# Proof 18: Root rules of algebra

#### Requirements:

Remember the sign rule: both ‘positive times positive gives positive’ and ‘negative times negative gives positive’. Only ‘negative times positive (or opposite) gives negative’. Same signs give positive. ‘A number times itself’ is two same-signed numbers multiplied and will thus always give a positive result.

Since the square root finds the number that was multiplied with itself, it is not possible to solve something like $\sqrt{-25}$. Because no number can be multiplied with itself to give this negative $-25$. Numbers multiplied with themselves three times can give a negative result, such as

$$(-3)^3=(-3)\cdot\underbrace{(-3)\cdot(-3)}_9 =-3\cdot 9=-27\qquad\text{ so }\qquad \sqrt{-27}=-3$$

But not four times. Five times is again possible, but not six times. The pattern is clearly that multiplying a number with itself an even number of times always gives a positive result. So, you cannot take an even root of a negative number because there is no answer to it: $\sqrt a$, $\sqrta$, $\sqrta$ etc. all require that $a\geq 0$.1 $\quad_\blacksquare$

#### Rules:

Roots can be split over multiplied and divided terms: $\sqrt{ab}=\sqrt a\cdot \sqrt b$ and $\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}$, because with numbers multiplied with themselves, $a=cc$ and $b=dd$, we get:

$$\sqrt{a\cdot b}=\sqrt{\underbrace{c\cdot c}_a \cdot \underbrace{d \cdot d}_b}= \sqrt{c\cdot d \cdot c \cdot d}=\sqrt{(c\cdot d)^2}=c\cdot d=\underbrace{\sqrt{c \cdot c}}_c \cdot \underbrace{\sqrt{d \cdot d}}_d=\sqrt a\cdot \sqrt b\quad_\blacksquare$$

$$\sqrt{\frac a b}=\sqrt{\frac{c\cdot c}{d \cdot d}}= \sqrt{\frac cd \cdot \frac cd}=\sqrt{\left(\frac cd\right)^2}=\frac cd=\frac{\sqrt{c\cdot c}}{\sqrt{d\cdot d}}=\frac{\sqrt a}{\sqrt b}\quad_\blacksquare$$

When roots are done on roots, they can be merged. For instance, think of a number like $a=d^6$:

$$\sqrt{\sqrta}= \sqrt{\sqrt{d\cdot d \cdot d \cdot d \cdot d \cdot d}}= \sqrt{\sqrt{(d\cdot d) \cdot (d \cdot d) \cdot (d \cdot d)}}= \sqrt{\sqrt{(d\cdot d)^{3}}}=\sqrt{d\cdot d}=d=\sqrt{d^6}= \sqrt{a}$$

This corresponds to simply multiplying the two indexes of the roots – the 2 and 3 from the square and cubic roots – into a 6th root. This is general for any numbers:

$$\sqrt[c]{\sqrt[b]a}=\sqrt[b\cdot c]a \quad_\blacksquare$$

## Proof 16: Fraction rules of algebra

Requirements: The denominator in a fraction tells ‘how many portions we split a value into’. Split it into 2 or 3 or 4 portions: $\frac12$,…