The World of Numbers

A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Answer

Given,

For 2 bags of spices, the merchant receives 15 copper ingots.

So, ingots received per bag = $\dfrac{15}{2}$

For 12 bags of spices :

⇒ $\dfrac{15}{2}$ x 12

⇒ 15 x 6

⇒ 90

Hence, the merchant will leave with 90 copper ingots.

Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Answer

The numbers 11, 13, 17 and 19 are all prime numbers (numbers that have exactly two factors — 1 and the number itself).

In fact, they are all the prime numbers between 10 and 20.

Continuing the pattern of prime numbers beyond 20 :

⇒ 23, 29 and 31 are the next three prime numbers.

Hence, the numbers have in common that they are all prime numbers, and the next three numbers in this pattern are 23, 29 and 31.

We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Answer

Natural numbers are not closed under subtraction because the result of subtracting two natural numbers may not be a natural number.

Example 1 : Consider 3 and 5, both are natural numbers.

⇒ 3 - 5 = -2, which is not a natural number (it is a negative integer).

Example 2 : Consider 7 and 7, both are natural numbers.

⇒ 7 - 7 = 0, which is not a natural number (zero is a whole number, not a natural number).

Hence, natural numbers are not closed under subtraction.

Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Answer

On one hand, there are 4 fingers (excluding the thumb), and each finger has 3 joints.

So, total number of joints that can be counted using the thumb :

⇒ 4 × 3

⇒ 12.

This is exactly equal to the base of the duodecimal (base-12) counting system.

This is why ancient civilisations developed base-12 counting systems — it could be naturally tracked using one hand, where the thumb counts the 12 finger joints. Many remnants of base-12 are still seen today, such as 12 hours on a clock, 12 months in a year and 12 inches in a foot.

Hence, we can count up to 12 on one hand, which corresponds to the base-12 counting system.

Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt. (Hint: If someone takes away (–) four of your debts that are each worth ₹3 (that is, –3), you are effectively ₹12 richer! Therefore, (–3) × (–4) = +12.)

Answer

A negative number represents a debt (an amount owed) and a negative multiplier represents the removal (taking away) of those debts.

Consider the following situation :

Suppose I have 4 debts, each worth ₹3.

Each debt is represented as -3.

If someone takes away (removes) these 4 debts, I am no longer in debt and my wealth effectively increases by ₹12.

Mathematically :

⇒ (-3) × (-4)

⇒ +12.

So, removing 4 debts of ₹3 each makes me ₹12 richer.

Hence, multiplying a negative number by a negative number gives a positive number, because removing debts increases wealth.

The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature?

Answer

Given,

Temperature at noon = 4 °C

Drop in temperature by midnight = 15 °C

Midnight temperature :

⇒ (4 - 15) °C

⇒ -11 °C

Hence, the midnight temperature in Ladakh is -11 °C.

A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.

Answer

Given,

Loan (debt) = -₹850 (negative as it is a debt)

Profit (fortune) = +₹1,200 (positive as it is a fortune)

Loss = -₹450 (negative as it is a loss)

Writing the equation using integers :

⇒ (-850) + 1200 + (-450)

⇒ -850 + 1200 - 450

⇒ 350 - 450

⇒ -100.

Hence, the spice trader's final financial standing is -₹100, which means he is in a debt of ₹100.

Calculate the following using Brahmagupta's laws:
(i) $(-12) \times 5$
(ii) $(-8) \times (-7)$
(iii) $0 - (-14)$
(iv) $(-20) \div 4$

Answer

(i) Given, (-12) × 5

By Brahmagupta's law, the product of a debt and a fortune is a debt.

⇒ (-12) × 5

⇒ -60.

Hence, (-12) × 5 = -60.

(ii) Given, (-8) × (-7)

By Brahmagupta's law, the product of two debts is a fortune.

⇒ (-8) × (-7)

⇒ +56.

Hence, (-8) × (-7) = 56.

(iii) Given, 0 - (-14)

Subtracting a negative number is the same as adding its positive value.

⇒ 0 - (-14)

⇒ 0 + 14

⇒ 14.

Hence, 0 - (-14) = 14.

(iv) Given, (-20) ÷ 4

The quotient of a debt and a fortune is a debt.

⇒ (-20) ÷ 4

⇒ $\dfrac{-20}{4}$

⇒ -5.

Hence, (-20) ÷ 4 = -5.

Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 - (-5) = 15).

Answer

Subtracting a negative number is equivalent to adding the corresponding positive number, because subtracting a debt means removing the debt, and removing debt makes you richer.

Real-world example :

Suppose you have ₹10 in your wallet and someone has previously recorded that you owe (debt) ₹5 to them.

Your effective wealth in this situation = ₹10 - ₹5 = ₹5.

Now, suppose your friend cancels (removes) the debt of ₹5 (that is, subtracts -5 from your account).

After cancellation :

⇒ 10 - (-5)

⇒ 10 + 5

⇒ 15.

So, removing the debt of ₹5 effectively increases your wealth by ₹5, making your total ₹15.

Hence, subtracting a negative number is the same as adding a positive number, because removing a debt is equivalent to gaining wealth.

Can you explain why we need $q \neq 0$ in the definition of a rational number?

Answer

In the definition of a rational number $\dfrac{p}{q}$, we need q ≠ 0 because division by zero is undefined in mathematics.

If we allow q = 0, then $\dfrac{p}{0}$ would have no meaningful value, since there is no number that gives p when multiplied by 0 (any number multiplied by 0 gives 0).

For example :

⇒ $\dfrac{5}{0}$ would mean "what number multiplied by 0 gives 5?"

But, since any number × 0 = 0, no such number exists.

Also, $\dfrac{0}{0}$ is indeterminate as 0 × any number = 0.

Hence, the condition q ≠ 0 is necessary so that the rational number $\dfrac{p}{q}$ has a well-defined value.

While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Answer

To add or subtract two rational numbers with different denominators, we make the denominators equal by finding the L.C.M. (Least Common Multiple) of the two denominators.

Steps to follow :

1. Find the L.C.M. of the two denominators.

2. Convert each rational number into an equivalent rational number whose denominator is the L.C.M.

3. Add or subtract the numerators while keeping the common denominator.

Example : Add $\dfrac{2}{3}$ and $\dfrac{1}{4}$.

L.C.M. of 3 and 4 = 12.

$\Rightarrow \dfrac{2}{3} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{1 \times 3}{4 \times 3} = \dfrac{3}{12} \\[1em] \Rightarrow \dfrac{8}{12} + \dfrac{3}{12} = \dfrac{11}{12}.$

Hence, we make the denominators equal by taking the L.C.M. of the denominators.

