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**NCERT Class 9 Mathematics Chapter 1 Number systems**

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**Solutions****NCERT Class 9 Mathematics Chapter 1 Number systems****Number systems**

**Number systems****Chapter – 1**

Exercise 1.1 |

**Q.1. Is zero a rational number? Can you write in the form p/q, where p and q are integers and q ≠ 0**

Ans: Yes, zero is a rational number and it can be written in the form of

p/q as 0 = 0/1 here p = 0 and q = 1

Notice that q can be any number you wish.

[∵ 0 = 0/2 = 0/4 etc.]

**Q.2. Find six rational numbers between 3 and 4**

Ans: Let a = 3, and b = 4

Rational number lying between 3 and 4 is

a+b/2 i.e. 3+4/2 = 7/2

Now rational number between 3 and

Then required six rational numbers between 3 and 4 are.

7/2, 13/4, 15/4, 27/8, 29/8, 31/8

ALITER

We want to find all six at one go. Since we want 6 numbers, so write.

3 = 21/6+1 = 21/7 and 4 = 28/6+1 = 28/7

Now six rational numbers between 3 and 4 are

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

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

Ans: To find five rational numbers between 3/5 and 4/5, we take 3/5 and 4/5

as a rational number with denominator 30. i.e. 3/5 = 3/5 × 6/6 = 18/30 and 4/5 = 4/5 × 6/6 = 24/30

Then, required six rational numbers between 3/5 and 4/5 are

19/30, 20/30, 21/30 and 23/30.

**Q.4. Are the following statements true or false? Give reasons for your answer.**

**(i) Every natural number is a whole number.**

Ans: Yes, because each natural number is a whole number.

**(ii) Every integer is a whole number.**

Ans: False, because all negative natural numbers (-1, -2, -3, …) are not whole numbers.

**(iii) Every rational number is a whole number.**

Ans: False, because the rational numbers of the type 2/5, 2/5, 3/5 etc are not whole numbers.

Exercise 1.2 |

**Q.1. State whether the following statements are true or false. Jus-tify your answer.**

**(i) Every irrational number is a real number.**

Ans: True, because the set of every rational and every irrational number is called a real number.

**(ii) Every point on the number line is of the form √m, where m is a natural number.**

Ans: False, no negative number can be the square root of any natural number.

**(iii) Every real number is an irrational number.**

Ans: False, because the real number is the set of every irrational and rational number. For example, 2 is real but not irrational.

**Q.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.**

Ans: No, the square roots of all positive integers are not always irrational.

For Example √4= 2 is a rational number.

**Q.3. Show how √5 can be represented on the number line.**

Ans: They represent √5 on the number line, taking 0 at zero on the number line. Again taking OA = 1 unit in the positive direction of the number line.

Then OB = √1² + 1² = √2

Construct OD of unit length perpendicular to OB.

Then OD= √(√2)² +1² = √3

Construct DE of unit length perpendicular to OD.

Then OE = √(√3)² + 1² = √4=2

Construct EF of unit length perpendicular to OE. Then

OF = √2²+1² = √5

Using a compass, with centre) and radius OF, draw an arc which inter- sects the number line in the point R. Then R corresponds to √5.

**Q.4. Classroom activity (Constructing the ‘Square root spiral’).**

Take a large sheet of paper and construct the ‘square spiral’ in the following fashion. Start with a point O and draw a line segment P ¹ P ² per- perpendicular to OP, of unit length. Now draw a line segment P ² P ³ perpendicular- lar to OP ². Then draw a line segment P ³ P ⁴ perpendicular to OP ³. Continuing in this manner, you can get the line segment Pn Pn by drawing a line segment of unit length perpendicular to Open. In this man-ner, you will have created the points P², P³,…, Pn, …, and join-ing them to a create beautiful spiral

depicting.

√2,√3, √4,…

Ans: For self-practice.

Exercise 1.3 |

**Q.1. Write the following in decimal form and say what kind of deci- mal expansion each has:**

(**i) 36/100**

Ans: 36/100 = 0.36 terminating decimal

**(ii) 1/11**

Ans: 1/11

Remainders: 1, 1, 1, 1………, Divisor: 11

∴ 1 11 =0.090909…. = 0.099

The decimal expansion is non-terminating.

