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# Is subtraction associative in rational numbers? Explain with an example.

Last updated date: 15th Jul 2024
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Hint: They are many different types of numbers starting with real and complex numbers and ending with natural numbers. One such type of numbers in the number system is called rational numbers. Rational numbers belong to the category of real numbers but they can always be expressed in a fractional form where the denominator of that fraction is never zero.

Complete step by step solution:
We need to identify if the given statement is true or false.
To understand if for rational numbers, subtraction becomes associative, we need to know the meanings of each term clearly. We should know what rational numbers are, what the operation subtraction does and we must also be clear about the associative property.
In mathematics, a rational number is any number that can be expressed in a specific form represented as $\dfrac{p}{q}$where $q \ne 0$. Furthermore, any fraction that has an integer denominator and a numerator that is not equal to zero falls into the group of rational numbers.
Subtraction is an operation that is used in Mathematics which describes the removal of values from another given value.
Then associative property is a grouping property wherein three terms will be given. If we group the first two terms among the three terms and if we find the value, then if we take the same three terms and group the last two terms and find the final value then the resultant value obtained by both the groupings should be equal. Then the terms for the given operation is said to be associative.
Now we can proceed to checking whether the subtraction of rational numbers is associative or not.
We make use of an example to carry out the checking:
We need three terms for checking associativity so let us take three simple rational numbers:
$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4}$
Now we need to check whether: $\left( {\dfrac{1}{2} - \dfrac{2}{3}} \right) - \dfrac{3}{4} = \dfrac{1}{2} - \left( {\dfrac{2}{3} - \dfrac{3}{4}} \right)$
We can perform the checking by calculating the left hand side and right hand side separately.
LHS:
$\Rightarrow \left( {\dfrac{1}{2} - \dfrac{2}{3}} \right) - \dfrac{3}{4}\;$
Solving we get:
$\Rightarrow \left( {\dfrac{1}{2} - \dfrac{2}{3}} \right) - \dfrac{3}{4}\;\; = \;\; - \dfrac{1}{6} - \dfrac{3}{4}$
Calculating the subtraction:
$\Rightarrow \left( {\dfrac{1}{2} - \dfrac{2}{3}} \right) - \dfrac{3}{4}\;\; = \;\; - \dfrac{{11}}{{12}}$
$\therefore$ LHS $= - \dfrac{{11}}{{12}}$
RHS:
$\Rightarrow \dfrac{1}{2} - \left( {\dfrac{2}{3} - \dfrac{3}{4}} \right)$
Solving as follows:
$\Rightarrow \dfrac{1}{2} - \left( {\dfrac{2}{3} - \dfrac{3}{4}} \right) = \dfrac{1}{2} - \left( { - \dfrac{1}{{12}}} \right)$
Completing the calculations:
$\Rightarrow \dfrac{1}{2} - \left( {\dfrac{2}{3} - \dfrac{3}{4}} \right) = \dfrac{7}{{12}}$
$\therefore$ RHS $= \dfrac{7}{{12}}$
Clearly the left hand side and right hand side operations do not match.
Therefore from the above example we proved that the associative property is not valid on subtraction of any rational numbers in general, since when groupings of three terms are different, the resulting values differ.

Note:
There is another property called commutative property where the values that are resulted from performing operations regardless of the order become equal. But subtraction of rational numbers is not commutative, this can be shown with an example. Let’s take two values $3,2$ to perform subtraction;
$\Rightarrow 3 - 2 \ne 2 - 3$
$\Rightarrow 1 \ne - 1$
So subtraction over rational numbers cannot be considered as commutative, on the other hand addition will be commutative.