Number Systems ⇒⇒ Exercise 1.3

Question 1(i)

Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36 / 100

Solution :

36 ${\div }$ 100 = 0.36

Decimal expression = 0.36 , This is terminating decimal expansion .

Question 1(ii)

Write the following in decimal form and say what kind of decimal expansion each has:
(ii) 1 / 11

Solution :

1 ${\div }$ 11
By long division method, we have ,

$$ \begin{array}{r} \mathbf{0.0909...} \\ 11\overline{)1.0000} \\ \underline{0}\phantom{0000} \\ 10\phantom{000} \\ \underline{0}\phantom{000} \\ 100\phantom{00} \\ \underline{99}\phantom{00} \\ 10\phantom{0} \\ \underline{0}\phantom{0} \\ 100 \\ \underline{99} \\ \mathbf{1}\phantom{.} \\ \end{array} $$

Decimal expression =0.09090909... or 0.09

Since decimal expansion repeats itself
Thus, it is non-terminating repeating decimal expansion .

Question 1(iii)

Write the following in decimal form and say what kind of decimal expansion each has:
(iii) 4${ 1 \over 8}$

Solution :

4${ 1 \over 8}$ = ${ 33 \over 8} $
By long division method, we have ,

$$ \begin{array}{r} \mathbf{4.125} \\ 8\overline{)33.000} \\ \underline{32}\phantom{0000} \\ 10\phantom{00} \\ \underline{8}\phantom{00} \\ 20\phantom{0} \\ \underline{16}\phantom{0} \\ 40\phantom{} \\ \underline{40}\phantom{} \\ \mathbf{0}\phantom{} \\ \end{array} $$

Decimal expression = 4.125.

Since decimal expansion has finite number of figures
Thus, it is terminating decimal expansion

Question 1(iv)

Write the following in decimal form and say what kind of decimal expansion each has:
(iv) ${ 3 \over 13}$

Solution :

${ 3 \over 13}$
By long division method, we have ,

$$ \begin{array}{r} 0.230769... \\ 13\overline{)3.000000} \\ \underline{0}\phantom{000000} \\ 30\phantom{00000} \\ \underline{26}\phantom{00000} \\ 40\phantom{0000} \\ \underline{39}\phantom{0000} \\ 10\phantom{000} \\ \underline{0}\phantom{000} \\ 100\phantom{00} \\ \underline{91}\phantom{00} \\ 90\phantom{0} \\ \underline{78}\phantom{0} \\ 120 \\ \underline{117} \\ \mathbf{3}\phantom{.} \\ \end{array} $$

Decimal expression =0.230769230... or 0.230769

Since decimal expansion repeats itself
Thus, it is non-terminating repeating decimal expansion .

Question 1(v)

Write the following in decimal form and say what kind of decimal expansion each has:
(v) ${ 2 \over 11}$

Solution :

${ 2 \over 11}$
By long division method, we have ,

$$ \begin{array}{r} \mathbf{0.1818...} \\ 11\overline{)2.0000} \\ \underline{0}\phantom{0000} \\ 20\phantom{000} \\ \underline{11}\phantom{000} \\ 90\phantom{00} \\ \underline{88}\phantom{00} \\ 20\phantom{0} \\ \underline{11}\phantom{0} \\ 90\phantom{} \\ \underline{88}\phantom{} \\ \mathbf{2}\phantom{.} \\ \end{array} $$

Decimal expression =0.181818... or 0.18

Since decimal expansion repeats itself
Thus, it is non-terminating repeating decimal expansion .

Question 1(vi)

Write the following in decimal form and say what kind of decimal expansion each has:
(vi) ${ 329 \over 400}$

Solution :

${ 329 \over 400}$
By long division method, we have ,

$$ \begin{array}{r} \mathbf{0.8225} \\ 400\overline{)329.0000} \\ \underline{0}\phantom{000000} \\ 3290\phantom{000} \\ \underline{3200}\phantom{000} \\ 900\phantom{00} \\ \underline{800}\phantom{00} \\ 1000\phantom{0} \\ \underline{800}\phantom{0} \\ 2000\phantom{} \\ \underline{2000}\phantom{} \\ \mathbf{0}\phantom{.} \\ \end{array} $$

Decimal expression = 0.8225.

