times 1 = 8)} \\[-3pt] \phantom{0}20 && \hbox{(10 - 8 = 2)} \\[-3pt] \underline{16\phantom{00}} && \hbox{(8 \times 2 = 16)} \\[-3pt] \phantom{0}40 && \hbox{(20 - 16 = 4)} \\[-3pt] \underline{\phantom{0}40} && \hbox{(8 \times 5 = 40)} \\[-3pt] \phantom{00}0 \end{array}$$Hence, $$4\frac{1}{8}$$ can be written as 4.125
Therefore, it is a terminating decimal.

iv) We have, 3/13
Dividing 3 by 13, we get, $$\require{enclose} \begin{array}{rll} 0.230769 && \hbox{(Explanations)} \\[-3pt] 13 \enclose{longdiv}{30}\kern-.2ex \\[-3pt] \underline{26\phantom{00}} && \hbox{(2 \times 13 = 26)} \\[-3pt] \phantom{0}40 && \hbox{(30 - 26 = 4)} \\[-3pt] \underline{39\phantom{00}} && \hbox{(3 \times 13 = 39)} \\[-3pt] \phantom{0}100 && \hbox{(40 - 39 = 1)} \\[-3pt] \underline{91\phantom{00}} && \hbox{(7 \times 13 = 91)} \\[-3pt] \phantom{0}90 && \hbox{(100 - 91 = 4)} \\[-3pt] \underline{78\phantom{00}} && \hbox{(6 \times 13 = 78)} \\[-3pt] \phantom{0}120 && \hbox{(90 - 78 = 12)} \\[-3pt] \underline{\phantom{0}117} && \hbox{(9 \times 13 = 117)} \\[-3pt] \phantom{00}3 \end{array}$$Hence, 3/13 can be written as 0.230769
Therefore, it is a non-terminating decimal.

v) We have, 2/11
Dividing 2 by 11, we get, $$\require{enclose} \begin{array}{rll} 0.1818 && \hbox{(Explanations)} \\[-3pt] 11 \enclose{longdiv}{20}\kern-.2ex \\[-3pt] \underline{11\phantom{00}} && \hbox{(1 \times 11 = 11)} \\[-3pt] \phantom{0}90 && \hbox{(20 - 11 = 9)} \\[-3pt] \underline{88\phantom{00}} && \hbox{(8 \times 11 = 11)} \\[-3pt] \phantom{0}20 && \hbox{(90 - 88 = 2)} \\[-3pt] \underline{11\phantom{00}} && \hbox{(1 \times 11 = 11)} \\[-3pt] \phantom{0}90 && \hbox{(20 - 11 = 9)} \\[-3pt] \underline{\phantom{0}88} && \hbox{(8 \times 11 = 88)} \\[-3pt] \phantom{00}2 \end{array}$$Hence, 2/11 can be written as 0.181818...
Therefore, it is a non-terminating decimal.

vi) We have, 329/400
Dividing 329 by 400, we get, $$\require{enclose} \begin{array}{rll} 0.8225 && \hbox{(Explanations)} \\[-3pt] 400 \enclose{longdiv}{3290}\kern-.2ex \\[-3pt] \underline{3200\phantom{00}} && \hbox{(8 \times 400 = 3200)} \\[-3pt] \phantom{0}900 && \hbox{(3290 - 3200 = 9)} \\[-3pt] \underline{800\phantom{00}} && \hbox{(2 \times 400 = 800)} \\[-3pt] \phantom{0}1000 && \hbox{(900 - 800 = 100)} \\[-3pt] \underline{800\phantom{00}} && \hbox{(2 \times 400 = 800)} \\[-3pt] \phantom{0}2000 && \hbox{(1000 - 800 = 200)} \\[-3pt] \underline{\phantom{0}2000} && \hbox{(5 \times 400 = 400)} \\[-3pt] \phantom{00}0 \end{array}$$Hence, 329/400 can be written as 0.8225
Therefore, it is a terminating decimal.

2. You know that $$1/7 = 0\overline{.142857.}$$ Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division. If so, how?
[Hint Study the remainders while finding the value 1/7 of carefully.]

