# Chapter – 1: Number Systems

# 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?**

Solution:

We know that, a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as, 0/1, 0/2, 0/3.

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

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

Solution:

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

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

Solution:

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

To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5

with 5+1=6 (or any number greater than 5)

i.e., (3/5) × (6/6) = 18/30

and, (4/5) × (6/6) = 24/30

The numbers in between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5.

Hence,19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/5

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

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

Solution:

**True**

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers= 1,2,3,4…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number.

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

Solution:

**False**

Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers.

i.e., integers= {…-4, -3, -2, -1,0,1,2,3,4…}

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

Every whole number is an integer; however, every integer is not a whole number.

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

Solution:

**False**

Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.

i.e., Rational numbers = 0, 19/30, 2, 9/-3, -12/7…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

All whole numbers are rational; however, all rational numbers are not whole numbers.

## Exercise 1.2

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

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

Solution:

**True**

Irrational Numbers – A number is said to be irrational, if it **cannot** be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Every irrational number is a real number, however, every real numbers are not irrational numbers.

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

Solution:

**False**

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 =3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

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

Solution:

**False**

The statement is false, the real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Irrational Numbers – A number is said to be irrational, if it **cannot** be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Every irrational number is a real number, however, every real number is not irrational.

**2. Are the square roots of all positive integer’s irrational? If not, give an example of the square root of a number that is a rational number.**

Solution:

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

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. (2 and 3, respectively).

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

Solution:

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

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

Step 3: Join CA

Step 4: Now, ABC is a right-angled triangle. Applying Pythagoras theorem,

AB^{2}+BC^{2} = CA^{2}

2^{2}+1^{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 lines. 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.

**4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP**_{1}** of unit length (see Fig. 1.9). Now draw a line segment P**_{2}**P**_{3}** perpendicular to OP**_{2}**. Then draw a line segment P**_{3}**P**_{4}** perpendicular to OP**_{3}**. Continuing in Fig. 1.9 :**

**Constructing this manner, you can get the line segment P**_{n-1}**Pn by square root spiral drawing a line segment of unit length perpendicular to OP**_{n-1}**. In this manner, you will have created the points P**_{2}**, P**_{3}**,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …**

Solution:

Step 1: Mark a point O on the paper. Here, O will be the centre of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….

## Exercise 1.3

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

(i) 36/100

Solution:

= 0.36 (Terminating)

(ii)1/11

Solution:

Solution:

(iv) 3/13

Solution:

(v) 2/11

Solution:

(vi) 329/400

Solution:

= 0.8225 (Terminating)

**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 division? If so, how?**

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

Solution:

**3. Express the following in the form p/q, where p and q are integers and q 0.**

(i)** **

Solution:

Assume that *x* = 0.666…

Then,10*x* = 6.666…

10*x* = 6 + *x*

9*x* = 6

*x* = 2/3

**(ii) ****0.4\overline {7}****0.47**

Solution:

0.4\overline {7} = 0. 4777..0.47=0. 4777..

= (4/10) + (0.777/10)

Assume that *x* = 0.777…

Then, 10*x* = 7.777…

10*x* = 7 + *x*

*x* = 7/9

(4/10) + (0. 777.. /10) = (4/10) +(7/90) (x = 7/9 and x = 0.777…0.777…/10 = 7/ (9×10) = 7/90)

= (36/90) +(7/90) = 43/90

Solution:

Assume that *x* = 0.001001…

Then, 1000*x* = 1.001001…

1000*x* = 1 + *x*

999*x* = 1

*x* = 1/999

**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.**

Solution:

Assume that *x* = 0.9999…. Eq (a)

Multiplying both sides by 10,

10*x* = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

10*x* = 9.9999

–*x* = -0.9999…

_____________

9*x* = 9

*x* = 1

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the 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.**

Solution:

1/17

Dividing 1 by 17:

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

**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?**

Solution:

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 observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

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

Solution:

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

- √3 = 1.732050807568
- √26 =5.099019513592
- √101 = 10.04987562112

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

Solution:

Three different irrational numbers are:

- 0.73073007300073000073…
- 0.75075007300075000075…
- 0.76076007600076000076…

**9. Classify the following numbers as rational or irrational according to their type:**

(i)√23

Solution:

√23 = 4.79583152331…

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

(ii)√225

Solution:

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

**(iii) 0.3796**

Solution:

Since the number,0.3796, is terminating, it is a rational number.

**(iv) 7.478478**

Solution:

The number,7.478478, is non-terminating but recurring, it is a rational number.

**(v) 1.101001000100001…**

Solution:

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

## Exercise 1.4

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

Solution:

## Exercise 1.5

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

**(i) 2 –√5**

Solution:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

**(ii) (3 +√23)- √23**

Solution:

(3 +**√**23) –√23 = 3+**√**23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (**3 +√23)- √23** is rational.

**(iii) 2√7/7√7**

Solution:

2√7/7√7 = (2/7) × (√7/√7)

We know that (√7/√7) = 1

Hence, (2/7) × (√7/√7) = (2/7) ×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

**(iv) 1/√2**

Solution:

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) × (√2/√2) = √2/2 (since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0. 7071..