**(iii) 4 ⅛ = 33/8**

Ans:

The decimal representation is terminating.

**(iv) 3/13**

Ans:

∴ 3/13 = 0.230769.

The decimal expansion is non-terminating repeating.

**(v) 2/11**

Ans:

Remainders: 9,2,9,2,…………

Divisor: 0..1818…………

We write

**(vi) 329/400**

Ans: 329/400

**Q.2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are without actually doing the long divi- sion? If so, how?**

Ans:

**Q.3. Express the following in the form p/q, where p and q are inte-gers and q≠0.**

Ans: (i) Let x = 0.6 or x=0.66666…(1)

Now, adding 6 both side in equation (1)

x + 6 = 0.66666….+ 6

or x + 6=6.6666… (2)

Again, multiplying by 10 both sides in equation (1)

10 x = 0.66666…x 10

or 10 x = 6.6666… (3)

Now, form equations (2) and (3)

x+6=10 x or 9 x = 6 ∴ x=6/9

Therefore, 0.6 = x = 2/3

**(ii) Let x = 0.47**

or x=0.47777 …(1)

Now, multiply 10th side in equation (1)

10 x = 0.47777…x10

or 10 x = 4.77777… (2)

Again, multiply 100 both side in equation (1)

100 = 0.477777 … X 100

or 100 x = 47.777 …(3)

Subtract equation (2) from equation (3)

90 x = 43.0000

or 90 x = 43 ∴ x = 43/90 Therefore 0.47 = x 43/90

Which is in the form of p/q

**(iii) Let x = 0.001**

or, x = 0.001001001… (1)

Now, add 1 both side in equation (1)

x + 1= 0.001001001…+1

or x + 1=1.001001001… (2)

Again, multiply 1000 both side in equation (1)

1000 × x =1000 x 0.001001001… (3)

or 1000x = 1.001001001…

From equations (2) and (3) we get

x + 1=1000x or 999x=1 ∴ x = 1/999

Therefore, 0.001 = x = 1/999′ Which is the form of p/q

**Q.4. Express 0.99999… in the form p/q. Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.**

Ans: Let x = 0.9999… (1)

Add 9 both side in equation (1)

x + 9 = 9 + 0.99999…

or x + 9 = 9.99999… (2)

Again, multiply 10 both side in equation (1)

10 x = 9.9999… (3)

From equations (2) and (3) we get 10 x = x+9

or 9 x = 9 or x = 1 Therefore, 0.9999 … = 1

It is because there is an infinite 9 coming after the point; which is very- very close to 1.

**Q.5. Look at several examples of rational numbers in the form p/q (q ≠ 0), where 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?**

Ans: **Some examples of rational number having terminating decimal representations are:**

(i) 1/2 = 0.5

(ii) 7/4 =1.75

(iii) 7/4 = 0.875

(iv) 2/5 = 0.4

It is clear that the prime factorisation of q has only power of 2 or power of 5 or both.

**Q.6. Write three numbers whose decimal expansions are non termi- nating non-recurring.**

Ans: We know that the decimal expansions of an irrational number is

non-terminating non-recurring. Three examples of such numbers are as

√2=1.4142135…

√3=1.722050807…

π = 3.1415926535…

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

Ans: To find an irrational number between 5/7 and 9/11 is non terminat- ing non recurring lying between them.

Therefore, the required three different irrational number which are ly-ing between 5/11 and 9/11 are:

0.720720072000 ….

0.730730073000… and 0.740740074000…

**Q.8. Classify the following number as rational or irrational:**

**(i) √23 **

Ans: (i) We have √23=4.795831523…

It is non terminating non recurring.

So, √23 is an irrational number.

**(ii) √225 **

Ans: (ii) We have √225 = 15 or √225= 15/1 which is a rational number.

**(iii) 0.3796 **

Ans:

So, 0.3796 is a rational number.

**(iv) 7.478478…**

Ans: (iv) We have, 7.478478… = 7.478 which is non terminating recurring.

Therefore, 7.478478…. is an irrational number.

**(v) 101001000100001**

Ans: We have 1.101001000100001…. which is non terminating non recurring.

Therefore, 1.101001000100001…. is an irrational number.

