Mathematics

State which of the following numbers will change into non-terminating, non-recurring decimals :
$\begin{matrix} \text{(i)} & - 3\sqrt{2} \\[1.5em] \text{(ii)} & \sqrt{\dfrac{256}{81}} \\[1.5em] \text{(iii)} & \sqrt{27 × 16} \\[1.5em] \text{(iv)} & \sqrt{\dfrac{5}{36}} \\[1.5em] \end{matrix}$

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$\text{(i) } -3\sqrt{2}$

it is an irrational number.

We know that $\sqrt2$ is a non-terminating, non-recurring decimal.
So, $\bold{-3\sqrt{2}}$ is also non-terminating, non-recurring decimal.

$\text{(ii) } \sqrt{\dfrac{256}{81}}$

Here, $\sqrt{\dfrac{256}{81}} = \dfrac{16}{9}$ it is a rational number.

$\text{(iii) } \sqrt{(27 × 16)} = \sqrt{27} × \sqrt{16} = 3\sqrt{3} × 4 = 12\sqrt{3}$

It is an irrational number.

We know that $\sqrt{3}$ is a non-terminating, non-recurring decimal.

So , $12\sqrt{3}$ is non-terminating, non-recurring decimal.

Hence, $\bold{\sqrt{27 × 16}}$ is also non-terminating, non-recurring decimal.

$\text{(iv) }\sqrt{\dfrac{5}{36}}$

Here, $\sqrt{\dfrac{5}{36}}$ = $\dfrac{\sqrt{5}}{6}$ , It is an irrational number.

As, $\dfrac{\sqrt{5}}{6}$ is non-terminating , non-recurring decimal So , $\sqrt{\dfrac{5}{36}}$ is also non-terminating, non-recurring decimal.

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