
As we know, we use
1,2,3,4,.... when we begin to count.

They come naturally when we start
counting.

Therefore, we call the counting numbers as natural numbers.
 Allnatural numbers are whole numbers, but all whole numbers are not natural numbers.
Natural Number:
The counting numbers 1,2, 3,4, are called natural numbers.
Whole Number:
0 is the smallest whole number. Whole numbers are 0, 1, 2, 3, ………
Properties of Whole Numbers:
 By Using Addition and multiplication of any 2 whole number give a whole number.
 But By using Subtraction and division of any 2 whole number may or may not give a whole number.
Number Line
 Draw a line.
 Mark a point on it and label it 0.
 Mark a second point to the right of 0 at the certain proper distance and label it 1.
 Then, the distance between the points labeled as 0 and 1 is called the unit distance. Now, mark another point on this line to the right of 1 at a unit distance from 1 and label it 2.
 Proceeding in this manner, we may find consecutive points and label them as 3, 4, 5,… in order.
 Thus, we can go to any whole number.
 This line is called the number line for whole numbers.
Use of Number Line
Addition
By using Number Line, we can add like this:
Let us add 4 and 3. We start from 4 on the number line and make 3 jumps to the right by unit distance each. We reach 5. So, 4 + 3 = 7.
Subtraction
By Using Number Line, we can subtract like this:
Let us find 4  3. We start from 4 on the number line and make 3 jumps to the left by unit distance each. We reach 1. So, 4 – 3 = 1.
Multiplication
By Using Number Line, multiplication is done like this:
Let us find 3 × 3. We start from 0 on the number line and move 3 units to the right at a time. We make 3 such moves. We reach 9. So. 3 × 3 = 9.
Division
By using Number line
For example 6 ÷ 2 = 3. Start from 6 and subtract 2 for a number of times till 0 is reached. The number of times 2 is subtracted gives the quotient.
Predecessor:
If we subtract 1 from a natural number, what we get is its predecessor.
For example, the predecessor of 11 is 11 – 1 = 10.
Successor:
If we add 1 to a natural number, what we get is its successor.
For example, the successor of 10 is 10 + 1 = 11
Properties of Number Line
Closure Property:
Whole numbers are closed under addition and also under multiplication. If a and b are any two whole numbers, then a+b, axb are also whole numbers.
Ex:
0 + 2 = 2 and 1 + 3 = 4
0 x 2 = 0 and 1 x 4 = 4
Commutative Property:
If a and b are any two whole numbers, then a + b = b + a and a × b = b × a
Ex:
0 + 3 = 3 + 0 = 3
0 × 2 = 0 or 2 × 0 = 0
Associate Property:
If a, b and c are any two whole numbers, then (a+b)+c = a+(b+c) and (a×b)×c = a×(b×c).
Ex:
0 + (2 + 3) = (0 + 2) + 3 = 5
0 × (2 × 3) = 0 or (0 × 2) × 3 = 0
10 − (2 − 1) = 9 but (10 − 2 ) − 1 = 7
Distributive Property:
If a, b and c are any two whole numbers, then a(b+c)=a × b + a × c
Ex:
2(3+4) = 2 × 3 + 2 × 4
Additive Identity:
If a is any whole number, then a + 0 = a = 0 + a.
Ex:
2 + 0 = 2
0 + 3 = 3
Multiplicative Identity:
If a is any whole number, then a × 1 = a = 1 × a
Ex:
1 × 1 = 1
5 × 1 = 5
Multiplication by zero:
If a is any whole number, then a × 0 =0 = 0 × a.
Ex:
1 × 0 = 0
5 × 0 = 0
Division by Zero:
If a is any whole number, then a ÷ 0 is not defined
NCERT SOLUTIONS
EXERCISE – 2.1
1. Write the next three natural numbers after 10999.
11000, 11001 and 11002
2. Write the three whole numbers occurring just before 10001.
10000, 9999 and 9998
3. Which is the smallest whole number?
The smallest whole number is 0
4. How many whole numbers are there between 32 and 53?
33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52
Hence, there are 20 whole numbers between 32 and 53.
5. Write the successor of:
(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670
The successors are
(a) 2440701 + 1 = 2440702
(b) 100199 + 1 = 100200
(c) 1099999 + 1 = 1100000
(d) 2345670 + 1 = 2345671
6. Write the predecessor of:
(a) 94 (b) 10000 (c) 208090 (d) 7654321
The predecessors are
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320
7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001
(a) 530, 503
Ans:
(a) 530 > 503
Clearly 503 is smaller than 530.
