Class 9 Maths Chapter 3 NCERT Solutions: The World of Numbers

3.1 THE DAWN OF MATHEMATICS: THE HUMAN NEED TO COUNT

EXERCISE SET 3.1

Fig. 3.1: Representation of the prime number tally groupings found on the Ishango bone

Question

1. 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?

Solution

We can solve this problem step-by-step using ratios or the unitary method.

Step 1: Understand the exchange rate given in the problem. The merchant gets $15$ copper ingots for a batch of $2$ bags of spices.
Step 2: Find out how many batches of $2$ bags are inside his total of $12$ bags. We divide the total bags by the batch size:

12 \div 2 = 6 b a t c h e s

Step 3: Since each batch earns him $15$ ingots, multiply the number of batches by $15$ :

6 \times 15 = 90 c o p p e r i n g o t s

Therefore, the merchant will leave the market with $90$ copper ingots.

Question

2. 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.

Solution

What they have in common: All of these numbers ( $11, 13, 17, 19$ ) are prime numbers lying between $10$ and $20$ . A prime number is a natural number greater than $1$ that cannot be divided evenly by any number other than $1$ and itself.
The next three numbers in the pattern: To continue this sequence of consecutive prime numbers after $19$ , we test the next natural numbers:
$20, 21, 22$ can all be divided by other numbers, so they are not prime.
$23$ can only be divided by $1$ and $23$ (First next prime).
$24, 25, 26, 27, 28$ can all be divided by other numbers.
$29$ can only be divided by $1$ and $29$ (Second next prime).
$30$ is divisible by $2, 3, 5,$ etc.
$31$ can only be divided by $1$ and $31$ (Third next prime).

Therefore, the next three numbers that fit this pattern are $23, 29,$ and $31$ .

Question

3. 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.

Solution

No, the set of Natural Numbers ( $ℕ = {1, 2, 3, 4, . . .}$ ) is not closed under subtraction.

What closure means: A set of numbers is "closed" under an operation (like subtraction) if performing that operation on any two numbers from the set always gives you an answer that is also a member of that same set.
Why it fails for Natural Numbers: When you subtract a smaller natural number from a larger one, the answer is a natural number (e.g., $10 - 3 = 7$ ). However, if you subtract a larger natural number from a smaller one, or subtract a number from itself, the answer falls outside the set of Natural Numbers!
Examples to justify this:
Example 1: Let us pick the natural numbers $4$ and $9$ . If we subtract $9$ from $4$ :

4 - 9 = - 5

The result, $- 5$ , is a negative integer. Since negative numbers do not exist in the set of Natural Numbers, closure fails.

Example 2: Let us pick the natural numbers $5$ and $5$ . If we subtract them:

5 - 5 = 0

The result is zero ( $0$ ). Since the Natural Numbers begin at $1$ , $0$ is not a natural number. Therefore, Natural Numbers are not closed under subtraction.

Question

4. 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?

Solution

How many you can count on one hand: Exclude the thumb (since the thumb acts as the pointer to touch the joints). You have $4$ fingers remaining on one hand (index, middle, ring, and little finger). Since each of these $4$ fingers has exactly $3$ segments or joints, we multiply them together:

4 f i n g e r s \times 3 j o i n t s e a c h = 12 j o i n t s

By touching the thumb to each joint sequentially, you can count up to $12$ on a single hand.

Relation to ancient base-12 systems: Today, we use a base-10 (decimal) system because humans have $10$ digits across both hands. Historically, many ancient cultures (including Vedic India and Mesopotamia) developed a base-12 (duodecimal) counting system precisely because of this finger-joint counting method. Because $12$ can be easily divided into halves ( $6$ ), thirds ( $4$ ), and quarters ( $3$ ), base-12 was exceptionally convenient for commerce, dividing time ( $24$ hours in a day, $12$ months in a year), and geometry!

3.3 INTEGERS: EXPANDING THE HORIZON

Fig. 3.2: Integers on the number line (illustrative)

Think and Reflect (Page 46)

Question

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) \times (- 4) = + 12

Solution

To understand why multiplying two negative numbers gives a positive answer, we can use Brahmagupta's ancient concepts of Fortunes (positive numbers) and Debts (negative numbers):

Let us say you owe ₹3 to a friend. In mathematical terms, your financial state is a debt of ₹3, which we write as $- 3$ .
Now, imagine a generous benefactor comes along and decides to cancel (take away) $4$ of these ₹3 debts that you owe.
In mathematics, "taking away" or removing something is represented by a negative sign ( $-$ ), and doing it $4$ times is written as $- 4$ .
When those debts are removed, you no longer have to pay out that money from your pocket. By having a ₹3 debt taken away $4$ times, your financial standing improves by $₹ 3 \times 4 = ₹ 12$ . You are effectively ₹12 richer!

Therefore, removing ( $-$ ) debts ( $-$ ) turns into a net fortune ( $+$ ):

(- 3) \times (- 4) = + 12

EXERCISE SET 3.2

Question

1. The temperature in the high-altitude desert of Ladakh is recorded as

4 \circ C

at noon. By midnight, it drops by

15 \circ C

. What is the midnight temperature?

Solution

Step 1: Identify the starting temperature at noon, which is $+ 4 \circ C$ .
Step 2: A "drop" in temperature means we must subtract from the starting value. The temperature drops by $15 \circ C$ .
Step 3: Write the mathematical equation using integers:

M i d n i g h t T e m p e r a t u r e = 4 - 15

Step 4: To subtract a larger number ( $15$ ) from a smaller number ( $4$ ), we find the difference between their absolute values ( $15 - 4 = 11$ ) and apply the negative sign because the drop is greater than the starting positive temperature:

4 - 15 = - 11 \circ C

Therefore, the midnight temperature in Ladakh is $- 11 \circ C$ .

Question

2. 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.

Solution

Using Brahmagupta's terminology, we represent financial gains (profits/fortunes) as positive integers and financial liabilities (loans/losses/debts) as negative integers:

Initial loan (debt) = $- 850$
Profit next day (fortune) = $+ 1200$
Loss following week (debt) = $- 450$

Step 1: Write the sequence as an integer equation:

F i n a l S t a n d i n g = (- 850) + 1200 + (- 450)

Step 2: Calculate step-by-step from left to right:
First, combine the initial debt with the profit:

(- 850) + 1200 = 1200 - 850 = + 350

(At this point, the trader has paid off the loan and has ₹350 left in his pocket).

Next, incorporate the loss of ₹450:

+ 350 + (- 450) = 350 - 450 = - 100

Therefore, the trader's final financial standing is $- 100$ , which means he is left with a net debt (or net loss) of ₹100.

Question

3. Calculate the following using Brahmagupta's laws:

(i)

(- 12) \times 5

(ii)

(- 8) \times (- 7)

(iii)

0 - (- 14)

(iv)

(- 20) \div 4

Solution

Let us solve each part step-by-step using Brahmagupta's arithmetic rules established in 628 CE:

(i) $(- 12) \times 5$
Rule: Brahmagupta stated that the product of a debt and a fortune is a debt (a negative number multiplied by a positive number gives a negative result).
Calculation: Multiply the numbers without signs first ( $12 \times 5 = 60$ ), then apply the negative sign:

(- 12) \times 5 = - 60

(ii) $(- 8) \times (- 7)$
Rule: Brahmagupta stated that the product of two debts is a fortune (a negative number multiplied by a negative number gives a positive result).
Calculation: $8 \times 7 = 56$ , and the two negative signs cancel out to become positive:

(- 8) \times (- 7) = + 56 (o r s i m p l y 56)

(iii) $0 - (- 14)$
Rule: Subtracting a debt (a negative number) is equivalent to adding a fortune (a positive number).
Calculation: The two consecutive minus signs turn into a plus sign:

0 - (- 14) = 0 + 14 = 14

(iv) $(- 20) \div 4$
Rule: Just like multiplication, dividing a debt by a positive number distributes that debt into equal parts, so the answer remains a debt (negative divided by positive is negative).
Calculation: Divide $20$ by $4$ to get $5$ , and keep the negative sign:

(- 20) \div 4 = - 5

Question

4. 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

Solution

Imagine you have a wallet containing a ₹10 currency note, but you also owe your local tea vendor ₹5 for tea you drank yesterday.

In mathematical terms, your current net worth is your cash minus your debt:

N e t W o r t h = ₹ 10 - ₹ 5 = ₹ 5

Now, imagine the tea vendor is feeling generous and says, "Don't worry about paying me back; I am cancelling your ₹5 debt!"
What just happened? The vendor subtracted your debt of ₹5. In our integer ledger, a debt of ₹5 is written as $- 5$ , and subtracting it is written as $- (- 5)$ .
Because you no longer have to hand over ₹5 from your wallet to pay the vendor, you get to keep your entire ₹10 currency note! In addition, compared to your old net worth of ₹5, your financial standing has jumped up by ₹5. Cancelling the debt had the exact same positive effect on your wealth as if someone had handed you an extra ₹5 note!

Therefore, taking away a debt is mathematically identical to receiving cash:

10 - (- 5) = 10 + 5 = 15

3.4 FILLING THE SPACES: FRACTIONS AND RATIONAL NUMBERS

Think and Reflect (Page 47)

Question

Can you explain why we need

𝑞 \neq 0

in the definition of a rational number?

Solution

A rational number is defined as any number that can be written in the form $𝑝 𝑞$ , where $𝑝$ and $𝑞$ are integers. The strict rule that the denominator $𝑞$ cannot equal zero ( $𝑞 \neq 0$ ) exists because division by zero is undefined in mathematics.

To understand why division by zero destroys mathematical logic, let us look at what division truly means:

When we say $123 = 4$ , it is because $4 \times 3 = 12$ . Division is simply the reverse of multiplication!
Now, imagine we allowed $𝑞$ to be $0$ , and we tried to evaluate $50$ . If $50$ equaled some number $𝑥$ , that would mean $𝑥 \times 0 = 5$ . But any number multiplied by zero always equals zero ( $𝑥 \times 0 = 0$ )! It is impossible to ever get $5$ . Thus, no answer exists.
What if we tried to evaluate $00$ ? If $00 = 𝑥$ , that would mean $𝑥 \times 0 = 0$ . Since every number multiplied by zero equals zero, $𝑥$ could be $1, 50, - 99,$ or anything else! We cannot assign a single, unique value to it.

To prevent these impossible and contradictory situations, mathematics strictly forbids dividing by zero. Therefore, $𝑞$ must never be $0$ in $𝑝 𝑞$ .

Think and Reflect (Page 49)

Question

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

Solution

To make different denominators equal before adding or subtracting, we follow a simple two-step process:

Step 1: Find the LCM (Least Common Multiple): Look at the different denominators and find their LCM. The LCM is the smallest positive number that can be divided cleanly by both denominators. This LCM will become our new, shared common denominator.
Step 2: Create Equivalent Fractions: For each fraction, ask: "By what integer do I need to multiply this denominator to turn it into the LCM?" Once you find that integer, multiply both the numerator and the denominator of the fraction by that exact same integer. This changes the appearance of the fractions so their bottom numbers match, without changing their mathematical value!

Example: To add $14 + 23$ :

The LCM of denominators $4$ and $3$ is $12$ .
Convert the first fraction: Multiply top and bottom by $3 \to 1 \times 3 4 \times 3 = 312$ .
Convert the second fraction: Multiply top and bottom by $4 \to 2 \times 4 3 \times 4 = 812$ .
Now that denominators are equal, simply add the numerators: $312 + 812 = 1112$ .

Question

2. Verify the distributive law for rational numbers.

Solution

The Distributive Law states that for any three rational numbers $𝑝, 𝑞,$ and $𝑟$ , multiplying a number by a sum is the same as multiplying each number separately and then adding the products:

𝑝 (𝑞 + 𝑟) = 𝑝 𝑞 + 𝑝 𝑟

Let us verify this law by picking three simple rational numbers: Let $𝑝 = 12$ , $𝑞 = 13$ , and $𝑟 = 14$ .