Since the number, 0. 7071… is non-terminating non-recurring, 1/√2 is an irrational number.

**(v) 2**

Solution:

We know that, the value of = 3.1415

Hence, 2 = 2×3. 1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

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

**(i) (3+√3) (2+√2)**

Solution:

(3+√3) (2+√2)

Opening the brackets, we get, (3×2) + (3×√2) + (√3×2) + (√3×√2)

= 6+3√2+2√3+√6

**(ii) (3+√3) (3-√3)**

Solution:

(3+√3) (3-√3) = 3^{2}-(√3)^{2} = 9-3

= 6

**(iii) (√5+√2)**^{2}

Solution:

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

= 5+2×√10+2 = 7+2√10

**(iv) (√5-√2) (√5+√2)**

Solution:

(√5-√2) (√5+√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?**

Solution:

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

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

Solution:

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

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

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, 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,

We get,

OD^{2}=BD^{2}+OB^{2}

⟹ (10.3/2)^{2} = BD^{2}+(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 = √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. Rationalize the denominators of the following:**

**(i) 1/√7**

Solution:

Multiply and divide 1/√7 by √7

(1×√7)/ (√7×√7) = √7/7

**(ii) 1/ (√7-√6)**

Solution:

Multiply and divide 1/ (√7-√6) by (√7+√6)

[1/ (√7-√6)] × (√7+√6)/ (√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√7^{2}-√6^{2 }[denominator is obtained by the property, (a + b) (a-b) = a^{2}-b^{2}]

= (√7+√6)/ (7-6)

= (√7+√6)/1

= √7+√6

**(iii) 1/ (√5+√2)**

Solution:

Multiply and divide 1/ (√5+√2) by (√5-√2)

[1/ (√5+√2)] × (√5-√2)/ (√5-√2) = (√5-√2)/ (√5+√2) (√5-√2)

= (√5-√2)/ (√5^{2}-√2^{2}) [denominator is obtained by the property, (a+b) (a-b) = a^{2}-b^{2}]

= (√5-√2) / 5-2)

= (√5-√2)/3

**(iv) 1/ (√7-2)**

Solution:

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7^{2}-2^{2}) [denominator is obtained by the property, (a+b)(a-b) = a^{2}-b^{2}]

= (√7+2)/(7-4)

= (√7+2)/3

## Exercise 1.6

**1. Find:**

**(i)64**^{1/2}

Solution:

64^{1/2} = (8×8)^{1/2}

= (8^{2})^{ ½}

= 8^{1} [⸪2×1/2 = 2/2 =1]

= 8

**(ii)32**^{1/5}

Solution:

32^{1/5 }= (2^{5})^{1/5}

= (2^{5})^{ ⅕}

= 2^{1} [⸪5×1/5 = 1]

= 2

**(iii)125**^{1/3}

Solution:

(125)^{1/3} = (5×5×5)^{1/3}

= (5^{3})^{ ⅓}

= 5^{1} (3×1/3 = 3/3 = 1)

= 5

**2. Find:**

**(i) 9**^{3/2}

Solution:

9^{3/2} = (3×3)^{3/2}

= (3^{2})^{3/2}

= 3^{3} [⸪2×3/2 = 3]

=27

**(ii) 32**^{2/5}

Solution:

32^{2/5 }= (2×2×2×2×2)^{2/5}

= (2^{5})^{2⁄5}

= 2^{2} [⸪5×2/5= 2]

= 4

**(iii)16**^{3/4}

Solution:

16^{3/4 }= (2×2×2×2)^{3/4}

= (2^{4})^{3⁄4}

= 2^{3} [⸪4×3/4 = 3]

= 8

**(iv) 125**^{-1/3}

125^{-1/3 }= (5×5×5)^{-1/3}

= (5^{3})^{-1⁄3}

= 5^{-1} [⸪3×-1/3 = -1]

= 1/5

**3. Simplify**:

**(i) 2**^{2/3}**×2**^{1/5}

Solution:

2^{2/3}×2^{1/5 }= 2^{(2/3) +(1/5)} [⸪Since, a^{m}×a^{n}=a^{m+n}____ Laws of exponents]

= 2^{13/15} [⸪2/3 + 1/5 = (2×5+3×1)/ (3×5) = 13/15]

**(ii) (1/3**^{3}**)**^{7}

Solution:

(1/3^{3})^{7 }= (3^{-3})^{7} [⸪Since, (a^{m})^{n }= a^{m x n}____ Laws of exponents]

= 3^{-21}

**(iii) 11**^{1/2}**/11**^{1/4}

Solution:

11^{1/2}/11^{1/4 }= 11^{(1/2) -(1/4)}

= 11^{1/4} [⸪ (1/2) – (1/4) = (1×4-2×1)/ (2×4) = 4-2)/8 = 2/8 = ¼]

**(iv) 7**^{1/2}**×8**^{1/2}

Solution:

7^{1/2}×8^{1/2} = (7×8)^{1/2} [⸪Since, (a^{m}×b^{m }= (a×b)^{m} ____ Laws of exponents]

= 56^{1/2}

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