Since decimal expansion has finite number of figures
Thus, it is terminating decimal expansion

Question 2

You know that ${ 1 \over 7}$ = 0.142857 . Can you predict what the decimal expansions of ${ 2 \over 7}, { 3 \over 7},{ 4 \over 7},{ 5 \over 7},{ 6 \over 7} $ are, without actually doing the long division? If so how ?

Solution :

Given that ${ 1 \over 7}$ = 0.142857 .

Thus , Decimal expansion of ${ 2 \over 7}$ = 2 × ${ 1 \over 7}$

= 2 × 0.142857

= 0.285714

In similar way,Decimal expansion of ${ 3 \over 7}$ = 3 × ${ 1 \over 7}$

= 3 × 0.142857

= 0.428571

Similarly,Decimal expansion of ${ 3 \over 7}$ = 3 × ${ 1 \over 7}$

= 3 × 0.142857

= 0.428571

In similar way,Decimal expansion of ${ 4 \over 7}$ = 4 × ${ 1 \over 7}$

= 4 × 0.142857

= 0.571428

Similarly,Decimal expansion of ${ 5 \over 7}$ = 5 × ${ 1 \over 7}$

= 5 × 0.142857

= 0.714285

Similarly,Decimal expansion of ${ 6 \over 7}$ = 6 × ${ 1 \over 7}$

= 6 × 0.142857

= 0.857142

Question 3 (i)

Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i) 0.6

Solution :

Given 0.6 = 0.6666666.....

Let x = 0.6666666..

By observation, We know that, only one digit is repeating, so multiplying both sides by 10

So, 10 × x = 10 × 0.6666..

Or, 10x = 6.666..

Or, 10x = 6 + 0.666..

( We know that, x = 0.6666.. )

10x = 6 + x

10x - x = 6

9x = 6

x = ${ 6 \over 9}$

x = ${ 2 \over 3}$

Hence, 0.6 = ${ 2 \over 3}$

Question 3 (ii)

Express the following in the form p/q, where p and q are integers and q ≠ 0.
(ii) 0.47

Solution :

Given 0.47 = 0.47777.....

Let x = 0.47777...

By observation, We know that, only one digit is repeating, so multiplying both sides by 10

So, 10 × x = 10 × 0.47777..

Or, 10x = 4.7777..

Or, 10x = 4.3 + 0.4777..

( We know that, x = 0.47777.... )

10x = 4.3 + x

10x - x = 4.3

9x = 4.3

x = ${ 4.3 \over 9}$

Or , x = ${ 43 \over 90}$

Hence, 0.47 = ${ 43 \over 90}$

Question 3 (iii)

Express the following in the form p/q, where p and q are integers and q ≠ 0.
(iii) 0.001

Solution :

Given 0.001 = 0.001001001.....

Let x = 0.001001...

By observation, We know that, only three digit is repeating, so multiplying both sides by 1000

So, 1000 × x = 1000 × 0.001001..

Or, 1000x = 1.001..

Or, 1000x = 1 + 0.001..

( We know that, x = 0.001001.. )

1000x = 1 + x

1000x - x = 1

999x = 1

x = ${ 1 \over 999}$

Or , x = ${ 1 \over 999}$

Hence, 0.001 = ${ 1 \over 999}$

Question 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 :

Given = 0.99999.....

Let x = 0.99999...

By observation, We know that, only three digit is repeating, so multiplying both sides by 10

So, 10 × x = 10 × 0.99999..

Or, 10x = 9.99999..

Or, 10x = 9 + 0.9999..

( We know that, x = 0.999. )

$ 10x = 9 + x $

$ 10x - x = 9 $

$ 9x = 9 $

$ x = { 9 \over 9}$

$Or , x = 1$

Thus, 0.9999. = 1

The difference between 1 and 0.999999 is o.000001 which is negligible.

Therefore, 1 as an answer can be justified and we are surprised with answer.

Another way to think about it is the difference: what is 1−0.99999...? It's 0.00000..., which is exactly 0. Because there is no difference between the two numbers, they must be equal.

This concept can be understood using the idea of a limit. The limit of the sequence 0.9, 0.99, 0.999, ... as the number of 9s approaches infinity is 1.

