Divide algebraic radicals

Consider the two mathematical objects $\frac{\sqrt{a}}{\sqrt{b}}$√a√b​ and $\sqrt{\frac{a}{b}}$√ab​. We can refer to them as "a fraction of surds" and "the radical of a fraction" respectively. These expression look very similar, but are they equal to each other?

Let's find out.

 

Exploration

Consider the expressions $\frac{\sqrt{12}}{\sqrt{3}}$√12√3​ and $\sqrt{\frac{12}{3}}$√123​.

For the first expression, we learned when simplifying radical expressions that we can write $\sqrt{12}$√12 as $2\sqrt{3}$2√3. This means that $\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}$√12√3​=2√3√3​ which simplifies nicely to $2$2.

For the second expression we can simplify the fraction in $\sqrt{\frac{12}{3}}$√123​ to get $\sqrt{4}$√4, which evaluates to $2$2 as well.

So for this example at least, we can see that a fraction of two surds is the same as the radical of the fraction. But does this work for all numbers?

 

To answer that question, we will return to the algebraic expressions $\frac{\sqrt{a}}{\sqrt{b}}$√a√b​ and $\sqrt{\frac{a}{b}}$√ab​.

To prove that these expressions are equal, we are going to first convert the surds to exponents and then use our knowledge of exponent laws to transform one expression into the other:

$\frac{\sqrt{a}}{\sqrt{b}}$√a√b​	$=$=	$\frac{a^{\frac{1}{2}}}{b^{\frac{1}{2}}}$a12​b12​​		(Convert the surds into exponential notation)
	$=$=	$\left(\frac{a}{b}\right)^{\frac{1}{2}}$(ab​)12​		(Apply the distributive property of exponents)
	$=$=	$\sqrt{\frac{a}{b}}$√ab​		(Convert back into surds)

In this way we have proven that the fraction of two surds is equal to the radical of the fraction, for all numbers allowed by the exponent law that we used. In particular, we have shown that $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√a√b​=√ab​ for any non-negative value of $a$a and any positive value of $b$b (note that $b$b cannot be zero since it is the denominator).

 

Practice question

Question 1

 

Surds of higher degree

We now know that the fraction of surds and the radical of a fraction will be equal for square roots, but what if we have surds of higher degree like in $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$³√a³√b​?

Fortunately, the proof that we used before works in the same way for cube roots as it did for square roots, by simply changing the $2$2s to $3$3s:

$\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$3√a3√b​	$=$=	$\frac{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}$a13​b13​​		(Convert the surds into exponential notation)
	$=$=	$\left(\frac{a}{b}\right)^{\frac{1}{3}}$(ab​)13​		(Apply the distributive property of exponents)
	$=$=	$\sqrt[3]{\frac{a}{b}}$3√ab​		(Convert back into surds)

 

In fact, we can use this kind of proof for any higher power by just the changing the power of the radical and its corresponding exponential form. This works no matter if we want to use $5$5, $10$10 or even $1000$1000.

Now that we know that we can use this rule for any power, we can write it like this:

 

 

 

Let's see this new rule in action.

 

WORKED EXAMPLE

Simplify the expression $\frac{\sqrt[3]{10x^4y}}{\sqrt[3]{5x}}$³√10x4y³√5x​ to a single radical.

Think: First we check that the powers of the surds are the same. Since both surds have a power of $3$3 we can apply the equivalence rule as stated above.

Do: Applying the rule gives us:

$\frac{\sqrt[3]{10x^4y}}{\sqrt[3]{5x}}=\sqrt[3]{\frac{10x^4y}{5x}}$³√10x4y³√5x​=³√10x4y5x​

Now that the expression is written as a single radical, all that is left is to simplify it.
We can start to simplify the fraction by dividing the numerator by $5x$5x. This gives us:

$\sqrt[3]{\frac{10x^4y}{5x}}=\sqrt[3]{2x^3y}$³√10x4y5x​=³√2x3y

Notice that we have $x^3$x3 inside of a cube root. So we can continue to simplify the expression by factoring out the cube root:

$\sqrt[3]{2x^3y}=\sqrt[3]{x^3}\sqrt[3]{2y}$³√2x3y=³√x3³√2y

Evaluating the cube root then gives us:

$\sqrt[3]{x^3}\sqrt[3]{2y}=x\sqrt[3]{2y}$³√x3³√2y=x³√2y

We can't simplify this expression any further, so we are done.

Reflect: When looking for a way to simplify the expression it was useful to first allow the terms in the different surds to interact with each other. This was achieved by combining the two surds into one using the equivalence rule. After this, simplifying the expression was much more straightforward.

 

This equivalence is a powerful tool for allowing us to simplify expressions in a way that we previously couldn't, as seen in the worked example, and the proof for it is a nice reminder that thinking outside the box can lead us to new concepts that, while unexpected, are extremely valuable.

 

Practice questions

Question 2

Question 3