Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 2 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Maths Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

Since, $5={2}^{0}×{5}^{1}$

$⇒$ The denominator is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal form of $\frac{13}{5}$ will be terminating type.

Since, $11={2}^{0}×{5}^{0}×{11}^{1}$

$⇒$ The denominator is not in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal form of $\frac{2}{11}$ will be non-terminating recurring type.

Since, $16={2}^{4}×{5}^{0}$

$⇒$ The denominator is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal form of $\frac{29}{16}$ will be terminating type.

Since, $125={2}^{0}×{5}^{3}$

$⇒$ The denominator is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal form of $\frac{17}{125}$ will be terminating type.

Since, $6={2}^{1}×{5}^{0}×{3}^{1}$

$⇒$ The denominator is not in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal form of $\frac{11}{6}$ will be non-terminating recurring type.

#### Question 2:

Write the following rational numbers in decimal form.

#### Question 3:

Write the following rational numbers in $\frac{p}{q}$ form

#### Question 1:

Show that 4$\sqrt{2}$ is an irrational number.

Let us assume that 4$\sqrt{2}$  is a rational number.

$⇒$4$\sqrt{2}$ = $\frac{p}{q}$, where p and q are the integers and q $\ne$ 0.

$⇒$$\sqrt{2}$ = $\frac{p}{4q}$

Since, pq and 4 are integers. So, $\frac{p}{4q}$ is a rational number.

$⇒$$\sqrt{2}$ is also a rational number.

but this contradicts the fact that $\sqrt{2}$ is an irrational number.

This contradiction has arisen due to the wrong assumption that 4$\sqrt{2}$ is a rational number.

Hence, 4$\sqrt{2}$ is an irrational number.

#### Question 2:

Prove that 3 + $\sqrt{5}$ is an irrational number.

Let us assume that 3 + $\sqrt{5}$  is a rational number.

$⇒$3 + $\sqrt{5}$ = $\frac{p}{q}$, where p and q are the integers and q $\ne$ 0.

$⇒$$\sqrt{5}$ = $\frac{p}{q}-3=\frac{p-3q}{q}$

Since, pq and 3 are integers. So, $\frac{p-3q}{q}$ is a rational number.

$⇒$$\sqrt{5}$ is also a rational number.

but this contradicts the fact that $\sqrt{5}$ is an irrational number.

This contradiction has arisen due to the wrong assumption that 3 + $\sqrt{5}$ is a rational number.

Hence, 3 + $\sqrt{5}$​ is an irrational number.

#### Question 3:

Represent the numbers $\sqrt{5}$ and  $\sqrt{10}$ on a number line .

(i) Steps of construction for $\sqrt{5}$:

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB $\perp$ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point P.

Thus, P is the point for $\sqrt{5}$ on the number line. (ii) Steps of construction for $\sqrt{10}$:

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB $\perp$ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point C.

Thus, C is the point for $\sqrt{10}$​ on the number line. #### Question 4:

Write any three rational numbers between the two numbers given below.

(i) 0.3 and -0.5
(ii) -2.3 and -2.33
(iii) 5.2 and 5.3
(iv) -4.5 and -4.6

(i) The three rational numbers between 0.3 and $-$0.5 are $-$0.4, 0 and, 0.1

(ii) The three rational numbers between $-$2.3 and $-$2.33 are $-$2.31, $-$2.32, and $-$2.325

(iii) The three rational numbers between 5.2 and 5.3 are 5.21, 5.24, and 5.28

(iv) The three rational numbers between $-$4.5 and $-$4.6 are $-$4.51, $-$4.55, and $-$4.59

#### Question 1:

State the order of the surds given below.

The order of the surd is 3.

The order of the surd is 2.

The order of the surd is 4.

The order of the surd is 2.

The order of the surd is 3.

#### Question 2:

State which of the following are surds. Justify.

(i)  (ii) $\sqrt{16}$ (iii) $\sqrt{81}$ (iv) $\sqrt{256}$ (v) $\sqrt{64}$ (vi) $\sqrt{\frac{22}{7}}$

(i) Since,

So,  is a surd.

(ii) Since, $\sqrt{16}=\sqrt{{2}^{4}}=2$

So, $\sqrt{16}$ is not a surd.

(iii) Since, $\sqrt{81}=\sqrt{{3}^{4}}={\left({3}^{4}\right)}^{\frac{1}{5}}={3}^{\frac{4}{5}}$

So, $\sqrt{81}$ is a surd.

(iv) Since, $\sqrt{256}=\sqrt{{16}^{2}}=16$

So, $\sqrt{256}$ is not a surd.

(v) Since, $\sqrt{64}=\sqrt{{4}^{3}}=4$

So, $\sqrt{64}$ is not a surd.