Question 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 :

${ 1 \over 17}$
By long division method, we have ,

$$ \begin{array}{r} \mathbf{0.0588235294117647...} \\ 17\overline{)1.0000000000000000} \\ \underline{0}\phantom{0000000000000000} \\ 10\phantom{000000000000000} \\ \underline{0}\phantom{000000000000000} \\ 100\phantom{00000000000000} \\ \underline{85}\phantom{00000000000000} \\ 150\phantom{0000000000000} \\ \underline{136}\phantom{0000000000000} \\ 140\phantom{000000000000} \\ \underline{136}\phantom{000000000000} \\ 40\phantom{00000000000} \\ \underline{34}\phantom{00000000000} \\ 60\phantom{0000000000} \\ \underline{51}\phantom{0000000000} \\ 90\phantom{000000000} \\ \underline{85}\phantom{000000000} \\ 50\phantom{00000000} \\ \underline{34}\phantom{00000000} \\ 160\phantom{0000000} \\ \underline{153}\phantom{0000000} \\ 70\phantom{000000} \\ \underline{68}\phantom{000000} \\ 20\phantom{00000} \\ \underline{17}\phantom{00000} \\ 30\phantom{0000} \\ \underline{17}\phantom{0000} \\ 130\phantom{000} \\ \underline{119}\phantom{000} \\ 110\phantom{00} \\ \underline{102}\phantom{00} \\ 80\phantom{0} \\ \underline{68}\phantom{0} \\ 120 \\ \underline{119} \\ 1 \\ \end{array} $$

Decimal expression =0.0588235294117647 ... or 0.0588235294117647

Since decimal expansion repeats itself
Thus, it is non-terminating repeating decimal expansion .

The maximum number of digits in the quotient while computing 1/ 17 = 16

Question 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 :

By observation, We know that, only when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

${ 5 \over 2}$ = 2.5 , denominator q = 2¹

${ 2 \over 5}$ = 0.4 , denominator q = 5¹

${ 7 \over 8}$ = 0.875 , denominator q = 2³

We observed that the denominators of above rational numbers are in the form of 2^m × 5ⁿ . Where, 'm' and 'n' are whole numbers.

A rational number p /q is a terminating decimal only, when prime factorisation of q must have only powers of 2 or 5 or both.

This is because our number system is base-10, and the prime factors of 10 are 2 and 5.

Therefore, any fraction that can be rewritten with a denominator that's a power of 10 = 2^m × 5^{n will have a finite number of decimal places.}

Question 7

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

Solution :

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

(i) 0.213221322213....

(ii)π = 3.14159265...

(iii) √2 = 1.414213562.....

Numbers with non-terminating and non-recurring decimal expansions are called irrational numbers. This means they cannot be expressed as a simple fraction p/q, where p and q are integers.

Question 8

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

Solution :

To find three irrational numbers between 5/7 and 9/11, first convert these fractions to their decimal forms.

We can write ${ 5 \over 7}$ = 0.7142857142857 . ., = 0.714285 .

And ${ 9 \over 11}$ = 0.8181818 . . . , = 0.81 .

To find irrational numbers between 0.714285... and 0.818181..., we can construct them by creating decimal numbers that do not repeat. We can start with a decimal that's greater than 0.714285 but less than 0.818181.

Therefore, three different irrational numbers between the rational numbers ${ 5 \over 7}$ and ${ 9 \over 11}$

(i) 0.723591787 ....

(ii) 0.756872357412....

(iii) 0.812385688423...

Question 9

Classify the following numbers as rational or irrational :
(i) √23
(ii) √225
(iii)0.3796
(iv) 7.478478..
(v) 1.101001000100001..

Solution :

(i) √23

Irrational. The number 23 is a prime number, so its square root is not an integer. The decimal expansion of √23 is non-terminating and non-repeating

(ii) √225 = 15 = 15 / 1

It is a rational number as it can be represented in the form of p/q

(iii) 0.3796

Rational. This number has a terminating decimal expansion, meaning it ends. It can be written as the fraction 3796/10000.

(iv) 7.478478... = 7.478

Rational. This number has a non-terminating but repeating decimal expansion (the block '478' repeats). Any number with a repeating decimal can be expressed as a fraction.

(v) 1.101001000100001..

Irrational. This number has a non-terminating and non-repeating decimal expansion. The pattern of an increasing number of zeros between the ones ensures the digits never repeat in a fixed block.