Write the two irrational numbers lying between 3 and 4.

Hint: In this question, we will first understand the basic definitions of rational and irrational numbers. As we know, there are infinite irrational numbers between two rational numbers. So, we can find any two of them. We also have a formula for writing irrational numbers that lie between two rational numbers given by
\[{{a}_{i}}=a+\dfrac{b-a}{2\left( n+1 \right)}\sqrt{2}\times i\]
Where, a and b are given rational numbers, n is number of irrational number required and i = 1, 2, 3 . . . . . . . . . n

Complete step-by-step solution:
Before solving this question, let us understand the basic definition of rational and irrational numbers.
Rational numbers are the numbers, which can be written in the form of $\dfrac{p}{q}$ where p and q both belong to integers and q is never equal to zero. It can be expressed as decimal also.
Irrational numbers are the non-terminating, non-recurring decimal which cannot be written in the form of $\dfrac{p}{q}$.
Now, if a and b are two distinct rational numbers then for a\[{{a}_{i}}=a+\dfrac{b-a}{2\left( n+1 \right)}\sqrt{2}\times i\] where i = 1, 2, 3 . . . . . . . . . n.
Here, we have to find two irrational numbers between 3 and 4, hence, a = 3, b = 4, and n = 2.
Putting all values in formula for i = 1, i = 2, we get:
\[\begin{align}
&{{a}_{1}}=a+\dfrac{b-a}{2\left( n+1 \right)}\sqrt{2}\times 1\text{ and }{{a}_{2}}=a+\dfrac{b-a}{2\left( n+1 \right)}\sqrt{2}\times 2 \\
&\Rightarrow {{a}_{1}}=3+\dfrac{4-3}{2\left( 2+1 \right)}\sqrt{2}\text{ and }{{a}_{2}}=3+\dfrac{2\left( 4-3 \right)}{2\left( 2+1 \right)}\sqrt{2} \\
 & \Rightarrow {{a}_{1}}=3+\dfrac{\sqrt{2}}{6}\text{ and }{{a}_{2}}=3+\dfrac{\sqrt{2}}{3} \\
\end{align}\]
Hence, two irrational numbers between 3 and 4 are $3+\dfrac{\sqrt{2}}{6}\text{ and }3+\dfrac{\sqrt{2}}{3}$.

Note: Students should note that these are irrational numbers because the addition of any rational number with the irrational number $\sqrt{2}$ is also an irrational number. Students can find irrational numbers in the following way also.
As the square of 3 is 9 and the square of 4 is 16, so, the square root of numbers between 9 and 16 are also irrational numbers.
Hence, \[\sqrt{10,}\sqrt{11},\sqrt{12},\sqrt{13},\sqrt{14},\sqrt{15}\] are also some of the irrational numbers between 3 and 4.