Here is a summary of some algebraic properties of exponents, with attempts at proof for each. In all cases below we are assuming that the variables $x, y, b, n, m$ etc are elements of some field. This would most normally be $\mathbf{R}$ or $\mathbf{C}$, but I think can be any field. Note that the inverse operation of exponentiation is taking the [[Logarithms|logarithm]]. ## Summary | Name | Formula | number | | ----------------------------------- | ---------------------------------------------- | ------------------------------------------------------- | | Distributive Property | $(ab)^x = a^xb^x$ | [[Exponents#1. Distributive property\|1]] | | Product Rule | $b^x \cdot b^y = b^{x+y}$ | [[Exponents#2. Product Rule\|2]] | | Power Rule | $\left( b^x \right)^y = b^{xy}$ | [[Exponents#3. Power Rule\|3]] | | Quotient Rule | $\frac{b^x}{b^y} = b^{x-y}$ | [[Exponents#4. Quotient Rule\|4]] | | Negative exponents | $\frac{1}{x^n} = x^{-n}$ | [[Exponents#5. Negative exponents\|5]] | | Distributive property for quotients | $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$ | [[Exponents#6. Distributive property for quotients\|6]] | | The zero exponent | $x^0 = 1$ | [[#7. The zero exponent\|7]] | | Fractional exponents | $x^{m/n} = \sqrt[n]{x^m}$ | [[Exponents#8. Fractional exponents\|8]] | ## 1. Distributive property $(ab)^x = a^xb^x \tag{1}$ #### Proof We seek to prove that $(ab)^x=a^x b^x$ for all real numbers $x$, but since I don’t know how to generalize induction to real numbers, for now we will simply prove it for natural numbers. Therefore let $P(n)$ be the proposition that $(ab)^n = a^n b^n$ for some natural number $n$. Then $P(1)$ is that $(ab)^1 = a^1b^1$. This is trivially true. Next we seek to prove that $P(k)\implies P(k+1)$. That is that if $(ab)^k=a^k b^k$ then $(ab)^{k+1}=a^{k+1} b^{k+1}$. $ \begin{align*} \text{Now suppose that }~ (ab)^k &= a^kb^k,~\text{then}\\ (ab)^{k+1} &= \left((ab)^kab\right)\\ &= (a^kb^k)ab &~\text{by induction hypothesis}\\ &= (a^ka)(b^kb) &~\text{by commutativity of multiplication}\\ &= a^{k+1}b^{k+1} \\ \text{therefore}~(ab)^k=a^k b^k &\implies (ab)^{k+1} = a^{k+1}b^{k+1}\\ \text{and thus since}~(ab)^1 &= a^1b^1\\ (ab)^n &= a^nb^\;\forall n \in \mathbb{N} \quad\square &~\text{by mathematical induction} \end{align*} $ ## 2. Product Rule $b^x \cdot b^y = b^{x+y} \tag{2}$ #### Sketch of intuition behind the proof $ \begin{align*} b^0 \cdot b^0 = 1 &= b^{0+0=0}\\ b^1 \cdot b^0 = b^0 \cdot b^1 &= b \\ b^\alpha &= b^{\alpha-1} \cdot b\\ b^x \cdot b^y &= b^{x-1}\cdot b \cdot b^{y-1} \cdot b \\ &= \underbrace{b \cdot b \cdots b}_\text{x times} \cdot \underbrace{b \cdot b \cdots b}_\text{y times} \\ &= \underbrace{b \cdot b \cdots b}_{{x+y}\text{ times}} \\ \text{therefore } b^x \cdot b^y &= b^{x+y} \quad \square \end{align*} $ #### Proof We seek to prove that $b^m b^n = b^{m+n}$ for all natural numbers $m$ and $n$. Let $P(n,m)$ be the proposition that $b^n b^m=b^{n+m}$ First we will show $ \begin{align*} b^0b^y &= 1\cdot b^y = b^{0+y} \\ b^xb^0 &= b^x\cdot 1 = b^{x+0} \\ \\ \text{Now suppose that}~b^mb^n &= b^{m+n},&~\text{then} \\ b^{m+1}b^n &= b^mb^nb \\ &= (b^mb^n)b&~\text{by