Chapter 2 Polynomials RS Aggarwal Solution for Class 9th Maths

Shivam Varshney

January 17, 2019

2410

Real Numbers

Question 1:

Exercise 1A

The numbers of the form , where p and q are integers and q ≠ 0 are known as rational numbers.

Ten examples of rational numbers are:

, , , , , , , , 1,

Question 2:

5
-3

(v) 1.3

(vi) -2.4

(vii)

Question 3:

A rational number lying between and is

Therefore, we have < < < < Or we can say that, < < < < That is, < < < <

Therefore, three rational numbers between and are

, and

Question 5:

Let and

Then, x < y because <

Or we can say that, That is, < .

We know that, 8 < 9 < 10 < 11 < 12 < 13 < 14 < 15.

Therefore, we have, < < < < < < < Thus, 5 rational numbers between, < are:

, , , and

Question 6:

Let x = 3 and y = 4

Then, x < y, because 3 < 4 We can say that, < .

We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28.

Therefore, we have, < < < < < < < Therefore, 6 rational numbers between 3 and 4 are:

, , , and

Question 7:

Let x = 2.1 and y = 2.2

Then, x < y because 2.1 < 2.2 Or we can say that, < Or,

That is, we have, <

We know that, 2100 < 2105 < 2110 < 2115 < 2120 < 2125 < 2130 < 2135 < 2140 <

2145 < 2150 < 2155 < 2160 < 2165 < 2170 < 2175 < 2180 < 2185 < 2190 < 2195 <

2200

Therefore, we can have,

Therefore, 16 rational numbers between, 2.1 and 2.2 are:

So, 16 rational numbers between 2.1 and 2.2 are:

2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17,

2.175, 2.18

Exercise 1B

Question 1:

(i)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since, 80 has prime factors 2 and 5, is a terminating decimal.

(ii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 12 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iv)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 35 has prime factors 5 and 7, and 7 is different from 2 and 5, is not a terminating decimal.

(v)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since 125 has prime factor 5 only is a terminating decimal.

Question 2:

(i)

= 0.625

(ii)

= 0.5625

(iii)

= 0.28

(iv)

= 0.458

(v)

= 2.41

Question 3:

Let x =

i.e x = 0.333 …. (i)

⇒ 10x = 3.333 …. (ii) Subtracting (i) from (ii), we get 9x = 3

⇒ x = = Hence, 0. =

Let x = 1.

i.e x = 1.333 …. (i)

⇒10x = 13.333 …. (ii)

Subtracting (i) from (ii) we get; 9x = 12

⇒ x = =

Hence, 1. =

Let x = 0.

i.e x = 0.3434 …. (i)

⇒ 100x = 34.3434 …. (ii)

Subtracting (i) from (ii), we get 99x = 34

⇒ x =

Hence, 0. =

Let x = 3.

i.e x = 3.1414 …. (i)

⇒ 100x = 314.1414 …. (ii)

Subtracting (i) from (ii), we get 99x = 311

⇒ x =

Hence, 3. =

Let x = 0.

i.e. x = 0.324324 ….(i)

⇒ 1000x = 324.324324….(ii)

Subtracting (i) from (ii), we get 999x = 324

⇒ x = =

Hence, 0. =

Let x = 0.

i.e. x = 0.177 …. (i)

⇒ 10x = 1.777 …. (ii)

and 100x = 17.777…. (iii) Subtracting (ii) from (iii), we get 90x = 16

⇒ x = =

Hence, 0. =

Let x = 0.

i.e. x = 0.544 …. (i)

⇒ 10 x = 5.44 …. (ii) and 100x = 54.44 ….(iii)

Subtracting (ii) from (iii), we get 90x = 49

⇒ x =

Hence, 0. =

(vii) Let x = Let x = 0.1 i.e. x = 0.16363 …. (i)

⇒ 10x = 1.6363 …. (ii)

and 1000 x = 163.6363 …. (iii) Subtracting (ii) from (iii), we get

990x = 162

⇒ x = = Hence, 0.1 =

Question 4:

True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers
False. Since 0 is whole number but it is not a natural number.
True. Every integer can be represented in a fraction form with denominator 1.
False. Since division of whole numbers is not closed under division, the value of , p and q are integers and q ≠ 0, may not be a whole number.
True. The prime factors of the denominator of the fraction form of terminating

decimal contains 2 and/or 5, which are integers and are not equal to zero.

