# Misconception About Multiplying Square Roots $\sqrt{ab} = \sqrt {a \vphantom b} \sqrt b$ does not hold true for all real numbers $a$ and $b$. Instead, it **holds true only when at least one of them is positive**. Put another way, the following holds true for all real numbers a and b: $\sqrt{|a|b} = \sqrt {|a|}\sqrt {b \vphantom |}$ [[Misconception About Roots and Fractions]] is a similar misconception. ## Proof Define $l=\sqrt{ab}$ and $r=\sqrt {a \vphantom b} \sqrt b$ $ \begin{align} l &= \sqrt{ab} & r &= \sqrt a \sqrt b \\ l^2 &= \left( \sqrt {ab} \right)^2 & r^2 &= \left( \sqrt a \right) ^2 \left( \sqrt b \right)^2 \\ l^2 &= ab & r^2 &= ab \end{align} $ Thus $l^2 = r^2$ for all real $a$, $b$. $ \begin{align} l^2 &= r^2 \\ l^2 - r^2 &= 0 \\ (l-r)(l+r) &= 0 \end{align} $ Therefore either: 1. $\sqrt{ab} = \sqrt {a \vphantom b} \sqrt b$ or 2. $\sqrt{ab} = -\sqrt {a \vphantom b} \sqrt b$ holds in any given case Need to determine which one applies, and when. Define three distinct situations: 1. Either both $a$ and $b$ are positive 2. One is positive while the other is negative 3. Both are negative. ### Both $a$ and $b$ are positive If so, the following can be said of $a$ and $b$: $\sqrt a$ is a positive real number [^alwaysPositive] $\sqrt b$ is a positive real number [^alwaysPositive] $-\sqrt a \sqrt b$ is a negative real number $\sqrt {ab}$ is a positive real number [^alwaysPositive]: In Algebra, the [[Square Root]] is defined as a mathematical [[Function (in Algebra)|function]] that is always positive. This is contrary to the common belief that it's an [[Inverse (in Algebra)|inverse]] of squaring, which doesn't exist as squaring is not a 1-to-1 function. As such, $\sqrt {ab} = -\sqrt a \sqrt b$ cannot be true because a negative real number cannot be equal to a positive real number. Therefore case 1 must be true: $\sqrt {ab} = \sqrt a \sqrt b$ ### One is positive and the other is negative If so (let’s label $a$ as the positive one), the following can be said of $a$ and $b$: $\sqrt a$ is a positive real number [^alwaysPositive] $\sqrt b$ is an imaginary number $y_b \cdot i$ and $y_b$ is a positive real number $\sqrt {ab}$ is an imaginary number $y_{ab} \cdot i$ and $y_{ab}$ is a positive real number. Lets assume case 2 to be true. If so then $ \begin{align} \sqrt {ab} &= -\sqrt a \sqrt b \\ y_{ab} \cdot i &= -\sqrt a \cdot y_b \cdot i \\ y_{ab} &= -\sqrt a \cdot y_b \end{align} $ This does not hold because a negative real number cannot be equal to a positive real number. Therefore case 1 must be true: $\sqrt {ab} = \sqrt a \sqrt b$ ### Both are negative If so, the following can be said of $a$ and $b$: $\sqrt a$ is an imaginary number $y_a \cdot i$ and $y_a$ is a positive real number $\sqrt b$ is an imaginary number $y_b \cdot i$ and $y_b$ is a positive real number $\sqrt {ab}$ is positive real number Lets assume case 1 to be true. If so then $ \begin{align} \sqrt {ab} &= \sqrt a \sqrt b \\ \sqrt {ab} &= y_a \cdot i \cdot y_b \cdot i \\ \sqrt {ab} &= y_a \cdot y_b \cdot i^2 \\ \sqrt {ab} &= -y_a \cdot y_b \end{align} $ This does not hold because a negative real number cannot be equal to a positive real number. Therefore case 2 must be true: $\sqrt {ab} = -\sqrt a \sqrt b$