{"id":545271,"date":"2021-10-04T08:35:38","date_gmt":"2021-10-04T08:35:38","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545271"},"modified":"2021-10-05T08:51:39","modified_gmt":"2021-10-05T08:51:39","slug":"rd-sharma-solutions-for-class-9-maths-chapter-1-number-system","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-1-number-system\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 1\u2013Number System"},"content":{"rendered":"\n

Class 9: Maths Chapter 1 solutions. Complete Class 9 Maths Chapter 1 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 1\u2013Number System<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 1, Class 9 Maths Chapter 1 solutions<\/p>\n\n\n\n

Exercise 1.1<\/h4>\n\n\n\n

Question 1: Is zero a rational number? Can you write it in the form p\/q, where p and q are integers and q \u2260 0?
Solution:<\/strong><\/p>\n\n\n\n

Yes, zero is a rational number.<\/p>\n\n\n\n

It can be written in p\/q form provided that q \u2260 0.<\/p>\n\n\n\n

For Example: 0\/1 or 0\/3 or 0\/4 etc.<\/p>\n\n\n\n

Question 2: Find five rational numbers between 1 and 2.
Solution:
<\/strong>
We know, one rational number between two numbers m and n = (m+n)\/2<\/p>\n\n\n\n

To find: 5 rational numbers between 1 and 2<\/p>\n\n\n\n

Step 1: Rational number between 1 and 2<\/p>\n\n\n\n

= (1+2)\/2<\/p>\n\n\n\n

= 3\/2<\/p>\n\n\n\n

Step 2: Rational number between 1 and 3\/2<\/p>\n\n\n\n

= (1+3\/2)\/2<\/p>\n\n\n\n

= 5\/4<\/p>\n\n\n\n

Step 3: Rational number between 1 and 5\/4<\/p>\n\n\n\n

= (1+5\/4)\/2<\/p>\n\n\n\n

= 9\/8<\/p>\n\n\n\n

Step 4: Rational number between 3\/2 and 2<\/p>\n\n\n\n

= 1\/2 [(3\/2) + 2)]<\/p>\n\n\n\n

= 7\/4<\/p>\n\n\n\n

Step 5: Rational number between 7\/4 and 2<\/p>\n\n\n\n

= 1\/2 [7\/4 + 2]<\/p>\n\n\n\n

= 15\/8<\/p>\n\n\n\n

Arrange all the results: 1 < 9\/8 < 5\/4 < 3\/2 < 7\/4 < 15\/8 < 2<\/p>\n\n\n\n

Therefore required integers are, 9\/8, 5\/4, 3\/2, 7\/4, 15\/8<\/p>\n\n\n\n

Question 3: Find six rational numbers between 3 and 4.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Steps to find n rational numbers between any two numbers:<\/p>\n\n\n\n

Step 1: Multiply and divide both the numbers by n+1.<\/p>\n\n\n\n

In this example, we have to find 6 rational numbers between 3 and 4. Here n = 6<\/p>\n\n\n\n

Multiply 3 and 4 by 7<\/p>\n\n\n\n

3 x 7\/7 = 21\/7 and<\/p>\n\n\n\n

4 x 7\/7 = 28\/7<\/p>\n\n\n\n

Step 2: Choose 6 numbers between 21\/7 and 28\/7<\/p>\n\n\n\n

3 = 21\/7 < 22\/7 < 23\/7 < 24\/7 < 25\/7 < 26\/7 < 27\/7 < 28\/7 = 4<\/p>\n\n\n\n

Therefore, 6 rational numbers between 3 and 4 are<\/p>\n\n\n\n

22\/7, 23\/7, 24\/7, 25\/7, 26\/7, 27\/7<\/p>\n\n\n\n

Question 4: Find five rational numbers between 3\/5 and 4\/5.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Steps to find n rational numbers between any two numbers:<\/p>\n\n\n\n

Step 1: Multiply and divide both the numbers by n+1.<\/p>\n\n\n\n

In this example, we have to find 5 rational numbers between 3\/5 and 4\/5. Here n = 5<\/p>\n\n\n\n