Step 1: Calculate the Left-Hand Side (LHS) $\to 𝑝 (𝑞 + 𝑟)$
First, solve inside the parentheses by finding a common denominator for $13$ and $14$ (LCM = $12$ ):

𝑞 + 𝑟 = 13 + 14 = 412 + 312 = 712

Now, multiply this sum by $𝑝$ :

L H S = 12 \times 712 = 1 \times 7 2 \times 12 = 724

Step 2: Calculate the Right-Hand Side (RHS) $\to 𝑝 𝑞 + 𝑝 𝑟$
First, multiply $𝑝 \times 𝑞$ :

𝑝 𝑞 = 12 \times 13 = 16

Next, multiply $𝑝 \times 𝑟$ :

𝑝 𝑟 = 12 \times 14 = 18

Now, add these two individual products together by finding a common denominator for $16$ and $18$ (LCM = $24$ ):

R H S = 16 + 18 = 424 + 324 = 724

Since the $L H S = 724$ and the $R H S = 724$ , we have successfully verified that $𝑝 (𝑞 + 𝑟) = 𝑝 𝑞 + 𝑝 𝑟$ . The distributive law holds true for rational numbers!

EXERCISE SET 3.3

Question

1. Prove that the following rational numbers are equal:

(i)

23

and

46

(ii)

54

and

108

(iii)

- 35

and

- 610

(iv)

93

and

3

Solution

We will use Brahmagupta's Equality Rule from the chapter: two rational numbers $𝑎 𝑏$ and $𝑐 𝑑$ are equal if and only if their cross-multiplied products are equal, meaning $𝑎 𝑑 = 𝑏 𝑐$ .

(i) $23$ and $46$
Here, $𝑎 = 2, 𝑏 = 3, 𝑐 = 4, 𝑑 = 6$ .
Cross-multiply: $𝑎 \times 𝑑 = 2 \times 6 = 12$
Cross-multiply: $𝑏 \times 𝑐 = 3 \times 4 = 12$
Since $12 = 12$ , we have proven that $23 = 46$ .

(ii) $54$ and $108$
Here, $𝑎 = 5, 𝑏 = 4, 𝑐 = 10, 𝑑 = 8$ .
Cross-multiply: $5 \times 8 = 40$
Cross-multiply: $4 \times 10 = 40$
Since $40 = 40$ , we have proven that $54 = 108$ .

(iii) $- 35$ and $- 610$
Keep the negative sign with the numerators: $𝑎 = - 3, 𝑏 = 5, 𝑐 = - 6, 𝑑 = 10$ .
Cross-multiply: $(- 3) \times 10 = - 30$
Cross-multiply: $5 \times (- 6) = - 30$
Since $- 30 = - 30$ , we have proven that $- 35 = - 610$ .

(iv) $93$ and $3$
Any integer can be written as a fraction over $1$ , so we write $3$ as $31$ . Now we compare $93$ and $31$ .
Cross-multiply: $9 \times 1 = 9$
Cross-multiply: $3 \times 3 = 9$
Since $9 = 9$ , we have proven that $93 = 3$ .

Question

2. Find the sum:

(i)

25 + 310

(ii)

712 + 58

(iii)

- 47 + 314

Solution

(i) $25 + 310$
The denominators are $5$ and $10$ . Their LCM is $10$ .
Convert $25$ to have a denominator of $10$ by multiplying top and bottom by $2$ :

2 \times 2 5 \times 2 = 410

Now add the numerators while keeping the common denominator:

410 + 310 = 4 + 3 10 = 710

(ii) $712 + 58$
The denominators are $12$ and $8$ . Their LCM is $24$ (since $12 \times 2 = 24$ and $8 \times 3 = 24$ ).
Convert both fractions to have a denominator of $24$ :

7 \times 2 12 \times 2 = 1424 a n d 5 \times 3 8 \times 3 = 1524

Add the numerators:

1424 + 1524 = 14 + 15 24 = 2924

(iii) $- 47 + 314$
The denominators are $7$ and $14$ . Their LCM is $14$ .
Convert $- 47$ by multiplying top and bottom by $2$ :

- 4 \times 2 7 \times 2 = - 814

Now add the numerators using integer addition rules:

- 814 + 314 = - 8 + 3 14 = - 514

Question

3. Find the difference:

(i)

56 - 14

(ii)

118 - 34

(iii)

- 79 - (- 23)

Solution

(i) $56 - 14$
Denominators are $6$ and $4$ . Their LCM is $12$ .
Convert equivalent fractions:

5 \times 2 6 \times 2 = 1012 a n d 1 \times 3 4 \times 3 = 312

Subtract the numerators:

1012 - 312 = 10 - 3 12 = 712

(ii) $118 - 34$
Denominators are $8$ and $4$ . Their LCM is $8$ .
Convert the second fraction:

3 \times 2 4 \times 2 = 68

Subtract:

118 - 68 = 11 - 6 8 = 58

(iii) $- 79 - (- 23)$
Remember that subtracting a negative fraction is the same as adding a positive fraction:

- 79 - (- 23) = - 79 + 23

The LCM of denominators $9$ and $3$ is $9$ . Convert $23$ :

2 \times 3 3 \times 3 = 69

Now combine the numerators:

- 79 + 69 = - 7 + 6 9 = - 19

Question

4. Find the product:

(i)

23 \times 310

(ii)

711 \times 58

(iii)

- 47 \times 514

Solution

To multiply rational numbers, multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator ( $𝑎 𝑏 \times 𝑐 𝑑 = 𝑎 𝑐 𝑏 𝑑$ ). Finally, simplify the fraction to its lowest terms if possible.

(i) $23 \times 310$
Multiply across:

2 \times 3 3 \times 10 = 630

Reduce to lowest terms by dividing both top and bottom by their greatest common factor ( $6$ ):

6 \div 6 30 \div 6 = 15

(Alternative method: You can cross-cancel the common factor of $3$ before multiplying: $23 \times 310 = 210 = 15$ ).

(ii) $711 \times 58$
Multiply across:

7 \times 5 11 \times 8 = 3588

(Since $35$ and $88$ share no common factors other than $1$ , this is already in simplest form).

(iii) $- 47 \times 514$
Multiply across (remembering that a negative times a positive gives a negative product):

- 4 \times 5 7 \times 14 = - 2098

Both numbers are even, so reduce by dividing top and bottom by $2$ :

- 20 \div 2 98 \div 2 = - 1049

Question

5. Find the quotient:

(i)

23 \div 310

(ii)

711 \div 58

(iii)

- 47 \div 514

Solution

To divide by a fraction, we flip the second fraction upside down (this is called taking the reciprocal) and change the division sign to multiplication ( $𝑎 𝑏 \div 𝑐 𝑑 = 𝑎 𝑏 \times 𝑑 𝑐$ ).

(i) $23 \div 310$
Flip $310$ to make it $103$ and multiply:

23 \times 103 = 2 \times 10 3 \times 3 = 209

(ii) $711 \div 58$
Flip $58$ to make it $85$ and multiply:

711 \times 85 = 7 \times 8 11 \times 5 = 5655

(iii) $- 47 \div 514$
Flip $514$ to make it $145$ and multiply:

- 47 \times 145 = - 4 \times 14 7 \times 5 = - 5635

Both $56$ and $35$ can be divided by $7$ , so we simplify:

- 56 \div 7 35 \div 7 = - 85

(Alternative method: Cross-cancel $14$ and $7$ before multiplying: $- 471 \times 1425 = - 85$ ).

Question

6. Show that:

(12 + 34) \times 83 = 12 \times 83 + 34 \times 83

Solution

This question asks us to prove that the equation is true by calculating the Left-Hand Side (LHS) and the Right-Hand Side (RHS) separately to see if they match. This is a practical test of the Distributive Property!

Step 1: Calculate the Left-Hand Side (LHS)

L H S = (12 + 34) \times 83

First, add the fractions inside the brackets. Convert $12$ to $24$ so denominators match:

12 + 34 = 24 + 34 = 54

Now multiply this result by $83$ :

L H S = 54 \times 83 = 4012

Reduce $4012$ by dividing top and bottom by $4$ :

L H S = 103

Step 2: Calculate the Right-Hand Side (RHS)

R H S = (12 \times 83) + (34 \times 83)

Solve the first multiplication block:

12 \times 83 = 86

Solve the second multiplication block:

34 \times 83 = 2412 = 126 (k e e p i n g d e n o m i n a t o r 6 f o r e a s y a d d i t i o n)

Now add the two blocks together:

R H S = 86 + 126 = 206

Reduce $206$ by dividing top and bottom by $2$ :

R H S = 103

Since $L H S = 103$ and $R H S = 103$ , we have shown that the two sides are exactly equal.

Question

7. Simplify the following using the distributive property:

79 (67 - 34)

Solution

The distributive property states that $𝑎 (𝑏 - 𝑐) = 𝑎 𝑏 - 𝑎 𝑐$ . Instead of subtracting inside the brackets first, we will multiply the outside number ( $79$ ) by each term inside the brackets:

Step 1: Distribute $79$ across both terms:

79 (67 - 34) = (79 \times 67) - (79 \times 34)

Step 2: Simplify each multiplication block separately:
First block: Notice that the $7$ in the numerator and the $7$ in the denominator cancel each other out!

79 \times 67 = 69 = 23

Second block: Multiply across:

79 \times 34 = 2136

Divide top and bottom by $3$ to reduce: $21 \div 3 36 \div 3 = 712$ .

Step 3: Subtract the simplified terms:

23 - 712

To subtract, we need a common denominator (LCM of $3$ and $12$ is $12$ ). Convert $23$ by multiplying top and bottom by $4$ :

2 \times 4 3 \times 4 = 812

Now perform the final subtraction:

812 - 712 = 112

Therefore, the simplified answer is $112$ .

Question

8. Find the rational number

𝑥

such that:

56 (𝑥 + 35) = 56 𝑥 + 12

Solution

Let us use basic algebra and the distributive property to simplify this equation and find $𝑥$ :

Step 1: Apply the distributive property on the left side of the equation:

(56 \times 𝑥) + (56 \times 35) = 56 𝑥 + 12

Step 2: Simplify the multiplication on the left side:
Notice how the $5$ s cancel out in $56 \times 35$ :

56 \times 35 = 36 = 12

Put this simplified fraction back into our equation:

56 𝑥 + 12 = 56 𝑥 + 12

Step 3: Analyze the result:

Look at both sides of the equation. The Left-Hand Side ( $56 𝑥 + 12$ ) is an exact mirror image of the Right-Hand Side ( $56 𝑥 + 12$ )! If we subtract $56 𝑥$ from both sides, we get:

12 = 12

Conclusion:

Because the variable $𝑥$ completely cancels out leaving a statement that is always true ( $12 = 12$ ), this equation is an algebraic identity. This means $𝑥$ can be ANY rational number whatsoever! Whether you plug in $𝑥 = 0$ , $𝑥 = 5$ , or $𝑥 = - 99100$ , the equation will always hold true.

Think and Reflect (Page 51)

Try and locate $- 74$ on a number line.

Solution

To locate $- 74$ accurately on a number line, we follow these clear steps:

Step 1: Understand the fraction's value: Since there is a negative sign, we will be moving to the left of the zero origin ( $0$ ). Notice that the top number ( $7$ ) is larger than the bottom number ( $4$ ). Let us convert it into a mixed fraction to see which integers it sits between:

- 74 = - 134

This tells us the number is located beyond $- 1$ , sitting between $- 1$ and $- 2$ .

Step 2: Prepare the number line: Draw a number line with an origin labeled $0$ . Mark the negative integers to the left: $- 1, - 2, - 3$ .
Step 3: Divide the intervals: Look at the denominator, which is $4$ . This tells us we must divide each unit interval (the space between $0$ and $- 1$ , between $- 1$ and $- 2$ , etc.) into $4$ equal sub-parts by drawing $3$ small tick marks between each integer.
Step 4: Count and mark the point: Starting from the origin $0$ , count exactly $7$ small tick marks to the left.
Step $4$ ticks left brings you to $- 44$ , which is exactly $- 1$ .
Step $3$ more ticks to the left brings you to $- 74$ .

This point lies exactly on the third tick mark between $- 1$ and $- 2$ (just one quarter-step before you hit $- 2$ ).

Think and Reflect (Page 52)

Question

Try to explain why the average of two rational numbers

𝑎

and

𝑏

, which equals

(𝑎 + 𝑏) 2

, is always a rational number between

𝑎

and

𝑏

Solution

There are two distinct mathematical facts we need to explain here: why the answer is rational, and why it is always trapped between the two original numbers.

Part 1: Why is the average always a rational number?
We know that rational numbers are closed under addition. This means if you take any two rational numbers $𝑎$ and $𝑏$ , their sum $(𝑎 + 𝑏)$ is guaranteed to be a rational number too.
We also know that rational numbers are closed under division (as long as you do not divide by zero). Since we are dividing the rational sum $(𝑎 + 𝑏)$ by the integer $2$ (which is not zero), the final quotient $𝑎 + 𝑏 2$ must be a rational number.