Hence, 503 will be on left side of 530 on number line.
(b) 370, 307
Ans:
307 < 370
Hence, 307 will be on the left side of 370 on number line.
(c) 98765, 56789
Ans:
98765 > 56789
Hence, 56789 is on the left side of 98765 on the number line.
(d) 9830415, 10023001
Ans:
9830415 < 10023001
Hence, 9830415 is on the left side of 10023001 on the number line.
8. Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
a. False b. False c. True d. True e. True f. False g. False h. False i. True j. False k.False l. True m. False
EXERCISE – 2.2
1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
2. Find the product by suitable rearrangement:
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25
Ans:
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000
(e) 285 x (5 x 60) = 285 x 300 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000
3. Find the value of the following:
(a) 297 × 17 + 297 × 3
= 297 × (17 + 3)
= 297 × 20
= 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8
= 54279 × (92 + 8)
= 54279 × 100
= 5427900
(c) 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 81265 × 100
= 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218
= 3845 × 5 × 782 + 3845 × 5 × 218
= 3845 × 5 × (782 + 218)
= 19225 × 1000
= 19225000
4. Find the product using suitable properties.
(a) 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3 (using distributive property)
= 73800 + 2214
= 76014
(b) 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 (using distributive property)
= 85400 + 1708
= 87108
(c) 258 × 1008
= 258 × (1000 + 8)
= 258 × 1000 + 258 × 8 (using distributive property)
= 258000 + 2064
= 260064
(d) 1005 × 168
= (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (using distributive property)
= 168000 + 840
= 168840
5. A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?
Ans:
Petrol filled on Monday = 40 litres
Cost of petrol = ₹44 per litre
Petrol filled on Tuesday = 50 litre
Cost of petrol = ₹44 pet litre
∴ Total money spent in all
= ₹(40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960
6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 45 per litre, how much money is due to the vendor per day?
Ans:
Milk supplied in the morning = 32 litres
Cost of milk = ₹45 per litre
Milk supplied in the evening = 68 litres
Total cost of milk per day = 45 × (32 + 68)
= 45 × 100
= ₹ 4500
7. Match the following:
(i) 425 × 136 = 425 × (6 + 30 + 100) 
(a) Commutativity under multiplication. 
(ii) 2 × 49 × 50 = 2 × 50 × 49 
(b) Commutativity under addition. 
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 
(c) Distributivity of multiplication over addition. 
Ans:
(i) = C,
(II) = A,
(III) = B
EXERCISE – 2.3
1. Which of the following will not represent zero:
(a) 1 + 0
(b) 0 × 0
(c) 0 / 2
(d) (10 – 10) / 2
Ans:
(a) 1 + 0 = 1
Hence, it does not represent zero
(b) 0 × 0 = 0
Hence,
it represents zero
(c) = 0/2 = 0
Hence, it represents zero
(d) = (10  10)/0 = 0
Hence, it represents zero
2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Ans:
If product of two whole numbers is zero, definitely one of them is zero
Example: 0 ×4 = 0 and 5 × 0 = 0
If product of two whole numbers is zero, both of them may be zero
Example: 0 × 0 = 0
Yes, if the product of two whole numbers is zero, then both of them will be zero
3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Ans:
This is only true, when each of the number are 1.
1 x 1 = 1
4. Find using distributive property:
(a) 728 × 101
= 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 72800 + 728
= 73528
(b) 5437 × 1001
= 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437
= 5442437
(c) 824 × 25
= (800 + 24) × 25
= (800 + 25 – 1) × 25
= 800 × 25 + 25 × 25 – 1 × 25
= 20000 + 625 – 25
= 20000 + 600
= 20600
(d) 4275 × 125
= (4000 + 200 + 100 – 25) × 125
= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)
= 500000 + 25000 + 12500 – 3125
= 534375
(e) 504 × 35
= (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17640
5. Study the pattern:
1 × 8 + 1 = 9 1234 × 8 + 4 = 9876
12 × 8 + 2 = 98 12345 × 8 + 5 = 98765
123 × 8 + 3 = 987
Write the next two steps. Can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1)
Ans:
Study the pattern:
1 x 8 + 1= 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Solution:
Step I: 123456 x 8 + 6 = 987654
Step II: 1234567 x 8 + 7 = 9876543
Working pattern:
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765
(123456) x 8 + 6 = (111111+11111 + 1111 + 111 + 11 + 1) x 8 + 6 = 987654
(1234567)x 8 + 7 = (1111111+ 111111+11111 + 1111 + 111 + 11 + 1) x 8 + 7 = 9876543