Question

Classify the following numbers as rational or irrational:
(i). $$2-\sqrt{5}$$
(ii). $$(3+\sqrt{23})-\sqrt{23}$$
(iii). $${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$$
(iv). $${\displaystyle \frac{1}{\sqrt{2}}}$$
(v). $$2\pi$$

Answer 1

Hint: Simplify the numbers and check if they can be represented in $${\displaystyle \frac{p}{q}}$$ form, where p and q are integers and $$q\ne 0$$ . If they can be represented, then they belong to rational numbers, if not, then they are irrational numbers.

Complete step-by-step answer:
A number that is in the form $${\displaystyle \frac{p}{q}}$$ or can be simplified to the form $${\displaystyle \frac{p}{q}}$$ where p and q are integers and $$q\ne 0$$ is called a rational number. They can also be represented in the decimal form as terminating decimals or non-terminating recurring decimals. Examples include $$22,{\displaystyle \frac{5}{7}},{\displaystyle \frac{1}{2}}$$ .
A real number that can’t be represented in the form $${\displaystyle \frac{p}{q}}$$ where both p and q are integers and $$q\ne 0$$ is called an irrational number. These numbers are non-terminating and non-recurring decimals.
Examples include $$\pi ,\sqrt{5},\sqrt[3]{2}$$ .
Now, having the knowledge of rational and irrational numbers, we can classify the numbers into these categories.
(i). The number $$2-\sqrt{5}$$ , is the difference between 2 and $$\sqrt{5}$$ .
2 is a rational number because it can be represented as $${\displaystyle \frac{2}{1}}$$ where 2 and 1 are integers.
$$\sqrt{5}$$ is an irrational number because it can’t be represented in the form of rational numbers.
We know that the sum or difference of a rational and an irrational number is an irrational number.
Hence, $$2-\sqrt{5}$$ is irrational.
(ii). The number $$(3+\sqrt{23})-\sqrt{23}$$ can be simplified as follows:
$$(3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23}$$
Cancelling $$\sqrt{23}$$ , we have:
$$(3+\sqrt{23})-\sqrt{23}=3$$
We know that 3 is a rational number since it can be represented as $${\displaystyle \frac{3}{1}}$$ where 3 and 1 are integers.
Hence, $$(3+\sqrt{23})-\sqrt{23}$$ is rational.
(iii). The number $${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$$ can be simplified by cancelling $$\sqrt{7}$$ as follows:
$${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}={\displaystyle \frac{2}{7}}$$
We can see that $${\displaystyle \frac{2}{7}}$$ is in the rational form since 2 and 7 are integers.
 Hence, $${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$$ is rational.
(iv). We can simplify the number $${\displaystyle \frac{1}{\sqrt{2}}}$$ by multiplying numerator and denominator by $$\sqrt{2}$$ .
$${\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}$$
$${\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{\sqrt{2}}{2}}$$
We know that $$\sqrt{2}$$ is irrational and 2 is rational.
Division of an irrational number by a rational number, results in an irrational number.
Hence, $${\displaystyle \frac{1}{\sqrt{2}}}$$ is irrational.
(v). In the number $$2\pi$$ , 2 is rational and $$\pi$$ is irrational.
Multiplication of a rational and an irrational number is an irrational number.
Hence, $$2\pi$$ is irrational.

Note: You might conclude that options (ii) and (iii) are irrational numbers because they contain the square root terms $$\sqrt{23}$$ and $$\sqrt{7}$$ respectively, but it is wrong. Simplify the number completely and then check for the $${\displaystyle \frac{p}{q}}$$ form.