Surds

UK Secondary (7-11)

Divide algebraic surds

Lesson

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

Let's find out.

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

For the first expression, we learned when simplifying surd 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 surd 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 $\sqrt{\frac{a}{b}}$√`a``b`.

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 |
$=$= | $\frac{a^{\frac{1}{2}}}{b^{\frac{1}{2}}}$a12b12 |
(Convert the surds into index 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 surd of the fraction, for all numbers allowed by the index law that we used. In particular, we have shown that $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√`a`√`b`=√`a``b` 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).

Simplify the expression $\frac{\sqrt{x}}{\sqrt{yz}}$√`x`√`y``z` to a single surd.

We now know that the fraction of surds and the surd 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}√`a`^{3}√`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}√a^{3}√b |
$=$= | $\frac{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}$a13b13 |
(Convert the surds into index 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 surd and its corresponding index 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:

Equivalence in surds and fractions

The fraction of surds $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$^{n}√`a`^{n}√`b` will always be equivalent to the surd of the fraction $\sqrt[n]{\frac{a}{b}}$^{n}√`a``b`.

Careful!

We can only use this rule if the powers of the surds in the fraction are the same. For example: $\frac{\sqrt[3]{a}}{\sqrt[4]{b}}$^{3}√`a`^{4}√`b` will **not** be equal to $\sqrt[3]{\frac{a}{b}}$^{3}√`a``b` or $\sqrt[4]{\frac{a}{b}}$^{4}√`a``b`.

Let's see this new rule in action.

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

**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}√10`x`4`y`^{3}√5`x`=^{3}√10`x`4`y`5`x`

Now that the expression is written as a single surd, all that is left is to simplify it.

We can start to simplify the fraction by dividing the numerator by $5x$5`x`. This gives us:

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

Notice that we have $x^3$`x`3 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}√2`x`3`y`=^{3}√`x`3^{3}√2`y`

Evaluating the cube root then gives us:

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

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.

Write the expression $\frac{\sqrt[5]{y^6}}{\sqrt[5]{y}}$^{5}√`y`6^{5}√`y` in its simplest form.

Write the expression $\frac{\sqrt{160pq}}{\sqrt{90p}}$√160`p``q`√90`p` as a single simplified surd.