Order of a Surd

Order of surds. Here we will learn about the order of a surd and its related terminologies. Some solved problems on the orders of surds are given at the end.

Definition of the order of a surd:

The index of the root involved in a surd is called the order of that surd.

Let us understand the definition mathematically. For a positive integer $n,$ we consider the surd $\sqrt[n]{a}.$ Here $n$ is the order of the surd $\sqrt[n]{a}.$

Examples: (i) The order of $\sqrt{2}$ is $2$

(ii) The order of $\sqrt[3]{2}$ is $3$

(iii) The order of $2^{\frac{2}{3}}$ is $3$ ♣

 

Some terminology related to the order of surds:

Quadratic Surd:

A surd of order $2$ is called a quadratic surd. It is also known as a second-order surd.

For example, $\sqrt{2}, \sqrt{3},$ $\sqrt{2^2},$ $5^{\frac{3}{2}},$ $\sqrt{a}$ are surds of order $2.$ So they are quadratic surds or second-order surds.

Cubic Surd:

A surd of order $3$ is called a cubic surd. This surd is also known as a third-order surd.

Example: The surds $\sqrt[3]{2}, \sqrt[3]{5},$ $\sqrt[3]{2^3},$ $7^{\frac{1}{3}},$ $10^{\frac{2}{3}},$ $\sqrt[3]{a},$ $\sqrt[3]{a^2-1}$ are of order $3.$ So these are the examples of cubic surds or third-order surds.

Quartic Surd or bi-quadratic:

A surd of order $4$ is called a quartic surd or a bi-quadratic surd. This is usually known as a fourth-order surd.

Example: The surds $\sqrt[4]{2},$ $\sqrt[4]{7},$ $9^{\frac{1}{4}},$ $\sqrt[4]{a}$ are of order $4.$ So by definition we can say that they are quartic surds or bi-quadratic surds.

A surd of order $n$ is called a $n$-th order surd. For example, $\sqrt[n]{2},$ $7^{\frac{1}{n}},$ $\sqrt[n]{11},$ $\sqrt[n]{a}$ are surds of order $n.$ So these surds are the examples of $n$-th order surds. ♣

 

Solved Problems on the order of surds:

Problem 1: Find the order of $5^{\frac{2}{3}}$

Solution.

As $5^{\frac{2}{3}}$ $=(5^2)^{\frac{1}{3}}$ $=(25)^{\frac{1}{3}}$ $=\sqrt[3]{25},$ the order of $5^{\frac{2}{3}}$ is $3.$ ♣

 

Problem 2: Simplify $\sqrt[4]{9}$ to a quadratic surd.

Solution.

Note that $9=3 \times 3$ $=3^2$

Now $\sqrt[4]{9}=9^{\frac{1}{4}}$ $[\because \sqrt[4]{a}=a^{\frac{1}{4}}]$

$=(3^2)^{\frac{1}{4}}$

$=3^{2 \times \frac{1}{4}}$ $[\because (a^m)^n=a^{m \times n}]$

$=3^{\frac{2}{4}}$ $=3^{\frac{1}{2}}$

$=\sqrt{3}$ ♣

 

Problem 3: Express $\sqrt{3}$ as a surd of order $12$

Solution.

$\sqrt{3}=3^{\frac{1}{2}}$ $[\because \sqrt{a}=a^{\frac{1}{2}}]$

$=3^{\frac{1 \times 6}{2 \times 6}}$      [As we have to express as a surd of order $12,$ we need to make the denominator of the power $12.$ So we are multiplying both the numerator and the denominator of $1/2$ by $6$]

$=3^{\frac{6}{12}}$ $=(3^6)^{\frac{1}{12}}$

$=\sqrt[12]{3^6}$

$=\sqrt[12]{729}$ ♣

 

Problem 4: Convert $\sqrt[5]{3}$ into a surd of order $15$

Solution.