Multiply 3\/5 and 4\/5 by 6<\/p>\n\n\n\n

3\/5 x 6\/6 = 18\/30 and<\/p>\n\n\n\n

4\/5 x 6\/6 = 24\/30<\/p>\n\n\n\n

Step 2: Choose 5 numbers between 18\/30 and 24\/30<\/p>\n\n\n\n

3\/5 = 18\/30 < 19\/30 < 20\/30 < 21\/30 < 22\/30 < 23\/30 < 24\/30 = 4\/5<\/p>\n\n\n\n

Therefore, 5 rational numbers between 3\/5 and 4\/5 are<\/p>\n\n\n\n

19\/30, 20\/30, 21\/30, 22\/30, 23\/30<\/p>\n\n\n\n

Question 5: Are the following statements true or false? Give reason for your answer.<\/strong><\/p>\n\n\n\n

(i) Every whole number is a natural number.<\/strong><\/p>\n\n\n\n

(ii) Every integer is a rational number.<\/strong><\/p>\n\n\n\n

(iii) Every rational number is an integer.<\/strong><\/p>\n\n\n\n

(iv) Every natural number is a whole number,<\/strong><\/p>\n\n\n\n

(v) Every integer is a whole number.<\/strong><\/p>\n\n\n\n

(vi) Every rational number is a whole number.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong> False.<\/p>\n\n\n\n

Reason: As 0 is not a natural number.<\/p>\n\n\n\n

(ii)<\/strong> True.<\/p>\n\n\n\n

(iii)<\/strong> False.<\/p>\n\n\n\n

Reason: Numbers such as 1\/2, 3\/2, 5\/3 are rational numbers but not integers.<\/p>\n\n\n\n

(iv)<\/strong> True.<\/p>\n\n\n\n

(v)<\/strong> False.<\/p>\n\n\n\n

Reason: Negative numbers are not whole numbers.<\/p>\n\n\n\n

(vi)<\/strong> False.<\/p>\n\n\n\n

Reason: Proper fractions are not whole numbers<\/p>\n\n\n\n

Exercise 1.2<\/h4>\n\n\n\n

Question 1: Express the following rational numbers as decimals.
(i) 42\/100 (ii) 327\/500 (iii) 15\/4<\/strong><\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Question 2: Express the following rational numbers as decimals.
(i) 2\/3 (ii) -4\/9 (iii) -2\/15 (iv) -22\/13 (v) 437\/999 (vi) 33\/26
Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong> Divide 2\/3 using long division:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii)<\/strong> Divide using long division: -4\/9<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Divide using long division: -2\/15<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv)<\/strong> Divide using long division: -22\/13<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v)<\/strong> Divide using long division: 437\/999<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi)<\/strong> Divide using long division: 33\/26<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: Look at several examples of rational numbers in the form p\/q (q \u2260 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations. Can you guess what property q must satisfy?<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

The decimal representation will be terminating, if the denominators have factors 2 or 5 or both. Therefore, p\/q is a terminating decimal, when prime factorization of q must have only powers of 2 or 5 or both.<\/p>\n\n\n\n

Exercise 1.3<\/h4>\n\n\n\n

Question 1: Express each of the following decimals in the form p\/q:<\/strong><\/p>\n\n\n\n

(i) 0.39<\/strong><\/p>\n\n\n\n

(ii) 0.750<\/strong><\/p>\n\n\n\n

(iii) 2.15<\/strong><\/p>\n\n\n\n

(iv) 7.010<\/strong><\/p>\n\n\n\n

(v) 9.90<\/strong><\/p>\n\n\n\n

(vi) 1.0001<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

0.39 = 39\/100<\/p>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

0.750 = 750\/1000 = 3\/4<\/p>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

2.15 = 215\/100 = 43\/20<\/p>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

