In comparison of surds we will discuss about the comparison of equiradical surds and comparison of non-equiradical surds.
I. Comparison of 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.
Comparison of surds can be done to compare the values of surd, which one has higher value and which has lower value. As the values of surds are irrational numbers and difficult to find out also as all have under root values, we have to follow some logic for comparing the values of surds.
A rational number x can be written in any order as given below.
x = \(\sqrt[2]{x^2}\) = \(\sqrt[3]{x^3}\) = \(\sqrt[4]{x^4}\) = \(\sqrt[n]{x^n}\)
Now if a > b, then \(\sqrt[n]{a}\) > \(\sqrt[n]{b}\).
So from the above two points, it is clear that a surd can be expressed in different orders in the multiple of its order of minimum value and two or more surds can be compared when the surds are in same order or equiradical.
II. Comparison of non-equiradical surds:
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.
If the surds are not in same order or non-equiradical, then we need to express the surds in the order of Lowest Common Multiple (LCM) of other surds. By this way, all the surds can be written in same order and we can compare their values by comparing the values of radicand.
For example we need to compare the following surds and arrange them in a descending order.
\(\sqrt[2]{3}\), \(\sqrt[3]{5}\),\(\sqrt[4]{12}\).
The surds are in the order of 2, 3, and 4 respectively. If we need to compare their values, we need to express them in same order. As the LCM of 2, 3, and 4 is 12, we should express the surds in order 12. We know surds can be expressed in any order in multiple of their lowest order.
\(\sqrt[2]{3}\) = \(3^{\frac{1}{2}}\) = \(3^{\frac{6}{12}}\)= \(729^{\frac{1}{12}}\) = \(\sqrt[12]{729}\)
\(\sqrt[3]{5}\) = \(5^{\frac{1}{3}}\) = \(5^{\frac{4}{12}}\)= \(625^{\frac{1}{12}}\) = \(\sqrt[12]{625}\)
\(\sqrt[4]{12}\) = \(12^{\frac{1}{4}}\) = \(12^{\frac{3}{12}}\) = \(1728^{\frac{1}{12}}\) = \(\sqrt[12]{1728}\)
Now all three surds are expressed in same order. As 1728 > 729 > 625
the descending order of the surds will be \(\sqrt[4]{12}\), \(\sqrt[2]{3}\), \(\sqrt[3]{5}\).
Examples of comparison of surds:
We will solve some similar problems to understand more on how to compare the values of surds.
1. 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
2. Arrange the following simple surds descending order.
\(\sqrt[2]{5}\), \(\sqrt[3]{7}\),\(\sqrt[6]{50}\)
Solution:
The given surds are \(\sqrt[2]{5}\), \(\sqrt[3]{7}\),\(\sqrt[6]{50}\).
The surds are in the order of 2, 3, and 6 respectively. As the LCM of 2, 3, and 6 is 6, we should express the surds in order 6.
\(\sqrt[2]{5}\) = \(5^{\frac{1}{2}}\) = \(5^{\frac{3}{6}}\)= \(125^{\frac{1}{6}}\) = \(\sqrt[6]{125}\)
\(\sqrt[3]{7}\) = \(7^{\frac{1}{3}}\) = \(7^{\frac{2}{6}}\) = \(49^{\frac{1}{6}}\) = \(\sqrt[6]{49}\)
\(\sqrt[6]{50}\) need not to be changed as it is already in order 6.
As 125, 49 and 50 are the radicands of the surds in same order that is 6, the descending order of the given surds is \(\sqrt[2]{5}\), \(\sqrt[6]{50}\), \(\sqrt[3]{7}\).
3. Arrange the following simple surds descending order.
\(2\sqrt[3]{10}\), \(3\sqrt[2]{7}\), \(5\sqrt[2]{3}\)
Solution:
If we need to compare the values of the given simple surds, we have to express them in the form of pure surds and after that we need to express them as equiradical surds. As the orders of the surds are 3, 2, 2 and LCM of the order is 6 we need change the order of the surds to 6.
\(2\sqrt[3]{10}\) = \(\sqrt[3]{2^{3}\times 10}\) = \(\sqrt[3]{8\times 10}\) = \(\sqrt[3]{80}\) = \(80^{\frac{1}{3}}\) = \(80^{\frac{2}{6}}\) = \(6400^{\frac{1}{6}}\) = \(\sqrt[6]{6400}\)
\(3\sqrt[2]{7}\) = \(\sqrt[2]{3^{2}\times 7}\) = \(\sqrt[2]{9\times 7}\) = \(\sqrt[2]{63}\) = \(63^{\frac{1}{2}}\) = \(63^{\frac{3}{6}}\) = \(250047^{\frac{1}{6}}\) = \(\sqrt[6]{250047}\)
\(5\sqrt[2]{3}\) = \(\sqrt[2]{5^{2}\times 3}\) = \(\sqrt[2]{25\times 3}\) = \(\sqrt[2]{75}\) =\(75^{\frac{1}{2}}\) = \(80^{\frac{3}{6}}\) = \(512000^{\frac{1}{6}}\) = \(\sqrt[6]{512000}\)
Hence the descending order of the given surds is \(5\sqrt[2]{3}\), \(3\sqrt[2]{7}\), \(2\sqrt[3]{10}\).
11 and 12 Grade Math
From Comparison of Surds to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Feb 23, 24 03:55 PM
Feb 23, 24 02:24 PM
Feb 23, 24 01:28 PM
Feb 22, 24 04:15 PM
Feb 22, 24 02:30 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.