Verify the distributive law for rational numbers.

Answer

The distributive law for rational numbers states that, for any three rational numbers p, q and r :

⇒ p × (q + r) = p × q + p × r.

Verification with example :

Let $p = \dfrac{1}{2}$, $q = \dfrac{2}{3}$ and $r = \dfrac{1}{4}$.

Using $p = \dfrac{1}{2}$, $q = \dfrac{2}{3}$ and $r = \dfrac{1}{4}$, we verify that $p \times (q + r) = p \times q + p \times r$.

Solving L.H.S. :

$\Rightarrow \dfrac{1}{2} \times \left(\dfrac{2}{3} + \dfrac{1}{4}\right) \\[1em] \Rightarrow \dfrac{1}{2} \times \left(\dfrac{8}{12} + \dfrac{3}{12}\right) \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{11}{12} \\[1em] \Rightarrow \dfrac{11}{24}.$

Solving R.H.S. :

$\Rightarrow \dfrac{1}{2} \times \dfrac{2}{3} + \dfrac{1}{2} \times \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{6} + \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{1}{3} + \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{8}{24} + \dfrac{3}{24} \\[1em] \Rightarrow \dfrac{11}{24}.$

Since L.H.S. = R.H.S.

Hence, the distributive law p × (q + r) = p × q + p × r is verified for rational numbers.

Prove that the following rational numbers are equal:

(i) $\dfrac{2}{3}$ and $\dfrac{4}{6}$

(ii) $\dfrac{5}{4}$ and $\dfrac{10}{8}$

(iii) $-\dfrac{3}{5}$ and $-\dfrac{6}{10}$

(iv) $\dfrac{9}{3}$ and $3$

Answer

Two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equal if ad = bc.

(i) Given, $\dfrac{2}{3}$ and $\dfrac{4}{6}$.

Cross-multiplying :

⇒ 2 × 6 = 12

⇒ 3 × 4 = 12

Since, 2 × 6 = 3 × 4.

Hence, $\dfrac{2}{3} = \dfrac{4}{6}$.

(ii) Given, $\dfrac{5}{4}$ and $\dfrac{10}{8}$.

Cross-multiplying :

⇒ 5 × 8 = 40

⇒ 4 × 10 = 40

Since, 5 × 8 = 4 × 10.

Hence, $\dfrac{5}{4} = \dfrac{10}{8}$.

(iii) Given, $-\dfrac{3}{5}$ and $-\dfrac{6}{10}$.

Cross-multiplying :

⇒ (-3) × 10 = -30

⇒ 5 × (-6) = -30

Since, (-3) × 10 = 5 × (-6).

Hence, $-\dfrac{3}{5} = -\dfrac{6}{10}$.

(iv) Given, $\dfrac{9}{3}$ and 3.

We can write 3 as $\dfrac{3}{1}$.

Cross-multiplying :

⇒ 9 × 1 = 9

⇒ 3 × 3 = 9

Since, 9 × 1 = 3 × 3.

Hence, $\dfrac{9}{3} = 3$.

Find the sum:

(i) $\dfrac{2}{5} + \dfrac{3}{10}$

(ii) $\dfrac{7}{12} + \dfrac{5}{8}$

(iii) $-\dfrac{4}{7} + \dfrac{3}{14}$

Answer

(i) Given, $\dfrac{2}{5} + \dfrac{3}{10}$

L.C.M. of 5 and 10 = 10.

$\Rightarrow \dfrac{2}{5} + \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2 \times 2}{5 \times 2} + \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{4}{10} + \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{4 + 3}{10} \\[1em] \Rightarrow \dfrac{7}{10}.$

Hence, $\dfrac{2}{5} + \dfrac{3}{10} = \dfrac{7}{10}$.

(ii) Given, $\dfrac{7}{12} + \dfrac{5}{8}$

L.C.M. of 12 and 8 = 24.

$\Rightarrow \dfrac{7}{12} + \dfrac{5}{8} \\[1em] \Rightarrow \dfrac{7 \times 2}{12 \times 2} + \dfrac{5 \times 3}{8 \times 3} \\[1em] \Rightarrow \dfrac{14}{24} + \dfrac{15}{24} \\[1em] \Rightarrow \dfrac{14 + 15}{24} \\[1em] \Rightarrow \dfrac{29}{24}.$

Hence, $\dfrac{7}{12} + \dfrac{5}{8} = \dfrac{29}{24}$.

(iii) Given, $-\dfrac{4}{7} + \dfrac{3}{14}$

L.C.M. of 7 and 14 = 14.

$\Rightarrow -\dfrac{4}{7} + \dfrac{3}{14} \\[1em] \Rightarrow -\dfrac{4 \times 2}{7 \times 2} + \dfrac{3}{14} \\[1em] \Rightarrow -\dfrac{8}{14} + \dfrac{3}{14} \\[1em] \Rightarrow \dfrac{-8 + 3}{14} \\[1em] \Rightarrow -\dfrac{5}{14}.$

Hence, $-\dfrac{4}{7} + \dfrac{3}{14} = -\dfrac{5}{14}$.

Find the difference:

(i) $\dfrac{5}{6} - \dfrac{1}{4}$

(ii) $\dfrac{11}{8} - \dfrac{3}{4}$

(iii) $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)$

Answer

(i) Given, $\dfrac{5}{6} - \dfrac{1}{4}$

L.C.M. of 6 and 4 = 12.

$\Rightarrow \dfrac{5}{6} - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{5 \times 2}{6 \times 2} - \dfrac{1 \times 3}{4 \times 3} \\[1em] \Rightarrow \dfrac{10}{12} - \dfrac{3}{12} \\[1em] \Rightarrow \dfrac{10 - 3}{12} \\[1em] \Rightarrow \dfrac{7}{12}.$

Hence, $\dfrac{5}{6} - \dfrac{1}{4} = \dfrac{7}{12}$.

(ii) Given, $\dfrac{11}{8} - \dfrac{3}{4}$

L.C.M. of 8 and 4 = 8.

$\Rightarrow \dfrac{11}{8} - \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{11}{8} - \dfrac{3 \times 2}{4 \times 2} \\[1em] \Rightarrow \dfrac{11}{8} - \dfrac{6}{8} \\[1em] \Rightarrow \dfrac{11 - 6}{8} \\[1em] \Rightarrow \dfrac{5}{8}.$

Hence, $\dfrac{11}{8} - \dfrac{3}{4} = \dfrac{5}{8}$.