associativity of multiplication}\\ &= (b^{m+n})b&~\text{by the induction hypothesis}\\ &= b^{m+n+1}&~\text{by the distributive property of multiplication}\\ \text{And since}~b^mb^n &= b^nb^m &~\text{by commutativity of multiplication}\\ b^mb^{n+1} &= (b^mb^n)b \\ &= b^{m+n+1} &~\text{in a similar fashion} \\ \text{therefore}~b^xb^y &= b^{x+y} \quad \square &~\text{by induction} \end{align*} $ ## 3. Power Rule $(b^x)^y = b^{xy} \tag{3}$ #### Proof $ \begin{align*} (b^x)^y &= \underbrace{b^x \cdot b^x \cdots b^x}_\text{y times}\\ &= b^{xy} \text{ given (2)} \\[15pt] \text{therefore } (b^x)^y &= b^{xy} \quad \square \end{align*} $ ## 4. Quotient Rule $\frac{b^x}{b^y} = b^{x-y} \tag{4}$ #### Derivation/intuition $ \begin{align*} \frac{b^x}{b^y} &= \frac{\underbrace{b \cdot b \cdots b}_\text{x times}}{\underbrace{b \cdot b \cdots b}_\text{y times}} \\ \end{align*} $ So $y$ factors of $b$ will cancel out top and bottom, leaving $b^{x-y}$. This feels unsatisfactory as a proof. What if $y > x$ for instance? #### Proof $ \begin{align*} \frac{b^x}{b^y} &= \frac{b^{x-y} \cdot b^y}{b^y} & \text{regrouping factors} \\ &= \frac{b^{x-y} \cdot b^y}{b^y} \cdot \frac{\frac{1}{b^y}}{\frac{1}{b^y}} & \text{multiplying top and bottom} \\ &= \frac{b^{x-y}}{1} & \text{simplifying}\\[15pt] &= b^{x-y} \\[15pt] \text{therefore } \frac{b^x}{b^y} &= b^{x-y} \quad \square \end{align*} $ ## 5. Negative exponents $\frac{1}{x^n} = x^{-n} \tag{5}$ #### Proof Recall from (4) that $\frac{x^a}{x^b} = x^{a-b}$ Therefore $\begin{align*} \frac{x^1}{x^1} &= x^{1-1} = x^0 \\ \text{But } \frac{\alpha}{\alpha} &= 1 \quad \forall \; \alpha \in \mathbf{F}\\ \text{so } x^0 &= 1. \\[12pt] \text{However, it follows also from (4) that } \frac{x^0}{x^n} &= x^{0-n} \\ \text{therefore } \frac{1}{x^n} &= x^{-n} \quad \square \end{align*} $ ## 6. Distributive property for quotients $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n} \tag{6}$ #### Proof $ \begin{align*} \frac{x}{y} &= x \cdot \frac{1}{y} \\ \frac{1}{y} &= y^{-1} & \text{from (5)} \\ \text{So} \; \frac{x}{y} &= x \cdot y^{-1} \\[15pt] ( x \cdot y^{-1} )^n &= x^n \cdot y^{-n} & \text{from (1)} \\ y^{-n} &= \frac{1}{y^n} & \text{from (5)} \\[15pt] \text{therefore } \left(\frac{x}{y}\right)^n &= \frac{x^n}{y^n} \quad \square \end{align*} $ ## 7. The zero exponent $ x^0 = 1 \tag{7} $ #### Proof $ \begin{align*} \text{Recall that }x^n \cdot x^{-n} &= x^{n-n} = x^0 & \text{from (2)} \\ \text{…and } x^{-n} &= \frac{1}{x^n} & \text{from (5)} \\ \text{…so } \frac{x^n}{x^n} &= x^0.\\ \\ \text{But } \frac{1}{x^n} &\text{ is the multiplicative inverse of } x^n, \text{ meaning} \\ \frac{x^n}{x^n} &= x^n \cdot \frac{1}{x^n} = 1 & \\[8pt] \text{So } x^0 &= 1 \quad \square \end{align*} $ **Note**: I am actually proving $x^0 = 1$ as a step in (5) at present. ## 8. Fractional exponents $x^{m/n} = \sqrt[n]{x^m} \tag{8}$ #### Proof $ \begin{align*} \text{Let } y &= \sqrt[n]{x^m} \\ \text{We know that } y^n &= x^m \\ \text{But } \left(x^{m/n}\right)^n &= x^{mn/n} = x^m & \text{from (3)} \\ \text{So } y &= x^{m/n} \\ \text{Therefore } x^{m/n} &= \sqrt[n]{x^m} \quad \square \end{align*} $ [[Exponential Models]]