7.010 = 7010\/1000 = 701\/100<\/p>\n\n\n\n

(v)<\/strong><\/p>\n\n\n\n

9.90 = 990\/100 = 99\/10<\/p>\n\n\n\n

(vi)<\/strong><\/p>\n\n\n\n

1.0001 = 10001\/10000<\/p>\n\n\n\n

Question 2: Express each of the following decimals in the form p\/q:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong> Let x = 0.4\u0305<\/p>\n\n\n\n

or x = 0.4\u0305 = 0.444 \u2026. (1)<\/p>\n\n\n\n

Multiplying both sides by 10<\/p>\n\n\n\n

10x = 4.444 \u2026..(2)<\/p>\n\n\n\n

Subtract (1) by (2), we get<\/p>\n\n\n\n

10x \u2013 x = 4.444\u2026 \u2013 0.444\u2026<\/p>\n\n\n\n

9x = 4<\/p>\n\n\n\n

x = 4\/9<\/p>\n\n\n\n

=> 0.4\u0305 = 4.9<\/p>\n\n\n\n

(ii)<\/strong> Let x = 0.3737.. \u2026. (1)<\/p>\n\n\n\n

Multiplying both sides by 100<\/p>\n\n\n\n

100x = 37.37\u2026 \u2026..(2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

100x \u2013 x = 37.37\u2026 \u2013 0.3737\u2026<\/p>\n\n\n\n

100x \u2013 x = 37<\/p>\n\n\n\n

99x = 37<\/p>\n\n\n\n

x = 37\/99<\/p>\n\n\n\n

(iii)<\/strong> Let x = 0.5454\u2026 (1)<\/p>\n\n\n\n

Multiplying both sides by 100<\/p>\n\n\n\n

100x = 54.5454\u2026. (2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

100x \u2013 x = 54.5454\u2026. \u2013 0.5454\u2026.<\/p>\n\n\n\n

99x = 54<\/p>\n\n\n\n

x = 54\/99<\/p>\n\n\n\n

(iv)<\/strong> Let x = 0.621621\u2026 (1)<\/p>\n\n\n\n

Multiplying both sides by 1000<\/p>\n\n\n\n

1000x = 621.621621\u2026. (2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

1000x \u2013 x = 621.621621\u2026. \u2013 0.621621\u2026.<\/p>\n\n\n\n

999x = 621<\/p>\n\n\n\n

x = 621\/999<\/p>\n\n\n\n

or x = 23\/37<\/p>\n\n\n\n

(v)<\/strong> Let x = 125.3333\u2026. (1)<\/p>\n\n\n\n

Multiplying both sides by 10<\/p>\n\n\n\n

10x = 1253.3333\u2026. (2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

10x \u2013 x = 1253.3333\u2026. \u2013 125.3333\u2026.<\/p>\n\n\n\n

9x = 1128<\/p>\n\n\n\n

or x = 1128\/9<\/p>\n\n\n\n

or x = 376\/3<\/p>\n\n\n\n

(vi)<\/strong> Let x = 4.7777\u2026. (1)<\/p>\n\n\n\n

Multiplying both sides by 10<\/p>\n\n\n\n

10x = 47.7777\u2026. (2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

10x \u2013 x = 47.7777\u2026. \u2013 4.7777\u2026.<\/p>\n\n\n\n

9x = 43<\/p>\n\n\n\n

x = 43\/9<\/p>\n\n\n\n

(vii)<\/strong> Let x = 0.47777\u2026.<\/p>\n\n\n\n

Multiplying both sides by 10<\/p>\n\n\n\n

10x = 4.7777\u2026. \u2026(1)<\/p>\n\n\n\n

Multiplying both sides by 100<\/p>\n\n\n\n

100x = 47.7777\u2026. (2)<\/p>\n\n\n\n

Subtract (1) from (2), we get<\/p>\n\n\n\n

100x \u2013 10x = 47.7777\u2026. \u2013 4.7777\u2026<\/p>\n\n\n\n

90x = 43<\/p>\n\n\n\n

x = 43\/90<\/p>\n\n\n\n

Exercise 1.4<\/h4>\n\n\n\n

Question 1: Define an irrational number.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

A number which cannot be expressed in the form of p\/q, where p and q are integers and q \u2260 0. It is non-terminating or non-repeating decimal.<\/p>\n\n\n\n