(iii) Given, $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right)$

L.C.M. of 9 and 3 = 9.

$\Rightarrow -\dfrac{7}{9} - \left(-\dfrac{2}{3}\right) \\[1em] \Rightarrow -\dfrac{7}{9} + \dfrac{2}{3} \\[1em] \Rightarrow -\dfrac{7}{9} + \dfrac{2 \times 3}{3 \times 3} \\[1em] \Rightarrow -\dfrac{7}{9} + \dfrac{6}{9} \\[1em] \Rightarrow \dfrac{-7 + 6}{9} \\[1em] \Rightarrow -\dfrac{1}{9}.$

Hence, $-\dfrac{7}{9} - \left(-\dfrac{2}{3}\right) = -\dfrac{1}{9}$.

Find the product:

(i) $\dfrac{2}{3} \times \dfrac{3}{10}$

(ii) $\dfrac{7}{11} \times \dfrac{5}{8}$

(iii) $-\dfrac{4}{7} \times \dfrac{5}{14}$

Answer

(i) Given, $\dfrac{2}{3} \times \dfrac{3}{10}$

$\Rightarrow \dfrac{2}{3} \times \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2 \times 3}{3 \times 10} \\[1em] \Rightarrow \dfrac{6}{30} \\[1em] \Rightarrow \dfrac{1}{5}.$

Hence, $\dfrac{2}{3} \times \dfrac{3}{10} = \dfrac{1}{5}$.

(ii) Given, $\dfrac{7}{11} \times \dfrac{5}{8}$

$\Rightarrow \dfrac{7}{11} \times \dfrac{5}{8} \\[1em] \Rightarrow \dfrac{7 \times 5}{11 \times 8} \\[1em] \Rightarrow \dfrac{35}{88}.$

Hence, $\dfrac{7}{11} \times \dfrac{5}{8} = \dfrac{35}{88}$.

(iii) Given, $-\dfrac{4}{7} \times \dfrac{5}{14}$

$\Rightarrow -\dfrac{4}{7} \times \dfrac{5}{14} \\[1em] \Rightarrow -\dfrac{4 \times 5}{7 \times 14} \\[1em] \Rightarrow -\dfrac{20}{98} \\[1em] \Rightarrow -\dfrac{10}{49}.$

Hence, $-\dfrac{4}{7} \times \dfrac{5}{14} = -\dfrac{10}{49}$.

Find the quotient:

(i) $\dfrac{2}{3} \div \dfrac{3}{10}$

(ii) $\dfrac{7}{11} \div \dfrac{5}{8}$

(iii) $-\dfrac{4}{7} \div \dfrac{5}{14}$

Answer

(i) Given, $\dfrac{2}{3} \div \dfrac{3}{10}$

$\Rightarrow \dfrac{2}{3} \div \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2}{3} \times \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2 \times 10}{3 \times 3} \\[1em] \Rightarrow \dfrac{20}{9}.$

Hence, $\dfrac{2}{3} \div \dfrac{3}{10} = \dfrac{20}{9}$.

(ii) Given, $\dfrac{7}{11} \div \dfrac{5}{8}$

$\Rightarrow \dfrac{7}{11} \div \dfrac{5}{8} \\[1em] \Rightarrow \dfrac{7}{11} \times \dfrac{8}{5} \\[1em] \Rightarrow \dfrac{7 \times 8}{11 \times 5} \\[1em] \Rightarrow \dfrac{56}{55}.$

Hence, $\dfrac{7}{11} \div \dfrac{5}{8} = \dfrac{56}{55}$.

(iii) Given, $-\dfrac{4}{7} \div \dfrac{5}{14}$

$\Rightarrow -\dfrac{4}{7} \div \dfrac{5}{14} \\[1em] \Rightarrow -\dfrac{4}{7} \times \dfrac{14}{5} \\[1em] \Rightarrow -\dfrac{4 \times 14}{7 \times 5} \\[1em] \Rightarrow -\dfrac{56}{35} \\[1em] \Rightarrow -\dfrac{8}{5}.$

Hence, $-\dfrac{4}{7} \div \dfrac{5}{14} = -\dfrac{8}{5}$.

Show that: $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$.

Answer

Given,

Equation : $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$

Solving L.H.S. of the equation :

$\Rightarrow \left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} \\[1em] \Rightarrow \left(\dfrac{2}{4} + \dfrac{3}{4}\right) \times \dfrac{8}{3} \quad \text{[L.C.M. of 2 and 4 is 4]} \\[1em] \Rightarrow \dfrac{5}{4} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{40}{12} \\[1em] \Rightarrow \dfrac{10}{3}.$

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{8}{6} + \dfrac{24}{12} \\[1em] \Rightarrow \dfrac{4}{3} + 2 \\[1em] \Rightarrow \dfrac{4}{3} + \dfrac{6}{3} \quad \text{[L.C.M. of 3 and 1 is 3]} \\[1em] \Rightarrow \dfrac{10}{3}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$.

Simplify the following using the distributive property: $\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right)$.

Answer

Given, $\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right)$

Using the distributive property : a(b - c) = ab - ac

$\Rightarrow \dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right) \\[1em] \Rightarrow \dfrac{7}{9} \times \dfrac{6}{7} - \dfrac{7}{9} \times \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{2}{1} - \dfrac{7}{3} \times \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{3} - \dfrac{7}{12} \\[1em] \Rightarrow \dfrac{8}{12} - \dfrac{7}{12} \quad \text{[L.C.M. of 3 and 12 is 12]} \\[1em] \Rightarrow \dfrac{1}{12}.$

Hence, $\dfrac{7}{9}\left(\dfrac{6}{7} - \dfrac{3}{4}\right) = \dfrac{1}{12}$.

Find the rational number $x$ such that: $\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}$.

Answer

Given,

Equation : $\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}$

Expanding L.H.S. using the distributive property :

$\Rightarrow \dfrac{5}{6}\left(x + \dfrac{3}{5}\right) \\[1em] \Rightarrow \dfrac{5}{6} \times x + \dfrac{5}{6} \times \dfrac{3}{5} \\[1em] \Rightarrow \dfrac{5}{6}x + \dfrac{15}{30} \\[1em] \Rightarrow \dfrac{5}{6}x + \dfrac{1}{2}.$

So, L.H.S. = $\dfrac{5}{6}x + \dfrac{1}{2}$ and R.H.S. = $\dfrac{5}{6}x + \dfrac{1}{2}$.

