Question

Find at least two irrational numbers between 2 and 3.

Answer 1

Hint: Irrational numbers are numbers that can not be represented in the form of \[\dfrac{p}{q}\]. The square roots and cube roots of most numbers are irrational. Use this to find the irrational numbers between 2 and 3.

Complete step-by-step answer:
Irrational numbers and rational numbers belong to the bigger domain of real numbers.
Rational numbers are numbers that can be represented in the form of \[\dfrac{p}{q}\] where \[q \ne 0\].
Irrational numbers are numbers that can not be represented in the form of \[\dfrac{p}{q}\]. Irrational numbers are non-recurring and non-terminating decimal numbers. The square roots of non-perfect square numbers are all irrational and the cube roots of non-perfect cubes are also irrational and so on.
Few examples of irrational numbers are \[\pi \], \[\sqrt 2 \], \[\sqrt[3]{7}\] and so on.
In this question, we need to find at least two irrational numbers between 2 and 3. We use the fact that the square root of non-perfect squares is irrational to solve it.
We know that the value of the square of 2 is 4 and the value of the square of 3 is 9.
We choose two numbers between 4 and 9, let's say 6 and 7. Then, we have the following:
\[4 < 6 < 9\]
\[4 < 7 < 9\]
Taking square root on all the three terms in each inequality, we get:
\[\sqrt 4 < \sqrt 6 < \sqrt 9 \]
\[\sqrt 4 < \sqrt 7 < \sqrt 9 \]
Simplifying the above expressions, we get:
\[2 < \sqrt 6 < 3\]
\[2 < \sqrt 7 < 3\]
Hence, the two irrational numbers between 2 and 3 are \[\sqrt 6 \] and \[\sqrt 7 \].

Note: There are infinite irrational numbers between two rational numbers. Hence, you can obtain as many irrational numbers as you want between the numbers 2 and 3. The answer is just one among them.