Let \(a \in \mathbb{R}\) and \(a \neq 0\) . Define \(a^0 = 1\), and for \(n \in \mathbb{Z}_+\), \(a^{-n} = 1 / a^n\) . Show that the laws of exponents hold for \(a, b \neq 0\) and \(n, m \in \mathbb{Z}\) .
Solution
Let \(a,b\ne0\), with \(a^0=1\) and \(a^{-n}=1/a^n\) for \(n>0\).
We first extend \[ a^ra^s=a^{r+s} \] to all integers \(r,s\). For \(r,s>0\) this is the addition law for positive exponents, E-MUN-4-6. If \(r,s>0\), then \[ a^{-r}a^{-s}=\frac1{a^ra^s}=\frac1{a^{r+s}}=a^{-(r+s)}. \] For mixed signs, say \(r\ge0\) and \(s=-q<0\), \[ a^ra^{-q}=\frac{a^r}{a^q}. \] If \(r\ge q\), E-MUN-4-6 gives \(a^r=a^{r-q}a^q\), hence the quotient is \(a^{r-q}\). If \(r<q\), then \(a^q=a^ra^{q-r}\), hence the quotient is \[ \frac1{a^{q-r}}=a^{r-q}. \] The other mixed case follows by commutativity. Thus the addition law holds for all integer exponents.
For the power law, if \(s>0\), repeated use of the addition law gives \[ (a^r)^s=a^{rs}. \] For \(s=0\) both sides equal \(1\), and for \(s=-q<0\), \[ (a^r)^s=\frac1{(a^r)^q}=\frac1{a^{rq}}=a^{-rq}=a^{rs}. \] Hence \[ \boxed{(a^r)^s=a^{rs}} \] for all integers \(r,s\).
Finally, E-MUN-4-6 gives \((ab)^m=a^mb^m\) for \(m>0\). For \(m=0\) both sides are \(1\), and for \(m=-q<0\), \[ (ab)^{-q}=\frac1{(ab)^q}=\frac1{a^qb^q}=a^{-q}b^{-q}. \] Thus all three laws of exponents hold for nonzero bases and arbitrary integer exponents.