(vi) Since, $\sqrt{\frac{22}{7}}={\left(\frac{22}{7}\right)}^{\frac{1}{2}}$

So, $\sqrt{\frac{22}{7}}$ is a surd.

#### Question 3:

Classify the given pair of surds into like surds and unlike surds.

(i)   (ii)  (iii)   (iv)   (v) $5\sqrt{22},7\sqrt{33}$  (vi)

(i)

Since, $\sqrt{52}=\sqrt{4×13}=2\sqrt{13}$

So,  is like surds.

(ii)

Since, $\sqrt{68}=\sqrt{4×17}=2\sqrt{17}$

So,  is unlike surds.

(iii)

Since, $4\sqrt{18}=4\sqrt{9×2}=4×3\sqrt{2}=12\sqrt{2}$

So,  is like surds.

(iv)

Since, $19\sqrt{12}=19\sqrt{4×3}=19×2\sqrt{3}=38\sqrt{3}$

So,  is like surds.

(v) $5\sqrt{22},7\sqrt{33}$

Since,

So, $5\sqrt{22},7\sqrt{33}$ is unlike surds.

(vi)

Since, $\sqrt{75}=\sqrt{25×3}=5\sqrt{3}$

So,  is like surds.

#### Question 4:

Simplify the following surds.

(i) $\sqrt{27}$  (ii) $\sqrt{50}$  (iii) $\sqrt{250}$  (iv) $\sqrt{112}$  (v) $\sqrt{168}$

(i) $\sqrt{27}=\sqrt{9×3}=3\sqrt{3}$

(ii) $\sqrt{50}=\sqrt{25×2}=5\sqrt{2}$

(iii) $\sqrt{250}=\sqrt{25×10}=5\sqrt{10}$

(iv) $\sqrt{112}=\sqrt{16×7}=4\sqrt{7}$

(v) $\sqrt{168}=\sqrt{4×42}=2\sqrt{42}$

#### Question 5:

Compare the following pair of surds.

(i)   (ii) $\sqrt{247},\sqrt{274}$  (iii) $2\sqrt{7},\sqrt{28}$   (iv)   (v)   (vi)   (vii)

(i)

Since,

So,

(ii) $\sqrt{247},\sqrt{274}$

(iii) $2\sqrt{7},\sqrt{28}$

Since, $2\sqrt{7}=\sqrt{4×7}=\sqrt{28}$

So,

(iv)

Since,

So,

(v)

Since,

So,

(vi)

Since,

So,

(vii)

Since,

So,

#### Question 6:

Simplify.

(i)    (ii)   (iii)   (iv)

(i) $5\sqrt{3}+8\sqrt{3}=13\sqrt{3}$

(ii)

(iii) $7\sqrt{48}-\sqrt{27}-\sqrt{3}=7\sqrt{16×3}-\sqrt{9×3}-\sqrt{3}=7×4\sqrt{3}-3\sqrt{3}-\sqrt{3}=28\sqrt{3}-3\sqrt{3}-\sqrt{3}=24\sqrt{3}$

(iv)

#### Question 7:

Multiply and write the answer in the simplest form.

(i)      (ii)

(iii)        (iv)

(i)

(ii)

(iii)

(iv)

#### Question 8:

Divide, and write the answer in simplest form.

(i)    (ii)   (iii)   (iv)

(i)

(ii)

(iii)

(iv)

#### Question 9:

Rationalize the denominator.

(i) $\frac{3}{\sqrt{5}}$   (ii) $\frac{1}{\sqrt{14}}$    (iii)  $\frac{5}{\sqrt{7}}$  (iv)  $\frac{6}{9\sqrt{3}}$  (v)  $\frac{11}{\sqrt{3}}$

(i) $\frac{3}{\sqrt{5}}$

$=\frac{3}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3\sqrt{5}}{{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{5}}{5}$

(ii) $\frac{1}{\sqrt{14}}$

$=\frac{1}{\sqrt{14}}×\frac{\sqrt{14}}{\sqrt{14}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{14}}{{\left(\sqrt{14}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{14}}{14}$

(iii) $\frac{5}{\sqrt{7}}$

$=\frac{5}{\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{5\sqrt{7}}{{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{5\sqrt{7}}{7}$

(iv)  $\frac{6}{9\sqrt{3}}$

$=\frac{2}{3\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}}{3×3}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}}{9}$

(v) $\frac{11}{\sqrt{3}}$

$=\frac{11}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{11\sqrt{3}}{{\left(\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{11\sqrt{3}}{3}$

#### Question 1:

Multiply

(i)    (ii)     (iii)

(i)

(ii)

(iii)

#### Question 2:

Rationalize the denominator.

(i)       (ii)   (iii)    (iv)

(i)

(ii)

(iii)

(iv)

#### Question 1:

Find the value.