Exercise 1.4 |

**Q.1. Visualise 3.765 on the number line, using successive magnification.**

Ans:

**Q.2 Visualise 4.26 on the number line, up to 4 decimal places.**

Ans:

Exercise 1.5 |

**Q.1. Classify the following numbers as rational irrational:**

**(i) 2 – √5**

Ans: We have 2 -√5

оr 2-2.236067977…

or -0.236067977….

which is non terminating non recurring. So, it is an irrational number.

**(ii) (3 + √23)-√23**

Ans: We have 3+√23-√23

or 3 + √23-√23 or 3

which is a rational number.

**(iii) (2√7/7√7**

Ans: We have, 2√7/7√7 or

which is a rational number.

**(iv) 1/√2**

Ans:

**(ν) 2 π**

Ans: We have, 2 π = 2×3.1415926535…..

= 6.2831853070….

which is non terminating non recurring.

Therefore, it is an irrational number.

**Q.2. Simplify each of the following expressions:**

**(i) (3+√3) (2+√2)**

Ans: We have, (3+ √3) (2+√2)

or 6+3√2+2√3+√6

**(ii) (3+√3) (3-√3)**

Ans: We have, (3+√3) (3-√3)

we know that (a+b)(a-b)=a² -b²

∴ (3+ √3)(3-√3)=(3)² -(√3)²=9-3=6

So, (3-√3) (3-√3)=6

**(iii) (√5+√2)²**

Ans: We have, (√5+ √2)²

we know that (a+b)² = a²+b²+2 ab

∴ (5+ √2)² = (√5)² + (√2)² +2√5.√2=5+2+2√10

or (√5+ √2)² = 7+2√10

**(iv) (√5-√2) (√5+ √2)**

Ans: We have, (√5-√2) (√5+√2)=(√5)² – (√2)²

[∵ (a+b)(a-b)=a²-b²=5-2]

∴ (√5-√2) (√5+√2)=3

**(v) (3√5-4√3)²**

Ans: (3√5-4√3)²

= (3√5)² -2.3√5.4√3+ (4√3)²

= 9 x 5-24 √15 +1 6 x 3

= 45 – 24 √15+48

= 93 – 24 √15

**(vi) (√7-6) (√3-√7)**

Ans: (√7-6) (√3-√7)

= √(√3-√7)-6(√3-√7)

= √21-√49-6√3+6√7

= √21-6√3-7+6√7

**(vii) (2+ √6) (4+ √6)**

Ans: (2+√6) (4+√6)

= 2(4+ √6) + √6 (4+ √6)

= 8+2√6+4√6+ √36

= 8+6+ 6√6 = 14 + 6√6

**Q.3. Recall, π is defined as the ratio of the circumference (say c) of **

**a circle to its diameter (say d). That is π=c/d. This seems to contradict **

**the fact that π is irrational/ How will you resolve this contradiction?**

Ans: Circumference = 2 π r (Irrational number) of the circle (c)

r → Radius of the circle

Diameter of circle (d) = 2 r (Rational number)

с/d = 2 π r/2 r Irrational number/Rational number = Irrational number

As we know, an irrational number divided by a rational number results in an Irrational number.

**Q.4. Represent √9.3 on the number line.**

Ans: To represent √9.3 on the number line. We mark a point B on the number line so that AB = 9.3 units. Again mark a point

C so that BC= 1 unit. Now take the mid point of AC and mark that point as 0. Draw a semicircle with centre 0 and radius OC. Draw the line perpendicular to

AC passing through B and in-

intersecting the semicircle at D.

Then, BD=√9.3.

Again taking BD as a radius and draw an arc which intersecting the number line atE. Then E represents √9.3 on the number line.

Take B at zero on the number line, then point E represents √9.3.

**Q.5. Rationalise the denominators of the following:**

(i)

Ans: We have

Multiply √7 both numerator and denominator

(ii)

Ans: We have,

Multiply √7+√6 both numerator and denominator

(iii)

Ans: We have,

Multiply

Both numerator and denominator

(iv)

Ans: We have,

(v)

Ans:

(vi)

Ans:

(vii)

Ans:

(viii)

Ans:

(ix)

Ans:

**Q.6. (i) **

Ans:

**(ii)**

Ans:

**(iii) **

Ans: