Question

is an irrational number.

• A. $2\sqrt{3}$
• B. $0.1\overline{32}$
• C. $5.\overline{46}$

Answer 1

The correct option is A $2\sqrt{3}$

i) $2\sqrt{3}$ is product of $2$ and $\sqrt{3}$. We know that $\sqrt{3}$ is an irrational number and cannot be expressed in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$. When we multiply a rational number with an irrational number, the result is an irrational number. Therefore, $2\sqrt{3}$ is an irrational number.

ii) $0.3\overline{32}$ is a recurring decimal
Let $x=0.3\overline{32}$. In this case we see that $1$ does not repeat itself but the block $32$ repeats itself. Since, three digits are not repeating, we multiply $x$ by $10$ to get

$10x=\mathrm{1.323232323232...}1000x=\mathrm{14332.3232..}$
Subtracting the two equations we get,
$990x=131x=\frac{131}{99000}$

which is in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$. Hence, it is a rational number.

iii) $5.\overline{46}$ is again a recurring decimal.
Let $x=\mathrm{5.464646464646...}100x=546.\overline{46}$

Subtracting the two equations, we get,
$99x=541$

Therefore, $x=\frac{541}{99}$

which is in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$.Hence, it is a rational number.