Practise Class 9 Maths Part 1 Chapter 3 – The World of Numbers with this Ganita Manjari Chapter 3 Worksheet. This worksheet includes important questions and revision exercises to strengthen understanding of numbers, improve mathematical practice and support effective preparation for class tests and examinations.
📘 The World of Numbers
Grade 9 Mathematics | Ganita Manjari | Part I
Easy Notes • Concepts • Complete Book Solutions
🌟 Easy Notes & Concepts
1. Natural Numbers
Natural Numbers are the basic counting numbers.
N = {1, 2, 3, 4, 5, …}
They are used for counting objects.
2. Zero — Śhūnya
Zero represents the absence of a quantity.
a + 0 = a
a − 0 = a
a × 0 = 0
a − 0 = a
a × 0 = 0
Brahmagupta gave rules for arithmetic involving zero.
3. Integers
Integers include positive numbers, negative numbers and zero.
Z = {…, −3, −2, −1, 0, 1, 2, 3, …}
Think of:
- Positive numbers → fortunes
- Negative numbers → debts
- Zero → neither fortune nor debt
4. Rational Numbers
A rational number is a number that can be written in the form:
p/q where p,q are integers and q ≠ 0
Examples:
- 1/2
- −3/5
- 7
- 0
- 0.25
- 0.333…
5. Irrational Numbers
Irrational numbers cannot be written as p/q.
Examples:
√2 √5 π
Their decimal expansions continue forever without repeating a fixed block.
6. Real Numbers
Real Numbers contain both rational and irrational numbers.
Real Numbers = Rational Numbers + Irrational Numbers
7. Decimal Expansion
| Decimal | Number Type |
|---|---|
| 0.25 | Rational — terminating |
| 0.333… | Rational — repeating |
| 0.142857142857… | Rational — repeating |
| 1.414213562… | Irrational |
8. Quick Rule for Terminating Decimals
If a rational number is in lowest form and its denominator has only the prime factors 2 and/or 5, its decimal expansion terminates.
Denominator = 2m × 5n
9. Dense Rational Numbers
There are infinitely many rational numbers between any two different rational numbers.
A very easy way to find one is to take their average:
(a + b) / 2
10. Number System in One Picture
🧠 Think and Reflect
Question
Book p. 47
Can you explain why we need q ≠ 0 in the definition of a rational number?
We cannot divide by zero. Therefore q cannot be 0 in p/q.
Question
Book p. 46
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.)
A negative sign can represent removing something.If we remove four debts of ₹3 each, we remove a total debt of ₹12.So we become ₹12 richer.Therefore:
(−3) × (−4) = +12
Question
Book p. 48
While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?
Find the LCM of the denominators and convert both fractions into equivalent fractions with that common denominator.
Question
Book p. 48
Verify the distributive law for rational numbers.
Take a simple example:
1/2 × (1/3 + 1/6)
1/3 + 1/6 = 1/2
So LHS = 1/2 × 1/2 = 1/4.
RHS = 1/2×1/3 + 1/2×1/6
= 1/6 + 1/12
= 1/4.
LHS = RHS. Therefore the distributive law is verified.
Question
Book p. 52
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.
Since a and b are rational, a+b is rational. Dividing by 2 still gives a rational number.Also, the average of two different numbers lies halfway between them.Therefore (a+b)/2 is a rational number between a and b.
Question
Book p. 53
Can √2 be written as a rational number p/q?
No. √2 is irrational and cannot be expressed as p/q where p and q are integers and q ≠ 0.
Question
Book p. 55
Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?
Yes. The same proof-by-contradiction idea works for √3, √5, √7 and √10.Assume the square root is rational, write it as p/q in lowest form, square both sides, and show that p and q would have a common factor. This contradicts the assumption that p/q was in lowest form.
