Reference

Properties of Radicals

Rule

$n^{th}$ Roots of $n^{th}$ Powers

To simplify an $n^{th}$ root, the radicand must first be expressed as a power. If the index of the radical and the power of the radicand are equal such that $n a^{n},$ the radical expression can be simplified as follows.

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

The absolute value of a number is always non-negative, so when $n$ is even, the result will always be non-negative. Consider a few example $n^{th}$ roots that can be simplified by using the formula.

$a$	$n$	$n a^{n}$	Is $n$ even?	Simplify
$4$	$3$	$3 4^{3}$	$\times$	$4$
$- 6$	$5$	$5 (- 6)^{5}$	$\times$	$- 6$
$2$	$6$	$6 2^{6}$	$✓$	$∣ 2 ∣ = 2$
$- 3$	$4$	$4 (- 3)^{4}$	$✓$	$∣ - 3 ∣ = 3$

Proof

Informal Justification

To write the expression for $n a^{n},$ there are two cases to consider.

$a \geq 0$
$a < 0$

Both cases will be considered one at a time.

$a \geq 0$

Start by noting that since

a

is non-negative,

a^{n}

is also non-negative. This means that the

n^{th}

root can be rewritten using a rational exponent.

n a^{n} = (a^{n})^{\frac{1}{n}}

Since both

n

and

\frac{1}{n}

are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.

(a^{n})^{\frac{1}{n}}

▼

Simplify

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

a^{n \cdot \frac{1}{n}}

$a \cdot \frac{1}{a} = 1$

a^{1}

ExponentOne

$a^{1} = a$

a

Therefore,

n a^{n}

is equal to

a

a

is non-negative.

$a < 0$

In case of a negative value of $a,$ there are also two cases two consider.

$n$ is even
$n$ is odd

Even $n$

Recall that a root with an even index

n_{e}

is defined only for non-negative numbers. Although

a

is negative,

a^{n_{e}}

is positive. Also, a power with a negative base and an even exponent can be rewritten as a power with a positive base.

a^{n_{e}} = (- a)^{n_{e}}

Now, since

- a

is positive, the Power of a Power Property for Rational Exponents can be applied again to simplify

n_{e} a^{n_{e}} .

n_{e} a^{n_{e}}

▼

Simplify

NegBaseToPosPow

$(- a)^{n_{e}} = a^{n_{e}}$

n_{e} (- a)^{n_{e}}

RootToPowD

$n a = a^{\frac{1}{n}}$

((- a)^{n_{e}})^{\frac{1}{n _{e}}}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

(- a)^{n_{e} \cdot \frac{1}{n _{e}}}

$a \cdot \frac{1}{a} = 1$

(- a)^{1}

ExponentOne

$a^{1} = a$

- a

Odd $n$

A root with an odd index

n_{o}

is defined for all real numbers. By the definition of the

n^{th}

root, the expression

n_{o} a^{n_{o}}

is the number

y

that, when multiplied by itself

n_{o}

times, will result in

a^{n_{o}} .

n_{o} times y \cdot y \cdot \dots \cdot y = a^{n_{o}}

Because

n_{o}

is odd and

a

is negative,

a^{n_{o}}

is also negative. This means that the best candidate for

y

is simply

a .

n a^{n} = a

Summary

If $a$ is non-negative, $n a^{n}$ is always equal to $a .$ However, in case of negative $a,$ the value of $n a^{n}$ depends on the parity of $n .$

	$a \geq 0$	$a < 0$
Even $n$	$n a^{n} = a$	$n a^{n} = - a$
Odd $n$	$n a^{n} = a$	$n a^{n} = a$

To conclude, for odd values of $n,$ the expression $n a^{n}$ is equal to $a .$ On the other hand, if $n$ is even, $n a^{n}$ can be written as $∣ a ∣ .$

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

Extra

Simplifying

n a^{m}

Depending on the index of the root and the power in the radicand, simplifying $n a^{m}$ may be problematic. Because real $n^{th}$ roots with an even index are defined only for non-negative numbers, the absolute value is sometimes needed.

If $n$ is even, $n a^{m}$ is defined only for non-negative $a^{m} .$

The following applet presents a decision tree to simplify

n a^{m} .

In this applet,

k

is the greatest common factor of

m

and

n .

Consider example

n^{th}

roots

n a^{m}

that can be simplified by using the decision tree.