We have, $$1/7 = 0\overline{.142857}$$
2/7 = 2 × 1/7
$$= 2 × 0\overline{.142857}$$
i.e. $$2/7 = 0\overline{.285714}$$

3/7 = 3 × 1/7
$$= 3 × 0\overline{.142857}$$
i.e. $$3/7 = 0\overline{.428571}$$

4/7 = 4 × 1/7
$$= 4 × 0\overline{.142857}$$
i.e. $$4/7 = 0\overline{.571428}$$

5/7 = 5 × 1/7
$$= 5 × 0\overline{.142857}$$
i.e. $$5/7 = 0\overline{.714285}$$

6/7 = 6 × 1/7
$$= 6 ×0 \overline{.142857}$$
i.e. $$6/7 = 0\overline{.857142}$$

3. Express the following in the form p/q where p and q are integers and $$q \ne 0$$ .
i)$$0\overline{.6}$$
ii)$$0.4\overline{7}$$
iii)$$0\overline{.001}$$

i) Let x = 0.66666... (i)
multiplying by eq. (i) by 10 ,
we get, 10 x = 6.666...... (ii)
On subtracting Eq. (ii) from Eq. (i),
we get (10 x – x) = (6.666…) – (0.666…)
i.e., 9x = 6
i.e., x = 6/9
Hence, x = 2/3

ii)Let x = 0.477777... (i)
multiplying by eq. (i) by 10 ,
we get, 10 x = 4.777....... (ii)
On subtracting Eq. (ii) from Eq. (i), we get, (10 x – x) = (4.777) - (0.4777)
i.e., 9x = 4.3
i.e., x = 4.3/9
Hence, x = 43/90

iii)Let x = 0.001001001...... (i)
multiplying by eq. (i) by 1000 , we get,
1000 x = 1.001001001....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(1000 x – x) = (1.001001001....) – (0.001001001....)
i.e., 999x = 1
Hence, x = 1/999

4.Express 0.99999… in the form p/q. Are you surprised by your answer?

Let,
x = 0.9999.... (i)
multiplying by eq. (i) by 10 , we get,
10 x = 9.9999......(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(10 x – x) = (9.99999…) – (0.99999…)
i.e., 9x = 9
Hence, x = 1

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ?

We know that, the maximum number of digits in the repeating block of digits in the decimal expansion of 1/17 is 17 – 1 = 16. $$\require{enclose} \begin{array}{rll} 0.0588235294117647 && \hbox{(Explanations)} \\[-3pt] 17 \enclose{longdiv}{100}\kern-.2ex \\[-3pt] \underline{85\phantom{00}} && \hbox{(5 \times 17 = 85)} \\[-3pt] \phantom{0}150 && \hbox{(100 - 85 = 15)} \\[-3pt] \underline{136\phantom{00}} && \hbox{(8 \times 17 = 136)} \\[-3pt] \phantom{0}140 && \hbox{(150 - 136 = 14)} \\[-3pt] \underline{136\phantom{00}} && \hbox{(8 \times 17 = 136)} \\[-3pt] \phantom{0}40 && \hbox{(140 - 136 = 4)} \\[-3pt] \underline{34\phantom{00}} && \hbox{(2 \times 17 = 34)} \\[-3pt] \phantom{0}60 && \hbox{(40 - 34 = 6)} \\[-3pt] \underline{51\phantom{00}} && \hbox{(3 \times 17 = 51)} \\[-3pt] \phantom{0}90 && \hbox{(60 - 51 = 9)} \\[-3pt] \underline{85\phantom{00}} && \hbox{(5 \times 17 = 85)} \\[-3pt] \phantom{0}50 && \hbox{(90 - 85 = 5)} \\[-3pt] \underline{34\phantom{00}} && \hbox{(2 \times 17 = 34)} \\[-3pt] \phantom{0}160 && \hbox{(50 - 34 = 16)} \\[-3pt] \underline{153\phantom{00}} && \hbox{(9 \times 17 = 153)} \\[-3pt] \phantom{0}70 && \hbox{(160 - 153 = 7)} \\[-3pt] \underline{68\phantom{00}} && \hbox{(4 \times 17 = 68)} \\[-3pt] \phantom{0}20 && \hbox{(70 - 68 = 2)} \\[-3pt] \underline{17\phantom{00}} && \hbox{(1 \times 17 = 17)} \\[-3pt] \phantom{0}30 && \hbox{(20 - 17 = 3)} \\[-3pt] \underline{28\phantom{00}} && \hbox{(2 \times 17 = 28)} \\[-3pt] \phantom{0}130 && \hbox{(30 - 17 = 13)} \\[-3pt] \underline{119\phantom{00}} && \hbox{(7 \times 17 = 119)} \\[-3pt] \phantom{0}110 && \hbox{(130 - 119 = 11)} \\[-3pt] \underline{102\phantom{00}} && \hbox{(6 \times 17 = 136)} \\[-3pt] \phantom{0}80 && \hbox{(110 - 102 = 8)} \\[-3pt] \underline{68\phantom{00}} && \hbox{(4 \times 17 = 68)} \\[-3pt] \phantom{0}40 && \hbox{(80 - 68 = 120)} \\[-3pt] \underline{\phantom{0}119} && \hbox{(7 \times 400 = 119)} \\[-3pt] \phantom{00}1 \end{array}$$ We have, Thus, 1/17 = $$= 0\overline{.0588235294117647}$$
i.e.a block of 16 digits is repeated.