Question 2: Explain, how irrational numbers differ from rational numbers?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers.<\/p>\n\n\n\n

It cannot be expressed as terminating or repeating decimal.<\/p>\n\n\n\n

For example, \u221a2 is an irrational number<\/p>\n\n\n\n

A rational number is a real number which can be written as a fraction and as a decimal i.e. it can be expressed as a ratio of integers.<\/p>\n\n\n\n

It can be expressed as terminating or repeating decimal.<\/p>\n\n\n\n

For examples: 0.10 and 5\/3 are rational numbers<\/p>\n\n\n\n

Question 3: Examine, whether the following numbers are rational or irrational:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> \u221a7<\/p>\n\n\n\n

Not a perfect square root, so it is an irrational number.<\/p>\n\n\n\n

(ii)<\/strong> \u221a4<\/p>\n\n\n\n

A perfect square root of 2.<\/p>\n\n\n\n

We can express 2 in the form of 2\/1, so it is a rational number.<\/p>\n\n\n\n

(iii)<\/strong> 2 + \u221a3<\/p>\n\n\n\n

Here, 2 is a rational number but \u221a3 is an irrational number<\/p>\n\n\n\n

Therefore, the sum of a rational and irrational number is an irrational number.<\/p>\n\n\n\n

(iv)<\/strong> \u221a3 + \u221a2<\/p>\n\n\n\n

\u221a3 is not a perfect square thus an irrational number.<\/p>\n\n\n\n

\u221a2 is not a perfect square, thus an irrational number.<\/p>\n\n\n\n

Therefore, sum of \u221a2 and \u221a3 gives an irrational number.<\/p>\n\n\n\n

(v)<\/strong> \u221a3 + \u221a5<\/p>\n\n\n\n

\u221a3 is not a perfect square and hence, it is an irrational number<\/p>\n\n\n\n

Similarly, \u221a5 is not a perfect square and also an irrational number.<\/p>\n\n\n\n

Since, sum of two irrational number, is an irrational number, therefore \u221a3 + \u221a5 is an irrational number.<\/p>\n\n\n\n

(vi)<\/strong> (\u221a2 \u2013 2)2<\/sup><\/p>\n\n\n\n

(\u221a2 \u2013 2)2<\/sup> = 2 + 4 \u2013 4 \u221a2<\/p>\n\n\n\n

= 6 \u2013 4 \u221a2<\/p>\n\n\n\n

Here, 6 is a rational number but 4\u221a2 is an irrational number.<\/p>\n\n\n\n

Since, the sum of a rational and an irrational number is an irrational number, therefore, (\u221a2 \u2013 2)2 is an irrational number.<\/p>\n\n\n\n

(vii)<\/strong> (2 \u2013 \u221a2)(2 + \u221a2)<\/p>\n\n\n\n

We can write the given expression as;<\/p>\n\n\n\n

(2 \u2013 \u221a2)(2 + \u221a2) = ((2)2<\/sup> \u2212 (\u221a2)2<\/sup>)[Since, (a + b)(a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= 4 \u2013 2 = 2 or 2\/1<\/p>\n\n\n\n

Since, 2 is a rational number, therefore, (2 \u2013 \u221a2)(2 + \u221a2) is a rational number.<\/p>\n\n\n\n

(viii)<\/strong> (\u221a3 + \u221a2)2<\/sup><\/p>\n\n\n\n

We can write the given expression as;<\/p>\n\n\n\n

(\u221a3 + \u221a2)2<\/sup> = (\u221a3)2<\/sup> + (\u221a2)2<\/sup> + 2\u221a3 x \u221a2<\/p>\n\n\n\n

= 3 + 2 + 2\u221a6<\/p>\n\n\n\n

= 5 + 2\u221a6[using identity, (a+b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

Since, the sum of a rational number and an irrational number is an irrational number, therefore, (\u221a3 + \u221a2)2<\/sup> is an irrational number.<\/p>\n\n\n\n

(ix)<\/strong> \u221a5 \u2013 2<\/p>\n\n\n\n

\u221a5 is an irrational number whereas 2 is a rational number.<\/p>\n\n\n\n

The difference of an irrational number and a rational number is an irrational number.<\/p>\n\n\n\n