L.H.S. = R.H.S. for any value of x. Therefore, the given equation is an identity (it is true for all rational numbers x).

Hence, every rational number x satisfies the given equation, i.e., the equation holds true for all rational values of x.

Try and represent $\dfrac{8}{5}$ and $-\dfrac{7}{4}$ on a number line.

Answer

To represent $\dfrac{8}{5}$ on the number line :

⇒ $\dfrac{8}{5} = 1\dfrac{3}{5}$, so it lies between 1 and 2.

⇒ Divide the unit interval between 1 and 2 into 5 equal parts and move 3 parts to the right of 1.

To represent $-\dfrac{7}{4}$ on the number line :

⇒ $-\dfrac{7}{4} = -1\dfrac{3}{4}$, so it lies between -2 and -1.

⇒ Divide the unit interval between -2 and -1 into 4 equal parts and move 3 parts to the left of -1 (toward -2).

Hence, $\dfrac{8}{5}$ lies between 1 and 2, and $-\dfrac{7}{4}$ lies between -2 and -1 on the number line.

Represent the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ on a single number line.

Answer

To represent the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ :

For $\dfrac{2}{3}$ :

⇒ It lies between 0 and 1. Divide the interval from 0 to 1 into 3 equal parts and mark the second division point.

For $-\dfrac{5}{4}$ :

⇒ $-\dfrac{5}{4} = -1\dfrac{1}{4}$, so it lies between -2 and -1. Divide the interval from -2 to -1 into 4 equal parts and mark the first division point from -1.

For $1\dfrac{1}{2}$ :

⇒ It lies between 1 and 2. Divide the interval from 1 to 2 into 2 equal parts and mark the midpoint.

Hence, the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ have been represented on a single number line.

Find three distinct rational numbers that lie strictly between $-\dfrac{1}{2}$ and $\dfrac{1}{4}$.

Answer

Given,

Two rational numbers : $-\dfrac{1}{2}$ and $\dfrac{1}{4}$.

Convert both to a common denominator of 8.

$\Rightarrow -\dfrac{1}{2} = -\dfrac{1 \times 4}{2 \times 4} = -\dfrac{4}{8} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{1 \times 2}{4 \times 2} = \dfrac{2}{8}.$

So we need to find three distinct rational numbers between $-\dfrac{4}{8}$ and $\dfrac{2}{8}$.

The integers between -4 and 2 are : -3, -2, -1, 0, 1.

Hence, three rational numbers between $-\dfrac{4}{8}$ and $\dfrac{2}{8}$ are :

$\Rightarrow -\dfrac{3}{8}, \\ -\dfrac{2}{8}, \\ \dfrac{1}{8} \\[1em] \Rightarrow -\dfrac{3}{8}, \\ -\dfrac{1}{4}, \\ \dfrac{1}{8}.$

Hence, three distinct rational numbers between $-\dfrac{1}{2}$ and $\dfrac{1}{4}$ are $-\dfrac{3}{8}, \\ -\dfrac{1}{4}$ and $\dfrac{1}{8}$.

Simplify the expression: $\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right)$.

Answer

Given, $\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right)$

L.C.M. of 4 and 12 = 12.

$\Rightarrow -\dfrac{1}{4} + \dfrac{5}{12} \\[1em] \Rightarrow -\dfrac{1 \times 3}{4 \times 3} + \dfrac{5}{12} \\[1em] \Rightarrow -\dfrac{3}{12} + \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{-3 + 5}{12} \\[1em] \Rightarrow \dfrac{2}{12} \\[1em] \Rightarrow \dfrac{1}{6}.$

Hence, $\left(-\dfrac{1}{4}\right) + \left(\dfrac{5}{12}\right) = \dfrac{1}{6}$.

A tailor has $15\dfrac{3}{4}$ metres of fine silk. If making one kurta requires $2\dfrac{1}{4}$ metres of silk, exactly how many kurtas can he make?

Answer

Given,

Total silk available = $15\dfrac{3}{4}$ metres = $\dfrac{63}{4}$ metres

Silk required for one kurta = $2\dfrac{1}{4}$ metres = $\dfrac{9}{4}$ metres

Number of kurtas the tailor can make :

$\Rightarrow \dfrac{63}{4} \div \dfrac{9}{4} \\[1em] \Rightarrow \dfrac{63}{4} \times \dfrac{4}{9} \\[1em] \Rightarrow \dfrac{63}{9} \\[1em] \Rightarrow 7.$

Hence, the tailor can make exactly 7 kurtas.

Find three rational numbers between 3.1415 and 3.1416.

Answer

Given,

Two rational numbers : 3.1415 and 3.1416.

To find rational numbers between them, we extend the decimal places.

⇒ 3.1415 = 3.14150

⇒ 3.1416 = 3.14160

Three rational numbers strictly between 3.14150 and 3.14160 are :

⇒ 3.14151, 3.14152, 3.14153.

Hence, three rational numbers between 3.1415 and 3.1416 are 3.14151, 3.14152 and 3.14153.

Can you think of other way(s) to find a rational number between any two rational numbers?

Answer

Yes, there are several ways to find a rational number between any two rational numbers a and b :

1. Average (Mean) method : The average of a and b, given by $\dfrac{a + b}{2}$, always lies between a and b.

For example, between $\dfrac{1}{2}$ and $\dfrac{3}{4}$ :

Average =

$= \dfrac{\dfrac{1}{2} + \dfrac{3}{4}}{2} \\[1em] = \dfrac{\dfrac{5}{4}}{2} \\[1em] = \dfrac{5}{4} \times \dfrac{1}{2} \\[1em] = \dfrac{5}{8}$

2. Common denominator method : Convert both rational numbers to a common denominator and select numerators in between. If the gap is too small, multiply numerator and denominator by a larger number to increase the gap.

For example, between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ :

⇒ $\dfrac{1}{3} = \dfrac{2}{6}$, $\dfrac{1}{2} = \dfrac{3}{6}$. To find more numbers, write as $\dfrac{20}{60}$ and $\dfrac{30}{60}$. Then $\dfrac{21}{60}, \dfrac{22}{60}, \ldots, \dfrac{29}{60}$ all lie between them.

3. Decimal method : Convert both rational numbers into decimals and pick decimal numbers between them by extending the decimal places.

4. Weighted average method : Use $\dfrac{2a + b}{3}$ or $\dfrac{a + 2b}{3}$, which divide the interval into thirds.

Hence, there are multiple ways to find a rational number between any two rational numbers — the average method, common denominator method, decimal method and weighted averages.

