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We Distribute, Yet Things Multiply - Some Properties of Multiplication

Grade 8CBSE

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

🔑Concepts

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The Distributive Property of Multiplication over Addition states that for any rational numbers aa, bb, and cc, the relation a(b+c)=ab+aca(b + c) = ab + ac holds true.

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The Distributive Property over Subtraction follows a similar rule: a(b−c)=ab−aca(b - c) = ab - ac.

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Commutative Property: Changing the order of the factors does not change the product, i.e., a×b=b×aa \times b = b \times a.

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Associative Property: The way in which factors are grouped in multiplication does not change the product: (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c).

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Multiplicative Identity: The number 11 is the multiplicative identity for rational numbers because a×1=1×a=aa \times 1 = 1 \times a = a.

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Multiplicative Inverse (Reciprocal): For any non-zero rational number ab\frac{a}{b}, its reciprocal is ba\frac{b}{a} such that ab×ba=1\frac{a}{b} \times \frac{b}{a} = 1.

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Multiplication by Zero: Any rational number multiplied by zero equals zero: a×0=0a \times 0 = 0.

📐Formulae

a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c) scores

a×(b−c)=(a×b)−(a×c)a \times (b - c) = (a \times b) - (a \times c)

a×b=b×aa \times b = b \times a

(a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c)

a×1=aa \times 1 = a

a×1a=1 (where a≠0)a \times \frac{1}{a} = 1 \text{ (where } a \neq 0 \text{)}

💡Examples

Problem 1:

Simplify using the distributive property: 12×10512 \times 105

Solution:

12×105=12×(100+5)12 \times 105 = 12 \times (100 + 5) =(12×100)+(12×5)= (12 \times 100) + (12 \times 5) =1200+60= 1200 + 60 1200+601260\begin{array}{r} 1200 \\ + 60 \\ \hline 1260 \\ \end{array}

Explanation:

The number 105105 is broken down into (100+5)(100 + 5) to make multiplication easier. We then apply the distributive property a(b+c)=ab+aca(b+c) = ab + ac.

Problem 2:

Verify the distributive property for a=12a = \frac{1}{2}, b=−23b = \frac{-2}{3}, and c=16c = \frac{1}{6}.

Solution:

LHS: a(b+c)=12(−23+16)a(b + c) = \frac{1}{2} (\frac{-2}{3} + \frac{1}{6}) =12(−4+16)=12(−36)=−312=−14= \frac{1}{2} (\frac{-4 + 1}{6}) = \frac{1}{2} (\frac{-3}{6}) = \frac{-3}{12} = -\frac{1}{4} RHS: (a×b)+(a×c)=(12×−23)+(12×16)(a \times b) + (a \times c) = (\frac{1}{2} \times \frac{-2}{3}) + (\frac{1}{2} \times \frac{1}{6}) =(−26)+(112)=−412+112=−312=−14= (\frac{-2}{6}) + (\frac{1}{12}) = \frac{-4}{12} + \frac{1}{12} = \frac{-3}{12} = -\frac{1}{4} Since LHS=RHSLHS = RHS, the property is verified.

Explanation:

We calculate the left-hand side (summing first, then multiplying) and the right-hand side (multiplying separately, then summing) to show they yield the same result.

Problem 3:

Find the product using properties: (35×−27)+(35×−57)(\frac{3}{5} \times \frac{-2}{7}) + (\frac{3}{5} \times \frac{-5}{7})

Solution:

35×(−27+−57)\frac{3}{5} \times (\frac{-2}{7} + \frac{-5}{7}) =35×(−2−57)= \frac{3}{5} \times (\frac{-2 - 5}{7}) =35×(−77)= \frac{3}{5} \times (\frac{-7}{7}) =35×(−1)=−35= \frac{3}{5} \times (-1) = -\frac{3}{5}

Explanation:

By identifying the common factor 35\frac{3}{5}, we can use the distributive property in reverse (ab+ac=a(b+c)ab + ac = a(b+c)) to simplify the calculation.