Therefore, \u221a5 \u2013 2 is an irrational number.<\/p>\n\n\n\n

(x)<\/strong> \u221a23<\/p>\n\n\n\n

Since, \u221a23 = 4.795831352331\u2026<\/p>\n\n\n\n

As decimal expansion of this number is non-terminating and non-recurring therefore, it is an irrational number.<\/p>\n\n\n\n

(xi)<\/strong> \u221a225<\/p>\n\n\n\n

\u221a225 = 15 or 15\/1<\/p>\n\n\n\n

\u221a225 is rational number as it can be represented in the form of p\/q and q not equal to zero.<\/p>\n\n\n\n

(xii)<\/strong> 0.3796<\/p>\n\n\n\n

As the decimal expansion of the given number is terminating, therefore, it is a rational number.<\/p>\n\n\n\n

(xiii)<\/strong> 7.478478\u2026\u2026<\/p>\n\n\n\n

As the decimal expansion of this number is non-terminating recurring decimal, therefore, it is a rational number.<\/p>\n\n\n\n

(xiv)<\/strong> 1.101001000100001\u2026\u2026<\/p>\n\n\n\n

As the decimal expansion of given number is non-terminating and non-recurring, therefore, it is an irrational number<\/p>\n\n\n\n

Question 4: Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong> \u221a4<\/p>\n\n\n\n

\u221a4 = 2, which can be written in the form of a\/b. Therefore, it is a rational number.<\/p>\n\n\n\n

Its decimal representation is 2.0.<\/p>\n\n\n\n

(ii)<\/strong> 3\u221a18<\/p>\n\n\n\n

3\u221a18 = 9\u221a2<\/p>\n\n\n\n

Since, the product of a rational and an irrational number is an irrational number.<\/p>\n\n\n\n

Therefore, 3\u221a18 is an irrational.<\/p>\n\n\n\n

Or 3 \u00d7 \u221a18 is an irrational number.<\/p>\n\n\n\n

(iii)<\/strong> \u221a1.44<\/p>\n\n\n\n

\u221a1.44 = 1.2<\/p>\n\n\n\n

Since, every terminating decimal is a rational number, Therefore, \u221a1.44 is a rational number.<\/p>\n\n\n\n

And, its decimal representation is 1.2.<\/p>\n\n\n\n

(iv)<\/strong> \u221a9\/27<\/p>\n\n\n\n

\u221a9\/27 = 1\/\u221a3<\/p>\n\n\n\n

Since, we know, quotient of a rational and an irrational number is irrational numbers, therefore, \u221a9\/27 is an irrational number.<\/p>\n\n\n\n

(v)<\/strong> \u2013 \u221a64<\/p>\n\n\n\n

\u2013 \u221a64 = \u2013 8 or \u2013 8\/1<\/p>\n\n\n\n

Therefore, \u2013 \u221a64 is a rational number.<\/p>\n\n\n\n

Its decimal representation is \u20138.0.<\/p>\n\n\n\n

(vi)<\/strong> \u221a100<\/p>\n\n\n\n

\u221a100 = 10<\/p>\n\n\n\n

Since, 10 can be expressed in the form of a\/b, such as 10\/1,<\/p>\n\n\n\n

Therefore, \u221a100 is a rational number.<\/p>\n\n\n\n

And it\u2019s decimal representation is 10.0.<\/p>\n\n\n\n

Question 5: In the following equation, find which variables x, y, z etc. represent rational or irrational numbers:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i)<\/strong> x2<\/sup> = 5<\/p>\n\n\n\n

Taking square root both the sides,<\/p>\n\n\n\n

x = \u221a5<\/p>\n\n\n\n

\u221a5 is not a perfect square root, so it is an irrational number.<\/p>\n\n\n\n

(ii)<\/strong> y2<\/sup> = 9<\/p>\n\n\n\n

y2<\/sup> = 9<\/p>\n\n\n\n

or y = 3<\/p>\n\n\n\n

3 can be expressed in the form of a\/b, such as 3\/1, so it a rational number.<\/p>\n\n\n\n

