Proof 17: Power rules of algebra


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

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


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

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.


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:

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

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