Question

Which of the following is an irrational number?

i) $2\sqrt{3}$

ii) $0.143\overline{32}$

iii) $5.\overline{46}$

iv) $\sqrt{5}$

• A. Only iv
• B. ii and iii
• C. i and iv
• D. Cannot be determined

Answer 1

The correct option is C

i and iv

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) $0.143\overline{32}$ is a recurring decimal
Let $x=0.143\overline{32}$. In this case we see that $143$ does not repeat itself but the block $32$ repeats itself. Since, three digits are not repeating, we multiply $x$ by $1000$ to get

$1000x=\mathrm{143.323232323232...}100000x=\mathrm{14332.3232..}$
Substracting the two equations we get,
$99000x=14159x=\frac{14159}{99000}$

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.

iv) $\sqrt{5}$ We know that $\sqrt{5}$ is an irrational number since it cannot be represented in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$. Its value is $\mathrm{2.23606797749979...}$ which is a non-terminating and non-recurring decimal which is the very definition of an irrational number.