####
**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*≠ 0?**Answer**

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

**Answer**

There are infinite rational numbers in between 3 and 4.

3 and 4 can be represented as 24/8 and 32/8 respectively.

Therefore, six rational numbers between 3 and 4 are

25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

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

**Answer**

There are infinite rational numbers in between 3/5 and 4/5

3/5 = 3×6/5×6 = 18/30

4/5 = 4×6/5×6 = 24/30

Therefore, five rational numbers between 3/5 and 4/5 are

19/30, 20/30, 21/30, 22/30, 23/30.

**4. State whether the following statements are true or false. Give reasons for your answers.**

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

► True, since the collection of whole numbers contains all natural numbers.

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

► False, as integers may be negative but whole numbers are always positive.

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

__Page No: 8__

####
**Exercise 1.2**

**1. State whether the following statements are true or false. Justify your answers.**

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

► True, since the collection of real numbers is made up of rational and irrational numbers.

► False, since positive number cannot be expressed as square roots.

► False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

**(ii) Every point on the number line is of the form√***m*, where m is a natural number.► False, since positive number cannot be expressed as square roots.

**(iii) Every real number is an irrational number.**► False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

**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.****Answer**

No, the square roots of all positive integers are not irrational. For example √4 = 2.

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

**Answer**

Step 1: Let AB be a line of length 2 unit on number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.

Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB

^{2}+ BC

^{2}= CA

^{2}

⇒ 2

^{2}+ 1

^{2}= CA

^{2}

⇒ CA

^{2}= 5

⇒ CA = √5

Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a centre draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose centre was A.

Thus, √5 is represented on the number line as shown in the figure.

__Page No: 14__

**Exercise 1.3**

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

(i) 36/100

(i) 36/100

= 0.36 (Terminating)

**(ii) 1/11**

0.09090909... = 0.9 (Non terminating repeating)

**(iii)**

= 33/8 = 4.125 (Terminating)

**(iv) 3/13**

= 0.230769230769... = 0.230769 (Non terminating repeating)

**(v) 2/11**

= 0.181818181818... = 0.18 (Non terminating repeating)

**(vi) 329/400**

= 0.8225 (Terminating)

**2. You know that 1/7 = 0.142857.Can you predict what the decimal expansion 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 of 1/7 carefully.]

[Hint: Study the remainders while finding the value of 1/7 carefully.]

**Answer**

Yes. We can be done this by:

**3. Express the following in the form**

*p*/*q*where*p*and*q*are integers and*q*≠ 0.**(i) 0.6**

**(ii) 0.47**

**(iii) 0.001**

**Answer**

(i) 0.6 = 0.666...

Let

10

10

9

(ii) 0.47 = 0.4777...

= 4/10 + 0.777/10

Let

10

10

Let

*x*= 0.666...10

*x*= 6.666...10

*x*= 6 +*x*9

*x*= 6*x*= 2/3(ii) 0.47 = 0.4777...

= 4/10 + 0.777/10

Let

*x*= 0.777…10

*x*= 7.777…10

*x*= 7 +*x**x*= 7/9

4/10 + 0.777.../10 = 4/10 + 7/90

= 36+7/90 = 43/90

(iii) 0.001 = 0.001001...

Let

1000*x*= 0.001001...*x*= 1.001001…

1000

*x*= 1 +

*x*

999

*x*= 1

*x*= 1/999

**4. Express 0.99999…in the form**

Answer

*p*/*q*. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.Answer

Let

*x*= 0.9999…

10

*x*= 9.9999…

10

*x*= 9 +

*x*

9

*x*= 9

*x*= 1

The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999 is too much near 1, Therefore, the 1 as answer can be justified.

**5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.**

**Answer**

1/17 = 0.0588235294117647

There are 16 digits in the repeating block of the decimal expansion of 1/17.

Division Check:

= 0.0588235294117647

**6. Look at several examples of rational numbers in the form**

*p*/*q*(*q*≠ 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?**Answer**

We observe that when q is 2, 4, 5, 8, 10... then the decimal expansion is terminating. For example:

1/2 = 0.5, denominator

*q*= 2

^{1}

7/8 = 0.875, denominator

*q*= 2

^{3}

4/5 = 0.8, denominator

*q*= 5

^{1}

We can observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

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

**Answer**

Three numbers whose decimal expansions are non-terminating non-recurring are:

0.303003000300003...

0.505005000500005...

