Question

Rewrite the following rational numbers in the simplest form:
(i) \[\dfrac{{ - 8}}{6}\]
(ii) \[\dfrac{{25}}{{45}}\]
(iii) \[\dfrac{{ - 44}}{{72}}\]

Answer 1

Hint: Here we write the prime factorization of each number in the numerator and denominator and then cancel out the same factors which exist in both numerator and denominator.
* Simplest form a fraction has no common factor between numerator and denominator.
* Prime factorization of a number is writing the number in multiples of its factors where all factors are prime numbers.

Complete step by step answer:
Since there are three parts of the question, we solve each part separately.
(i)\[\dfrac{{ - 8}}{6}\]
Numerator of the above fraction is -8
Denominator of the above fraction is 6
We write prime factorization of the numbers 8 and 6:
\[8 = 2 \times 2 \times 2\]
\[6 = 2 \times 3\]
Now substitute the values of 8 and 6 from prime factorization in the fraction.
\[ \Rightarrow \dfrac{{ - 8}}{6} = \dfrac{{ - (2 \times 2 \times 2)}}{{(2 \times 3)}}\]
Cancel out same factor from the numerator and denominator
\[ \Rightarrow \dfrac{{ - 8}}{6} = \dfrac{{ - (2 \times 2)}}{3}\]
\[ \Rightarrow \dfrac{{ - 8}}{6} = \dfrac{{ - 4}}{3}\]

\[\therefore \]The simplest form of fraction \[\dfrac{{ - 8}}{6} = \dfrac{{ - 4}}{3}\].

(ii)\[\dfrac{{25}}{{45}}\]
Numerator of the above fraction is 25
Denominator of the above fraction is 45
We write prime factorization of the numbers 25 and 45:
\[25 = 5 \times 5\]
\[6 = 3 \times 3 \times 5\]
Now substitute the values of 25 and 45 from prime factorization in the fraction.
\[ \Rightarrow \dfrac{{25}}{{45}} = \dfrac{{(5 \times 5)}}{{(3 \times 3 \times 5)}}\]
Cancel out same factor from the numerator and denominator
\[ \Rightarrow \dfrac{{25}}{{45}} = \dfrac{5}{{3 \times 3}}\]
\[ \Rightarrow \dfrac{{25}}{{45}} = \dfrac{5}{9}\]

\[\therefore \]The simplest form of fraction \[\dfrac{{25}}{{45}} = \dfrac{5}{9}\]

(iii)\[\dfrac{{ - 44}}{{72}}\]
Numerator of the above fraction is -44
Denominator of the above fraction is 72
We write prime factorization of the numbers 44 and 72:
\[44 = 2 \times 2 \times 11\]
\[72 = 2 \times 2 \times 2 \times 3 \times 3\]
Now substitute the values of 8 and 6 from prime factorization in the fraction.
\[ \Rightarrow \dfrac{{ - 44}}{{72}} = \dfrac{{ - (2 \times 2 \times 11)}}{{(2 \times 2 \times 2 \times 3 \times 3)}}\]
Cancel out same factor from the numerator and denominator
\[ \Rightarrow \dfrac{{ - 44}}{{72}} = \dfrac{{ - (11)}}{{(2 \times 3 \times 3)}}\]
\[ \Rightarrow \dfrac{{ - 44}}{{72}} = \dfrac{{ - 11}}{{18}}\]

\[\therefore \] The simplest form of fraction \[\dfrac{{ - 44}}{{72}} = \dfrac{{ - 11}}{{18}}\].

Note:
It is always recommended to write the prime factorization as those are the smallest factors of a number and we don’t miss out on cancelling any factor if we write all the prime factors of that number, otherwise if we write general factors we are likely to miss out factors. Keep in mind we can check the simplest form is correct if there is no common factor between numerator and denominator of the fraction.