Question

If a number P is a positive integer but not a perfect square, then $\sqrt{P}$ is definitely _________.

• A. An integer
• B. A rational number
• C. An irrational number
• D. A terminating decimal number

Answer 1

The correct option is C An irrational number

A perfect square is a number which can be represented as a product of two same numbers.
Accordingly, a non perfect square is a number which cannot be represented as the product of two same numbers.
Let’s consider a non perfect square number, P = 48.
Then, $\sqrt{P}=\sqrt{48}$.
$\Rightarrow \sqrt{P}=\mathrm{6.9282...}$
Here, 6.9282... is a non-terminating non-recurring decimal number, and non-terminating non-recurring numbers are irrational numbers.
Hence, $\sqrt{P}$ will be an irrational number if P is not a perfect square.
If P had been 25, which can be expressed as $5\times 5$, then $\sqrt{P}$ would be an integer.
If P had been $\frac{25}{144}$, which can be expressed as $\frac{5}{12}\times \frac{5}{12}$ , then
$\sqrt{P}$ would be a rational number.
If P had been $\frac{25}{144}$, which can be expressed as $\frac{5}{12}\times \frac{5}{12}$ , then $\sqrt{P}$ would be a rational number.
By considering any value of $\sqrt{P}$ as a positive non perfect square integer, P will always be an irrational number.