Part 2: Why does the average always lie strictly between $𝑎$ and $𝑏$ ?
Let us assume that $𝑎$ is the smaller number and $𝑏$ is the larger number, so $𝑎 < 𝑏$ .
Proof that the average is greater than $𝑎$ :

Take the inequality $𝑎 < 𝑏$ . If we add the same number $𝑎$ to both sides, the inequality remains true:

𝑎 + 𝑎 < 𝑎 + 𝑏 ⟹ 2 𝑎 < 𝑎 + 𝑏

Now, divide both sides by $2$ :

2 𝑎 2 < 𝑎 + 𝑏 2 ⟹ 𝑎 < 𝑎 + 𝑏 2

This proves the average is strictly greater than $𝑎$ .

Proof that the average is less than $𝑏$ :

Take our starting inequality $𝑎 < 𝑏$ again. This time, add $𝑏$ to both sides:

𝑎 + 𝑏 < 𝑏 + 𝑏 ⟹ 𝑎 + 𝑏 < 2 𝑏

Divide both sides by $2$ :

𝑎 + 𝑏 2 < 2 𝑏 2 ⟹ 𝑎 + 𝑏 2 < 𝑏

This proves the average is strictly less than $𝑏$ .

By combining both proven inequalities, we get the complete chain:

𝑎 < 𝑎 + 𝑏 2 < 𝑏

This proves that the average $𝑎 + 𝑏 2$ is securely sandwiched between $𝑎$ and $𝑏$ , proving the infinite density of rational numbers on the number line!

EXERCISE SET 3.4

Number line for Exercise Set 3.4, Question 1 (common denominator 12)

Question

1. Represent the rational numbers

23

- 54

, and

112

on a single number line.

Number line locating 2/3, −5/4 and 1½

Solution

To plot different fractions accurately on the same number line without guessing their positions, we must first convert them so they all share a common denominator.

Step 1: Convert all numbers to improper fractions with a common denominator:
Our numbers are $23$ , $- 54$ , and $112$ (which is $32$ ).
Look at the denominators: $3, 4,$ and $2$ . Their Least Common Multiple (LCM) is $12$ . Let us convert each fraction to an equivalent fraction with a denominator of $12$ :
$23 \times 44 = 𝟖 𝟏 𝟐$
$- 54 \times 33 = - 𝟏 𝟓 𝟏 𝟐$
$32 \times 66 = 𝟏 𝟖 𝟏 𝟐$ (which is equal to $1612$ )

Step 2: Prepare the number line:
Draw a long horizontal line and mark an origin labeled $0$ in the center.
Mark integer boundaries to the right ( $+ 1, + 2$ ) and to the left ( $- 1, - 2$ ).
Because our common denominator is $12$ , divide every single integer unit interval into $12$ equal sub-intervals by drawing $11$ small tick marks between each whole number. Every tick mark represents a step of $112$ .

Step 3: Locate and mark each point:
For $23$ (or $812$ ): Start at $0$ and count $8$ tick marks to the right. Place a solid dot here and label it $23$ .
For $- 54$ (or $- 1512$ ): Start at $0$ and move to the left. Moving $12$ ticks left brings you to $- 1$ . Count $3$ more tick marks past $- 1$ to the left (for a total of $15$ ticks left). Place a solid dot here and label it $- 54$ .
For $112$ (or $1812$ ): Start at $0$ and move to the right. Moving $12$ ticks right brings you to $+ 1$ . Count $6$ more tick marks past $+ 1$ to the right (exactly halfway between $+ 1$ and $+ 2$ ). Place a solid dot here and label it $112$ .

Question

2. Find three distinct rational numbers that lie strictly between

- 14

and

- 12

Solution

Step 1: Ensure denominators match:

We want numbers between $- 12$ (the smaller number) and $- 14$ (the larger number). Let us convert them to a common denominator of $4$ :

- 12 = - 24 a n d - 14

Right now, looking at the numerators $- 2$ and $- 1$ , there are no whole integers between them!

Step 2: Magnify the gap by expanding the fractions:

To create plenty of integer steps between the numerators, let us multiply both top and bottom of our fractions by a large number, say $10$ :

- 2 \times 10 4 \times 10 = - 𝟐 𝟎 𝟒 𝟎 a n d - 1 \times 10 4 \times 10 = - 𝟏 𝟎 𝟒 𝟎

Because we multiplied top and bottom by the same amount, their mathematical values remain completely unchanged.

Step 3: Pick three intermediate numerators:

Now we simply look at the integer numerators between $- 20$ and $- 10$ . We can freely choose any three distinct integers from this range (such as $- 19, - 18, - 17, . . ., - 11$ ). Let us pick $- 15, - 14,$ and $- 13$ :

O u r n u m b e r s a r e : - 1540, - 1440, a n d - 1340

Step 4: Simplify to lowest terms (optional but good practice):
$- 1540$ divides by $5 \to - 𝟑 𝟖$
$- 1440$ divides by $2 \to - 𝟕 𝟐 𝟎$
$- 1340$ cannot be reduced $\to - 𝟏 𝟑 𝟒 𝟎$

Therefore, three distinct rational numbers lying strictly between $- 14$ and $- 12$ are $- 38$ , $- 720$ , and $- 1340$ .

Question

3. Simplify the expression:

(- 14) + (512)

Solution

Step 1: Find a common denominator:

The denominators are $4$ and $12$ . Since $12$ can be cleanly divided by $4$ , the LCM is $12$ .

Step 2: Convert the first fraction:

Multiply the numerator and denominator of $- 14$ by $3$ :

- 1 \times 3 4 \times 3 = - 312

Step 3: Add the numerators:

Now combine the fractions:

- 312 + 512 = - 3 + 5 12 = 212

Step 4: Reduce to lowest terms:

Divide both top and bottom by their common factor of $2$ :

2 \div 2 12 \div 2 = 16

Therefore, the simplified value is $16$ .

Question

4. A tailor has

1534

metres of fine silk. If making one kurta requires

214

metres of silk, exactly how many kurtas can he make?

Solution

To find out how many kurtas can be made, we need to divide the total length of silk available by the length of silk required for a single kurta.

N u m b e r o f K u r t a s = T o t a l S i l k \div S i l k p e r K u r t a

Step 1: Convert both mixed fractions into improper fractions:
Total silk: $1534 = (15 \times 4) + 3 4 = 𝟔 𝟑 𝟒 m e t r e s$
Silk per kurta: $214 = (2 \times 4) + 1 4 = 𝟗 𝟒 m e t r e s$

Step 2: Perform the fraction division:

N u m b e r o f K u r t a s = 634 \div 94

Remember our rule for fraction division: flip the second fraction upside down (reciprocal) and multiply:

N u m b e r o f K u r t a s = 634 \times 49

Step 3: Simplify and multiply:

Notice that the $4$ in the numerator and the $4$ in the denominator completely cancel each other out!

634 \times 49 = 639

Now simply divide $63$ by $9$ :

63 \div 9 = 7

Therefore, the tailor can make exactly $7$ kurtas with no fabric left over.

Question

5. Find three rational numbers between

3.1415

and

3.1416

Solution

Step 1: Understand decimal places:

At first glance, $3.1415$ and $3.1416$ seem like consecutive numbers with no space between them. However, we can add trailing zeroes to the right of the decimal point without changing their mathematical value! Let us append a zero to both numbers:

3.1415 = 𝟑 . 𝟏 𝟒 𝟏 𝟓 𝟎 a n d 3.1416 = 𝟑 . 𝟏 𝟒 𝟏 𝟔 𝟎

Step 2: Pick intermediate decimal numbers:

Now it is easy to see the gap! We can pick any numbers ending in $51, 52, 53,$ etc. Let us choose three simple terminating decimals strictly between them:

𝟑 . 𝟏 𝟒 𝟏 𝟓 𝟏, 𝟑 . 𝟏 𝟒 𝟏 𝟓 𝟐, a n d 𝟑 . 𝟏 𝟒 𝟏 𝟓 𝟓

Step 3: Why are these rational?

Every terminating decimal can be written as a fraction over a power of $10$ . For example, $3.14151 = 314151100000$ , which is a ratio of integers with a non-zero denominator.

Therefore, three rational numbers between $3.1415$ and $3.1416$ are $3.14151$ , $3.14152$ , and $3.14155$ .

Question

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

Solution

Yes! Aside from the Averaging Method ( $𝑎 + 𝑏 2$ ) and the Common Denominator Expansion method we used in earlier problems, mathematicians use a clever and lightning-fast technique called the Farey Addition (or Mediant) Method:

How the Mediant Method works:

If you have two rational numbers written as positive fractions $𝑎 𝑏$ and $𝑐 𝑑$ (where $𝑎 𝑏 < 𝑐 𝑑$ ), you can form a new number between them by simply adding their numerators together and adding their denominators together:

M e d i a n t = 𝑎 + 𝑐 𝑏 + 𝑑

In normal fraction addition, adding straight across is a famous mistake. However, if your goal is solely to find an intermediate number lying trapped between two fractions, this "mistake" is actually a guaranteed mathematical theorem! The mediant $𝑎 + 𝑐 𝑏 + 𝑑$ will always be a valid rational number lying strictly between $𝑎 𝑏$ and $𝑐 𝑑$ .

Example to prove how well it works:

Let us find a rational number between $13$ and $23$ :

Using Farey Addition: Add top numbers ( $1 + 2 = 3$ ) and add bottom numbers ( $3 + 3 = 6$ ).
The resulting mediant is $36 = 12$ .
Since $13 \approx 0.333$ and $23 \approx 0.667$ , the number $12 (0.5)$ lies perfectly between them! This method allows you to generate endless rational numbers between any two fractions without ever calculating an LCM.

3.5 IRRATIONAL NUMBERS

Proof that √2 is irrational (textbook illustration)

Fig. 3.11: Constructing the length √2 on the number line

Think and Reflect (Page 53)

Question

Can

\sqrt 2

be written as a rational number?

Solution

No, $\sqrt 2$ cannot be written as a rational number. As you learned in Grade 8 and as the chapter proves using Hippasus's ancient method, it is mathematically impossible to find any two integers $𝑝$ and $𝑞$ such that their ratio $𝑝 𝑞$ equals $\sqrt 2$ . Because it defies fractional representation and leaves an unfillable gap among rational numbers on the number line, $\sqrt 2$ is classified as an Irrational Number.

Think and Reflect (Page 55)

Question

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

Solution

We will prove that $\sqrt 3$ is irrational by following the exact same logical steps Hippasus used for $\sqrt 2$ in Section 3.5.1.

Step 1: The Assumption (The Contradiction setup)

Assume the exact opposite of what we want to prove. Assume $\sqrt 3$ is a rational number. This means it can be written as a fraction in its simplest, lowest terms:

\sqrt 3 = 𝑝 𝑞 (𝑞 \neq 0)

(Important: "Simplest terms" means $𝑝$ and $𝑞$ are co-prime integers that share no common factors other than $1$ ).

Step 2: Square both sides of the equation

3 = 𝑝 2 𝑞 2

Step 3: Rearrange to clear the denominator

Multiply both sides by $𝑞 2$ :

3 𝑞 2 = 𝑝 2

Step 4: Make a deduction about $𝑝$

Since $𝑝 2$ is equal to $3$ multiplied by some integer ( $𝑞 2$ ), $𝑝 2$ must be a multiple of $3$ (divisible by $3$ ). A fundamental property of prime numbers states: if a prime number (like $3$ ) divides a square number ( $𝑝 2$ ), it must also divide the base number itself ( $𝑝$ ). Therefore, $𝑝$ is a multiple of $3$ . We can write $𝑝$ as:

𝑝 = 3 𝑘 (w h e r e 𝑘 i s s o m e i n t e g e r)

Step 5: Substitute $𝑝 = 3 𝑘$ back into our equation from Step 3

3 𝑞 2 = (3 𝑘) 2

3 𝑞 2 = 9 𝑘 2

Now, divide both sides of the equation by $3$ :

𝑞 2 = 3 𝑘 2

Step 6: Make a deduction about $𝑞$

Now we see that $𝑞 2$ is equal to $3$ multiplied by an integer ( $𝑘 2$ ). This means $𝑞 2$ is a multiple of $3$ . Following the same logic as before, if $𝑞 2$ is divisible by $3$ , then $𝑞$ must also be a multiple of $3$ !

Step 7: The Fatal Contradiction!

