Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 7. [M23] Given that $x$ and $y$ are integers, prove the laws of exponents, starting from the definition given by Eq. (4).

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Assume throughout that $b\neq 0$. The recursive definition of integer powers is

$$ b^0=1, $$

$$ b^n=b^{n-1}b \qquad (n>0), $$

$$ b^n=\frac{b^{n+1}}{b} \qquad (n<0). $$

We shall prove the exponent laws

$$ b^{x+y}=b^x b^y, $$

and

$$ (b^x)^y=b^{xy}, $$

for all integers $x,y$.

First we establish several consequences of the recursive definition that will be used later.

Lemma 1

For every integer $n$,

$$ b^{n+1}=b^n b. $$

Proof

If $n\ge 0$, then $n+1>0$, so the defining recursion gives directly

$$ b^{n+1}=b^n b. $$

Now suppose $n<0$. Then $n+1\le 0$. Since $n<0$, the negative recursion gives

$$ b^n=\frac{b^{n+1}}{b}. $$

Multiplying both sides by $b$, we obtain

$$ b^{n+1}=b^n b. $$

Thus the identity holds for every integer $n$. ∎

Lemma 2

For every integer $n$,

$$ b^{n-1}=\frac{b^n}{b}. $$

Proof

If $n\le 0$, then $n-1<0$, so the negative recursion gives directly

$$ b^{n-1}=\frac{b^n}{b}. $$

Now suppose $n>0$. By Lemma 1,

$$ b^n=b^{n-1}b. $$

Dividing by $b$, we obtain

$$ b^{n-1}=\frac{b^n}{b}. $$

Thus the identity holds for every integer $n$. ∎

Lemma 3

For every integer $n\ge 0$,

$$ b^{-n}=\frac1{b^n}. $$

Proof

We use induction on $n$.

If $n=0$, then

$$ b^{-0}=b^0=1=\frac1{b^0}. $$

Now assume

$$ b^{-n}=\frac1{b^n}. $$

Since $-(n+1)<0$, the negative recursion gives

$$ b^{-(n+1)} = \frac{b^{-n}}{b}. $$

Using the induction hypothesis,

$$ b^{-(n+1)} = \frac{1/b^n}{b} = \frac1{b^{n+1}}. $$

Therefore the formula holds for $n+1$. Hence

$$ b^{-n}=\frac1{b^n} $$

for all $n\ge 0$. ∎

Proof of the addition law

We prove that

$$ b^{x+y}=b^x b^y $$

for all integers $x,y$.

Fix an integer $x$. Define

$$ P(y): \qquad b^{x+y}=b^x b^y. $$

We prove $P(y)$ for all integers $y$.

Base case

When $y=0$,

$$ b^{x+0}=b^x=b^x\cdot 1=b^x b^0. $$

Thus $P(0)$ holds.

Upward induction

Assume $P(y)$ holds. Then

$$ b^{x+y}=b^x b^y. $$

Using Lemma 1,

$$ b^{x+(y+1)} = b^{(x+y)+1} = b^{x+y}b. $$

Substituting the induction hypothesis,

$$ b^{x+(y+1)} = (b^x b^y)b. $$

By associativity of multiplication,

$$ (b^x b^y)b = b^x(b^y b). $$

Using Lemma 1 again,

$$ b^y b=b^{y+1}. $$

Therefore

$$ b^{x+(y+1)}=b^x b^{y+1}. $$

Hence $P(y+1)$ follows from $P(y)$.

Downward induction

Assume $P(y)$ holds. Then

$$ b^{x+y}=b^x b^y. $$

Using Lemma 2,

$$ b^{x+(y-1)} = b^{x+y-1} = \frac{b^{x+y}}{b}. $$

Substituting the induction hypothesis,

$$ b^{x+(y-1)} = \frac{b^x b^y}{b}. $$

Associativity gives

$$ \frac{b^x b^y}{b} = b^x\frac{b^y}{b}. $$

Applying Lemma 2,

$$ \frac{b^y}{b}=b^{y-1}. $$

Therefore

$$ b^{x+(y-1)}=b^x b^{y-1}. $$

Hence $P(y-1)$ follows from $P(y)$.

Since $P(0)$ is true, upward induction proves $P(y)$ for all $y\ge0$, and downward induction proves $P(y)$ for all $y\le0$. Therefore

$$ b^{x+y}=b^x b^y $$

for all integers $x,y$. ∎

Proof of the power law

We now prove

$$ (b^x)^y=b^{xy} $$

for all integers $x,y$.

Fix an integer $x$. Define

$$ Q(y): \qquad (b^x)^y=b^{xy}. $$

We prove $Q(y)$ for all integers $y$.

Base case

When $y=0$,

$$ (b^x)^0=1=b^0=b^{x\cdot 0}. $$

Thus $Q(0)$ holds.

Upward induction

Assume $Q(y)$ holds. Then

$$ (b^x)^y=b^{xy}. $$

Applying Lemma 1 with base $b^x$,

$$ (b^x)^{y+1}=(b^x)^y b^x. $$

Substituting the induction hypothesis,

$$ (b^x)^{y+1}=b^{xy}b^x. $$

Using the addition law already proved,

$$ b^{xy}b^x=b^{xy+x}. $$

Since

$$ xy+x=x(y+1), $$

we obtain

$$ (b^x)^{y+1}=b^{x(y+1)}. $$

Thus $Q(y+1)$ holds.

Downward induction

Assume $Q(y)$ holds. Then

$$ (b^x)^y=b^{xy}. $$

Applying Lemma 2 with base $b^x$,

$$ (b^x)^{y-1}=\frac{(b^x)^y}{b^x}. $$

Substituting the induction hypothesis,

$$ (b^x)^{y-1} = \frac{b^{xy}}{b^x}. $$

By the addition law,

$$ b^{x(y-1)}b^x = b^{x(y-1)+x} = b^{xy}. $$

Dividing by $b^x$, we obtain

$$ b^{x(y-1)} = \frac{b^{xy}}{b^x}. $$

Hence

$$ (b^x)^{y-1}=b^{x(y-1)}. $$

Thus $Q(y-1)$ follows from $Q(y)$.

Starting from $Q(0)$, upward induction proves the formula for all $y\ge0$, and downward induction proves it for all $y\le0$. Therefore

$$ (b^x)^y=b^{xy} $$

for all integers $x,y$. ∎