(iii)<\/strong> z2<\/sup> = 0.04<\/p>\n\n\n\n

z2<\/sup> = 0.04<\/p>\n\n\n\n

Taking square root both the sides, we get<\/p>\n\n\n\n

z = 0.2<\/p>\n\n\n\n

0.2 can be expressed in the form of a\/b such as 2\/10, so it is a rational number.<\/p>\n\n\n\n

(iv)<\/strong> u2<\/sup> = 17\/4<\/p>\n\n\n\n

Taking square root both the sides, we get<\/p>\n\n\n\n

u = \u221a17\/2<\/p>\n\n\n\n

Since, quotient of an irrational and a rational number is irrational, therefore, u is an Irrational number.<\/p>\n\n\n\n

(v)<\/strong> v2<\/sup> = 3<\/p>\n\n\n\n

Taking square root both the sides, we get<\/p>\n\n\n\n

v = \u221a3<\/p>\n\n\n\n

Since, \u221a3 is not a perfect square root, so v is irrational number.<\/p>\n\n\n\n

(vi)<\/strong> w2<\/sup> = 27<\/p>\n\n\n\n

Taking square root both the sides, we get<\/p>\n\n\n\n

w = 3\u221a3<\/p>\n\n\n\n

Since, the product of a rational and irrational is an irrational number. Therefore, w is an irrational number.<\/p>\n\n\n\n

(vii)<\/strong> t2<\/sup> = 0.4<\/p>\n\n\n\n

Taking square root both the sides, we get<\/p>\n\n\n\n

t = \u221a(4\/10)<\/p>\n\n\n\n

t = 2\/\u221a10<\/p>\n\n\n\n

Since, quotient of a rational and an irrational number is irrational number. Therefore, t is an irrational number.<\/p>\n\n\n\n

Exercise 1.5<\/h4>\n\n\n\n

Question 1: Complete the following sentences:<\/strong><\/p>\n\n\n\n

(i) Every point on the number line corresponds to a \u2026\u2026 number which many be either \u2026\u2026 or \u2026\u2026.<\/strong><\/p>\n\n\n\n

(ii) The decimal form of an irrational number is neither \u2026.. nor \u2026\u2026<\/strong><\/p>\n\n\n\n

(iii) The decimal representation of a rational number is either ..\u2026 or \u2026..<\/strong><\/p>\n\n\n\n

(iv) Every real number is either \u2026 number or \u2026 number.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> Every point on the number line corresponds to a real<\/strong> number which many be either rational<\/strong> or irrational<\/strong>.<\/p>\n\n\n\n

(ii)<\/strong> The decimal form of an irrational number is neither terminating<\/strong> nor repeating<\/strong>.<\/p>\n\n\n\n

(iii)<\/strong> The decimal representation of a rational number is either terminating<\/strong> or non-terminating recurring<\/strong>.<\/p>\n\n\n\n

(iv)<\/strong> Every real number is either rational<\/strong> number or an irrational<\/strong> number.<\/p>\n\n\n\n

Question 2: Represent \u221a6, \u221a7, \u221a8 on the number line.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Find the equivalent values of \u221a6, \u221a7, \u221a8<\/p>\n\n\n\n

\u221a6 = 2.449<\/p>\n\n\n\n

\u221a7 = 2.645<\/p>\n\n\n\n

\u221a8 = 2.828<\/p>\n\n\n\n

We can see that, all the given numbers lie between 2 and 3.<\/p>\n\n\n\n

Draw on number line:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 3: Represent \u221a3.5, \u221a9.4, \u221a10.5 and on the real number line.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Represent \u221a3.5 on number line<\/strong><\/p>\n\n\n\n

Step 1: Draw a line segment AB = 3.5 units<\/p>\n\n\n\n

Step 2: Produce B till point C, such that BC = 1 unit<\/p>\n\n\n\n

Step 3: Find the mid-point of AC, say O.<\/p>\n\n\n\n

Step 4: Taking O as the centre draw a semi circle, passing through A and C.<\/p>\n\n\n\n