In Step $4$ , we proved that $𝑝$ is divisible by $3$ . In Step $6$ , we proved that $𝑞$ is divisible by $3$ . This means both $𝑝$ and $𝑞$ share a common factor of $3$ . However, in Step $1$ , we explicitly stated that $𝑝 𝑞$ was in its simplest form, sharing no common factors other than $1$ ! Because our step-by-step algebra was flawless, our very first assumption (" $\sqrt 3$ is rational") must be completely false. Therefore, $\sqrt 3$ is an irrational number.

Will this same approach work for $\sqrt 5$ , $\sqrt 7$ , or $\sqrt 10$ ?

Yes! This exact proof by contradiction works seamlessly for $\sqrt 5, \sqrt 7, \sqrt 10$ , and in fact for the square root of any positive integer that is not a perfect square. For instance, with $\sqrt 5$ , your equations would show that $𝑝 2 = 5 𝑞 2$ , proving both $𝑝$ and $𝑞$ are multiples of $5$ , creating the exact same fatal contradiction!

Think and Reflect (Page 55)

Question

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

Solution

We obtain line segments of precise irrational lengths using geometric construction based on the Baudhāyana-Pythagoras Theorem.

While you cannot measure an irrational length like $\sqrt 2$ using fractions on a standard ruler, geometry provides a perfect solution:

The Baudhāyana-Pythagoras Theorem states that in any right-angled triangle, the square of the hypotenuse ( $𝑐$ ) equals the sum of the squares of the two perpendicular base legs ( $𝑎$ and $𝑏$ ):

𝑐 2 = 𝑎 2 + 𝑏 2 ⟹ 𝑐 = \sqrt 𝑎 2 + 𝑏 2

If we deliberately draw a right-angled triangle where the two perpendicular legs have clean, integer lengths of $1$ unit each ( $𝑎 = 1, 𝑏 = 1$ ), let us see what happens to the hypotenuse:

𝑐 = \sqrt 12 + 12 = \sqrt 1 + 1 = \sqrt 2

By simply connecting the two ends of our integer legs, the resulting diagonal hypotenuse line is magically forced to have an exact, perfect length of $\sqrt 2$ units (as shown in Fig. 3.10)! We can then use a compass to transfer this exact irrational length down onto our straight number line.

Think and Reflect (Page 56)

Question

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.

Solution

Let us see how to construct $\sqrt 3$ and $\sqrt 5$ step-by-step on a number line, and then discover the universal rule for $\sqrt 𝑛$ .

Part 1: Constructing Length $\sqrt 3$

To get $\sqrt 3$ , we build a new right-angled triangle directly on top of the $\sqrt 2$ line segment we already created in Fig. 3.11:

Step 1: On your number line, locate the point labeled $𝑃$ which sits at distance $\sqrt 2$ from the origin $0$ (so segment $𝑂 𝑃 = \sqrt 2$ ).
Step 2: Use your ruler and compass to construct a vertical line segment exactly $1$ unit high, standing perpendicular to the number line at point $𝑃$ . Call the top of this vertical line point $𝐷$ .
Step 3: Connect the origin $𝑂$ to point $𝐷$ with a straight line. By the Baudhāyana-Pythagoras Theorem, the length of this new hypotenuse $𝑂 𝐷$ is:

𝑂 𝐷 = \sqrt (𝑂 𝑃) 2 + (𝑃 𝐷) 2 = \sqrt (\sqrt 2) 2 + 12 = \sqrt 2 + 1 = \sqrt 𝟑

Step 4: Place your compass needle at origin $𝑂$ , open the pencil out to point $𝐷$ (radius $\sqrt 3$ ), and swing an arc down to intersect the number line. That exact intersection point represents the irrational number $\sqrt 3$ .

Part 2: Constructing Length $\sqrt 5$ (The Shortcut!)

Instead of laboriously building $\sqrt 2, \sqrt 3,$ and $\sqrt 4$ first, we can get $\sqrt 5$ in a single, brilliant step by choosing smarter integer legs:

Notice that $5$ can be written as the sum of two integer squares: $5 = 4 + 1 = 22 + 12$ .
Step 1: On your number line, measure a base line segment from origin $𝑂$ to integer point $2$ (so base length = $2$ units).
Step 2: At point $2$ , draw a vertical perpendicular line segment exactly $1$ unit high. Call the top point $𝐸$ .
Step 3: Connect origin $𝑂$ to point $𝐸$ . By the theorem, the length of hypotenuse $𝑂 𝐸$ is:

𝑂 𝐸 = \sqrt 22 + 12 = \sqrt 4 + 1 = \sqrt 𝟓

Step 4: Swing an arc with your compass from origin $𝑂$ with radius $𝑂 𝐸$ down onto the number line to mark $\sqrt 5$ .

Part 3: Generalizing for any length $\sqrt 𝑛$

To construct a line segment of length $\sqrt 𝑛$ for any positive integer $𝑛$ , you can use the Sequential Step Method:

Assume you have already constructed a segment of length $\sqrt 𝑛 - 1$ along the base.
Draw a perpendicular line segment of length $1$ unit at the end of that base.
When you join the origin to the top of that perpendicular leg, the hypotenuse will always equal $\sqrt 𝑛$ :

H y p o t e n u s e = \sqrt (\sqrt 𝑛 - 1) 2 + 12 = \sqrt (𝑛 - 1) + 1 = \sqrt 𝐧

(Note: Whenever $𝑛$ can be broken down into the sum of two perfect squares like $𝑎 2 + 𝑏 2$ , you can bypass the sequence entirely and simply draw a right triangle with legs of length $𝑎$ and $𝑏$ !)

3.6 REAL NUMBERS: DECIMALS AND CYCLIC PATTERNS

Fig. 3.13: Classification of Real Numbers

Section 3.6.1 Rational Decimals: Terminating and Repeating

Example 2 (Page 57)

Question

38 = 0.375

. (Can you tell for which rational numbers the decimal will be terminating?)

Solution

A rational number written in its simplest form ( $𝑝 𝑞$ , where $𝑝$ and $𝑞$ share no common factors) will have a terminating decimal expansion if and only if the prime factorization of its denominator $𝑞$ contains only the prime numbers 2, 5, or both.

In mathematical notation, the denominator must take the form:

𝑞 = 2 𝑚 \times 5 𝑛 (w h e r e 𝑚 a n d 𝑛 a r e n o n - n e g a t i v e i n t e g e r s)

Why does this work for $38$ ? Look at the denominator $8$ . Its prime factorization is $8 = 2 \times 2 \times 2 = 23$ . Since it contains only powers of $2$ and no other prime numbers, it is guaranteed to terminate!

Think and Reflect (Page 57)

Question

Try to find the decimal expansions of

103

and

1112

. What do you observe about the repetition of the digits after the decimal point?

Solution

Let us perform long division for both fractions to observe how their decimal digits behave:

Part 1: Decimal expansion of $103$
Divide $10$ by $3$ : $10 \div 3 = 3$ with a remainder of $1$ .
Add a decimal point and bring down a zero ( $10$ ). Divide by $3$ again: $3$ times with a remainder of $1$ .
Because the remainder $1$ keeps appearing endlessly at every single step, the digit $3$ in the quotient repeats forever:

103 = 3.33333 . . . = 𝟑 . ―― 𝟑

Part 2: Decimal expansion of $1112$
Divide $11.0$ by $12$ : $110 \div 12 = 9$ (since $12 \times 9 = 108$ ), leaving a remainder of $2$ .
Bring down a zero ( $20$ ). Divide by $12$ : $20 \div 12 = 1$ (since $12 \times 1 = 12$ ), leaving a remainder of $8$ .
Bring down a zero ( $80$ ). Divide by $12$ : $80 \div 12 = 6$ (since $12 \times 6 = 72$ ), leaving a remainder of $8$ .
Bring down a zero ( $80$ ). Notice that the remainder $8$ has appeared again! From this point forward, the division loops endlessly, producing $6$ s forever:

1112 = 0.916666 . . . = 𝟎 . 𝟗 𝟏 ―― 𝟔

What do we observe about the repetition?

We observe two distinct styles of repeating patterns:

For $103 = 3 . ―― 3$ , the repetition begins immediately after the decimal point. This is called a pure repeating decimal.
For $1112 = 0.91 ―― 6$ , there are two initial digits ( $9$ and $1$ ) after the decimal point that do not repeat, and only after them does the digit $6$ begin repeating. This is called a general (or mixed) repeating decimal.

Think and Reflect (Page 58)

Question

The decimal expansion of

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?

Solution

The secret lies in our standard base-10 (decimal) number system.

When a decimal terminates (stops), such as $0.375$ or $0.15$ , what does it truly mean as a fraction?

0.375 = 3751000 a n d 0.15 = 15100

Any terminating decimal is simply a fraction whose denominator is an exact power of $10$ ( $10, 100, 1000, 104,$ etc.)!

Now, let us look at the prime building blocks of the number $10$ :

10 = 2 \times 5

Because $10$ is made entirely of $2$ s and $5$ s, any power of $10$ is also made exclusively of $2$ s and $5$ s ( $100 = 22 \times 52$ , $1000 = 23 \times 53$ , etc.).

Therefore, if a fraction's simplified denominator $𝑞$ contains only $2$ s and $5$ s, we can easily "complete the pairs" by multiplying both the top and bottom of the fraction by whatever matching powers of $2$ or $5$ are missing! This transforms the denominator into a perfect power of $10$ , causing the decimal to terminate cleanly.
What happens if any other prime number is present? If the denominator contains a prime factor like $3, 7,$ or $11$ , it is mathematically impossible to multiply it by any integer to turn it into a power of $10$ (since powers of $10$ can never be divided by $3, 7,$ or $11$ ). Because it can never reach a base of $10$ , the long division can never terminate and is forced to loop into a repeating decimal!

Worked Examples 4 to 9 (Pages 58–60): Full Step-by-Step Solutions

Here are the complete, step-by-step conversions for all six worked examples in the chapter, written out clearly using algebraic rules without shortcut omissions.

Example 4: Convert $0.35$ into the form $𝑝 𝑞$ .

Solution

This is a Terminating Decimal (it ends after two decimal places).

Step 1: Write the decimal as a fraction over a power of $10$ . Because there are two digits after the decimal point ( $3$ and $5$ ), the denominator will be $102 = 100$ :

0.35 = 35100

Step 2: Reduce the fraction to its lowest terms by finding the greatest common factor between $35$ and $100$ , which is $5$ . Divide both numerator and denominator by $5$ :

35 \div 5 100 \div 5 = 720

Therefore, $0.35$ in the form $𝑝 𝑞$ is $720$ .

Question

Example 5: Convert

0 . ―― 6

into the form

𝑝 𝑞

Solution

This is a Pure Repeating Decimal where one digit ( $6$ ) loops infinitely ( $0.6666 . . .$ ).

Step 1: Let our variable $𝑥$ equal the decimal number:

𝑥 = 0.6666 . . . - - - (E q u a t i o n 1)

Step 2: Count the number of repeating digits. Here, exactly $1$ digit repeats. Multiply both sides of Equation 1 by $101 = 10$ to shift one repeating block to the left of the decimal point:

10 𝑥 = 6.6666 . . . - - - (E q u a t i o n 2)

Step 3: Subtract Equation 1 from Equation 2. Notice how the infinite tail of decimal $6$ s matches perfectly on both equations, completely canceling out to zero!

10 𝑥 = 6.6666 . . . - 𝑥 = 0.6666 . . . 9 𝑥 = 6.0000 . . .

Step 4: Solve for $𝑥$ by dividing both sides by $9$ :

9 𝑥 = 6 ⟹ 𝑥 = 69

Step 5: Reduce $69$ to lowest terms by dividing top and bottom by $3$ :

𝑥 = 6 \div 3 9 \div 3 = 23

Therefore, $0 . ―― 6$ in the form $𝑝 𝑞$ is $23$ .

Question

Example 6: Convert

0 . ――― 45

into the form

𝑝 𝑞

Solution

This is a Pure Repeating Decimal where a block of two digits ( $45$ ) loops infinitely ( $0.454545 . . .$ ).

Step 1: Let $𝑥$ equal the decimal number:

𝑥 = 0.454545 . . . - - - (E q u a t i o n 1)

Step 2: Since $2$ digits repeat, multiply both sides by $102 = 100$ to shift one entire two-digit cycle to the left of the decimal point:

100 𝑥 = 45.454545 . . . - - - (E q u a t i o n 2)

Step 3: Subtract Equation 1 from Equation 2 to eliminate the infinite decimal tail:

100 𝑥 = 45.454545 . . . - 𝑥 = 0.454545 . . . 99 𝑥 = 45

Step 4: Solve for $𝑥$ :

𝑥 = 4599

Step 5: Reduce to lowest terms by dividing top and bottom by $9$ :

𝑥 = 45 \div 9 99 \div 9 = 511

Therefore, $0 . ――― 45$ in the form $𝑝 𝑞$ is $511$ .

