# Misconception About Roots and Fractions $\sqrt \frac{a}{b} = \frac {\sqrt a} {\sqrt b}$ does not hold true for all real numbers $a$ and $b$. Instead, it **holds true only when $b$ is positive, or when both $a$ and $b$ are negative**. Put another way, the following holds true for all real $a$ and $b$: $\sqrt \frac{a}{|b|} = \frac {\sqrt a} {\sqrt {|b|}}$ If you succumb to this misconception you can come to the conclusion that $-1=1$.[^1] $ \begin{align} \frac{-1}{1} &= \frac{1}{-1} && \text {take a square root of both sides} \\ \sqrt {\frac{-1}{1}} &= \sqrt {\frac{1}{-1}} \\ \frac {\sqrt{-1}}{\sqrt 1} &= \frac {\sqrt 1}{\sqrt {-1}} && \text {(!)} \\ \frac {i}{1} &= \frac {1}{i} && /\cdot i \\ {i^2} &= 1 \\ -1 &= 1 \end{align} $ [^1]: From [Brilliant](https://brilliant.org/practice/why-complete-this-exploration/?p=6). [[Misconception About Multiplying Square Roots]] is a similar misconception worth checking out. ## Proof Define $l=\sqrt \frac{a}{b}$ and $r=\frac{\sqrt a} {\sqrt b}$ $ \begin{align} l&=\sqrt \frac{a}{b} & r&=\frac{\sqrt a} {\sqrt b}\\ l^2 &= \left( \sqrt \frac{a}{b} \right)^2 & r^2 &= \frac{\left( \sqrt a \right )^2} {\left ( \sqrt b \right )^2} \\ l^2 &= \frac{a}{b} & r^2 &= \frac{a}{b} \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. $\sqrt \frac{a}{b} = \frac {\sqrt a} {\sqrt b}$ or 2. $\sqrt \frac{a}{b} = -\frac {\sqrt a} {\sqrt b}$ holds in any given case Need to determine which one applies, and when. Four distinct situations exist: 1. Both $a$ and $b$ are positive 2. $a$ is negative and $b$ is positive 3. $a$ is positive and $b$ is negative 4. 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] $\sqrt \frac {a}{b}$ is a positive real number [^alwaysPositive] $\frac {\sqrt a}{\sqrt b}$ 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, $\sqrt \frac{a}{b} = -\frac {\sqrt a} {\sqrt b}$ cannot be true because a negative real number cannot be equal to a positive real number. Therefore case 1 must be true: $\sqrt \frac{a}{b} = \frac {\sqrt a} {\sqrt b}$. ### $a$ is negative and $b$ is positive 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 a positive real number [^alwaysPositive] $\sqrt \frac {a}{b}$ 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 \frac{a}{b} &= -\frac {\sqrt a} {\sqrt b} \\ y_{ab} \cdot i &= -\frac{y_a \cdot i}{\sqrt b} \\ y_{ab} &= -\frac{y_a}{\sqrt 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: $\sqrt \frac{a}{b} = \frac {\sqrt a} {\sqrt b}$. ### $a$ is positive and $b$ is negative If so, 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 $\sqrt \frac {a}{b}$ is an imaginary number $y_{ab} \cdot i$ and $y_{ab}$ is a positive real number. Lets assume case 1 to be true. If so then $ \begin{align} \sqrt \frac{a}{b} &= \frac {\sqrt a} {\sqrt b} \\ y_{ab} \cdot i &= \frac{\sqrt a}{y_b \cdot i} &/\cdot i\\ -y_{ab} &= \frac{\sqrt a}{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: $\sqrt \frac{a}{b} = -\frac {\sqrt a} {\sqrt b}$. ### 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 $\sqrt {\frac {a} {b}}$ is positive real number Lets assume case 2 to be true. If so then $ \begin{align} \sqrt \frac{a}{b} &= -\frac {\sqrt a} {\sqrt b} \\ \sqrt \frac{a}{b} &= -\frac{y_a \cdot i}{y_b \cdot i} \\ \sqrt \frac{a}{b} &= -\frac{y_a}{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: $\sqrt \frac{a}{b} = \frac {\sqrt a} {\sqrt b}$.