6.Look at several examples of rational numbers in the form p/q ($$q \ne 0$$). 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?

Considering some rational numbers in the form p/q (q?0) with no common factors other than 1 and having terminating decimal representations(expansions).
we can say, the various such rational numbers are 1\2, 7/125, 1/4, 19/20, etc.
$$1/2 = (1 × 5)/(2 × 5) = 5/10 = 0.5.$$
$$7/125 = (7 × 8)/(125 × 8) = 56/1000 = 0.056$$
$$1/4 = (1 × 25)/(25 × 4) = 25/100 = 0.25$$
$$19/25 = (19 × 4)/(25 × 4) = 76/100 = 0.76$$
In all the cases mentioned above, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10.
Thus, we find that, the decimal expansion of above numbers are terminating.
Along with, we see that the denominator of above numbers i.e. q is in the form has only powers of 2 or power of 5 or both of them.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

The required numbers are as follows :

i) 4.020020002000002

ii)0.36366366636666

iii) 0.101100110001

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
So, for 9/11,
we get, $$\require{enclose} \begin{array}{rll} 0.8181 && \hbox{(Explanations)} \\[-3pt] 11 \enclose{longdiv}{90}\kern-.2ex \\[-3pt] \underline{88\phantom{00}} && \hbox{(8 \times 11 = 88)} \\[-3pt] \phantom{0}20 && \hbox{(90 - 88 = 2)} \\[-3pt] \underline{11\phantom{00}} && \hbox{(1 \times 11 = 11)} \\[-3pt] \phantom{0}90 && \hbox{(20 - 11 = 9)} \\[-3pt] \underline{\phantom{0}88} && \hbox{(8 \times 11 = 88)} \\[-3pt] \phantom{00}2 \end{array}$$ Now, for 5/11,
we get,$$\require{enclose} \begin{array}{rll} 0.714285 && \hbox{(Explanations)} \\[-3pt] 7 \enclose{longdiv}{50}\kern-.2ex \\[-3pt] \underline{49\phantom{00}} && \hbox{(7 \times 7 = 49)} \\[-3pt] \phantom{0}10 && \hbox{(50 - 49 = 1)} \\[-3pt] \underline{7\phantom{00}} && \hbox{(1 \times 7 = 7)} \\[-3pt] \phantom{0}30 && \hbox{(10 - 7 = 3)} \\[-3pt] \underline{28\phantom{00}} && \hbox{(4 \times 7 = 28)} \\[-3pt] \phantom{0}20 && \hbox{(30 - 28 = 2)} \\[-3pt] \underline{14\phantom{00}} && \hbox{(2 \times 7 = 14)} \\[-3pt] \phantom{0}60 && \hbox{(20 - 14 = 6)} \\[-3pt] \underline{56\phantom{00}} && \hbox{(8 \times 7 = 56)} \\[-3pt] \phantom{0}40 && \hbox{(60 - 56 = 4)} \\[-3pt] \underline{\phantom{0}35} && \hbox{(5 \times 7 = 35)} \\[-3pt] \phantom{00}5 \end{array}$$ Thus, $$5/7 = 0\overline{.714285} = 0.71428571428...$$
$$9/11 = 0\overline{.818181} = 0.8181818.....$$

Therefore, the three rational numbers between these will be

0.74074007400074000074...

0.7750775007750007750000...

0.80800800080000...

9. Classify the following numbers as rational or irrational :
i)$$\sqrt{23}$$
ii)$$\sqrt{225}$$
iii) 0.3796
iv)sqrt7.478478.....
v)1.101001000100001.....

i) Irrational (Since, it is not a perfect square).

ii)Rational (Since, it is a perfect square).

iii)0.3796 = Rational (terminating)

iv) 7.478478… = Rational (non-terminating repeating.)

v)1.101001000100001… = Irrational (non-terminating non-repeating.)