Question

Example 7: Convert

0.1 ―― 6

into the form

𝑝 𝑞

Solution

This is a General (Mixed) Repeating Decimal ( $0.16666 . . .$ ) because the digit $1$ does not repeat, and only the digit $6$ repeats after it.

Step 1: Let $𝑥$ equal the decimal number:

𝑥 = 0.16666 . . . - - - (E q u a t i o n 1)

Step 2: First, we must shift the decimal point to place the repeating block immediately after the decimal point. Since there is $1$ non-repeating digit ( $1$ ), multiply Equation 1 by $101 = 10$ :

10 𝑥 = 1.6666 . . . - - - (E q u a t i o n 2)

Step 3: Next, we need a second equation where one full repeating cycle is shifted to the left of the decimal point. Since the repeating block is $1$ digit long ( $6$ ), multiply Equation 2 by $10$ again (which means multiplying the original $𝑥$ by $100$ ):

100 𝑥 = 16.6666 . . . - - - (E q u a t i o n 3)

Step 4: Subtract Equation 2 from Equation 3. Notice that both equations now have the exact same infinite tail ( $0.6666 . . .$ ), allowing them to cancel cleanly:

100 𝑥 = 16.6666 . . . - 10 𝑥 = 1.6666 . . . 90 𝑥 = 15

Step 5: Solve for $𝑥$ :

𝑥 = 1590

Step 6: Reduce to lowest terms by dividing top and bottom by $15$ :

𝑥 = 15 \div 15 90 \div 15 = 16

Therefore, $0.1 ―― 6$ in the form $𝑝 𝑞$ is $16$ .

Question

Example 8: Convert

2.35 ―― 7

into the form

𝑝 𝑞

Solution

This is a General Repeating Decimal ( $2.357777 . . .$ ) with integers before the decimal point, two non-repeating digits ( $35$ ), and one repeating digit ( $7$ ).

Step 1: Let $𝑥$ equal the decimal number:

𝑥 = 2.357777 . . . - - - (E q u a t i o n 1)

Step 2: Shift the $2$ non-repeating digits ( $35$ ) before the decimal point by multiplying Equation 1 by $102 = 100$ :

100 𝑥 = 235.7777 . . . - - - (E q u a t i o n 2)

Step 3: Shift one cycle of the $1$ repeating digit ( $7$ ) past the decimal point by multiplying Equation 2 by $10$ (which is $1000 \times 𝑥$ ):

1000 𝑥 = 2357.7777 . . . - - - (E q u a t i o n 3)

Step 4: Subtract Equation 2 from Equation 3 to eliminate the infinite repeating tail:

1000 𝑥 = 2357.7777 . . . - 100 𝑥 = 235.7777 . . . 900 𝑥 = 2122

Step 5: Solve for $𝑥$ :

𝑥 = 2122900

Step 6: Reduce to lowest terms by dividing top and bottom by $2$ (since both are even):

𝑥 = 2122 \div 2 900 \div 2 = 1061450

Therefore, $2.35 ―― 7$ in the form $𝑝 𝑞$ is $1061450$ .

Question

Example 9: Convert

2.45 ――― 37

into the form

𝑝 𝑞

Solution

This is a General Repeating Decimal ( $2.45373737 . . .$ ) with two non-repeating digits ( $45$ ) and a repeating block of two digits ( $37$ ).

Step 1: Let $𝑥$ equal the decimal number:

𝑥 = 2.45373737 . . . - - - (E q u a t i o n 1)

Step 2: Shift the $2$ non-repeating digits ( $45$ ) to the left of the decimal point by multiplying Equation 1 by $102 = 100$ :

100 𝑥 = 245.373737 . . . - - - (E q u a t i o n 2)

Step 3: Shift one full cycle of the $2$ repeating digits ( $37$ ) past the decimal point by multiplying Equation 2 by $102 = 100$ (which equals $10000 \times 𝑥$ ):

10000 𝑥 = 24537.373737 . . . - - - (E q u a t i o n 3)

Step 4: Subtract Equation 2 from Equation 3 to eliminate the infinite repeating decimal tail:

10000 𝑥 = 24537.373737 . . . - 100 𝑥 = 245.373737 . . . 9900 𝑥 = 24292

Step 5: Solve for $𝑥$ :

𝑥 = 242929900

Step 6: Reduce to lowest terms by dividing top and bottom by their common factor of $4$ :

𝑥 = 24292 \div 4 9900 \div 4 = 60732475

Therefore, $2.45 ――― 37$ in the form $𝑝 𝑞$ is $60732475$ .

EXERCISE SET 3.5

Question

1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating:

720, 415

and

13250

. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Solution

Part 1: Prediction using Prime Factorization

A rational number in lowest terms terminates if and only if the prime factorization of its denominator contains ONLY powers of $2$ and $5$ .

For $720$ : Look at denominator $20$ . Its prime factorization is $20 = 2 \times 2 \times 5 = 22 \times 5$ . Because it contains only primes $2$ and $5$ , it will be a Terminating Decimal.
For $415$ : Look at denominator $15$ . Its prime factorization is $15 = 3 \times 5$ . Because it contains the prime factor $3$ (which is not $2$ or $5$ ), it will be a Repeating Decimal.
For $13250$ : Look at denominator $250$ . Its prime factorization is $250 = 2 \times 5 \times 5 \times 5 = 2 \times 53$ . Because it contains only primes $2$ and $5$ , it will be a Terminating Decimal.

Part 2: Verification by Explicit Long Division

Let us perform actual long division to check our predictions:

Check $720$ :
Divide $7.00$ by $20$ :
$20$ goes into $70$ three times ( $20 \times 3 = 60$ ), remainder $10$ .
Bring down a zero to make $100$ . $20$ goes into $100$ five times ( $20 \times 5 = 100$ ), remainder $0$ .

720 = 𝟎 . 𝟑 𝟓 - - - V e r i f i e d : I t t e r m i n a t e s!

Check $415$ :
Divide $4.000$ by $15$ :
$15$ goes into $40$ two times ( $15 \times 2 = 30$ ), remainder $10$ .
Bring down zero ( $100$ ). $15$ goes into $100$ six times ( $15 \times 6 = 90$ ), remainder $10$ .
Bring down zero ( $100$ ). $15$ goes into $100$ six times ( $90$ ), remainder $10$ . The remainder $10$ loops endlessly!

415 = 0.2666 . . . = 𝟎 . 𝟐 ―― 𝟔 - - - V e r i f i e d : I t r e p e a t s!

Check $13250$ :
Divide $13.000$ by $250$ :
$250$ cannot divide $13$ or $130$ , so write $0.0$ .
Look at $1300$ . $250$ goes into $1300$ five times ( $250 \times 5 = 1250$ ), remainder $50$ .
Bring down zero ( $500$ ). $250$ goes into $500$ exactly two times ( $250 \times 2 = 500$ ), remainder $0$ .

13250 = 𝟎 . 𝟎 𝟓 𝟐 - - - V e r i f i e d : I t t e r m i n a t e s!

Question

2. Perform the long division for

113

. Identify the repeating block of digits. Does it show cyclic properties if you evaluate

213

? Now compute

313, 413

etc. What do you notice?

Solution

Step 1: Perform Long Division for $113$

Let us divide $1.0000000$ by $13$ :

$10 \div 13 = 0$ , remainder $10$
$100 \div 13 = 𝟕$ ( $13 \times 7 = 91$ ), remainder $9$
$90 \div 13 = 𝟔$ ( $13 \times 6 = 78$ ), remainder $12$
$120 \div 13 = 𝟗$ ( $13 \times 9 = 117$ ), remainder $3$
$30 \div 13 = 𝟐$ ( $13 \times 2 = 26$ ), remainder $4$
$40 \div 13 = 𝟑$ ( $13 \times 3 = 39$ ), remainder $1$
Notice that our remainder is now $1$ , which is the exact number we started with! From here, the entire 6-digit division cycle will repeat indefinitely.

113 = 0.076923076923 . . . = 𝟎 . ――――― 𝟎 𝟕 𝟔 𝟗 𝟐 𝟑

The repeating block of digits is 076923 (a 6-digit block).

Step 2: Evaluate $213$ and check for cyclic properties

213 = 0.153846153846 . . . = 𝟎 . ――――― 𝟏 𝟓 𝟑 𝟖 𝟒 𝟔

Let us divide $2$ by $13$ : Observation: Notice that the repeating block for $213$ is 153846. This is completely different from the block 076923! Unlike $17$ (where every single multiple was just a shifted circle of the same digits 142857), $213$ does not share the same cyclic circle as $113$ .

Step 3: Compute other multiples and state what we notice

Let us look at the decimal blocks for the first few multiples of $113$ :

$113 = 0 . ――――― 𝟎 𝟕 𝟔 𝟗 𝟐 𝟑$
$213 = 0 . ――――― 𝟏 𝟓 𝟑 𝟖 𝟒 𝟔$
$313 = 0 . ――――― 𝟐 𝟑 𝟎 𝟕 𝟔 𝟗$
$413 = 0 . ――――― 𝟑 𝟎 𝟕 𝟔 𝟗 𝟐$
$513 = 0 . ――――― 𝟑 𝟖 𝟒 𝟔 𝟏 𝟓$
$613 = 0 . ――――― 𝟒 𝟔 𝟏 𝟓 𝟑 𝟖$

Question

What do we notice?

We discover a fascinating internal pattern! The multiples of $113$ split into two distinct cyclic families (or groups):

Family 1 (The 076923 circle): The fractions $113, 313, 413, 913, 1013,$ and $1213$ all share the exact same cyclic block 076923, simply starting from different digit positions! (For example, look at $313 = 0 . ――――― 230769$ ; it is simply the digits $076923$ starting from the digit $2$ !).
Family 2 (The 153846 circle): The remaining fractions $213, 513, 613, 713, 813,$ and $1113$ all share the second cyclic block 153846, again shifting seamlessly in a circle!

Question

3. Classify the following numbers as rational or irrational:

(i)

\sqrt 81

(ii)

\sqrt 12

(iii)

0.33333 . . .

(iv)

0.123451234512345 . . .

(v)

1.01001000100001 . . .

(Notice the pattern: Is it repeating a single block?)

(vi)

23.560185612239874790120

Find the explicit fractions in case they are rational.

Solution

Let us classify each number using the golden rule of decimals: Rational numbers always terminate or repeat a fixed block; Irrational numbers never terminate and never repeat a fixed block.

(i) $\sqrt 81$
Since $81$ is a perfect square ( $9 \times 9 = 81$ ), we simplify: $\sqrt 81 = 9$ .
Classification: Rational Number.
Explicit fraction: $𝟗 𝟏$ .

(ii) $\sqrt 12$
$12$ is not a perfect square. We can simplify it to $\sqrt 4 \times 3 = 2 \sqrt 3$ . Because $\sqrt 3$ is irrational, multiplying it by $2$ leaves it irrational.
Classification: Irrational Number (Cannot be written as an explicit fraction).

(iii) $0.33333 . . .$
This is a non-terminating decimal, but notice that the single digit $3$ repeats endlessly ( $0 . ―― 3$ ). Any repeating decimal is rational.
Classification: Rational Number.
Explicit fraction: Let $𝑥 = 0 . ―― 3 ⟹ 10 𝑥 = 3 . ―― 3 ⟹ 9 𝑥 = 3 ⟹ 𝑥 = 𝟏 𝟑$ .

(iv) $0.123451234512345 . . .$
Notice that the 5-digit block 12345 repeats continuously ( $0 . ―――― 12345$ ).
Classification: Rational Number.
Explicit fraction: Let $𝑥 = 0 . ―――― 12345$ . Multiply by $105$ : $100000 𝑥 = 12345 . ―――― 12345$ . Subtract: $99999 𝑥 = 12345 ⟹ 𝑥 = 1234599999$ . Reduce by dividing top and bottom by their GCD $3$ : $𝟒 𝟏 𝟏 𝟓 𝟑 𝟑 𝟑 𝟑 𝟑$ .

(v) $1.01001000100001 . . .$
Look at the pattern: we have one zero between $1$ s, then two zeroes, then three zeroes, then four zeroes! Because the number of zeroes keeps growing forever, this decimal never settles into a single fixed repeating block. It is non-terminating and non-repeating.
Classification: Irrational Number (Cannot be written as a fraction).

