# Proof 17: Power rules of algebra

#### Definitions:

The pattern looks as if reducing the exponent means removing an $a$ term:

$$\vdots\\ a^4=a\cdot a \cdot a \cdot a\\ a^3=a\cdot a \cdot a\\ a^2=a\cdot a \\ a^1=a$$

As if $a^3=a^4/a$, $a^2=a^3/a$ and $a^1=a^2/a$. To keep this pattern, let’s set $a^0=a^1/a$:1

$$a^0=1\quad_\blacksquare$$

To continue the pattern even further with negative exponents, let’s set:

$$a^{-1}=\frac1a\\ a^{-2}=\frac1{a\cdot a}\\ a^{-3}=\frac1{a\cdot a \cdot a }\\ \vdots$$

So, we have made a general rule for negative exponents saying:

$$a^{-b}=\frac 1{a^b}\quad_\blacksquare$$

Lastly, what about exponents that aren’t whole numbers? Like a half or a third? If we can include those, we have covered all possible numbers. Let’s decide to convert those into roots:

$$a^{1/2}=\sqrt[2]{a}\qquad a^{1/3}=\sqrt[3]{a} \qquad a^{1/4}=\sqrt[4]{a} \qquad \cdots$$

In general, the rule is thus:

$$a^{1/b}=\sqrt[b]{a}\quad_\blacksquare$$

Those were some important definitions. They may seem a bit randomly chosen, especially the latter two. But in fact, they are chosen like this so that the below calculations rules for powers work for all types of exponents.

#### Rules:

Exponents can be added up or merged:

$$a^b\cdot a^c=\underbrace{(a\cdot a \cdot a \cdots)}_{b\text{ many terms}} \cdot \underbrace{(a\cdot a \cdot a \cdots)}_{c\text{ many terms}}= \underbrace{a\cdot a \cdot a \cdot a \cdot a \cdot a \cdots}_{\text{In total }b+c\text{ terms}}=a^{b+c}\quad_\blacksquare$$

$$\left(a^b\right)^c=\underbrace{(a\cdot a \cdot a \cdots)^c}_{b\text{ many terms}} = \underbrace{ \underbrace{(a\cdot a \cdot a \cdots)}_{b\text{ many terms}}\cdot \underbrace{(a\cdot a \cdot a \cdots)}_{b\text{ many terms}} \cdot \underbrace{(a\cdot a \cdot a \cdots)}_{b\text{ many terms}} \cdots }_{c\text{ many portions of }b\text{ many terms}}=a^{b\cdot c}\quad_\blacksquare$$

Exponents can be split out on multiplied and divided terms:

$$(ab)^c=\underbrace{(a\cdot b)\cdot (a\cdot b)\cdot (a\cdot b)\cdots}_{c\text{ many }a\cdot b \text{ terms}} =\underbrace{a\cdot b\cdot a\cdot b\cdot a\cdot b\cdots}_{c\text{ many }a\cdot b \text{ terms}} = \underbrace{\underbrace{a\cdot a\cdot a\cdots}_{c \text{ many } a\text{ terms}} \;\cdot \underbrace{b\cdot b\cdot b\cdots} _{c \text{ many } b\text{ terms}} }_{c\text{ many }a\cdot b \text{ terms}} = a^c\cdot b^c\quad_\blacksquare$$

$$\left(\frac ab\right)^c=\underbrace{\frac ab\cdot \frac ab \cdot \frac ab \cdots}_{ c\text{ many }\frac ab \text{ terms} } = \underbrace{a\frac 1b\cdot a\frac 1b \cdot a\frac 1b \cdots}_{ c\text{ many }\frac ab \text{ terms} }= \underbrace{\underbrace{a\cdot a\cdot a\cdots}_{c \text{ many terms}} \;\cdot \underbrace{\frac 1b\cdot \frac 1b \cdot \frac 1b \cdots}_{c \text{ many terms}}}_{c\text{ many }\frac ab \text{ terms}}=a^c\cdot \frac1{\underbrace{b\cdot b\cdot b\cdots}_{c \text{ many terms}}}=a^c\cdot \frac 1{b^c}=\frac{a^c}{b^c} \quad_\blacksquare$$

The ‘adding up exponents’ rule and the choices for the definitions mean that $a^{3-4}=\frac{a^3}{a^4}$, since:

$$a^{b-c}=a^b\cdot a^{-c}=a^b\cdot \frac 1{a^c}=\frac{a^b}{a^c}\quad_\blacksquare$$

## Proof 18: Root rules of algebra

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## Proof 19: Logarithm rules of algebra

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## Proof 16: Fraction rules of algebra

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## Proof 26: Parallel-axis theorem

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