Question

Which one of the following is a rational number
[a] $\sqrt{2}$
[b] $\sqrt{\pi }$
[c] $\sqrt{7}$
[d] $\sqrt{\dfrac{5}{25}}$
[e] $\sqrt{\dfrac{64}{49}}$

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]:

Since 2 is not a perfect square, we have $\sqrt{2}$ is not a rational number.

Checking option [b]:

Claim: $\sqrt{\pi }$ is irrational.

Proof: Suppose not and let $\sqrt{\pi }$ be rational

Since the product of two rational numbers is rational, we have $\sqrt{\pi }\times \sqrt{\pi }=\pi $ is a rational number.

But we know that $\pi $ is an irrational number.

We reached a contradiction.

Hence our assumption is incorrect.

Hence $\sqrt{\pi }$ is irrational

Checking option [c]:

Since 7 is not a perfect square, we have $\sqrt{7}$ is an irrational number.

Checking option [d]:

Claim: $\sqrt{\dfrac{5}{25}}$ is irrational.

Proof: Suppose not, and let $\sqrt{\dfrac{5}{25}}$ is rational.

Since the product of two rational numbers is rational, we have $\sqrt{\dfrac{5}{25}}\times 5=\sqrt{\dfrac{5}{25}\times 25}=\sqrt{5}$ is rational.

But since 5 is not a perfect square $\sqrt{5}$ is not a rational number.

We reached a contradiction.

Hence our assumption is incorrect.

Hence $\sqrt{\dfrac{5}{25}}$ is irrational.

Checking option [d]:

We have $\sqrt{\dfrac{64}{49}}=\sqrt{\dfrac{{{8}^{2}}}{{{7}^{2}}}}=\dfrac{8}{7}$

Hence $\sqrt{\dfrac{64}{49}}=\dfrac{p}{q}$ where p = 8 and q = 4 are integers and q is non-zero. Hence $\sqrt{\dfrac{64}{49}}$ is a rational number.

Hence option [e] is correct.


Note: Alternatively, we have:

[i] Square root of a prime number is irrational

[ii] Product of a rational number and an irrational number is irrational provided that the rational number is non-zero.

[iii] Sum of a rational and an irrational number is irrational

[iv] Difference of a rational number and an irrational number is an irrational number.

[v] Dividing an irrational number by a rational number yields an irrational number.

[vi] Square root of an irrational number is an irrational number.

Keeping all the above-mentioned points in mind, we have options [a], [b], [c] and [d] are all irrational.

And since option [e] can be expressed in the form of $\dfrac{p}{q},p,q\in \mathbb{Z},q\ne

0$, option [e] is a rational number.