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Whole Numbers - Introduction to Whole Numbers

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Natural Numbers and Whole Numbers: The counting numbers 1,2,3,1, 2, 3, \dots are called natural numbers. When we include 00 to the set of natural numbers, we get the collection of Whole Numbers 0,1,2,3,0, 1, 2, 3, \dots. Visually, this can be represented as a set where natural numbers are a subset inside the larger circle of whole numbers.

Predecessor and Successor: Given any whole number nn, the number that comes immediately after it is its successor, calculated as n+1n + 1. The number that comes immediately before it (except for 00) is its predecessor, calculated as n1n - 1. For example, the successor of 1919 is 2020, and the predecessor of 1919 is 1818. Note that the whole number 00 has no predecessor.

The Number Line: A number line for whole numbers is a horizontal line starting from a point marked 00 and extending infinitely to the right. Points are marked at equal intervals and labeled 1,2,3,1, 2, 3, \dots. The distance between any two consecutive points is called a unit distance. Numbers increase as we move to the right and decrease as we move to the left.

Operations on the Number Line: Addition is represented by moving to the right; for 4+34 + 3, start at 44 and move 33 units right to land on 77. Subtraction is represented by moving to the left; for 727 - 2, start at 77 and move 22 units left to reach 55. Multiplication is shown as equal jumps from 00; for 2×32 \times 3, make 33 jumps of 22 units each starting from 00.

Closure Property: Whole numbers are 'closed' under addition and multiplication. This means if aa and bb are whole numbers, then (a+b)(a + b) and (a×b)(a \times b) will always be whole numbers. However, they are not closed under subtraction or division, as 353 - 5 or 5÷25 \div 2 do not result in whole numbers.

Commutative and Associative Properties: Addition and multiplication are commutative (a+b=b+aa + b = b + a) and associative ((a+b)+c=a+(b+c))( (a + b) + c = a + (b + c) ). This means the order or grouping of numbers does not change the result. Visually, a grid of 3×43 \times 4 dots contains the same number of dots as a grid of 4×34 \times 3 dots.

Identity Elements: The number 00 is the additive identity because a+0=aa + 0 = a for any whole number aa. The number 11 is the multiplicative identity because a×1=aa \times 1 = a. Adding zero or multiplying by one leaves the original value unchanged.

Distributive Property: Multiplication distributes over addition. This is written as a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c). Visually, this is like calculating the total area of two adjacent rectangles with the same height aa but different widths bb and cc.

📐Formulae

Successor: n+1n + 1

Predecessor: n1n - 1 (for n>0n > 0)

Commutative Property (Addition): a+b=b+aa + b = b + a

Commutative Property (Multiplication): a×b=b×aa \times b = b \times a

Associative Property (Addition): (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)

Associative Property (Multiplication): (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c)

Distributive Property: a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c)

Additive Identity: a+0=aa + 0 = a

Multiplicative Identity: a×1=aa \times 1 = a

Division by Zero: a÷0a \div 0 is not defined

💡Examples

Problem 1:

Find the value of 125×8×40×25125 \times 8 \times 40 \times 25 using suitable rearrangement.

Solution:

Step 1: Group numbers that make multiplication easier (look for pairs that result in multiples of 100100 or 10001000). (125×8)×(40×25)(125 \times 8) \times (40 \times 25) Step 2: Calculate the products inside the brackets. 1000×10001000 \times 1000 Step 3: Multiply the results. 1,000,0001,000,000 Final Answer: 1,000,0001,000,000

Explanation:

This solution uses the Associative and Commutative properties of multiplication to rearrange the numbers so that we get products ending in multiple zeros, making the final calculation simpler.

Problem 2:

Find the value of 738×103738 \times 103 using the distributive property.

Solution:

Step 1: Break 103103 into a sum of 100100 and 33. 738×(100+3)738 \times (100 + 3) Step 2: Apply the Distributive Property a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c). (738×100)+(738×3)(738 \times 100) + (738 \times 3) Step 3: Calculate the individual products. 73800+221473800 + 2214 Step 4: Add the results. 7601476014 Final Answer: 76,01476,014

Explanation:

The distributive property allows us to break down a difficult multiplication into two simpler multiplications and then add the results together.