Can $\sqrt{2}$ be written as a rational number $\dfrac{p}{q}$?

Answer

No, $\sqrt{2}$ cannot be written as a rational number $\dfrac{p}{q}$.

Proof by contradiction :

Assume $\sqrt{2}$ is rational. Then it can be expressed as $\dfrac{p}{q}$, where p and q are integers with q ≠ 0 and gcd(p, q) = 1 (lowest form).

Squaring both sides :

$\Rightarrow \sqrt{2} = \dfrac{p}{q} \\[1em] \Rightarrow 2 = \dfrac{p^2}{q^2} \\[1em] \Rightarrow 2q^2 = p^2.$

This means p² is even, and so p must also be even.

Let p = 2k for some integer k. Then :

$\Rightarrow 2q^2 = (2k)^2 \\[1em] \Rightarrow 2q^2 = 4k^2 \\[1em] \Rightarrow q^2 = 2k^2.$

This means q² is even, so q is also even.

But, both p and q being even contradicts our assumption that gcd(p, q) = 1.

Hence, our assumption is wrong.

Hence, $\sqrt{2}$ cannot be written as $\dfrac{p}{q}$. It is an irrational number.

Try to prove the irrationality of $\sqrt{3}$ using the approach of proof by contradiction. Will the same approach work for $\sqrt{5}$, $\sqrt{7}$, or $\sqrt{10}$?

Answer

Proof of irrationality of $\sqrt{3}$ by contradiction :

Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{p}{q}$, where p and q are integers with q ≠ 0 and gcd(p, q) = 1.

Squaring both sides :

$\Rightarrow \sqrt{3} = \dfrac{p}{q} \\[1em] \Rightarrow 3 = \dfrac{p^2}{q^2} \\[1em] \Rightarrow 3q^2 = p^2.$

So, p² is divisible by 3, which means p is also divisible by 3.

Let p = 3k for some integer k. Then :

$\Rightarrow 3q^2 = (3k)^2 \\[1em] \Rightarrow 3q^2 = 9k^2 \\[1em] \Rightarrow q^2 = 3k^2.$

So, q² is divisible by 3, which means q is also divisible by 3.

Both p and q being divisible by 3 contradicts gcd(p, q) = 1.

Hence, $\sqrt{3}$ is irrational.

Yes, the same approach works for $\sqrt{5}$, $\sqrt{7}$ and $\sqrt{10}$, since 5, 7 and 10 are not perfect squares.

The proof relies on the fact that if a prime number p divides n², then p divides n. This works for any non-perfect square integer.

Hence, $\sqrt{3}$ is irrational, and the same approach proves $\sqrt{5}$, $\sqrt{7}$ and $\sqrt{10}$ are also irrational.

We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?

Answer

To obtain a line segment of irrational length, we use the Baudhāyana–Pythagoras theorem and a geometric construction with ruler and compass.

For example, to construct a line segment of length $\sqrt{2}$ :

Step 1 : Draw a line segment OA of length 1 unit.

Step 2 : At point A, draw a perpendicular AB of length 1 unit.

Step 3 : Join O to B. By the Pythagoras theorem :

⇒ OB² = OA² + AB²

⇒ OB² = 1² + 1²

⇒ OB² = 2

⇒ OB = $\sqrt{2}$.

Step 4 : With O as centre and OB as radius, draw an arc cutting the number line at point P. Then OP = $\sqrt{2}$.

This way, we can construct line segments of any irrational length of the form $\sqrt{n}$ using the Pythagoras theorem repeatedly.

Hence, we can obtain lines of irrational length by using right-angled triangles and applying the Pythagoras theorem.

Try to extend this method for constructing line segments of lengths $\sqrt{3}$ and $\sqrt{5}$ using a ruler and a compass. Generalise this method to construct a line segment of any length of the form $\sqrt{n}$, where $n$ is a positive integer.

Answer

Construction of $\sqrt{3}$ :

Step 1 : Construct OB = $\sqrt{2}$ (as shown previously).

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

Step 3 : Join O to D. By the Pythagoras theorem :

⇒ OD² = OB² + BD²

⇒ OD² = ($\sqrt{2}$)² + 1²

⇒ OD² = 2 + 1 = 3

⇒ OD = $\sqrt{3}$.

Step 4 : With O as the centre and OD as the radius, draw an arc cutting the number line at point Q. Then OQ = $\sqrt{3}$.

Construction of $\sqrt{5}$ :

Step 1: On the number line, measure a horizontal segment OA = 2 units.

Step 2: At point A, draw a perpendicular line segment AC of length 1 unit.

Step 3: Join O to C. The length OC is:

⇒ OC² = OA² + AC²

⇒ OC² = (2)² + 1²

⇒ OC² = 4 + 1 = 5

⇒ OC = $\sqrt{5}$.

Step 4 : With O as the centre and OC as the radius, draw an arc cutting the number line at point R. Then OR = $\sqrt{5}$.

Generalisation to construct $\sqrt{n}$ :

Suppose we have a line segment of length $\sqrt{n - 1}$. Then at one end of this segment, draw a perpendicular of length 1 unit. The hypotenuse of the resulting right triangle will have length :

$\Rightarrow \sqrt{(\sqrt{n - 1})^2 + 1^2} \\[1em] \Rightarrow \sqrt{(n - 1) + 1} \\[1em] \Rightarrow \sqrt{n}.$

This method, when extended repeatedly starting from a unit segment, gives the famous "square root spiral".

Hence, by repeatedly applying the Pythagoras theorem with a unit perpendicular to a known segment of length $\sqrt{n - 1}$, we can construct a line segment of any length of the form $\sqrt{n}$ for any positive integer n.

Try to find the decimal expansions of $\dfrac{10}{3}$ and $\dfrac{11}{12}$. What do you observe about the repetition of the digits after the decimal point?

Answer

Decimal expansion of $\dfrac{10}{3}$ :

By long division :

$\begin{array}{l} \phantom{3)\quad1}{3.333\ldots} \\ 3\overline{\smash{\big)}\quad 10.000\ldots} \\ \phantom{3)}\phantom{)1}\underline{-9} \\ \phantom{3)\quad))}10 \\ \phantom{3)\quad,}\underline{-9} \\ \phantom{3)\quad1))}10 \\ \phantom{3)\quad10}\underline{-9} \\ \phantom{3)\quad))10}10 \\ \phantom{3)\quad100}\underline{-9} \\ \phantom{3)\quad)1000}1\ldots \end{array}$

⇒ $\dfrac{10}{3} = 3.3333\ldots = 3.\overline{3}$

The digit 3 repeats infinitely after the decimal point.