(i)   (ii)   (iii)

(i)

(ii)

(iii)

#### Question 2:

Solve.

(i)   (ii)    (iii)    (iv)

(i) $\left|3x-5\right|=1$

(ii)

(iii)

(iv)

#### Question 1:

Choose the correct alternative answer for the questions given below.

(i) Which one of the following is an irrational number ?
(A)   (B) $\sqrt{5}$  (C)  $\frac{3}{9}$   (D) $\sqrt{196}$

(ii) Which of the following is an irrational number?

(A) 0.17 (B)  $1.\overline{)513}$   (C)  $0.27\overline{)46}$    (D) 0.101001000.....

(iii) Decimal expansion of which of the following is non-terminating recurring ?

(A) $\frac{2}{5}$    (B)  $\frac{3}{16}$  (C)  $\frac{3}{11}$ (D) $\frac{137}{25}$

iv) Every point on the number line represent, which of the following numbers?
(A) Natural numbers (B) Irrational numbers   (C) Rational numbers (D) Real numbers.

(v) The number $0.\stackrel{°}{4}$ in  $\frac{p}{q}$form is .....
(A) $\frac{4}{9}$ (B) $\frac{40}{9}$  (C)$\frac{3.6}{9}$ (D) $\frac{36}{9}$

(vi) What is  $\sqrt{n}$ , if n is not a perfect square number ?
(A) Natural number  (B) Rational number
(C) Irrational number  (D) Options A, B, C all are correct.

(vii) Which of the following is not a surd ?
(A) $\sqrt{7}$    (B) $\sqrt{17}$   (C)  $\sqrt{64}$  (D)  $\sqrt{193}$

(viii) What is the order of the surd $\sqrt{\sqrt{5}}$ ?
(A) 3  (B) 2  (C) 6  (D) 5

(ix) Which one is the conjugate pair of   ?
(A) $-$    (B)   (C)     (D)

(x) The value of   is...............
(A) $-$68   (B) 68  (C) $-$32   (D) 32

(i) Since,

= $\frac{4}{5}$ is a rational number; $\frac{3}{9}$ is a rational number; $\sqrt{196}$ = 14 is a rational number; and $\sqrt{5}$ is an irrational number.

Hence, the correct option is (B).

(ii) Since,

0.17 has a terminating decimal expansion, so, it is rational number;

$1.\overline{)513}$ has non-terminating recurring decimal expansion, so, it is rational number;

$0.27\overline{)46}$ has non-terminating recurring decimal expansion, so, it is rational number;

0.101001000..... has non-terminating non-recurring decimal expansion, so, it is irrational number;

Hence, the correct option is (D).

(iii)

(A) Since, 5 = ${2}^{0}×{5}^{1}$, which is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal expansion of $\frac{2}{5}$ is terminating.

(B) Since, 16 = ${2}^{4}×{5}^{0}$, which is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal expansion of $\frac{3}{16}$ is terminating.

(C) Since, 11 = ${2}^{0}×{5}^{0}×{11}^{1}$, which is not in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal expansion of $\frac{3}{11}$​ is non-terminating recurring.

(D) Since, 25 = ${2}^{0}×{5}^{2}$, which is in the form of ${2}^{m}×{5}^{n}$, where m and n are non-negative integers.

So, the decimal expansion of $\frac{137}{25}$ is terminating.

Hence, the correct option is (C).

(iv) Since, every point on the number line represents a real number.

Hence, the correct option is (D).

(v)

Hence, the correct option is (A).

(vi) If n is not a perfect square number, then $\sqrt{n}$ is an irrational number.

Hence, the correct option is (C).

(vii) Since, $\sqrt{64}$ = 4

Hence, the correct option is (C).

(viii) Since, $\sqrt{\sqrt{5}}={\left(\sqrt{5}\right)}^{\frac{1}{3}}={\left[{\left(5\right)}^{\frac{1}{2}}\right]}^{\frac{1}{3}}={5}^{\frac{1}{6}}=\sqrt{5}$

So, the order of the surd $\sqrt{\sqrt{5}}$ is 6.

Hence, the correct option is (C).

(ix) Since, the conjugate pair of () is ($-$).

​Hence, the correct option is (A).

(x) Since, $\left|12-\left(13+7\right)×4\right|=\left|12-20×4\right|=\left|12-80\right|=\left|-68\right|=68$

So, the value of  is  68 .

​Hence, the correct option is (B).

#### Question 2:

Write the following numbers in $\frac{p}{q}$ form
(i ) 0.555 (ii)  $29.\overline{)568}$     (iii) 9.315 315 ...       (iv) 357.417417...
(v) $30.\overline{)219}$

(i ) 0.555

$=\frac{555}{1000}\phantom{\rule{0ex}{0ex}}=\frac{111}{200}$

(ii) $29.\overline{)568}$

(iii) 9.315 315 ...