#### Solution for exercise 1.4

1. Visualize $$3\overline{.765}$$ on the number line, using successive magnification.

We know that, 3.765... lies between 3 and 4.
So, let us divide the part of the number line between and 3 and 4 into 10 equal parts, mark each point of division and look at the portion between 3.7 and 3.8 through a magnifying glass.
Now 3.765..... lies between 3.7 and 3.8 Figure (i). Now, we imagine to divide this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and soon.
To see this clearly, we magnify this as shown in Figure (ii). Again 3.765.... lie between 3.76 and 3.77 Figure (ii).
So, let us focus on this portion of the number line Figure (iii) and imagine dividing it again into ten equal parts Figure (iii).
Here, we can visualize that 3.761 is the first mark and 3.765...... is the 5th mark in these subdivisions.
We call this process of visualization of representation of number on the number line through a magnifying glass as the process of successive magnification.
So, we get seen that it is possible by sufficient successive magnification of visualize the position (or representation) of a real number with a terminating decimal expansion on the number line.

2. Visualize $$4\overline{.26}$$ on the number line, upto 4 Decimal places.

We know that, by process of successive magnification and successively decrease the length of the portion of the number line in which 4.26... is located.
Since 4.26... is located between 4 and 5 and is divided into 10 equal parts Figure (i). In further, we locate 4.26... between 4.2 and 4.3 Figure (ii).
To get more accurate visualization of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualize that 4.26.... lies between 4.26 and 4.27.
To visualize 4.26... more clearly we divide again between 4.26 and 4.27 into 10 equal parts and visualize the representation of 4.26... between 4.262 and 4.263 Figure (iii).

Now, for a much better visualization between 4.262 and 4.263 is again divided into 10 equal parts Figure (iv). Notice that 4.26..... is located closer to 4.263 then to 4.262 at 4.2627.

#### Solution for Exercise 1.5

1.Classify the following numbers as rational or irrational :
i)$$2 - \sqrt{5}$$
ii)$$(3 + \sqrt{23}) - \sqrt{23}$$
iii)$$2\sqrt{7} / 7\sqrt{7}$$
iv)$$1 / \sqrt{2}$$
v)$$2\pi$$

i) Irrational number (Since, 2 is a rational number and 5 is an irrational number).
Therefore, $$2 - \sqrt{5}$$ is an irrational number.

ii)$$(3 + \sqrt{23}) - \sqrt{23}= 3$$ i.e. rational number .

iii) $$2\sqrt{7} / 7\sqrt{7} = 2/7$$ i.e. rational number

iv) $$1 / \sqrt{2}$$ is an irrational number because, as numerator is a rational number but denominator is an irrational number.

v) $$2\pi$$ is a Irrational number , since 2 is a rational number and $$\pi$$ is an irrational number.

2. Simplify each of the following expressions :
i) $$(3 + \sqrt{3})(2 + \sqrt{2})$$
ii) $$(3 + \sqrt{3})(3 - \sqrt{3})$$
iii) $$(\sqrt{5} + \sqrt{2})(\sqrt{5} + \sqrt{2})$$
iv)$$(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})$$

i)= $$(3 + \sqrt{3})(2 + \sqrt{2})$$
Therefore, $$(3 + \sqrt{3})(2 + \sqrt{2})$$ =$$6 + 3\sqrt{2} + 2\sqrt{3} + 6$$

ii)= $$(3 + \sqrt{3})(3 - \sqrt{3})$$)
$$= 3^2 - (\sqrt{3})^2$$
Because, We know that, ($$a + b)(a - b) = a^2 - b^2$$
i.e., $$= 9 - 3$$
Therefore, $$(3 + \sqrt{3})(3 - \sqrt{3}) = 6$$

iii)$$= (\sqrt{5} + \sqrt{2})^2$$
$$= (\sqrt{5})^2 + 2 × \sqrt{5}\sqrt{2} + (\sqrt{2})^2$$
Because, We know that, $$(a + b)^2 = a^2 + 2 × a × b + b^2$$
i.e., $$= 5 + 2\sqrt{10} + 2$$
Therefore, $$(\sqrt{5} + \sqrt{2})(\sqrt{5} + \sqrt{2}) = 7 + 2\sqrt{10}$$

iv)= $$(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})$$
$$= (\sqrt{5})^2 - (\sqrt{2})^2$$
Because, We know that, $$(a + b)(a - b) = a^2 - b^2$$
i.e., $$= 5 - 2$$
Therefore, $$(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 3$$

3. Recall, $$\pi$$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d).
That is, $$\pi$$ = c/d. This seems to contradict the fact that $$\pi$$ is irrational.
How will you resolve this contradiction?