Decimal expansion of $\dfrac{11}{12}$ :

By long division:

$\begin{array}{l} \phantom{12\overline{)},1}0.9166\ldots \\ 12\overline{\smash{\big)}\phantom{)}11.0000\ldots} \\ \phantom{12\overline{}}\underline{-108} \\ \phantom{12\overline{)}111}20\phantom{00\ldots} \\ \phantom{12\overline{)},0}\underline{-12}\phantom{00\ldots} \\ \phantom{12\overline{)}1111}80\phantom{0\ldots} \\ \phantom{12\overline{)},00}\underline{-72}\phantom{0\ldots} \\ \phantom{12\overline{)}11111}80\phantom{\ldots} \\ \phantom{12\overline{)},000}\underline{-72}\phantom{\ldots} \\ \phantom{12\overline{)}111111}8\ldots \end{array}$

⇒ $\dfrac{11}{12} = 0.91666\ldots = 0.91\overline{6}$

After the decimal point, "91" appears once and then the digit 6 repeats infinitely.

Observations :

1. Both decimal expansions are non-terminating but repeating.

2. In $\dfrac{10}{3}$, the repetition starts immediately after the decimal point (pure repeating decimal).

3. In $\dfrac{11}{12}$, there are some non-repeating digits before the repeating block begins (general/mixed repeating decimal).

This happens because the denominator 3 (in $\dfrac{10}{3}$) has only the prime factor 3, and the denominator 12 = 2² × 3 (in $\dfrac{11}{12}$) has both 2 and 3 as factors. The presence of a prime factor other than 2 and 5 in the denominator makes the decimal repeating.

Hence, $\dfrac{10}{3} = 3.\overline{3}$ and $\dfrac{11}{12} = 0.91\overline{6}$. Both have repeating digits after the decimal point.

The decimal expansion of $\dfrac{p}{q}$ will be terminating precisely when the prime factors of $q$ are only 2, only 5 or both 2 and 5. Can you explain why?

Answer

The decimal expansion of $\dfrac{p}{q}$ terminates if and only if the denominator q (in lowest terms) has only 2 and/or 5 as its prime factors.

Reason :

Our number system is base-10, and 10 = 2 × 5. So, any denominator that is a power of 10 produces a terminating decimal.

If q has only the prime factors 2 and/or 5, we can multiply both the numerator and the denominator by a suitable number to make the denominator a power of 10. This gives a terminating decimal.

Example 1 :

⇒ $\dfrac{3}{20} = \dfrac{3}{2^2 \times 5} = \dfrac{3 \times 5}{2^2 \times 5 \times 5} = \dfrac{15}{100} = 0.15$ (terminating).

Example 2 :

⇒ $\dfrac{7}{8} = \dfrac{7}{2^3} = \dfrac{7 \times 5^3}{2^3 \times 5^3} = \dfrac{875}{1000} = 0.875$ (terminating).

But, if q has any prime factor other than 2 or 5 (such as 3, 7, 11, etc.), no multiplication can make the denominator a power of 10. So, the long division will never give a remainder of 0, and the decimal expansion will be non-terminating.

Example 3 :

⇒ $\dfrac{1}{3} = 0.3333\ldots$ (non-terminating, repeating). Here 3 is a prime factor other than 2 or 5.

Hence, the decimal expansion of $\dfrac{p}{q}$ terminates only when the denominator q (in lowest form) has 2 and/or 5 as its only prime factors, because only then can it be expressed as a fraction with a power of 10 in the denominator.

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\dfrac{7}{20}$, $\dfrac{4}{15}$ and $\dfrac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Answer

To determine if a rational number $\dfrac{p}{q}$ in lowest form has a terminating decimal expansion, the prime factorisation of q must contain only 2's and/or 5's.

(i) $\dfrac{7}{20}$

Prime factorisation of 20 = 2² × 5.

Since the denominator has only 2's and 5's, it is a terminating decimal.

Performing long division :

$\begin{array}{l} \phantom{20\overline{)},}0.35 \\ 20\overline{\smash{\big)}\phantom{)}7.00} \\ \phantom{20\overline{}}\underline{-60}\phantom{0} \\ \phantom{20\overline{)})}100 \\ \phantom{20\overline{}}\underline{-100} \\ \phantom{20\overline{)},00}0 \end{array}$

⇒ $\dfrac{7}{20}$ = 0.35

(ii) $\dfrac{4}{15}$

Prime factorisation of 15 = 3 × 5.

Since the denominator has 3 (a prime other than 2 or 5), it is a repeating decimal.

Performing long division :

$\begin{array}{l} \phantom{15\overline{)},}0.2666\ldots \\ 15\overline{\smash{\big)}\phantom{)}4.0000\ldots} \\ \phantom{15\overline{}}\underline{-30} \\ \phantom{15\overline{)})}100\phantom{0\ldots} \\ \phantom{15\overline{)}}\underline{-90}\phantom{0\ldots} \\ \phantom{15\overline{)}1)}100\phantom{\ldots} \\ \phantom{15\overline{)}1}\underline{-90}\phantom{\ldots} \\ \phantom{15\overline{)}11)}100 \\ \phantom{15\overline{)}00}\underline{-90} \\ \phantom{15\overline{)})111}10\ldots \end{array}$

⇒ $\dfrac{4}{15} = 0.2666\ldots = 0.2\overline{6}$

(iii) $\dfrac{13}{250}$

Prime factorisation of 250 = 2 × 5³.

Since the denominator has only 2's and 5's, it is a terminating decimal.

Performing long division :

$\begin{array}{l} \phantom{250\overline{)},1}0.052 \\ 250\overline{\smash{\big)}\phantom{)}13.000} \\ \phantom{250\overline{)}}\underline{-0} \\ \phantom{250\overline{)})}130\phantom{00} \\ \phantom{250\overline{1},}\underline{-0} \\ \phantom{250\overline{)})}1300\phantom{0} \\ \phantom{250\overline{}}\underline{-1250} \\ \phantom{250\overline{)}11)}500 \\ \phantom{250\overline{)}))}\underline{-500} \\ \phantom{250\overline{)},0000}0 \end{array}$

⇒ $\dfrac{13}{250}$ = 0.052

Hence, $\dfrac{7}{20} = 0.35$ (terminating), $\dfrac{4}{15} = 0.2\overline{6}$ (repeating) and $\dfrac{13}{250} = 0.052$ (terminating).