(vi) $23.560185612239874790120$
While this looks like a long and chaotic string of digits, notice that there are no "dots" ( $. . .$ ) at the end! This means the decimal terminates (stops completely) after exactly $21$ decimal places. Every terminating decimal is rational.
Classification: Rational Number.
Explicit fraction: Write the entire number without the decimal point over $1021$ (a $1$ followed by twenty-one zeroes):

𝟐 𝟑 𝟓 𝟔 𝟎 𝟏 𝟖 𝟓 𝟔 𝟏 𝟐 𝟐 𝟑 𝟗 𝟖 𝟕 𝟒 𝟕 𝟗 𝟎 𝟏 𝟐 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎

Question

4. The number

0 . ―― 9

(which means

0.99999 . . .

) is a rational number. Using algebra (let

𝑥 = 0 . ―― 9

, multiply by 10, and subtract), explain why

0 . ―― 9

is exactly equal to 1.

Solution

Let us follow the standard algebraic conversion steps for repeating decimals to reveal this surprising mathematical truth:

Step 1: Set the variable $𝑥$ equal to the repeating decimal:

𝑥 = 0.99999 . . . - - - (E q u a t i o n 1)

Step 2: Since exactly $1$ digit ( $9$ ) repeats, multiply both sides of Equation 1 by $101 = 10$ :

10 𝑥 = 9.99999 . . . - - - (E q u a t i o n 2)

Step 3: Subtract Equation 1 from Equation 2. Look at the right side: the infinite tails of $9$ s cancel each other out completely:

10 𝑥 = 9.99999 . . . - 𝑥 = 0.99999 . . . 9 𝑥 = 9.00000 . . .

Step 4: Solve for $𝑥$ by dividing both sides by $9$ :

9 𝑥 = 9 ⟹ 𝑥 = 99 = 𝟏

Conclusion:

Since we started by defining $𝑥 = 0 . ―― 9$ and mathematically proved that $𝑥 = 1$ , it is an inescapable mathematical fact that $0 . ―― 9 = 1$ . They are not two different numbers where one is "slightly less" than the other; they are simply two different written representations of the exact same number on the Real number line!

Question

5. We have seen that the repeating block of

17

is a cyclic number. Try to find more numbers (

𝑛

) whose reciprocals (

1 𝑛

) produce decimals with repeating blocks that are cyclic.

Solution

A prime number $𝑝$ is called a full-repetend prime (or long prime) if its reciprocal $1 𝑝$ produces a repeating block of digits that has the maximum possible length, which is always $𝑝 - 1$ digits long.

Whenever a prime number has this maximum repeating length, that repeating block of digits is guaranteed to be a cyclic number! Just like $17$ produces a $6$ -digit cyclic circle ( $142857$ ), let us list the next few numbers $𝑛$ whose reciprocals generate cyclic numbers:

$𝑛 = 17$ (Generates a 16-digit cyclic number):

117 = 0 . ――――――――――― 𝟎 𝟓 𝟖 𝟖 𝟐 𝟑 𝟓 𝟐 𝟗 𝟒 𝟏 𝟏 𝟕 𝟔 𝟒 𝟕

If you multiply this 16-digit block by any number from $1$ to $16$ , you will get the exact same 16 digits shifting in a perfect circle!

$𝑛 = 19$ (Generates an 18-digit cyclic number):

119 = 0 . ―――――――――――― 𝟎 𝟓 𝟐 𝟔 𝟑 𝟏 𝟓 𝟕 𝟖 𝟗 𝟒 𝟕 𝟑 𝟔 𝟖 𝟒 𝟐 𝟏

$𝑛 = 23$ (Generates a 22-digit cyclic number):

123 = 0 . ――――――――――――――― 𝟎 𝟒 𝟑 𝟒 𝟕 𝟖 𝟐 𝟔 𝟎 𝟖 𝟔 𝟗 𝟓 𝟔 𝟓 𝟐 𝟏 𝟕 𝟑 𝟗 𝟏 𝟑

Other cyclic-generating prime numbers ( $𝑛$ ):

Following $23$ , the next numbers that create single cyclic circles are $29, 47, 59, 61, 97, 109, a n d 113$ .

3.7 CONCLUSION: THE NEVER-ENDING JOURNEY

Think and Reflect (Page 64)

Question

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 our Real number line.

Solution

This puzzle highlights the final conceptual frontier of numbers discussed in the chapter!

As Brahmagupta's laws prove, multiplying two positive numbers gives a positive result ( $1 \times 1 = 1$ ), and multiplying two negative numbers also gives a positive result ( $(- 1) \times (- 1) = + 1$ ). Because any real number squared is always non-negative ( $𝑥 2 \geq 0$ ), it is impossible to find a number on our 1-dimensional Real Number Line that squares to give $- 1$ .
To solve this dilemma and answer equations like $𝑥 2 = - 1$ , mathematicians realized they had to step completely off the real number line into a new mathematical dimension! They invented a new number denoted by the lowercase letter $𝑖$ (standing for Imaginary Unit), which is defined precisely by the property:

𝑖 = \sqrt - 1 m e a n i n g 𝑖 2 = - 1

When you combine Real numbers with Imaginary numbers (e.g., $3 + 4 𝑖$ ), you form the set of Complex Numbers. While the name "imaginary" sounds like fiction, these numbers are absolutely real in their utility; they form the mathematical backbone of modern electrical engineering, quantum mechanics, and the digital signal processing that powers mobile phones and Wi-Fi networks!

END-OF-CHAPTER EXERCISES

Question

1. 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)

350

(ii)

29

Solution

(i) $350$
Set up long division: divide $3.00$ by $50$ .
$50$ cannot go into $3$ or $30$ , so write $0.0$ in the quotient.
Look at $300$ : $50$ goes into $300$ exactly six times ( $50 \times 6 = 300$ ), leaving a remainder of $0$ .

350 = 𝟎 . 𝟎 𝟔

This is a Terminating Decimal.

(ii) $29$
Set up long division: divide $2.000$ by $9$ .
$9$ goes into $20$ two times ( $9 \times 2 = 18$ ), leaving a remainder of $2$ .
Bring down zero to make $20$ : $9$ goes into $20$ two times ( $18$ ), leaving a remainder of $2$ .
Because the remainder $2$ keeps appearing endlessly, the digit $2$ repeats forever:

29 = 0.2222 . . . = 𝟎 . ―― 𝟐

This is a Non-terminating and Repeating Decimal.

Question

2. Prove that

\sqrt 5

is an irrational number.

Solution

We will prove this using Proof by Contradiction, following the same logical structure used for $\sqrt 2$ and $\sqrt 3$ in the chapter.

Step 1: The Assumption

Assume that $\sqrt 5$ is a rational number. Therefore, it can be written as a simplified fraction of two co-prime integers $𝑝$ and $𝑞$ (meaning $𝑝$ and $𝑞$ share no common factors other than $1$ ):

\sqrt 5 = 𝑝 𝑞 (𝑞 \neq 0)

Step 2: Square both sides and clear denominator

5 = 𝑝 2 𝑞 2 ⟹ 5 𝑞 2 = 𝑝 2

Step 3: Deduce divisibility for $𝑝$

Since $𝑝 2$ equals $5$ multiplied by an integer ( $𝑞 2$ ), $𝑝 2$ is divisible by $5$ . By the fundamental theorem of arithmetic, if a prime number ( $5$ ) divides a square ( $𝑝 2$ ), it must also divide the base number ( $𝑝$ ). Therefore, $𝑝$ is divisible by $5$ . We can write $𝑝$ as:

𝑝 = 5 𝑘 (w h e r e 𝑘 i s a n i n t e g e r)

Step 4: Substitute and deduce divisibility for $𝑞$

Substitute $𝑝 = 5 𝑘$ back into our equation from Step 2:

5 𝑞 2 = (5 𝑘) 2 ⟹ 5 𝑞 2 = 25 𝑘 2

Divide both sides by $5$ :

𝑞 2 = 5 𝑘 2

Now we see that $𝑞 2$ is a multiple of $5$ . Therefore, following the exact same logic, $𝑞$ must also be divisible by $5$ !

Step 5: The Contradiction

We have deduced that $𝑝$ is divisible by $5$ and $𝑞$ is divisible by $5$ . This means both $𝑝$ and $𝑞$ share a common factor of $5$ . However, this contradicts our initial statement in Step 1 that $𝑝 𝑞$ was in simplest terms with no common factors other than $1$ ! Because this logical contradiction is unavoidable, our assumption that $\sqrt 5$ is rational must be false. Hence, $\sqrt 5$ is an irrational number.

Question

3. Convert the following decimal numbers in the form of

𝑝 𝑞

(i)

12.6

(ii)

0.0120

(iii)

3.0 ――― 52

(iv)

1.235

(v)

0 . ――― 23

(vi)

2.0 ―― 5

(vii)

2.125

(viii)

3.125

(ix)

2.1625

Solution

Let us convert each decimal into its simplest fractional form $𝑝 𝑞$ :

(i) $12.6$ (Terminating)

12.6 = 12610

Divide top and bottom by $2$ : $𝟔 𝟑 𝟓$ .

(ii) $0.0120$ (Terminating)

Drop the trailing zero ( $0.0120 = 0.012$ ). Write over $1000$ :

0.012 = 121000

Divide top and bottom by $4$ : $𝟑 𝟐 𝟓 𝟎$ .

(iii) $3.0 ――― 52$ (General Repeating)
Let $𝑥 = 3.0 ――― 52$ .
Shift $1$ non-repeating digit ( $0$ ) by multiplying by $10$ : $10 𝑥 = 30 . ――― 52$ .
Shift $2$ repeating digits ( $52$ ) by multiplying by $1000$ : $1000 𝑥 = 3052 . ――― 52$ .
Subtract: $1000 𝑥 - 10 𝑥 = 3052 - 30 ⟹ 990 𝑥 = 3022$ .
$𝑥 = 3022990$ . Reduce by dividing top and bottom by $2$ : $𝟏 𝟓 𝟏 𝟏 𝟒 𝟗 𝟓$ .

(iv) $1.235$ (Terminating)

1.235 = 12351000

Divide top and bottom by $5$ : $𝟐 𝟒 𝟕 𝟐 𝟎 𝟎$ .

(v) $0 . ――― 23$ (Pure Repeating)
Let $𝑥 = 0 . ――― 23$ .
Multiply by $100$ : $100 𝑥 = 23 . ――― 23$ .
Subtract: $100 𝑥 - 𝑥 = 23 ⟹ 99 𝑥 = 23$ .
Fraction: $𝟐 𝟑 𝟗 𝟗$ .

(vi) $2.0 ―― 5$ (General Repeating)
Let $𝑥 = 2.0 ―― 5$ .
Shift $1$ non-repeating digit ( $0$ ): $10 𝑥 = 20 . ―― 5$ .
Shift $1$ repeating digit ( $5$ ): $100 𝑥 = 205 . ―― 5$ .
Subtract: $100 𝑥 - 10 𝑥 = 205 - 20 ⟹ 90 𝑥 = 185$ .
$𝑥 = 18590$ . Reduce by dividing top and bottom by $5$ : $𝟑 𝟕 𝟏 𝟖$ .

(vii) $2.125$ (Terminating)

2.125 = 21251000

Divide top and bottom by $125$ : $𝟏 𝟕 𝟖$ .

(viii) $3.125$ (Terminating)

3.125 = 31251000

Divide top and bottom by $125$ : $𝟐 𝟓 𝟖$ .

(ix) $2.1625$ (Terminating)

2.1625 = 2162510000

Divide top and bottom by $625$ : $𝟏 𝟕 𝟑 𝟖 𝟎$ .

Question

4. Locate the following rational numbers on the number line.

(i)

0.532

(ii)

1.1 ―― 5

Solution

Because these decimals involve thousandths and infinite repetition, we locate them on the number line using Successive Magnification (zoom in step by step).

(i) Locating $0.532$
Step 1 (Tenths): The first decimal digit is $5$ , so $0.532$ lies between $0.5$ and $0.6$ .
Step 2 (Hundredths): Magnify $[0.5, 0.6]$ into $10$ equal parts. The second digit is $3$ , so the number lies between $0.53$ and $0.54$ .
Step 3 (Thousandths): Magnify $[0.53, 0.54]$ into $10$ equal parts. Count $2$ tick marks to the right of $0.530$ . That tick is exactly $0.532$ .