$\sqrt[5]{3}=3^{\frac{1}{5}}$ $[\because \sqrt[5]{a}=a^{\frac{1}{5}}]$

$=3^{\frac{1 \times 3}{5 \times 3}}$      [Because, to convert $3^{1/5}$ of order $15$, we will transform the denominator of  $1/5$ to $15$]

$=3^{\frac{3}{15}}$ $=(3^3)^{\frac{1}{15}}$

$=\sqrt[15]{3^3}$

$=\sqrt[15]{27}$ ♣

 

Problem 5: Convert $\sqrt{5}$ and $\sqrt[3]{7}$ into surds of the same but smallest order.

Solution.

The order of $\sqrt{5}$ is $2$ and the order of $\sqrt[3]{7}$ is $3.$

Note that the LCM of $2$ and $3$ is $6.$ So we will convert both $\sqrt{5}$ and $\sqrt[3]{7}$ as surds of order $6.$ This will be their surd forms of the same but smallest order.

Now,

$\sqrt{5}$ $=5^{\frac{1}{2}}$ $=5^{\frac{1 \times 3}{2 \times 3}}$ $=5^{\frac{3}{6}}$ $=\sqrt[6]{5^3}$ $=\sqrt[6]{125}$

and

$\sqrt[3]{7}$ $=7^{\frac{1}{3}}$ $=7^{\frac{1 \times 2}{3 \times 2}}$ $=7^{\frac{2}{6}}$ $=\sqrt[6]{7^2}$ $=\sqrt[6]{49}$ ♣

 

Problem 6: Which is greater between $\sqrt{2}$ and $\sqrt[3]{3}$

Solution. 

The order of $\sqrt{2}$ is $2$ and the order of $\sqrt[3]{3}$ is $3.$

Note that the LCM of $2$ and $3$ is $6.$

So we will convert both $\sqrt{2}$ and $\sqrt[3]{3}$ as surds of order $6.$

$\sqrt{2}$ $=2^{\frac{1}{2}}$ $=2^{\frac{1 \times 3}{2 \times 3}}$ $=2^{\frac{3}{6}}$ $=\sqrt[6]{2^3}$ $=\sqrt[6]{8}$

and

$\sqrt[3]{3}$ $=3^{\frac{1}{3}}$ $=3^{\frac{1 \times 2}{3 \times 2}}$ $=3^{\frac{2}{6}}$ $=\sqrt[6]{3^2}$ $=\sqrt[6]{9}$

So we obtain that

$\sqrt{2}=\sqrt[6]{8}$ and $\sqrt[3]{3}$ $=\sqrt[6]{9}$

As $9>8,$ we have $\sqrt[6]{9}>\sqrt[6]{8}$

∴ we conclude that $\sqrt[3]{3}>\sqrt{2}.$ ♣

 

Problem 7: Arrange in the ascending order: $\sqrt[3]{3},$ $\sqrt[6]{11},$ $\sqrt[12]{120}$

Solution. 

To arrange in ascending order, we need to convert them as surds of the same order. Note that the orders of $\sqrt[3]{3},$ $\sqrt[6]{11},$ $\sqrt[12]{120}$ respectively are $3,$ $6$ and $12.$

Observe that the LCM of $3,$ $6$ and $12$ is $12.$

So we will make them as surds of order $12.$

Now,

$\sqrt{3}$ $=3^{\frac{1}{3}}$ $=3^{\frac{1 \times 4}{3 \times 4}}$ $=2^{\frac{4}{12}}$ $=\sqrt[12]{3^4}$ $=\sqrt[12]{81},$

$\sqrt[6]{11}$ $=11^{\frac{1}{6}}$ $=11^{\frac{1 \times 2}{2 \times 6}}$ $=11^{\frac{2}{12}}$ $=\sqrt[12]{11^2}$ $=\sqrt[12]{121}$

and

$\sqrt[12]{120}$ is already a surd of order $12.$

So we need to arrange in ascending order of

$\sqrt{3}$ $=\sqrt[12]{81},$ $\sqrt[6]{11}$ $=\sqrt[12]{121}$ and $\sqrt[12]{120}$

As $81<120<121,$ the following is the desired ascending order:

$\sqrt[3]{3},$ $\sqrt[12]{120}$ and $\sqrt[6]{11}.$ ♣

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