Perform the long division for $\dfrac{1}{13}$. Identify the repeating block of digits. Does it show cyclic properties if you evaluate $\dfrac{2}{13}$? Now compute $\dfrac{3}{13}$, $\dfrac{4}{13}$, etc. What do you notice?

Answer

Performing long division for $\dfrac{1}{13}$ :

$\begin{array}{l} \phantom{13\overline{)},}0.076923\ldots \\ 13\overline{\smash{\big)}\phantom{)}1.000000\ldots} \\ \phantom{13\overline{}}\underline{-0} \\ \phantom{13\overline{)})}10\phantom{0000\ldots} \\ \phantom{13\overline{)}}\underline{-0} \\ \phantom{13\overline{)},0}100\phantom{000\ldots} \\ \phantom{13\overline{)}1}\underline{-91}\phantom{000\ldots} \\ \phantom{13\overline{)},000}90\phantom{00\ldots} \\ \phantom{13\overline{)}11}\underline{-78}\phantom{00\ldots} \\ \phantom{13\overline{)},000}120\phantom{0\ldots} \\ \phantom{13\overline{)})1}\underline{-117}\phantom{0\ldots} \\ \phantom{13\overline{)},00000}30\phantom{\ldots} \\ \phantom{13\overline{)}1111}\underline{-26}\phantom{\ldots} \\ \phantom{13\overline{)},000000}40 \\ \phantom{13\overline{)}11111}\underline{-39} \\ \phantom{13\overline{)},0000000}1\ldots \end{array}$

⇒ $\dfrac{1}{13} = 0.076923 \\ 076923 \ldots = 0.\overline{076923}$

The repeating block is 076923 (6 digits).

Computing other multiples of $\dfrac{1}{13}$ :

Long division for $\dfrac{2}{13}$:

$\begin{array}{l} \phantom{13\overline{)},}0.153846\ldots \\ 13\overline{\smash{\big)}\phantom{)}2.000000\ldots} \\ \phantom{13\overline{}}\underline{-0} \\ \phantom{13\overline{)}1}20\phantom{0000\ldots} \\ \phantom{13\overline{,}}\underline{-13} \\ \phantom{13\overline{)},00}70\phantom{000\ldots} \\ \phantom{13\overline{)}1}\underline{-65}\phantom{000\ldots} \\ \phantom{13\overline{)},111}50\phantom{00\ldots} \\ \phantom{13\overline{)}11}\underline{-39}\phantom{00\ldots} \\ \phantom{13\overline{)},0000}110\phantom{0\ldots} \\ \phantom{13\overline{)}111}\underline{-104}\phantom{0\ldots} \\ \phantom{13\overline{)}))11111}60\phantom{\ldots} \\ \phantom{13\overline{)}11111}\underline{-52}\phantom{\ldots} \\ \phantom{13\overline{)}))111111}80 \\ \phantom{13\overline{)},000000}\underline{-78} \\ \phantom{13\overline{)})11111111}2\ldots \end{array}$

⇒ $\dfrac{2}{13} = 0.\overline{153846}$

Long division for $\dfrac{3}{13}$:

$\begin{array}{l} \phantom{13\overline{)},}0.230769\ldots \\ 13\overline{\smash{\big)}\phantom{)}3.000000\ldots} \\ \phantom{13\overline{}}\underline{-0} \\ \phantom{13\overline{)})}30\phantom{0000\ldots} \\ \phantom{13\overline{}}\underline{-26} \\ \phantom{13\overline{)})1}40\phantom{000\ldots} \\ \phantom{13\overline{}1}\underline{-39} \\ \phantom{13\overline{)}111}10\phantom{00\ldots} \\ \phantom{13\overline{)})11}\underline{-0}\phantom{00\ldots} \\ \phantom{13\overline{)},0000}100\phantom{0\ldots} \\ \phantom{13\overline{)}1111}\underline{-91}\phantom{0\ldots} \\ \phantom{13\overline{)}))11111}90\phantom{\ldots} \\ \phantom{13\overline{)}11111}\underline{-78}\phantom{\ldots} \\ \phantom{13\overline{)}))11111}120 \\ \phantom{13\overline{)},00000}\underline{-117} \\ \phantom{13\overline{)})11111111}3\ldots \end{array}$

⇒ $\dfrac{3}{13} = 0.\overline{230769}$

Similarly, we have:

⇒ $\dfrac{4}{13} = 0.\overline{307692}$

⇒ $\dfrac{5}{13} = 0.\overline{384615}$

⇒ $\dfrac{6}{13} = 0.\overline{461538}$

⇒ $\dfrac{7}{13} = 0.\overline{538461}$

⇒ $\dfrac{8}{13} = 0.\overline{615384}$

⇒ $\dfrac{9}{13} = 0.\overline{692307}$

⇒ $\dfrac{10}{13} = 0.\overline{769230}$

⇒ $\dfrac{11}{13} = 0.\overline{846153}$

⇒ $\dfrac{12}{13} = 0.\overline{923076}$

Observations :

1. The decimal expansions of $\dfrac{n}{13}$ form two distinct cyclic groups of 6 digits each, unlike $\dfrac{1}{7}$ which has one cyclic group.

2. Group 1 : $\dfrac{1}{13}, \dfrac{3}{13}, \dfrac{4}{13}, \dfrac{9}{13}, \dfrac{10}{13}, \dfrac{12}{13}$ all share cyclic permutations of "076923".

3. Group 2 : $\dfrac{2}{13}, \dfrac{5}{13}, \dfrac{6}{13}, \dfrac{7}{13}, \dfrac{8}{13}, \dfrac{11}{13}$ all share cyclic permutations of "153846".

Hence, $\dfrac{1}{13} = 0.\overline{076923}$ shows cyclic behaviour, but unlike $\dfrac{1}{7}$, it produces two different cyclic blocks among its multiples.

Classify the following numbers as rational or irrational:
(i) $\sqrt{81}$
(ii) $\sqrt{12}$
(iii) $0.33333\ldots$
(iv) $0.123451234512345\ldots$
(v) $1.01001000100001\ldots$ (Notice the pattern: Is it repeating a single block?)
(vi) $23.560185612239874790120$

Find the explicit fractions in case they are rational.

Answer

(i) $\sqrt{81}$

⇒ $\sqrt{81} = 9 = \dfrac{9}{1}$

Since 9 can be written as $\dfrac{p}{q}$ with q ≠ 0,

$\sqrt{81}$ is rational.