$a$	$m$	$n$	$n a^{m}$	Simplify
$2$	$8$	$4$	$4 2^{8}$	$2^{\frac{8}{4}} = 2^{2} = 4$
$- 3$	$6$	$2$	$(- 3)^{6}$	$∣ ∣ ∣ ∣ (- 3)^{\frac{6}{2}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ (- 3)^{3} ∣ ∣ ∣ = ∣ - 27 ∣ = 27$
$x$	$6$	$3$	$3 x^{6}$	$x^{\frac{6}{3}} = x^{2}$
$- 3$	$2$	$8$	$8 (- 3)^{2}$	$4 ∣ - 3 ∣ = 43$
$2$	$3$	$6$	$6 2^{3}$	$2$
$- 2$	$3$	$9$	$9 (- 2)^{3}$	$3 - 2$
$2$	$6$	$8$	$8 2^{6}$	$4 2^{3} = 48$
$- 2$	$6$	$8$	$8 (- 2)^{6}$	$4 ∣ - 2 ∣^{3} = 4 2^{3} = 48$

Rule

Product Property of Radicals

Given two non-negative numbers $a$ and $b,$ the $n^{th}$ root of their product equals the product of the $n^{th}$ root of each number.

$n a b = n a \cdot n b,$ for $a \geq 0$ and $b \geq 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Product Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

b

is equal to

0 .

By the Zero Property of Multiplication, the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} n a \cdot n b

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and one of the factors on the right-hand side equal

0 .

Again, by the Zero Property of Multiplication, the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = \underline{Right - Hand Side} 0

Therefore, the property is true when

a

b

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n a b .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = a b (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be multiplied by

y^{n},

which is equal to

b .

x^{n} = a ⇓ x^{n} y^{n} = a y^{n}

Now, substitute Equations (II) and (III) into the obtained equation.

x^{n} y^{n} = a y^{n}

▼

Substitute values and simplify

Substitute

$y^{n} = b$

x^{n} y^{n} = a b

Substitute

$a b = z^{n}$

x^{n} y^{n} = z^{n}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

(x y)^{n} = z^{n}

RearrangeEqn

Rearrange equation

z^{n} = (x y)^{n}

Since

x y

and

z

are of the same sign, the final equation implies that

z = x y .

z^{n} = (x y)^{n} \Rightarrow z = x y

The last step is substituting

z = n a b,

x = n a,

and

y = n b

into this equation.

z = x y \Leftrightarrow n a b = n a \cdot n b

Rule

Quotient Property of Radicals

Let $a$ be a non-negative number and $b$ be a positive number. The $n^{th}$ root of the quotient $\frac{a}{b}$ equals the quotient of the $n^{th}$ roots of $a$ and $b .$

$n \frac{a}{b} = \frac{n a}{n b},$ for $a \geq 0,$ $b > 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Quotient Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

is equal to

0 .

Because

\frac{0}{b} = 0,

the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} \frac{n 0}{n b}

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and the numerator on the right-hand side equal

0 .

Again, since dividing

0

by a nonzero number results in

0,

the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = ✓ \underline{Right - Hand Side} 0

Therefore, the property is true when

a

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n \frac{a}{b} .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = \frac{a}{b} (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be divided by

y^{n},

which is equal to

b .

x^{n} = a ⇓ \frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

Now, substitute Equations (II) and (III) into the obtained equation.

\frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

▼

Substitute values and simplify

Substitute

$y^{n} = b$

\frac{x ^{n}}{y ^{n}} = \frac{a}{b}

Substitute

$\frac{a}{b} = z^{n}$

\frac{x ^{n}}{y ^{n}} = z^{n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

(\frac{x}{y})^{n} = z^{n}

RearrangeEqn

Rearrange equation

z^{n} = (\frac{x}{y})^{n}

Since

\frac{x}{y}

and

z

are of the same sign, the final equation implies that

z = \frac{x}{y} .

z^{n} = (\frac{x}{y})^{n} \Rightarrow z = \frac{x}{y}

The last step is substituting

z = n \frac{a}{b},

x = n a,

and

y = n b

into this equation.

z = \frac{x}{y} \Leftrightarrow n \frac{a}{b} = \frac{n a}{n b}

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Proof

a≥0

a<0

Even n

Odd n

Summary

Extra

Proof

Proof

$a \geq 0$

$a < 0$

Even $n$

Odd $n$