0.7207200720007200007200000…**8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.**

**Answer**

5/7 = 0.714285

9/11 = 0.81

9/11 = 0.81

Three different irrational numbers are:

0.73073007300073000073…
0.75075007300075000075…

0.76076007600076000076…

**9. Classify the following numbers as rational or irrational:**

**(i) √23**

**(ii) √225**

**(iii) 0.3796**

(iv) 7.478478

(iv) 7.478478

**(v) 1.101001000100001…**

**Answer**

(i) √23 = 4.79583152331...

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii) √225 = 15 = 15/1

Since the number is rational number as it can represented in

Since the number is rational number as it can represented in

*p*/*q*form.
(iii) 0.3796

Since the number is terminating therefore, it is an rational number.

(iv) 7.478478 = 7.478

Since the this number is non-terminating recurring, therefore, it is a rational number.
(v) 1.101001000100001…

Since the number is non-terminating non-repeating, therefore, it is an irrational number.__Page No: 18__

####
**Exercises 1.4**

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

**Answer**

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

**Answer**

4.26 = 4.2626

__Page No: 24__

####
**Exercise 1.5**

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

**(i) 2 - √5**

**(ii) (3 + √23) - √23**

**(iii) 2√7/7√7**

**(iv) 1/√2**

**(v) 2π**

**Answer**

(i) 2 - √5 = 2 - 2.2360679… = - 0.2360679…

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3 = 3/1

Since the number is rational number as it can represented in

(iii) 2√7/7√7 = 2/7

Since the number is rational number as it can represented in

(iv) 1/√2 = √2/2 = 0.7071067811...

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3 = 3/1

Since the number is rational number as it can represented in

*p*/*q*form.(iii) 2√7/7√7 = 2/7

Since the number is rational number as it can represented in

*p*/*q*form.(iv) 1/√2 = √2/2 = 0.7071067811...

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…

Since the number is is non-terminating non-recurring therefore, it is an irrational number.

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

**(i) (3 + √3) (2 + √2)**

**(ii) (3 + √3) (3 - √3)**

**(iii) (√5 + √2)**

(iv) (√5 - √2) (√5 + √2)

^{2}(iv) (√5 - √2) (√5 + √2)

**Answer**

(i) (3 + √3) (2 + √2)

⇒ 3 × 2 + 2 + √3 + 3√2+ √3 ×√2
⇒ 6 + 2√3 +3√2 + √6

(ii) (3 + √3) (3 - √3) [∵ (

*a*+*b*) (*a*-*b*) =*a*^{2}-*b*^{2}]
⇒ 3

⇒ 9 - 3

⇒ 6^{2}- (√3)^{2}⇒ 9 - 3

(iii) (√5 + √2)

⇒ (√5)

⇒ 5 + 2 + 2 × √5× 2

^{2 }[∵ (*a*+*b*)^{2}=*a*^{2}+*b*^{2}+ 2*ab*]⇒ (√5)

^{2}+ (√2)^{2}+ 2 ×√5 × √2⇒ 5 + 2 + 2 × √5× 2

⇒ 7 +2√10

(iv) (√5 - √2) (√5 + √2) [∵ (

(iv) (√5 - √2) (√5 + √2) [∵ (

*a*+*b*) (*a*-*b*) =*a*^{2}-*b*^{2}]
⇒ (√5)

^{2}- (√2)^{2}
⇒ 5 - 2

⇒ 3

**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?**

**Answer**

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

**Answer**

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.

Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.

Step 3: Draw a semi circle with radius OC and centre O.

Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.

Step 5: Now, OBD is a right angled triangle.

Here, OD = 10.3/2 (radius of semi circle), OC = 10.3/2, BC = 1

OB = OC – BC = (10.3/2) – 1 = 8.3/2

Using Pythagoras theorem,

OD

^{2}= BD

^{2}+ OB

^{2}

⇒ (10.3/2)

^{2}= BD2 + (8.3/2)

^{2}

⇒ BD

^{2}= (10.3/2)

^{2}- (8.3/2)

^{2}

⇒ BD

^{2}= (10.3/2 – 8.3/2) (10.3/2 + 8.3/2)

⇒ BD

^{2}= 9.3

⇒ BD

^{2}= √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

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

**(i) 1/√7**

**(ii) 1/√7-√6**

**(iii) 1/√5+√2**

**(iv) 1/√7-2**

**Answer**

__Page No: 26__

####
**Exercise 1.6**

**1. Find:**

(i) 64

(ii) 32

(iii) 125

(i) 64

^{1/2}(ii) 32

^{1/5}(iii) 125

^{1/3}

**Answer**

**2. Find:**

**(i) 9**

^{3/2}**(ii) 32**

^{2/5}**(iii) 16**

^{3/4}**(iv) 125**

^{-1/3}**Answer**

**3. Simplify:**

**(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}**Answer**

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