(iv) 357.417417...

(v) $30.\overline{)219}$

#### Question 3:

Write the following numbers in its decimal form. .

(i)  $\frac{-5}{7}$  (ii)  $\frac{9}{11}$ (iii)  $\sqrt{5}$   (iv) $\frac{121}{13}$  (v)  $\frac{29}{8}$

(i) $\frac{-5}{7}=-0.\overline{)714285}$

(ii) $\frac{9}{11}=0.\overline{)81}$

(iii) $\sqrt{5}=2.23606797...$

(iv) $\frac{121}{13}=9.\overline{)307692}$

(v) $\frac{29}{8}=3.625$

#### Question 4:

Show that 5 + $\sqrt{7}$ is an irrational number.

Let us assume that 5 + $\sqrt{7}$ is a rational number.

Since, .

$⇒$ $\sqrt{7}$ is also a rational number.

But this contradicts the fact that $\sqrt{7}$ is an irrational number.

This contradiction has arisen due to our assumption that 5 + $\sqrt{7}$ is a rational number.

Hence, 5 + $\sqrt{7}$ is an irrational number.

#### Question 5:

Write the following surds in simplest form.

(i) $\frac{3}{4}\sqrt{8}$    (ii)  $-\frac{5}{9}\sqrt{45}$

(i) $\frac{3}{4}\sqrt{8}=\frac{3}{4}×\sqrt{4×2}=\frac{3}{4}×2\sqrt{2}=\frac{3}{2}\sqrt{2}$

(ii) $-\frac{5}{9}\sqrt{45}=-\frac{5}{9}\sqrt{9×5}=-\frac{5}{9}×3\sqrt{5}=-\frac{5}{3}\sqrt{5}$

#### Question 6:

Write the simplest form of rationalising factor for the given surds.

(i)

#### Question 7:

Simplify.

(i)

(ii)  $5\sqrt{3}+2\sqrt{27}+\frac{1}{\sqrt{3}}$

(iii) $\sqrt{216}-5\sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}$

(iv)$4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

(v)  $2\sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}$

(i)

(ii)  $5\sqrt{3}+2\sqrt{27}+\frac{1}{\sqrt{3}}$

$=5\sqrt{3}+2\sqrt{9×3}+\frac{1}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=5\sqrt{3}+6\sqrt{3}+\frac{\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=11\sqrt{3}+\frac{\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{33\sqrt{3}+\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{34\sqrt{3}}{3}$

(iii) $\sqrt{216}-5\sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}$

$=\sqrt{36×6}-5\sqrt{6}+\sqrt{49×6}-\frac{3}{\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}\phantom{\rule{0ex}{0ex}}=6\sqrt{6}-5\sqrt{6}+7\sqrt{6}-\frac{3\sqrt{6}}{6}\phantom{\rule{0ex}{0ex}}=8\sqrt{6}-\frac{\sqrt{6}}{2}\phantom{\rule{0ex}{0ex}}=\frac{18\sqrt{6}-\sqrt{6}}{2}\phantom{\rule{0ex}{0ex}}=\frac{17\sqrt{6}}{2}$

(iv) $4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

$=4\sqrt{4×3}-\sqrt{25×3}-7\sqrt{16×3}\phantom{\rule{0ex}{0ex}}=8\sqrt{3}-5\sqrt{3}-28\sqrt{3}\phantom{\rule{0ex}{0ex}}=-25\sqrt{3}$

(v) $2\sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}$

$=2\sqrt{16×3}-\sqrt{25×3}-\frac{1}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=8\sqrt{3}-5\sqrt{3}-\frac{\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=3\sqrt{3}-\frac{\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{9\sqrt{3}-\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=\frac{8\sqrt{3}}{3}$

#### Question 8:

Rationalize the denominator.

(i) $\frac{1}{\sqrt{5}}$   (ii)  $\frac{2}{3\sqrt{7}}$   (iii)  $\frac{1}{\sqrt{3}-\sqrt{2}}$   (iv)  $\frac{1}{3\sqrt{5}+2\sqrt{2}}$   (v)   $\frac{12}{4\sqrt{3}-\sqrt{2}}$

(i) $\frac{1}{\sqrt{5}}$

$=\frac{1}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{5}}{5}$

(ii) $\frac{2}{3\sqrt{7}}$

$=\frac{2}{3\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{7}}{3×7}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{7}}{21}$

(iii) $\frac{1}{\sqrt{3}-\sqrt{2}}$

(iv) $\frac{1}{3\sqrt{5}+2\sqrt{2}}$

(v) $\frac{12}{4\sqrt{3}-\sqrt{2}}$

View NCERT Solutions for all chapters of Class 9