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. All of the above

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 two digits are 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 $x=5.4646464646\overline{46}100x=546.\overline{46}99x=541$

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 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.