Question

From the following find the smallest and the greatest numbers:

$\frac{-3}{4},\frac{14}{-18},\frac{2}{3},\frac{5}{-6},\frac{-9}{-36}$

Answer 1

Step 1: Converting given fraction into equivalent fractions.

To convert them in equivalent fraction we have to make $\frac{14}{-18},\frac{5}{-6},\frac{-9}{-36}$ in standard form.

$\Rightarrow \frac{14}{-18}=\frac{-14}{18},\frac{5}{-6}=\frac{-5}{6}$ (Multiplying numerator and denominator by $-1$)

$\Rightarrow \frac{-9}{-36}=\frac{9}{36}$ (minus-minus sine gets cancelled)

So, LCM of $4,18,3,6$ and $36$ is $72$.

Now convert the given rational numbers in equivalent fraction with denominator $72$.

$$\begin{gathered} \frac{-3\times 18}{4\times 18}=\frac{-54}{72}, \\ \frac{-14\times 4}{18\times 4}=\frac{-56}{72}, \\ \frac{2\times 24}{3\times 24}=\frac{48}{72}, \\ \frac{-5\times 12}{6\times 12}=\frac{-60}{72} \end{gathered}$$

and $\frac{9\times 2}{36\times 2}=\frac{18}{72}$

Step 2: Comparing the fractions.

We know that in equivalent fractions, fractions with smaller numerators are smaller and vice-versa.

Here, $48>18>-54>-56>-60$

So, $\frac{48}{72}>\frac{18}{72}>\frac{-54}{72}>\frac{-56}{72}>\frac{-60}{72}$

$\Rightarrow \frac{2}{3}>\frac{-9}{-36}>\frac{-3}{4}>\frac{14}{-18}>\frac{5}{-6}$

Hence $\frac{5}{-6}$ is the smallest integer and $\frac{2}{3}$ is the greatest integer.