Write any two irrational number lying between $3$ and $4$ .

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Solution

To find: Irrational numbers between 3 and 4.
$3^{2} = 9$ $9 \leq x \leq 16$
$4^{2} = 16$ $3 \leq \sqrt{x} \leq 4$
where $\sqrt{x}$ is an irrational number between 3 and 4.
$∴$ the irrational numbers can be $\sqrt{10}, \sqrt{11}, \sqrt{12} . . . .$

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