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The rational number whose absolute value is $\left( {\dfrac{4}{3}} \right)$ is:
(A) $\left( {\dfrac{8}{6}} \right)$
(B) $\left( {\dfrac{{16}}{9}} \right)$
(C) $1$
(D) $ - 1$

Answer
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Hint: In this question, the numerator of the given rational number is $4$ and the denominator is equal to $3$ and we have to find a rational number whose absolute value is same as $\left( {\dfrac{4}{3}} \right)$ . When the numerator and the denominator are in the form of prime factors and don’t have any common factor, then the rational number is said to be simplified. We can find whether the given rational number is simplified or not by writing both the numerator and the denominator as a product of their prime factors, and thus simplify the fraction if it is not in the simplified form.

Complete step by step answer:
We can easily find the rational number whose absolute value is $\left( {\dfrac{4}{3}} \right)$ by seeking for the absolute value of the rational numbers given to us in the options.
So, we are given the rational number $\left( {\dfrac{8}{6}} \right)$ in the Option (A).
We know that prime factorization of \[6\] is:
$6 = 2 \times 3$
Also, prime factorization of 8 is:
$8 = 2 \times 2 \times 2$
We see that $2$ is a prime factor of both the numerator and the denominator, so we have $2$ as a common factor, and thus we divide the numerator and the denominator by $2$.
So, we get, \[\left( {\dfrac{8}{6}} \right) = \left[ {\dfrac{{\left( {\dfrac{8}{2}} \right)}}{{\left( {\dfrac{6}{2}} \right)}}} \right] = \dfrac{4}{3}\]
Now, the numerator and the denominator are both prime numbers and don’t have any common factor, so it cannot be simplified further.

Hence, the simplified form of $\left( {\dfrac{6}{8}} \right)$ is $\left( {\dfrac{3}{4}} \right)$. So, $\left( {\dfrac{6}{8}} \right)$ is the rational number whose absolute value is same as $\left( {\dfrac{3}{4}} \right)$.

Note: The process of writing a number as a product of the prime factors is known as its prime factorization. The rational number whose absolute value is $\left( {\dfrac{4}{3}} \right)$ can also be found by multiplying the numerator and denominator by $2$.
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