Question
Book p. 55
We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
A right-angled triangle can be used.Take two perpendicular sides of length 1 unit. By the Baudhāyana–Pythagoras Theorem, the hypotenuse is √2 units.That gives a line segment of irrational length.
Question
Book p. 57
Try to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?
10/3 = 3.333…
11/12 = 0.91666…
In both cases, the decimal continues and a digit or block of digits repeats.
11/12 = 0.91666…
In both cases, the decimal continues and a digit or block of digits repeats.
Question
Book p. 58
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?
Because powers of 10 contain only the prime factors 2 and 5:
10 = 2 × 5
So a denominator containing only 2s and 5s can be changed into a power of 10. Then the decimal terminates.Question
Book p. 64
Consider this puzzle: What is the square root of –1? We know that 1 × 1 = 1. We also know that (–1) × (–1) = 1. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, −1 cannot exist on number line.
There is no real number whose square is −1.Mathematicians introduce a new number i such that:
i² = −1
These are called imaginary numbers.📗 Exercise Set 3.1
1.
Book p. 43
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?
2 bags → 15 ingots
12 bags = 6 × 2 bags
So ingots = 6 × 15 = 90
✅ Answer: 90 copper ingots.
2.
Book p. 43
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.
11, 13, 17 and 19 are prime numbers.The next three prime numbers are:23, 29 and 31.
3.
Book p. 43
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.
No. Natural Numbers are not closed under subtraction.Example:5 − 2 = 3 → Natural Number.But:2 − 5 = −3 → Not a Natural Number.Therefore subtraction does not always give a Natural Number.
*4.
Book p. 43
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?
There are 4 fingers other than the thumb.
Each finger has 3 joints.
4 × 3 = 12
So one hand can count 12.This connects with the ancient base-12 counting system.
📗 Exercise Set 3.2
1.
Book p. 46
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?
4 − 15 = −11
✅ Midnight temperature = −11 °C.
2.
Book p. 46
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.
−850 + 1200 − 450 = −100
The trader has a final debt of ₹100.
3.
Book p. 46
Calculate the following using Brahmagupta’s laws:
(i) (–12) × 5
(ii) (–8) × (–7)
(iii) 0 – (–14)
(iv) (–20) ÷ 4
(i) (−12) × 5 = −60
(ii) (−8) × (−7) = 56
(iii) 0 − (−14) = 14
(iv) (−20) ÷ 4 = −5
(ii) (−8) × (−7) = 56
(iii) 0 − (−14) = 14
(iv) (−20) ÷ 4 = −5
4.
Book p. 46
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).
Suppose you have a debt of ₹10.If a debt of ₹5 is removed, your debt becomes ₹5 less.So subtracting a negative amount is like adding a positive amount.
10 − (−5) = 10 + 5 = 15
📗 Exercise Set 3.3
1.
Book p. 49
Prove that the following rational numbers are equal:
(i) 2/3 and 4/6
(ii) 5/4 and 10/8
(iii) −3/5 and −6/10
(iv) 9/3 and 3/2
(i)2/3 = 2×2 / 3×2 = 4/6 ✓
(ii)5/4 = 5×2 / 4×2 = 10/8 ✓
(iii)−3/5 = −3×2 / 5×2 = −6/10 ✓
(iv)As printed in the uploaded PDF:9/3 = 3but3/2 = 1.5Therefore these two are not equal as written.
(ii)5/4 = 5×2 / 4×2 = 10/8 ✓
(iii)−3/5 = −3×2 / 5×2 = −6/10 ✓
(iv)As printed in the uploaded PDF:9/3 = 3but3/2 = 1.5Therefore these two are not equal as written.
The question has been kept unchanged. The solution does not alter the printed question.
2.