Number line for (i) 0.532 using successive magnification

(ii) Locating $1.1 ―― 5$ (that is, $1.1555 \dots$ )
Step 1 (Whole & Tenths): The number lies between $1$ and $2$ , specifically between $1.1$ and $1.2$ .
Step 2 (Hundredths): Magnify $[1.1, 1.2]$ into $10$ parts. The next digit is $5$ , so it lies between $1.15$ and $1.16$ .
Step 3 (Thousandths & beyond): Magnify $[1.15, 1.16]$ . Because $5$ keeps repeating, the point is at $1.1555 \dots$ — slightly more than halfway between $1.155$ and $1.156$ .

Number line for (ii) 1.1̅5 using successive magnification

Question

5. Find 6 rational numbers between 3 and 4.

Solution

To easily find $6$ numbers between any two consecutive integers, we use a simple formula: write both integers as fractions with a common denominator of $(𝑛 + 1)$ , where $𝑛$ is the number of fractions you want to find!

Here we want $𝑛 = 6$ numbers, so our target denominator is $6 + 1 = 𝟕$ .
Step 1: Convert both integers to equivalent fractions with denominator $7$ :

3 = 3 \times 7 1 \times 7 = 𝟐 𝟏 𝟕 a n d 4 = 4 \times 7 1 \times 7 = 𝟐 𝟖 𝟕

Step 2: List the integer numerators between $21$ and $28$ . There are exactly six of them:

𝟐 𝟐 𝟕, 𝟐 𝟑 𝟕, 𝟐 𝟒 𝟕, 𝟐 𝟓 𝟕, 𝟐 𝟔 𝟕, a n d 𝟐 𝟕 𝟕

These are six distinct rational numbers lying strictly between $3$ and $4$ .

Question

6. Find 5 rational numbers between

25

and

35

Solution

We want $𝑛 = 5$ numbers between two fractions that already share a denominator ( $5$ ). Let us expand their gap using our $(𝑛 + 1)$ rule: multiply top and bottom by $5 + 1 = 𝟔$ :

Step 1: Expand both fractions:

2 \times 6 5 \times 6 = 𝟏 𝟐 𝟑 𝟎 a n d 3 \times 6 5 \times 6 = 𝟏 𝟖 𝟑 𝟎

Step 2: List five consecutive numerators between $12$ and $18$ :

1330, 1430, 1530, 1630, a n d 1730

Step 3: Simplify to lowest terms where possible:

𝟏 𝟑 𝟑 𝟎, 𝟕 𝟏 𝟓, 𝟏 𝟐, 𝟖 𝟏 𝟓, a n d 𝟏 𝟕 𝟑 𝟎

Question

7. Find 5 rational numbers between

16

and

25

Solution

Step 1: Find a common denominator:

The LCM of denominators $6$ and $5$ is $30$ . Convert both fractions:

1 \times 5 6 \times 5 = 𝟓 𝟑 𝟎 a n d 2 \times 6 5 \times 6 = 𝟏 𝟐 𝟑 𝟎

Step 2: Check the integer gap:

Look at the numerators $5$ and $12$ . The whole integers lying strictly between them are $6, 7, 8, 9, 10,$ and $11$ . Since there are six integers available and we only need five numbers, we can simply pick the first five directly without needing any further expansion!

Step 3: Write and simplify five numbers:

630 = 𝟏 𝟓, 730 = 𝟕 𝟑 𝟎, 830 = 𝟒 𝟏 𝟓, 930 = 𝟑 𝟏 𝟎, a n d 1030 = 𝟏 𝟑

Question

8. If

𝑥 3 + 𝑥 5 = 1615

, find the rational number

𝑥

Solution

Let us solve this linear fraction equation step-by-step:

Step 1: Combine the fractions on the Left-Hand Side (LHS):

Find a common denominator for $𝑥 3$ and $𝑥 5$ (LCM is $15$ ):

5 𝑥 15 + 3 𝑥 15 = 1615

Add the numerators:

8 𝑥 15 = 1615

Step 2: Clear the denominators:

Since both sides of the equation are divided by the exact same denominator ( $15$ ), we can multiply both sides by $15$ to cancel it out completely:

8 𝑥 = 16

Step 3: Solve for $𝑥$ :

Divide both sides by $8$ :

𝑥 = 168 ⟹ 𝐱 = 𝟐

Therefore, the rational number $𝑥$ is $2$ .

Question

9. Let

𝑎

and

𝑏

be two non-zero rational numbers such that

𝑎 + 1 𝑏 = 0

. Without assigning any numerical values, determine whether

𝑎 𝑏

is positive or negative. Justify your answer.

Solution

We can determine the exact algebraic sign of the product $𝑎 𝑏$ by isolating variables:

Step 1: Start with the given algebraic equation:

𝑎 + 1 𝑏 = 0

Step 2: Subtract $1 𝑏$ from both sides of the equation to isolate $𝑎$ :

𝑎 = - 1 𝑏

Step 3: Since we are given that $𝑏$ is non-zero ( $𝑏 \neq 0$ ), we can safely multiply both sides of the equation by $𝑏$ :

𝑎 \times 𝑏 = - 1 𝑏 \times 𝑏 ⟹ 𝐚 𝐛 = - 𝟏

Conclusion and Justification:

Without plugging in a single numerical value for $𝑎$ or $𝑏$ , algebra proves that their product $𝑎 𝑏$ is permanently locked to the exact value of $- 1$ . Because $- 1$ is strictly less than zero ( $- 1 < 0$ ), the product $𝑎 𝑏$ is always negative.

Question

10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form

𝑝 104

where

𝑝

is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by

24

54

? Give reasons.

Solution

Let us break this conceptual proof down into two clear parts:

Part 1: Proving it can be written as $𝑝 104$ with $𝑝$ not divisible by 10

If a decimal terminates at exactly the 4th decimal place, it looks like:

N u m b e r = 𝑑 0 . 𝑑 1 𝑑 2 𝑑 3 𝑑 4 (w h e r e 𝑑 4 i s t h e 4 t h d i g i t a f t e r t h e d e c i m a l p o i n t)

Because we are explicitly told that the last non-zero digit occurs at the 4th place, we know for a fact that $𝑑 4 \neq 0$ .
When we convert any 4-decimal place number into a fraction, we remove the decimal point and place the entire integer over $104$ ( $10000$ ):

F r a c t i o n = 𝑑 0 𝑑 1 𝑑 2 𝑑 3 𝑑 4 10000 = 𝐩 𝟏 𝟎 𝟒

Here, $𝑝$ is the integer formed by all the digits ( $𝑑 0 𝑑 1 𝑑 2 𝑑 3 𝑑 4$ ).

Why is $𝑝$ not divisible by 10? An integer is divisible by $10$ if and only if its last digit is $0$ . Since the last digit of our integer $𝑝$ is $𝑑 4$ , and we know $𝑑 4 \neq 0$ , $𝑝$ cannot end in zero, and therefore is not divisible by 10.

Part 2: Is the lowest-form denominator necessarily divisible by $24$ or $54$ ?

Yes, it is absolutely necessary that the lowest-form denominator is divisible by either $24$ (which is $16$ ) or $54$ (which is $625$ ). Here is the reason:

Look at our prime-factored denominator:

104 = (2 \times 5) 4 = 𝟐 𝟒 \times 𝟓 𝟒

To reduce $𝑝 24 \times 54$ to its lowest form $𝑝 𝑞$ , we must cancel out any factors that $𝑝$ shares with the denominator ( $24 \times 54$ ).
Because we proved in Part 1 that $𝑝$ is not divisible by $10$ ( $2 \times 5$ ), $𝑝$ cannot be simultaneously divisible by both 2 and 5!
If $𝑝$ is even (divisible by $2$ ), it is NOT divisible by $5$ . Therefore, while we might cancel some or all of the $24$ term, the $54$ term in the denominator remains completely untouched.
If $𝑝$ is a multiple of $5$ , it is NOT even. Therefore, while we might cancel some of the $54$ term, the $24$ term in the denominator remains completely untouched.
If $𝑝$ is neither even nor a multiple of $5$ , then neither term cancels, and the denominator keeps both $24$ and $54$ .

Therefore, when reduced to lowest terms, the final denominator $𝑞$ is guaranteed to retain either the full $24$ factor ( $16$ ), the full $54$ factor ( $625$ ), or both!

Question

11. Without performing division, determine whether the decimal expansion of

18125

is terminating or non-terminating. If it terminates, state the number of decimal places.

Solution

Step 1: Simplify to lowest terms:

Check if $18$ and $125$ share any common factors. $18 = 2 \times 32$ , and $125 = 53$ . They share no common factors, so the fraction is already in lowest terms.

Step 2: Check prime factorization of the denominator:

The denominator is $125$ . Its prime factorization is:

125 = 𝟓 𝟑

Because the only prime factor present is $5$ (with no $3, 7,$ etc.), the decimal expansion is Terminating.

Step 3: Find the number of decimal places without dividing:

To terminate, we need to turn the denominator into an exact power of $10$ ( $10 𝑛$ ). Since we have $53$ , we multiply top and bottom by matching powers of two ( $23 = 8$ ):

18 \times 23 53 \times 23 = 18 \times 8 (5 \times 2) 3 = 144 𝟏 𝟎 𝟑 = 1441000

Because the denominator is $103$ ( $1000$ ), dividing by it shifts the decimal point exactly $3$ places to the left ( $0.144$ ). Therefore, it terminates after 3 decimal places. (Shortcut Rule: For any terminating fraction in lowest terms with denominator $2 𝑚 \times 5 𝑛$ , the number of decimal places is simply the higher exponent between $𝑚$ and $𝑛$ . Here, between $20$ and $53$ , the higher exponent is $3$ !)

Question

12. A rational number in its lowest form has denominator

23 \times 5

. How many decimal places will its decimal expansion have? Explain your answer.

Solution

The decimal expansion will have exactly 3 decimal places.

Explanation:

The number of decimal places in a terminating fraction equals the number of tens ( $10 𝑛$ ) required in the denominator to make a complete decimal transformation.
Our denominator is currently:

D e n o m i n a t o r = 23 \times 51

To create powers of $10$ ( $2 \times 5$ ), every $2$ needs a matching $5$ as a partner. Right now, we have three $2$ s ( $23$ ), but only one $5$ ( $51$ ).
To complete the pairs without changing the fraction's value, we must multiply both the top and bottom by $52$ ( $25$ ):

N e w D e n o m i n a t o r = (23 \times 51) \times 52 = 23 \times 53 = (2 \times 5) 3 = 𝟏 𝟎 𝟑

Because the smallest power of $10$ that can accommodate the denominator is $103$ ( $1000$ ), converting the fraction to a decimal will always result in dividing by $1000$ . Dividing any integer by $1000$ shifts the decimal point exactly three steps to the left, producing exactly $3$ decimal places.

Question

13. Let

𝑎 = 712

and

𝑏 = 56

. Express both

𝑎

and

𝑏

in the form

𝑘 1 𝑚

and

𝑘 2 𝑚

where

𝑘 1, 𝑘 2

and

𝑚

are integers and

𝑘 2 - 𝑘 1 > 6

. Using the same denominator

𝑚

, write exactly five distinct rational numbers lying between

𝑎

and

𝑏

keeping an integer numerator. Explain why the condition

𝑘 2 - 𝑘 1 > 𝑛 + 1

is necessary to find

𝑛

such rational numbers between the two rational numbers

𝑎

and

𝑏

using this method.

Solution

Let us solve this multi-part problem step-by-step:

Part 1: Express $𝑎$ and $𝑏$ with a shared denominator $𝑚$ such that $𝑘 2 - 𝑘 1 > 6$

Right now, $𝑎 = 712$ and $𝑏 = 56$ . If we use a common denominator of $12$ , we get:

𝑎 = 712 a n d 𝑏 = 1012

Here, the difference between numerators is $10 - 7 = 3$ , which is NOT greater than $6$ .

To make the gap larger than $6$ , let us multiply both top and bottom of these fractions by $3$ :

𝑎 = 7 \times 3 12 \times 3 = 𝟐 𝟏 𝟑 𝟔 a n d 𝑏 = 10 \times 3 12 \times 3 = 𝟑 𝟎 𝟑 𝟔

Let us check our values: our shared denominator is $𝑚 = 36$ , $𝑘 1 = 21$ , and $𝑘 2 = 30$ . The difference between numerators is:

𝑘 2 - 𝑘 1 = 30 - 21 = 𝟗

Since $9 > 6$ , we have successfully met the condition!