(ii) $\sqrt{12}$

⇒ $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

Since $\sqrt{3}$ is irrational, $2\sqrt{3}$ is also irrational.

So, $\sqrt{12}$ is irrational.

(iii) 0.33333…

This is a repeating decimal with the digit 3 repeating.

Let x = 0.3333…

⇒ 10x = 3.3333…

⇒ 10x - x = 3.3333… - 0.3333… = 3

⇒ 9x = 3

⇒ x = $\dfrac{3}{9} = \dfrac{1}{3}$.

Since 0.33333… can be written as $\dfrac{p}{q}$ with q ≠ 0,

So, 0.33333… is rational and equals $\dfrac{1}{3}$.

(iv) 0.123451234512345…

This is a repeating decimal with the block "12345" repeating.

Let x = 0.12345 12345…

⇒ 100000x = 12345.12345…

⇒ 100000x - x = 12345.12345… - 0.12345… = 12345

⇒ 99999x = 12345

⇒ x = $\dfrac{12345}{99999} = \dfrac{4115}{33333}$.

Since 0.123451234512345… can be written as $\dfrac{p}{q}$ with q ≠ 0,

So, 0.123451234512345… is rational and equals $\dfrac{4115}{33333}$.

(v) 1.01001000100001…

The pattern increases the number of zeros each time (one 0, then two 0's, then three 0's, etc.). This means there is no single repeating block.

So, 1.01001000100001… is non-terminating and non-repeating.

Hence, it is irrational.

(vi) 23.560185612239874790120

This is a terminating decimal with 21 decimal places.

It can be written as :

⇒ $\dfrac{23560185612239874790120}{10^{21}}$.

Since 23.560185612239874790120 can be written as $\dfrac{p}{q}$ with q ≠ 0,

So, 23.560185612239874790120 is rational.

The number $0.\overline{9}$ (which means $0.99999\ldots$) is a rational number. Using algebra (let $x = 0.\overline{9}$, multiply by 10, and subtract), explain why $0.\overline{9}$ is exactly equal to 1.

Answer

Let, x = $0.\overline{9}$ = 0.9999…

Multiplying both sides by 10 :

⇒ 10x = 9.9999…

Subtracting the first equation from the second :

⇒ 10x - x = 9.9999... - 0.9999...

⇒ 9x = 9

⇒ x = 1

So, $0.\overline{9}$ = 1.

This confirms that the recurring decimal $0.\overline{9}$ is exactly equal to 1, not slightly less than 1. Although it may seem counter-intuitive, the infinite chain of 9's collectively equals 1.

Hence, $0.\overline{9}$ is exactly equal to 1.

We have seen that the repeating block of $\dfrac{1}{7}$ is a cyclic number. Try to find more numbers (n) whose reciprocals $\left(\dfrac{1}{n}\right)$ produce decimals with repeating blocks that are cyclic.

Answer

Numbers n whose reciprocals $\dfrac{1}{n}$ produce fully cyclic repeating blocks are called full reptend primes. These are prime numbers p for which the period of $\dfrac{1}{p}$ has exactly p - 1 digits.

Examples of such numbers :

1. n = 7 : $\dfrac{1}{7} = 0.\overline{142857}$ (6-digit cyclic block).

2. n = 17 : $\dfrac{1}{17} = 0.\overline{0588235294117647}$ (16-digit cyclic block).

3. n = 19 : $\dfrac{1}{19} = 0.\overline{052631578947368421}$ (18-digit cyclic block).

4. n = 23 : $\dfrac{1}{23} = 0.\overline{0434782608695652173913}$ (22-digit cyclic block).

5. n = 29 : $\dfrac{1}{29}$ has a 28-digit cyclic block.

6. Other examples : 47, 59, 61, 97, 109, etc.

The cyclic property : If n is a full reptend prime, then multiplying the repeating block of $\dfrac{1}{n}$ by integers 1, 2, …, n - 1 gives cyclic permutations of the same digits.

For example, for n = 7 :

⇒ 142857 × 1 = 142857

⇒ 142857 × 2 = 285714

⇒ 142857 × 3 = 428571 (and so on, all permutations of the same digits).

Hence, prime numbers like 7, 17, 19, 23, 29, 47, 59, 61, 97, etc. (full reptend primes) produce reciprocals with cyclic repeating blocks.

Consider this puzzle: What is the square root of –1? We know that $1 \times 1 = 1$. We also know that $(-1) \times (-1) = 1$. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, $\sqrt{-1}$ cannot exist on number line.

Answer

It is true that $\sqrt{-1}$ cannot exist on the real number line.

Reason :

By Brahmagupta's laws :

⇒ A positive number multiplied by a positive number gives a positive number : (+) × (+) = (+).

⇒ A negative number multiplied by a negative number gives a positive number : (-) × (-) = (+).

So, the square of any real number (positive, negative or zero) is always non-negative. There is no real number whose square is -1.

Hence, $\sqrt{-1}$ has no place on the real number line.

To handle expressions like $\sqrt{-1}$, mathematicians invented a new kind of number called the Imaginary Unit, denoted by i, where :

⇒ i² = -1

⇒ i = $\sqrt{-1}$.

This led to the development of Imaginary Numbers, which extend the real number line into a new dimension and form the basis of Complex Numbers (numbers of the form a + bi, where a and b are real numbers).

Imaginary numbers are essential in modern electrical engineering, quantum mechanics, signal processing and several other fields.

Hence, $\sqrt{-1}$ does not exist on the real number line; it is denoted by i, the imaginary unit, which forms the foundation of complex numbers.

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) $\dfrac{3}{50}$

(ii) $\dfrac{2}{9}$

Answer

(i) Given, $\dfrac{3}{50}$

Prime factorisation of 50 = 2 × 5².

Since the denominator has only 2's and 5's, it has a terminating decimal expansion.

By long division :

$\begin{array}{l} \phantom{50\overline{)},}0.06 \\ 50\overline{\smash{\big)}\phantom{)}3.00} \\ \phantom{50\overline{}}\underline{-0}\phantom{.00} \\ \phantom{50\overline{)}1}30\phantom{.0} \\ \phantom{50\overline{)})}\underline{-0}\phantom{.0} \\ \phantom{50\overline{)}1}300 \\ \phantom{50\overline{,}}\underline{-300} \\ \phantom{50\overline{)},00}0 \end{array}$

⇒ $\dfrac{3}{50} = \dfrac{3 \times 2}{50 \times 2} = \dfrac{6}{100} = 0.06.$