Question

Verify Associative property of multiplication using the rational numbers 3/ 4 , 4 /5 and 5/6

Answer 1

★ How to do :-

Here, we are given with three fractions in which we are asked to verify the associative property with all those three numbers. For associative property we always need correctly three numbers or fractions. Here, we are given with all the required fractions. In one side, we group the first two numbers with brackets and on the other side, we group the last two numbers. We can also interchange the brackets on both LHS and RHS at a time so that it's value doesn't change. So, let's solve!!

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➤ Solution :-

Associative property says that,

${\tt \leadsto \boxed{\tt \bigg( \dfrac{3}{4} \times \dfrac{4}{5} \bigg) \times \dfrac{5}{6}} = \boxed{\tt \dfrac{3}{4} \times \bigg( \dfrac{4}{5} \times \dfrac{5}{6} \bigg)}}$

Let's solve LHS and RHS separately.

LHS :-

${\tt \leadsto \bigg( \dfrac{3}{4} \times \dfrac{4}{5} \bigg) \times \dfrac{5}{6}}$

First, we should solve the numbers given in bracket.

${\tt \leadsto \bigg( \dfrac{3 \times 4}{4 \times 5} \bigg) \times \dfrac{5}{6}}$

Multiply the numerator and denominator given in bracket.

${\tt \leadsto \dfrac{12}{20} \times \dfrac{5}{6}}$

Now, let's multiply the remaining numbers.

${\tt \leadsto \dfrac{12 \times 5}{20 \times 6} = \dfrac{60}{120}}$

Write the obtained fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{60}{120} = \dfrac{1}{2} \: - - - \sf LHS}$

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RHS :-

${\tt \leadsto \dfrac{3}{4} \times \bigg( \dfrac{4}{5} \times \dfrac{5}{6} \bigg)}$

First, we should solve the numbers given in bracket.

${\tt \leadsto \dfrac{3}{4} \times \bigg( \dfrac{4 \times 5}{5 \times 6} \bigg)}$

Multiply the numerator and denominator given in the bracket.

${\tt \leadsto \dfrac{3}{4} \times \dfrac{20}{30}}$

Now, let's multiply the remaining numbers.

${\tt \leadsto \dfrac{3 \times 20}{4 \times 30} = \dfrac{60}{120}}$

Write the obtained fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{60}{120} = \dfrac{1}{2} \: - - - \sf RHS}$

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Now, we can compare the fractions by seeing that whether they are equal or not.

${\tt \leadsto \dfrac{1}{2} = \dfrac{1}{2}}$

So,

$\sf \leadsto LHS = RHS$

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Hence verified !!