Question

Verify the associative property of subtraction for the following rational numbers, [i.e., $x - (y - z) \neq (x - y) - z$ ].

$\frac{9}{8}, \frac{- 3}{4} and \frac{7}{12}$

Accepted Answer

We are given to verify the associative property of subtraction for the rational numbers $\frac{9}{8}, \frac{- 3}{4} and \frac{7}{12}$ .

We know that rational numbers cannot be subtraction in any order. So, if $x, y$ and $z$ are three rational numbers, then $x - (y - z) \neq (x - y) - z$ is true.

Now, considering $x = \frac{9}{8}, y = \frac{- 3}{4}$ and $z = \frac{7}{12}$ , we obtain,

$x - (y - z) = \frac{9}{8} - (\frac{- 3}{4} - \frac{7}{12}) = \frac{9 \times 3}{8 \times 3} - (\frac{- 3 \times 6}{4 \times 6} - \frac{7 \times 2}{12 \times 2}) = \frac{27}{24} - (\frac{- 18}{24} - \frac{14}{24})$

$= \frac{27}{24} - \frac{- 18 - 14}{24} = \frac{27}{24} - \frac{- 32}{24} = \frac{27 - (- 32)}{24} = \frac{59}{24}$ .

And, $(x - y) - z = (\frac{9}{8} - \frac{- 3}{4}) - \frac{7}{12} = (\frac{9 \times 3}{8 \times 3} - \frac{- 3 \times 6}{4 \times 6}) - \frac{7 \times 2}{12 \times 2} = (\frac{27}{24} - \frac{- 18}{24}) - \frac{14}{24}$

$= \frac{27 - (- 18)}{24} - \frac{14}{24} = \frac{45}{24} - \frac{14}{24} = \frac{45 - 14}{24} = \frac{31}{24}$ .

We observe that $x - (y - z) \neq (x - y) - z$ .

Hence, the associative property of subtraction is verified for the given rational numbers.

Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Operations on Rational Numbers, Exercise 2: EXERCISE 3.2

Dr Pooranmal Mathematics Solutions for Exercise - Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Operations on Rational Numbers, Exercise 2: EXERCISE 3.2

Questions from Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Operations on Rational Numbers, Exercise 2: EXERCISE 3.2 with Hints & Solutions

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