Book p. 49
Find the sum:
(i) 2/5 + 3/10
(ii) 7/12 + 5/8
(iii) −4/7 + 3/14
(i)2/5 = 4/10Therefore:4/10 + 3/10 = 7/10
(ii)LCM of 12 and 8 = 24.7/12 = 14/245/8 = 15/24Therefore:14/24 + 15/24 = 29/24
(iii)−4/7 = −8/14−8/14 + 3/14 = −5/14
(ii)LCM of 12 and 8 = 24.7/12 = 14/245/8 = 15/24Therefore:14/24 + 15/24 = 29/24
(iii)−4/7 = −8/14−8/14 + 3/14 = −5/14
3.
Book p. 49
Find the difference:
(i) 5/6 − 1/4
(ii) 11/8 − 3/4
(iii) −7/9 − (−2/3)
(i)LCM = 1210/12 − 3/12 = 7/12
(ii)3/4 = 6/811/8 − 6/8 = 5/8
(iii)−7/9 − (−2/3)= −7/9 + 2/3= −7/9 + 6/9= −1/9
(ii)3/4 = 6/811/8 − 6/8 = 5/8
(iii)−7/9 − (−2/3)= −7/9 + 2/3= −7/9 + 6/9= −1/9
4.
Book p. 49
Find the product:
(i) 2/3 × 3/10
(ii) 7/11 × 5/8
(iii) −4/7 × 5/14
(i)2/3 × 3/10 = 1/5
(ii)7/11 × 5/8 = 35/88
(iii)−4/7 × 5/14= −20/98= −10/49
(ii)7/11 × 5/8 = 35/88
(iii)−4/7 × 5/14= −20/98= −10/49
5.
Book p. 49
Find the quotient:
(i) 2/3 ÷ 3/10
(ii) 7/11 ÷ 5/8
(iii) −4/7 ÷ 5/14
Rule: To divide fractions, multiply by the reciprocal.
(i)2/3 × 10/3 = 20/9
(ii)7/11 × 8/5 = 56/55
(iii)−4/7 × 14/5 = −8/5
(i)2/3 × 10/3 = 20/9
(ii)7/11 × 8/5 = 56/55
(iii)−4/7 × 14/5 = −8/5
6.
Book p. 50
Show that: (1 2/3 + 3/4) × 8/3 = 1 2/3 × 8/3 + 3/4 × 8/3.
1 2/3 = 5/3
LHS = (5/3 + 3/4) × 8/3
5/3 + 3/4 = 20/12 + 9/12 = 29/12
LHS = 29/12 × 8/3 = 58/9
RHS = 5/3×8/3 + 3/4×8/3
= 40/9 + 2 = 58/9
LHS = RHS. Hence proved.
7.
Book p. 50
Simplify the following using the distributive property: 7/9 − (6/7 − 3/4).
6/7 − 3/4 = 24/28 − 21/28 = 3/28
7/9 − 3/28
LCM = 252
7/9 = 196/252
3/28 = 27/252
196/252 − 27/252 = 169/252
8.
Book p. 50
Find the rational number x such that: 5/6 (3/5 + x) = 5/6 + 1/2.
5/6(3/5+x)=5/6+1/2
5/6+1/2 = 5/6+3/6 = 8/6 = 4/3
3/5+x = 4/3 × 6/5 = 8/5
x = 8/5 − 3/5 = 1
✅ x = 1
📗 Exercise Set 3.4
1.
Book p. 52
Represent the rational numbers 2/3, –5/4 and 11/2 on a single number line.
2/3 lies between 0 and 1.
−5/4 = −1 1/4 lies between −2 and −1.
11/2 = 5 1/2 lies between 5 and 6.
−5/4 = −1 1/4 lies between −2 and −1.
11/2 = 5 1/2 lies between 5 and 6.
2.
Book p. 52
Find three distinct rational numbers that lie strictly between –1/2 and 1/4.
One possible set is:−1/4, 0, 1/8.All three lie strictly between −1/2 and 1/4.
3.
Book p. 53
Simplify the expression: −1/4 + 5/12.
LCM of 4 and 12 = 12.
−1/4 = −3/12.