True. The prime factors of the denominator of the fraction form of repeating decimal contains integers, which are not equal to zero.
True. 0 can considered as a fraction , which is a rational number.

Question 1:

Exercise 1C

Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot

be expressed in the fraction form, , p and q are integers and q ≠ 0

For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.

Also, etc are examples of irrational numbers.

Question 2:

(i)

We know that, if n is a perfect square, then is a rational number. Here, 4 is a perfect square and hence, = 2 is a rational number. So, is a rational number.

(ii)

We know that, if n is a perfect square, then is a rational number. Here, 196 is a perfect square and hence is a rational number. So, is rational.

(iii)

We know that, if n is a not a perfect square, then is an irrational number. Here, 21 is a not a perfect square number and hence, is an irrational number. So, is irrational.

(iv)

We know that, if n is a not a perfect square, then is an irrational number. Here, 43 is not a perfect square number and hence, is an irrational number. So, is irrational.

(v)

, is the sum of a rational number 3 and irrational number .

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, , is an irrational number.

(vi)

= + (-2) is the sum of a rational number and an irrational number.

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, + (-2) , is an irrational number.

So, is irrational.

(vii)

= × is the product of a rational number and an irrational number .

Theorem: The product of a non-zero rational number and an irrational number is an irrational number.

Thus, by the above theorem, × is an irrational number.

So, is an irrational number.

Every rational number can be expressed either in the terminating form or in the non- terminating, recurring decimal form.

Therefore, 0. = 0.6666

Question 3:

Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.

Take OA = 1 unit and draw BA ⊥ OA such that AB = 1 unit, join OB. Then,

With O as centre and OB as radius, drawn an arc, meeting OX at P. Then, OP = OB = units

Thus the point P represents on the real line.

Now draw BC ⊥ OB such that BC = 1 units Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. The, OQ = OC = units

Thus, the point Q represents on the real line. Now draw CD ⊥ OC such that CD = 1 units

Join OD. Then,

Now draw DE ⊥ OD such that DE = 1 units Join OE. Then,

With O as centre and OE as radius draw an arc, meeting OX at R. Then, OR = OE = units

Thus, the point R represents on the real line.

Question 4:

Draw horizontal line X’OX taken as the x-axis Take O as the origin to represent 0.

Let OA = 2 units and let AB ⊥ OA such that AB = 1 units

Join OB. Then,

With O as centre and OB as radius draw an arc meeting OX at P. Then, OP = OB =

Now draw BC ⊥ OB and set off BC = 1 unit Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. Then, OQ = OC =

Thus, Q represents on the real line. Now, draw CD ⊥ OC as set off CD = 1 units Join OD. Then,

With O as centre and OD as radius, draw an arc, meeting OX at R. Then OR = OD =

Thus, R represents on the real line.

Question 5:

(i)

Since 4 is a rational number and is an irrational number.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(ii)

Since – 3 is a rational number and is irrational.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(iii)

Since 5 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(iv)

Since -3 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(v)

is irrational because it is the product of a rational number and the irrational number .

(vi)

is an irrational number because it is the product of rational number and irrational number .

Question 6:

True
False
True
False
True
False
False
True
True

Exercise 1D

Question 1:

(i)

We have:

(ii)

We have:

Question 2:

Question 3:

(i) by

(ii) by

(iii) by

Question 4:

Question 5:

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 units. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle. Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With B as centre and BD as radius, draw an arc meeting AC produced at E. Then, BE = BD = units.

Question 6:

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With D as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD = units.

Question 7:

Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for al real numbers a, b, c.

Commutative Law: a + b = b + a, for all real numbers a and b.

Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (-a) such that

a + (-a) = (-a) + a= 0

a and (-a) are called the additive inverse of each other. Existence of inverse of multiplication:

For each non zero real number a, there exists a real number such that