Comparison of Surds

In comparison of surds we will discuss about the comparison of equiradical surds and comparison of non-equiradical surds.

In case of equiradical surds (i.e., surds of the same order) $$\sqrt[n]{a}$$ and $$\sqrt[n]{b}$$, we have $$\sqrt[n]{a}$$ > $$\sqrt[n]{b}$$ when x > y.

For example,

(i) √5 > √3, since 5 > 3

(ii) ∛21 < ∛28, since 21 < 28.

(iii) ∜10 > ∜6, since 10 > 6.

In case of comparison between two or more non-equiradical surds (i.e., surds of different orders) we express them to surds of the same order (i.e., equiradical surds). Thus, to compare between ∛7 and ∜5 we express them to surds of the same order as follows:

Clearly, the orders of the given surds are 3 and 4 respectively and LCM Of 3 and 4 is 12.

Therefore, ∛7 = 7$$^{1/3}$$ = 7$$^{4/12/}$$ = $$\sqrt[12]{7^{4}}$$ = $$\sqrt[12]{2401}$$ and

∜5 = 5$$^{1/4}$$ = 5$$^{3/12}$$ = $$\sqrt[12]{5^{3}}$$ = $$\sqrt[12]{125}$$

Clearly, we see that 2401 > 125

Therefore, ∛7 > ∜5.

Example of comparison of surds:

Convert each of the following surds into equiradical surds of the lowest order and then arrange them in ascending order.

∛2, ∜3 and $$\sqrt[12]{4}$$

Solution:

∛2, ∜3 and $$\sqrt[12]{4}$$

We see that the orders of the given surds are 3, 4 and 12 respectively.

Now we need to find the lowest common multiple of 3, 4 and 12.

The lowest common multiple of 3, 4 and 12 = 12

Therefore, the given surds are expressed as equiradical surds of the lowest order (i.e. 12th order) as follows:

∛2 = 2$$^{1/3}$$ = 2$$^{4/12}$$ = $$\sqrt[12]{2^{4}}$$ = $$\sqrt[12]{16}$$

∜3 = 3$$^{1/4}$$ = 3$$^{3/12}$$ = $$\sqrt[12]{3^{3}}$$ = $$\sqrt[12]{27}$$

$$\sqrt[12]{4}$$ = 4$$^{1/12}$$ = $$\sqrt[12]{4^{1}}$$ = $$\sqrt[12]{4}$$

Therefore, equiradical surds of the lowest order ∛2, ∜3 and $$\sqrt[12]{4}$$ are $$\sqrt[12]{16}$$, $$\sqrt[12]{27}$$ and $$\sqrt[12]{4}$$ respectively.

Clearly, 4 < 16 < 27; hence the required ascending order of the given surds is:

$$\sqrt[12]{4}$$, ∛2, ∜3