−3/12 + 5/12 = 2/12.
✅ 1/6
4.
Book p. 53
A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can he make?
15 3/4 = 63/4
2 1/4 = 9/4
63/4 ÷ 9/4 = 63/9 = 7
✅ The tailor can make 7 kurtas.
5.
Book p. 53
Find three rational numbers between 3.1415 and 3.1416.
Possible answers:3.14151, 3.14155, 3.14159.
*6.
Book p. 53
Can you think of other way(s) to find a rational number between any two rational numbers?
Yes.Apart from taking the average, we can write both numbers with a common denominator and choose a numerator between their numerators.There are actually infinitely many rational numbers between any two different rational numbers.
📗 Exercise Set 3.5
1.
Book p. 61
Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
| Number | Decimal | Type |
|---|---|---|
| 7/20 | 0.35 | Terminating |
| 4/15 | 0.2666… | Repeating |
| 13/250 | 0.052 | Terminating |
20 = 2²×5 and 250 = 2×5³, so those decimals terminate. 15 contains the prime factor 3, so 4/15 repeats.
2.
Book p. 62
Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?
| Fraction | Decimal |
|---|---|
| 1/13 | 0.076923076923… |
| 2/13 | 0.153846153846… |
| 3/13 | 0.230769230769… |
| 4/13 | 0.307692307692… |
| 5/13 | 0.384615384615… |
| 6/13 | 0.461538461538… |
The decimal expansions repeat. The digits show a cyclic pattern as the numerator changes.
3.
Book p. 62
Classify the following numbers as rational or irrational:
(i) √81
(ii) √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.
| Number | Answer |
|---|---|
| (i) √81 | 9 → Rational |
| (ii) √12 | Irrational |
| (iii) 0.33333… | Rational = 1/3 |
| (iv) 0.123451234512345… | Rational = 4115/33333 |
| (v) 1.01001000100001… | Irrational |
| (vi) 23.560185612239874790120 | Rational — terminating decimal |
For (vi), because the decimal terminates, it can be written over a suitable power of 10 and reduced.
4.
Book p. 62
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.
Let x = 0.999…
10x = 9.999…
Subtract:
10x − x = 9.999… − 0.999…
9x = 9
Therefore:x = 1So:0.999… = 1.
*5.
Book p. 62
We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.
Examples include denominators such as:7, 13, 17 and 19.Their reciprocals have repeating decimal blocks with interesting cyclic patterns.
🏁 End-of-Chapter Exercises
1.
Book p. 64
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) 3/50 (ii) 2/9
(i)3/50 = 0.06→ terminating.
(ii)2/9 = 0.2222…→ non-terminating and repeating.
(ii)2/9 = 0.2222…→ non-terminating and repeating.
2.
Book p. 64
Prove that √5 is an irrational number.
Assume √5 is rational.
Then √5 = p/q, where p and q have no common factor and q ≠ 0.
Squaring:5 = p²/q²
Therefore:p² = 5q²
So p² is divisible by 5, which means p is divisible by 5.
Let p = 5k.
Substitution gives q² = 5k².
Therefore q is also divisible by 5.
So p and q have a common factor 5. This contradicts our assumption that p/q was in lowest form.Therefore:√5 is irrational.
3.
Book p. 64
Convert the following decimal numbers in the form of p/q.
(i) 12.6 (ii) 0.0120 (iii) 3.052
(iv) 1.235 (v) 0.23 (vi) 2.05
(vii) 2.125 (viii) 3.125 (ix) 2.1625
| Decimal | Fraction |
|---|---|
| 12.6 | 63/5 |
| 0.0120 | 3/250 |
| 3.052 | 763/250 |
| 1.235 | 247/200 |
| 0.23 | 23/100 |
| 2.05 | 41/20 |
| 2.125 | 17/8 |
| 3.125 | 25/8 |
| 2.1625 | 173/80 |
4.
