Comparison between Two Irrational Numbers

As we know that the numbers which can’t be written in $\frac{p}{q}$ form or fraction form are known as irrational numbers. These are non- recurring decimal numbers. The square roots, cute roots of numbers which are not perfect roots are examples of irrational numbers. In such cases in which perfect square roots or cube roots can’t be found out, it is difficult to compare them without knowing their approximate or actual value.

For comparing them, we should always keep in mind that if square or cube roots of two numbers (‘a’ and ‘b’) are to be compared, such that ‘a’ is greater than ‘b’, then a$^{2}$ will be greater than b$^{2}$ and a$^{3}$ will be greater than b$^{3}$ and so on, i.e., nth power of ‘a’ will be greater than nth power of ‘b’.

1. Compare $\sqrt{2}$ and $\sqrt{3}$

Solution:

We know that if ‘a’ and ‘b’ are two numbers such that ‘a’ is greater than ‘b’, then a$^{2}$ will be greater than b$^{2}$. Hence, for $\sqrt{2}$ and $\sqrt{3}$, let us square both the numbers and then compare them:

$(\sqrt{2})^{2}$ = $\sqrt{2}$ × $\sqrt{2}$ = 2,

$(\sqrt{3})^{2}$ = $\sqrt{3}$ × $\sqrt{3}$ = 3

Since, 2 is less than 3.

Hence, $\sqrt{2}$ will be less than $\sqrt{3}$.

2. Compare $\sqrt{17}$ and $\sqrt{15}$.

Solution:

Let us find out the square of both the numbers and then compare them. So,

$(\sqrt{17})^{2}$ = $\sqrt{17}$ × $\sqrt{17}$ = 17,

$(\sqrt{15})^{2}$ = $\sqrt{15}$ × $\sqrt{15}$ = 15

Since, 17 is greater than 15.

So, $\sqrt{17}$ will be greater than $\sqrt{15}$.

3. Compare 2$\sqrt{3}$ and $\sqrt{5}$.

Solution:

To compare the given numbers let us first find the square of both the numbers and then carry out the comparison process. So,

$2(\sqrt{3})^{2}$ = 2$\sqrt{3}$ x 2$\sqrt{3}$ = 2 × 2 × $\sqrt{3}$ × $\sqrt{3}$ = 4 × 3 = 12,

$(\sqrt{5})^{2}$ = $\sqrt{5}$ × $\sqrt{5}$ = 5

Since, 12 is greater than 5.

So, 2$\sqrt{3}$ is greater than $\sqrt{5}$.

4. Arrange the following in ascending order:

$\sqrt{5}$, $\sqrt{3}$, $\sqrt{11}$, $\sqrt{21}$, $\sqrt{13}$.

Solution:

Arranging in ascending order stands for arrangement of series from smaller value to the larger value. To arrange the given series in ascending order let us find the square of every element of the series. So,

$(\sqrt{5})^{2}$ = $\sqrt{5}$ × $\sqrt{5}$ = 5.

$(\sqrt{3})^{2}$ = $\sqrt{3}$ × $\sqrt{3}$ = 3.

$(\sqrt{11})^{2}$ = $\sqrt{11}$ × $\sqrt{11}$ = 11.

$(\sqrt{21})^{2}$ = $\sqrt{21}$ × $\sqrt{21}$ = 21.

$(\sqrt{13})^{2}$ = $\sqrt{13}$ × $\sqrt{13}$ = 13.

Since, 3 < 5 < 11 < 13 < 21. Hence, the required order of the series is:

$\sqrt{3}$ < $\sqrt{5}$ < $\sqrt{11}$ < $\sqrt{13}$ < $\sqrt{21}$.

5. Arrange the following in descending order:

$\sqrt[3]{5}$, $\sqrt[3]{7}$, $\sqrt[3]{15}$, $\sqrt[3]{2}$, $\sqrt[3]{39}$.

Solution:

Descending order stands for arrangement of given series in larger value to the smaller value. To find the required series, let us find the cube of each element of the series. So,

$(\sqrt[3]{5})^{3}$ = $\sqrt[3]{5}$ × $\sqrt[3]{5}$ × $\sqrt[3]{5}$ = 5.

$(\sqrt[3]{7})^{3}$ = $\sqrt[3]{7}$ × $\sqrt[3]{7}$ × $\sqrt[3]{7}$ = 7.

$(\sqrt[3]{15})^{3}$ = $\sqrt[3]{15}$ × $\sqrt[3]{15}$ × $\sqrt[3]{15}$ = 15.

$(\sqrt[3]{2})^{3}$ = $\sqrt[3]{2}$ × $\sqrt[3]{2}$ x $\sqrt[3]{2}$ = 2.

$(\sqrt[3]{39})^{3}$ = $\sqrt[3]{39}$ × $\sqrt[3]{39}$ × $\sqrt[3]{39}$ = 39.

Since, 39 > 15 > 7 > 5 > 2.

So, the required order of the series is:

$\sqrt[3]{39}$ > $\sqrt[3]{15}$ > $\sqrt[3]{7}$ > $\sqrt[3]{5}$ > $\sqrt[3]{2}$

Irrational Numbers

Definition of Irrational Numbers

Representation of Irrational Numbers on The Number Line

Comparison between Two Irrational Numbers