TAOCP 1.2.2 Exercise 9
Let x=\frac pq,\qquad y=\frac rs, where $p,r\in\mathbb Z$ and $q,s\in\mathbb Z_{>0}$.
Section 1.2.2: Numbers, Powers, and Logarithms
Exercise 9. [M23] Given that $x$ and $y$ are rational, prove the laws of exponents under the assumption that the laws hold when $x$ and $y$ are integers.
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Solution
Let
$$ x=\frac pq,\qquad y=\frac rs, $$
where $p,r\in\mathbb Z$ and $q,s\in\mathbb Z_{>0}$. By Eq. (6),
$$ b^{p/q}=\sqrt[q]{,b^p,}, $$
so rational powers are defined in terms of positive roots.
We shall prove
$$ b^{x+y}=b^x b^y, \qquad (b^x)^y=b^{xy}, $$
assuming the laws of exponents are already known for integer exponents.
1. Proof that $b^{x+y}=b^x b^y$
Since
$$ x+y=\frac{ps+rq}{qs}, $$
we have
$$ b^{x+y} = b^{(ps+rq)/(qs)} = \sqrt[qs]{,b^{ps+rq},}. $$
Using the integer exponent law,
$$ b^{ps+rq}=b^{ps}b^{rq}, $$
hence
$$ b^{x+y} = \sqrt[qs]{,b^{ps}b^{rq},}. $$
Now compute the $qs$-th power of $b^x b^y$:
$$ (b^x b^y)^{qs} = \left(\sqrt[q]{b^p}\right)^{qs} \left(\sqrt[s]{b^r}\right)^{qs}. $$
Since
$$ \left(\sqrt[q]{b^p}\right)^q=b^p, $$
it follows that
$$ \left(\sqrt[q]{b^p}\right)^{qs} = \left(\left(\sqrt[q]{b^p}\right)^q\right)^s = (b^p)^s = b^{ps}, $$
where the last step uses the integer exponent laws. Similarly,
$$ \left(\sqrt[s]{b^r}\right)^{qs} = b^{rq}. $$
Therefore
$$ (b^x b^y)^{qs} = b^{ps}b^{rq} = b^{ps+rq}. $$
Also,
$$ (b^{x+y})^{qs} = \left(\sqrt[qs]{b^{ps+rq}}\right)^{qs} = b^{ps+rq}. $$
Thus $b^{x+y}$ and $b^xb^y$ are positive numbers with the same $qs$-th power. By uniqueness of the positive $qs$-th root,
$$ b^{x+y}=b^xb^y. $$
2. A lemma: $(b^x)^m=b^{mx}$ for every integer $m$
This was used implicitly in the flawed proof and must be established.
Case $m>0$
By repeated multiplication,
$$ (b^x)^m = \underbrace{b^x b^x\cdots b^x}_{m\text{ factors}}. $$
Using the addition law already proved,
$$ b^x b^x=b^{x+x}, $$
and repeating this,
$$ (b^x)^m = b^{x+\cdots+x} = b^{mx}. $$
Case $m=0$
$$ (b^x)^0=1=b^0=b^{0x}. $$
Case $m<0$
Write $m=-n$ with $n>0$. Then
$$ (b^x)^m = (b^x)^{-n} = \frac1{(b^x)^n}. $$
By the positive case,
$$ (b^x)^n=b^{nx}, $$
hence
$$ (b^x)^m = \frac1{b^{nx}} = b^{-nx} = b^{mx}. $$
The last equality follows from the addition law:
$$ b^{nx}b^{-nx}=b^0=1. $$
Therefore
$$ (b^x)^m=b^{mx} $$
for every integer $m$.
3. Proof that $(b^x)^y=b^{xy}$
Since
$$ y=\frac rs, $$
the definition of rational powers gives
$$ (b^x)^y = \sqrt[s]{(b^x)^r}. $$
By the lemma,
$$ (b^x)^r=b^{rx}, $$
so
$$ (b^x)^y = \sqrt[s]{,b^{rx},}. $$
Since $x=p/q$,
$$ rx=\frac{rp}{q}, $$
hence
$$ (b^x)^y = \sqrt[s]{,b^{rp/q},}. $$
Let
$$ A=(b^x)^y, \qquad B=b^{xy}=b^{pr/(qs)}. $$
We compare their $s$-th powers.
From the preceding expression for $A$,
$$ A^s=b^{rp/q}. $$
Now
$$ B = b^{pr/(qs)} = \sqrt[qs]{,b^{pr},}. $$
To compute $B^s$, we must justify the root identity that was missing in the flawed proof.
Let
$$ C=\left(\sqrt[qs]{a}\right)^s, \qquad a>0. $$
Then
$$ C^q = \left(\left(\sqrt[qs]{a}\right)^s\right)^q = \left(\sqrt[qs]{a}\right)^{qs} = a. $$
Since $C>0$, $C$ is the unique positive $q$-th root of $a$. Therefore
$$ \left(\sqrt[qs]{a}\right)^s=\sqrt[q]{a}. $$
Applying this with $a=b^{pr}$,
$$ B^s = \left(\sqrt[qs]{b^{pr}}\right)^s = \sqrt[q]{b^{pr}} = b^{pr/q}. $$
Hence
$$ B^s=b^{pr/q}=b^{rp/q}=A^s. $$
Both $A$ and $B$ are positive, and their $s$-th powers are equal. By uniqueness of positive $s$-th roots,
$$ A=B. $$
Therefore
$$ (b^x)^y=b^{xy}. $$
We have proved
$$ b^{x+y}=b^xb^y, \qquad (b^x)^y=b^{xy}, $$
for all rational $x$ and $y$, assuming the exponent laws for integers.
$$ \boxed{\text{The laws of exponents hold for rational exponents.}} $$