Book p. 64
Locate the following rational numbers on the number line.
(i) 0.532
(ii) 1.15
0.532 lies between 0 and 1.
1.15 lies between 1 and 2.
1.15 lies between 1 and 2.
5.
Book p. 65
Find 6 rational numbers between 3 and 4.
One possible answer:
3.1, 3.2, 3.3, 3.4, 3.5, 3.6
6.
Book p. 65
Find 5 rational numbers between 2/5 and 3/5.
2/5 = 12/30
3/5 = 18/30
Therefore five possible numbers are:
13/30, 14/30, 15/30, 16/30, 17/30
7.
Book p. 65
Find 5 rational numbers between1/6 and 2/5.
1/6 = 5/30
2/5 = 12/30
Five possible answers:
6/30, 7/30, 8/30, 9/30, 10/30
8.
Book p. 65
If x/3 + x/5 = 16/15, find the rational number x.
x/3 + x/5 = 16/15
5x/15 + 3x/15 = 16/15
8x/15 = 16/15
x = 2
9.
Book p. 65
Let a and b be two non-zero rational numbers such that a/b + 1 = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
a/b + 1 = 0
a/b = −1
a = −b
ab = −b²
Since b² is positive for non-zero b, −b² is negative.Therefore ab is negative.
10.
Book p. 65
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 or 5⁴? Give reasons.
A terminating decimal with four decimal places can be written over 10⁴.
For example:0.1234 = 1234/10000
No. It is not necessary that the lowest-form denominator contain 2⁴ or 5⁴.Some factors may cancel when the fraction is reduced.
11.
Book p. 65
Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.
125 = 5³.
The denominator contains only the prime factor 5.
Therefore the decimal terminates.18/125 = 0.144.So it has 3 decimal places.
12.
Book p. 65
A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Denominator = 2³ × 5.
Multiply by 5² to make the denominator 10³.
2³ × 5 × 5² = 2³ × 5³ = 10³
Therefore the decimal expansion will have at most 3 decimal places.
*13.
Book p. 65
Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ – k₁ > 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 k₂ – k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
a = 7/12
b = 5/6 = 10/12
Multiply both by 3:a = 21/36b = 30/36
Five possible rational numbers are:
22/36, 23/36, 24/36, 25/36, 26/36
We need enough integers strictly between k₁ and k₂. To find n such numbers, the difference k₂−k₁ must be greater than n+1.
*14.
Book p. 65
Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
From x+y+z=0:z = −(x+y)
Substitute this in xy+yz+zx=0.
We get:x² + xy + y² = 0
Multiply by 4:(2x+y)² + 3y² = 0
Both terms are non-negative. Their sum can be zero only when both are zero.
Therefore y=0, x=0 and hence z=0.So:x = y = z = 0.
*15.
Book p. 66
Show that the rational number (a + b)/2 lies between the rational numbers a and b.
Assume a < b.
Then:2a < a+b < 2b
Divide by 2:a < (a+b)/2 < b
Therefore (a+b)/2 lies between a and b.
16.
Book p. 66
Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
Using the Baudhāyana–Pythagoras Theorem, the successive hypotenuses in the square root spiral are:
√2, √3, √4, √5, √6, …
Each new triangle adds one unit side to the previous hypotenuse.Picture idea: each new right triangle has one side of length 1 unit.
⭐ Quick Revision
Number System
N ⊂ Z ⊂ Q ⊂ R
- N → Natural Numbers
- Z → Integers
- Q → Rational Numbers
- I → Irrational Numbers
- R → Real Numbers
Remember These Rules
- Never divide by zero.
- Negative × negative = positive.
- Negative × positive = negative.
- Rational numbers are dense.
- There are infinitely many rational numbers between two different rational numbers.
- Terminating decimals have denominators involving only 2 and/or 5 after reduction.
- Repeating decimals are rational.
- Non-terminating non-repeating decimals are irrational.
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