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Reference

Properties of Square Roots

Rule

Product Property of Square Roots

Given two non-negative numbers $a$ and $b,$ the square root of their product equals the product of the square root of each number.

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

Proof

Let

x,

y,

and

z

be three non-negative numbers such that

x = a,

y = b,

and

z = a b .

By the definition of a square root, each of these numbers squared is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{2} = a y^{2} = b z^{2} = a b (I) (II) (III)

Next, multiply Equation (I) by

y^{2} .

x^{2} = ⇓ x^{2} y^{2} = a a y^{2}

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

x^{2} y^{2} = a y^{2}

Substitute values and simplify

$y^{2} = b$

x^{2} y^{2} = a b

$a b = z^{2}$

x^{2} y^{2} = z^{2}

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

(x y)^{2} = z^{2}

Rearrange equation

z^{2} = (x y)^{2}

Since

x,

y,

and

z

are non-negative, the final equation implies that

z = x y .

z^{2} = (x y)^{2} \Rightarrow z = x y

The last step is substituting

z = a b,

x = a,

and

y = b

into this equation.

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

Rule

Quotient Property of Square Roots

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

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

Proof

Let

x,

y,

and

z

be non-negative numbers such that

x = a,

y = b,

and

z = \frac{a}{b} .

By the definition of a square root, each of these numbers squared is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ x^{2} = a y^{2} = b z^{2} = \frac{a}{b} (I) (II) (III)

Since

b

is a positive number,

y^{2}

is also positive. Therefore, Equation (I) can be divided by

y^{2} .

x^{2} = ⇓ \frac{x ^{2}}{y ^{2}} = a \frac{a}{y ^{2}}

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

\frac{x ^{2}}{y ^{2}} = \frac{a}{y ^{2}}

Substitute values and simplify

$y^{2} = b$

\frac{x ^{2}}{y ^{2}} = \frac{a}{b}

$\frac{a}{b} = z^{2}$

\frac{x ^{2}}{y ^{2}} = z^{2}

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

(\frac{x}{y})^{2} = z^{2}

Rearrange equation

z^{2} = (\frac{x}{y})^{2}

Since

x,

y,

and

z

are non-negative, the final equation implies that

z = \frac{x}{y} .

z^{2} = (\frac{x}{y})^{2} \Rightarrow z = \frac{x}{y}

The last step is substituting

z = \frac{a}{b},

x = a,

and

y = b

into this equation.

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

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