Question

Which one of the following is a rational number
[a] ${{\left( \sqrt{7} \right)}^{2}}$
[b] $211\sqrt{7}$
[c] $8+\sqrt{7}$
[d] $\dfrac{\sqrt{7}}{9}$

Answer 1

Hint: Recall the definition of a rational number. Check option wise which of the numbers satisfy the definition and which don’t. Hence determine which of the numbers are rational numbers and which of the numbers are not.

Complete step-by-step answer:
Rational Numbers: Numbers which can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q is non-zero are called rational numbers. Rational numbers have terminating or non-terminating but recurring decimal representation.
Property 1: If p is not a perfect square, then $\sqrt{p}$ is irrational.
Property 2: sum, Difference, product and division of two rational numbers is rational provided the divisor is non-zero.
We will use these two properties of rational numbers to determine which of the options are rational numbers and which are irrational numbers.
Checking option [a]:
We have ${{\left( \sqrt{7} \right)}^{2}}=\dfrac{7}{1}$.
Here p = 7 and q = 1. Since both p and q are integers and q is non-zero, we have ${{\left( \sqrt{7} \right)}^{2}}$ is a rational number
Checking option [b]:
Claim: $211\sqrt{7}$ is irrational.
Proof: Suppose not, and let $211\sqrt{7}$ be rational
Since the division of two rational numbers is a rational number, we have $\dfrac{211\sqrt{7}}{211}=\sqrt{7}$ is rational. But since 7 is not a perfect square $\sqrt{7}$ is not rational.
We reached a contradiction.
Hence our assumption is incorrect.
Hence $211\sqrt{7}$ is irrational
Checking option [c]:
Claim: $8+\sqrt{7}$ is irrational
Proof: Suppose not, and let $8+\sqrt{7}$ is rational.
Since the difference of two rational numbers is rational, we have $8+\sqrt{7}-8=\sqrt{7}$ is rational.
But since 7 is not a perfect square $\sqrt{7}$ is not rational.
We reached a contradiction.
Hence our assumption is incorrect.
Hence $8+\sqrt{7}$ is irrational.
Checking option [d]:
Claim: $\dfrac{\sqrt{7}}{9}$ is irrational.
Proof: Suppose not, and let $\dfrac{\sqrt{7}}{9}$ is rational.
Since the product of two rational numbers is rational, we have $\dfrac{\sqrt{7}}{9}\times 9=\sqrt{7}$ is rational.
But since 7 is not a perfect square $\sqrt{7}$ is not rational.
We reached a contradiction.
Hence our assumption is incorrect.
Hence $\dfrac{\sqrt{7}}{9}$ is irrational.
Hence option [a] is correct.

Note: Alternatively, we have:
[i] Product of a rational number and an irrational number is irrational provided that the rational number is non-zero.
[ii] Sum of a rational and an irrational number is irrational
[iii] Difference of a rational number and an irrational number is an irrational number.
[iv] Dividing an irrational number by a rational number yields an irrational number.
Keeping all the above-mentioned points in mind, we have options [b],[c] and [d] are all irrational.
And since option [a] can be expressed in the form of $\dfrac{p}{q},p,q\in \mathbb{Z},q\ne 0$, option [a] is a rational number.