Question

Find two rational and irrational numbers between $\sqrt 2 $ and $\sqrt 7 $

Answer 1

Hint: Here, in the given question, we need to find two rational and irrational numbers between $\sqrt 2 $ and $\sqrt 7 $. Let us first understand what rational and irrational numbers are. Rational number is a type of real number, which is in the form of $\dfrac{p}{q}$ where $q \ne 0$. Any fraction with non-zero denominator is a rational number. Every rational number can be expressed as a decimal. Rational numbers are those decimal numbers which are either terminating or repeating. Irrational numbers are the real numbers that cannot be represented as a simple fraction , i.e., irrational numbers cannot be represented in the form of $\dfrac{p}{q}$. For example, $\sqrt 3 $ is an irrational number. Also, the decimal expansion of an irrational number is neither terminating nor recurrent. The square root numbers except the perfect square root numbers are irrational numbers.

Complete step-by-step answer:
Let us find rational numbers between $\sqrt 2 $ and $\sqrt 7 $.
Let us first write the values of $\sqrt 2 $ and $\sqrt 7 $.
 $ \Rightarrow \sqrt 2 = 1.414$
$ \Rightarrow \sqrt 7 = 2.64$
As we know, rational numbers are those decimal numbers which are either terminating or repeating.
Thus, we can write $2$ (it can be expressed in the form of $\dfrac{p}{q}$ as $\dfrac{2}{1}$) and $2.5$ as our answer.
Therefore, $2,2.5$ are rational numbers between $\sqrt 2 $ and $\sqrt 7 $.

Now, let us find irrational numbers between $\sqrt 2 $ and $\sqrt 7 $.
As we know, the square root numbers except the perfect square root numbers are irrational numbers. So, now we will write square root numbers between $\sqrt 2 $ and $\sqrt 7 $.
$\sqrt 3 ,\sqrt 4 ,\sqrt 5 ,\sqrt 6 $
As we know, $\sqrt 4 = 2$. Therefore, it is not an irrational number and the perfect square of $\sqrt 3 ,\sqrt 5 ,\sqrt 6 $ numbers doesn’t exist. Thus, $\sqrt 3 ,,\sqrt 5 ,\sqrt 6 $ are irrational numbers. But we need to write only two numbers,
Therefore, $\sqrt 3 ,\sqrt 5 $ are irrational numbers between $\sqrt 2 $ and $\sqrt 7 $.

Note: Remember that rational and irrational numbers are contradictory numbers because rational number is represented in the form of $\dfrac{p}{q}$, but irrational number can’t be represented in the form of $\dfrac{p}{q}$. Remember that $0$ is also a rational number, as we can represent it in many forms such as $\dfrac{0}{1}$, $\dfrac{0}{2}$, $\dfrac{0}{3}$ etc.