Part 2: Write five distinct rational numbers between $𝑎$ and $𝑏$ using denominator $𝑚 = 36$

𝟐 𝟐 𝟑 𝟔, 𝟐 𝟑 𝟑 𝟔, 𝟐 𝟒 𝟑 𝟔, 𝟐 𝟓 𝟑 𝟔, a n d 𝟐 𝟔 𝟑 𝟔

Now we simply pick five consecutive integer numerators lying between $21$ and $30$ , keeping the denominator $36$ :

Part 3: Why is the condition $𝑘 2 - 𝑘 1 > 𝑛 + 1$ (or $\geq 𝑛 + 1$ ) necessary to find $𝑛$ numbers?

Imagine you want to fit exactly $𝑛$ distinct integer steps strictly between two boundary integers $𝑘 1$ and $𝑘 2$ .
The intermediate integers after $𝑘 1$ would be:

1 s t n u m b e r : 𝑘 1 + 1, 2 n d n u m b e r : 𝑘 1 + 2, \dots, 𝑛 - t h n u m b e r : 𝑘 1 + 𝑛

For this final, $𝑛$ -th integer ( $𝑘 1 + 𝑛$ ) to still remain strictly inside the boundary and not crash into the upper limit $𝑘 2$ , it must be strictly less than $𝑘 2$ :

𝑘 1 + 𝑛 < 𝑘 2

Rearranging this inequality by subtracting $𝑘 1$ from both sides gives:

𝐤 𝟐 - 𝐤 𝟏 > 𝐧

Because $𝑘 1$ and $𝑘 2$ are whole integers, stating that their difference is strictly greater than $𝑛$ is mathematically identical to saying the difference must be at least $𝑛 + 1$ ( $𝑘 2 - 𝑘 1 \geq 𝑛 + 1$ ). Without this minimum gap of $(𝑛 + 1)$ integer steps between the numerators, there simply would not be enough physical room on the number line to fit $𝑛$ distinct integer numerators between them!

Question

14. Three rational numbers

𝑥, 𝑦, 𝑧

satisfy

𝑥 + 𝑦 + 𝑧 = 0

and

𝑥 𝑦 + 𝑦 𝑧 + 𝑧 𝑥 = 0

. Show that all the rational numbers

𝑥, 𝑦, 𝑧

must be simultaneously zero.

Solution

We can prove this elegantly using a famous algebraic identity:

Step 1: State the identity for the square of a trinomial:

(𝑥 + 𝑦 + 𝑧) 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 + 2 (𝑥 𝑦 + 𝑦 𝑧 + 𝑧 𝑥)

Step 2: Substitute our given zero values into the identity:

We are given that $(𝑥 + 𝑦 + 𝑧) = 0$ and $(𝑥 𝑦 + 𝑦 𝑧 + 𝑧 𝑥) = 0$ . Plug these zeroes into our formula:

(0) 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 + 2 (0)

0 = 𝑥 2 + 𝑦 2 + 𝑧 2

Step 3: Analyze the properties of squared rational numbers:

In the set of rational numbers (and all real numbers), any number multiplied by itself is always positive or zero. A square number can never be negative!

𝑥 2 \geq 0, 𝑦 2 \geq 0, a n d 𝑧 2 \geq 0

Step 4: The Logical Conclusion:

How can you add three positive or non-negative numbers together ( $𝑥 2 + 𝑦 2 + 𝑧 2$ ) and get a total sum of exactly $0$ ? If even one of the variables (say $𝑥$ ) was anything other than zero (like $0.5$ or $- 3$ ), its square ( $𝑥 2$ ) would be strictly positive ( $+ 0.25$ or $+ 9$ ). Because none of the other squared terms can ever be negative to cancel it out, the total sum would be greater than zero! Therefore, the only mathematical way the sum of three squares can equal zero is if every single square is individually equal to zero:

𝑥 2 = 0 ⟹ 𝐱 = 𝟎, 𝑦 2 = 0 ⟹ 𝐲 = 𝟎, a n d 𝑧 2 = 0 ⟹ 𝐳 = 𝟎

Thus, we have proven that $𝑥, 𝑦,$ and $𝑧$ must all be simultaneously zero.

Question

15. Show that the rational number

(𝑎 + 𝑏) 2

lies between the rational numbers

𝑎

and

𝑏

Solution

(Note: This is the formal algebraic proof for the same concept we explored in Think and Reflect on page 52).

Let us assume without loss of generality that $𝑎$ is the smaller number and $𝑏$ is the larger number, meaning $𝑎 < 𝑏$ . We must prove two things: that the average is greater than $𝑎$ , and less than $𝑏$ .

Part 1: Prove that $𝑎 + 𝑏 2 > 𝑎$
Start with our true statement:

𝑎 < 𝑏

Add the number $𝑎$ to both sides of the inequality:

𝑎 + 𝑎 < 𝑎 + 𝑏 ⟹ 2 𝑎 < 𝑎 + 𝑏

Divide both sides by the positive number $2$ :

2 𝑎 2 < 𝑎 + 𝑏 2 ⟹ 𝐚 < 𝐚 + 𝐛 𝟐

This proves the average sits above the lower boundary $𝑎$ .

Part 2: Prove that $𝑎 + 𝑏 2 < 𝑏$
Start again with our true statement:

𝑎 < 𝑏

This time, add the number $𝑏$ to both sides of the inequality:

𝑎 + 𝑏 < 𝑏 + 𝑏 ⟹ 𝑎 + 𝑏 < 2 𝑏

Divide both sides by $2$ :

𝑎 + 𝑏 2 < 2 𝑏 2 ⟹ 𝐚 + 𝐛 𝟐 < 𝐛

This proves the average sits below the upper boundary $𝑏$ .

Final Conclusion:

By combining the two proven inequalities, we get the complete statement:

𝐚 < 𝐚 + 𝐛 𝟐 < 𝐛

This rigorously proves that the average $𝑎 + 𝑏 2$ always lies trapped directly between $𝑎$ and $𝑏$ .

Question

16. Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

Solution

The Square Root Spiral (also known as the Spiral of Theodorus) is constructed by taking a right-angled triangle, and then using its hypotenuse as the base leg for a brand new right triangle, adding a perpendicular leg of $1$ unit each time!

Let us use the Baudhāyana-Pythagoras Theorem ( $H y p o t e n u s e = \sqrt B a s e 2 + H e i g h t 2$ ) to calculate the length of every hypotenuse in sequence from the center outward:

Triangle 1 (Innermost): Both base leg and perpendicular height are $1$ unit.

H y p o t e n u s e ℎ 1 = \sqrt 12 + 12 = \sqrt 1 + 1 = \sqrt 𝟐

Triangle 2: The base leg is now $\sqrt 2$ , and height is $1$ .

H y p o t e n u s e ℎ 2 = \sqrt (\sqrt 2) 2 + 12 = \sqrt 2 + 1 = \sqrt 𝟑

Triangle 3: The base leg is $\sqrt 3$ , and height is $1$ .

H y p o t e n u s e ℎ 3 = \sqrt (\sqrt 3) 2 + 12 = \sqrt 3 + 1 = \sqrt 4 = 𝟐

Triangle 4: The base leg is $\sqrt 4$ (or $2$ ), and height is $1$ .

H y p o t e n u s e ℎ 4 = \sqrt (\sqrt 4) 2 + 12 = \sqrt 4 + 1 = \sqrt 𝟓

Triangle 5: The base leg is $\sqrt 5$ , and height is $1$ .

H y p o t e n u s e ℎ 5 = \sqrt (\sqrt 5) 2 + 12 = \sqrt 5 + 1 = \sqrt 𝟔

Triangle 6: The base leg is $\sqrt 6$ , and height is $1$ .

H y p o t e n u s e ℎ 6 = \sqrt (\sqrt 6) 2 + 12 = \sqrt 6 + 1 = \sqrt 𝟕

Triangle 7: The base leg is $\sqrt 7$ , and height is $1$ .

H y p o t e n u s e ℎ 7 = \sqrt (\sqrt 7) 2 + 12 = \sqrt 7 + 1 = \sqrt 8 = 𝟐 \sqrt 𝟐

Triangle 8: The base leg is $\sqrt 8$ , and height is $1$ .

H y p o t e n u s e ℎ 8 = \sqrt (\sqrt 8) 2 + 12 = \sqrt 8 + 1 = \sqrt 9 = 𝟑

Triangle 9: The base leg is $\sqrt 9$ (or $3$ ), and height is $1$ .

H y p o t e n u s e ℎ 9 = \sqrt (\sqrt 9) 2 + 12 = \sqrt 9 + 1 = \sqrt 𝟏 𝟎

Triangle 10: The base leg is $\sqrt 10$ , and height is $1$ .

H y p o t e n u s e ℎ 10 = \sqrt (\sqrt 10) 2 + 12 = \sqrt 10 + 1 = \sqrt 𝟏 𝟏

Triangle 11 (Outermost in Fig. 3.14): The base leg is $\sqrt 11$ , and height is $1$ .

H y p o t e n u s e ℎ 11 = \sqrt (\sqrt 11) 2 + 12 = \sqrt 11 + 1 = \sqrt 12 = 𝟐 \sqrt 𝟑

Summary of Hypotenuse Lengths: The lengths form a beautiful, perfect sequence of square roots of consecutive natural numbers:

\sqrt 𝟐, \sqrt 𝟑, 𝟐, \sqrt 𝟓, \sqrt 𝟔, \sqrt 𝟕, \sqrt 𝟖, 𝟑, \sqrt 𝟏 𝟎, \sqrt 𝟏 𝟏, a n d \sqrt 𝟏 𝟐

3.1 THE DAWN OF MATHEMATICS: THE HUMAN NEED TO COUNT

EXERCISE SET 3.1

3.3 INTEGERS: EXPANDING THE HORIZON

Think and Reflect (Page 46)

EXERCISE SET 3.2

3.4 FILLING THE SPACES: FRACTIONS AND RATIONAL NUMBERS

Think and Reflect (Page 47)

Think and Reflect (Page 49)

EXERCISE SET 3.3

Think and Reflect (Page 51)

Think and Reflect (Page 52)

EXERCISE SET 3.4

3.5 IRRATIONAL NUMBERS

Think and Reflect (Page 53)

Think and Reflect (Page 55)

Think and Reflect (Page 55)

Think and Reflect (Page 56)

Part 1: Constructing Length √3

Part 2: Constructing Length √5 (The Shortcut!)

Part 3: Generalizing for any length √𝑛

3.6 REAL NUMBERS: DECIMALS AND CYCLIC PATTERNS

Section 3.6.1 Rational Decimals: Terminating and Repeating

Example 2 (Page 57)

Think and Reflect (Page 57)

Think and Reflect (Page 58)

Worked Examples 4 to 9 (Pages 58–60): Full Step-by-Step Solutions

EXERCISE SET 3.5

Part 1: Prediction using Prime Factorization

Part 2: Verification by Explicit Long Division

Step 1: Perform Long Division for 113

Step 2: Evaluate 213 and check for cyclic properties

Step 3: Compute other multiples and state what we notice

3.7 CONCLUSION: THE NEVER-ENDING JOURNEY

Think and Reflect (Page 64)

END-OF-CHAPTER EXERCISES

Part 1: Proving it can be written as 𝑝104 with 𝑝 not divisible by 10

Part 2: Is the lowest-form denominator necessarily divisible by 24 or 54?

Part 1: Express 𝑎 and 𝑏 with a shared denominator 𝑚 such that 𝑘2 −𝑘1 >6

Part 2: Write five distinct rational numbers between 𝑎 and 𝑏 using denominator 𝑚 =36

Part 3: Why is the condition 𝑘2 −𝑘1 >𝑛 +1 (or ≥𝑛 +1) necessary to find 𝑛 numbers?

Part 1: Constructing Length $\sqrt 3$

Part 2: Constructing Length $\sqrt 5$ (The Shortcut!)

Part 3: Generalizing for any length $\sqrt 𝑛$

Step 1: Perform Long Division for $113$

Step 2: Evaluate $213$ and check for cyclic properties

Part 1: Proving it can be written as $𝑝 104$ with $𝑝$ not divisible by 10

Part 2: Is the lowest-form denominator necessarily divisible by $24$ or $54$ ?

Part 1: Express $𝑎$ and $𝑏$ with a shared denominator $𝑚$ such that $𝑘 2 - 𝑘 1 > 6$

Part 2: Write five distinct rational numbers between $𝑎$ and $𝑏$ using denominator $𝑚 = 36$

Part 3: Why is the condition $𝑘 2 - 𝑘 1 > 𝑛 + 1$ (or $\geq 𝑛 + 1$ ) necessary to find $𝑛$ numbers?