Step 5: Draw a line passing through B perpendicular to OB, and cut semicircle at D.<\/p>\n\n\n\n

Step 6: Consider B as a centre and BD as radius draw an arc cutting OC produced at E.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, from right triangle OBD,<\/p>\n\n\n\n

BD2<\/sup> = OD2<\/sup> \u2013 OB2<\/sup><\/p>\n\n\n\n

= OC2<\/sup> \u2013 (OC \u2013 BC)2<\/sup><\/p>\n\n\n\n

(As, OD = OC)<\/p>\n\n\n\n

BD2<\/sup> = 2OC x BC \u2013 (BC)2<\/sup><\/p>\n\n\n\n

= 2 x 2.25 x 1 \u2013 1<\/p>\n\n\n\n

= 3.5<\/p>\n\n\n\n

=> BD = \u221a3.5<\/p>\n\n\n\n

Represent \u221a9.4 on number line<\/strong><\/p>\n\n\n\n

Step 1: Draw a line segment AB = 9.4 units<\/p>\n\n\n\n

Follow step 2 to Step 6 mentioned above.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

BD2<\/sup> = 2OC x BC \u2013 (BC)2<\/sup><\/p>\n\n\n\n

= 2 x 5.2 x 1 \u2013 1<\/p>\n\n\n\n

= 9.4<\/p>\n\n\n\n

=> BD = \u221a9.4<\/p>\n\n\n\n

Represent \u221a10.5 on number line<\/strong><\/p>\n\n\n\n

Step 1: Draw a line segment AB = 10.5 units<\/p>\n\n\n\n

Follow step 2 to Step 6 mentioned above, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

BD2<\/sup> = 2OC x BC \u2013 (BC)2<\/sup><\/p>\n\n\n\n

= 2 x 5.75 x 1 \u2013 1<\/p>\n\n\n\n

= 10.5<\/p>\n\n\n\n

=> BD = \u221a10.5<\/p>\n\n\n\n

Question 4: Find whether the following statements are true or false:<\/strong><\/p>\n\n\n\n

(i) Every real number is either rational or irrational.<\/strong><\/p>\n\n\n\n

(ii) \u03c0 is an irrational number.<\/strong><\/p>\n\n\n\n

(iii) Irrational numbers cannot be represented by points on the number line.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

(i) True.<\/p>\n\n\n\n

(ii) True.<\/p>\n\n\n\n

(ii) False.<\/p>\n\n\n\n

Exercise 1.6<\/h4>\n\n\n\n

Question 1: Visualise 2.665 on the number line, using successive magnification.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

2.665 is lies between 2 and 3 on the number line.<\/p>\n\n\n\n

Divide selected segment into 10 equal parts and mark each point of division as 2.1, 2.2, \u2026.,2.9, 2.10<\/p>\n\n\n\n

2.665 is lies between 2.6 and 2.7<\/p>\n\n\n\n

Divide line segment between 2.6 and 2.7 in 10 equal parts such as 2.661, 2.662, and so on.<\/p>\n\n\n\n

Here we can see that 5th point will represent 2.665.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 2: Visualise the representation of 5.37\u0305 on the number line upto 5 decimal places, that is upto 5.37777.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Clearly 5.37\u0305 is located between 5 and 6.<\/p>\n\n\n\n

Again by successive magnification, and successively decrease 5.37\u0305 located between 5.3 and 5.4.<\/p>\n\n\n\n

For more clarity, divide 5.3 and 5.4 portion of the number line into 10 equal parts and we can see 5.37\u0305 lies between 5.37 and 5.38.<\/p>\n\n\n\n

To visualize 5.37\u0305 more accurately, divide line segment between 5.37 and 5.38 in ten equal parts.<\/p>\n\n\n\n

5.37\u0305 lies between 5.377 and 5.378.<\/p>\n\n\n\n

Again divide above portion between 5.377 and 5.378 into 10 equal parts, which shows 5.37\u0305 is located closer to 5.3778 than to 5.3777<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 1: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 1\u2013Number System<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 1\u2013Number System PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n