First of all, there is not any contradiction. $$\pi$$ defined here has an approximate value of 22/7 i.e. by comparing with c/d.
Also, any value measured on scale or physically is an approx value.

4. Represent $$\sqrt{9.3}$$ on the number line.

Firstly we draw AB = $$\sqrt{9.3}$$ units. Now, from B, mark a distance of 1 unit.
Let this point be C. Let O be the mid-point of AC. Now, draw a semi-circle with centre O and radius OA.
Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
Therefore, the distance BD = $$\sqrt{9.3}$$ units.

Draw an arc with centre B and radius BD which intersects the number line at point E, then the point E represents $$\sqrt{9.3}$$ .

5. Rationalize the denominator of the following :
i)$$1/\sqrt{7}$$
ii)$$1/{\sqrt{7} - \sqrt{6}}$$
iii)$$1/{\sqrt{5} + \sqrt{2}}$$
iv)$$1/{\sqrt{7} - 2}$$

To rationalize the denominator, we need to multiply the irrational denominator by it's conjugate.
i) = $$1/\sqrt{7}$$ × $$\sqrt{7}/\sqrt{7}$$
= $$\sqrt{7}/7$$

ii) = $$1/(\sqrt{7} - \sqrt{6})$$ × $$(\sqrt{7} + \sqrt{6})/(\sqrt{7} + \sqrt{6})$$
= $$(\sqrt{7} + \sqrt{6})/(7-6)$$
= $${\sqrt{7} + \sqrt{6}}$$

iii) = $$1/(\sqrt{5} + \sqrt{2})$$ × $$(\sqrt{5} - \sqrt{2})/(\sqrt{5} - \sqrt{2})$$
$$= (\sqrt{5} - \sqrt{2})/(5 - 2)$$
= $$(\sqrt{5} - \sqrt{2})/3$$

iv) = $$1/(\sqrt{7} - 2)$$ × $$(\sqrt{7} + 2)/(\sqrt{7} + 2)$$
=$$(\sqrt{7} + 2)/(7 - 4)$$
=$${\sqrt{7} - 2}/3$$

#### Solution for Exersice 1.6

1.Find :
i)$$64^{1/2}$$
ii)$$32^{1/5}$$
iii)$$125^{1/3}$$

i) $$64^{1/2}$$ = $$(8 × 8)^{1/2}$$ = $$8^{1/2 × 2}$$ =$$8$$

ii) $${32}^{1/5}$$ = $$(2 × 2 × 2 × 2 × 2)^{1/5}$$ = $$2^{1/5 × 5}$$ =$$2$$

iii) $${125}^{1/3}$$ = $$(5 × 5 × 5)^{1/3}$$ = $$5^{3 × 1/3}$$ = $$5$$

2.Find :
i)$$9^{3/2}$$
ii)$$32^{2/5}$$
iii)$$16^{3/4}$$
iv)$$125^{-1/3}$$

i) $$9^{3/2}$$ = $$(3 × 3)^{3/2}$$ = $$3^{2 × 3/2}$$ = $$3^{3}$$ = $$27$$

ii) $$32^{2/5}$$ = $$(2 × 2 × 2 × 2 × 2)^{2/5}$$ = $$2^{2/5 × 5}$$ = $$2^2$$ = $$4$$

iii) $$16^{3/4}$$ = $$(2 × 2 × 2 × 2)^{3/4}$$ = $$2^{4 × 3/4}$$ = $$2^{3}$$ =$$8$$

iv) $$125^{-1/3}$$ = $$(5 × 5 × 5)^{-1/3}$$ = $$5^{3 × (-1/3)}$$ = $$5^{-1}$$ = $$1/5$$

3.Find :
i)$$2^{2/3} × 2^{1/5}$$
ii)$$(1/{3^{3}})^7$$
iii)$${11^{1/2}}/11^{1/4}$$
iv)$$7^{1/2} × 8^{1/2}$$

i)$$2^{2/3} × 2^{1/5}$$ = $$2^{2/3 + 1/5}$$ = $$2^{(10+3)/15}$$ = $$2^{13/15}$$
ii)$$(1/{3^{3}})^7$$ = $$1/{3^{3 × 7}}$$ = $$1/{3^{21}}$$ = $$3^{-21}$$
iii)$${11^{1/2}}/11^{1/4}$$ = $${11^{1/2}} × 11^{-1/4}$$ = $$11^{(2-1)/4}$$ = $$11^{1/4}$$
iv)$$7^{1/2} × 8^{1/2}$$ = $${7 × 8}^{1/2}$$ = $$56^{1/2}$$