Proving exponent law for negative integers

Question

I'm trying to prove exponent law: $x^nx^m = x^{n+m}$ for $n,m\in \mathbb Z$ using the exponent laws for natural numbers.

In the case $m\in \mathbb N$ and $0>n\in \mathbb Z $ we get:

$x^mx^n = x^m \frac 1 {x^{-n}} = \frac {x^m}{x^{-n}} $

but I don't know how to proceed from here (I can't use the quotient rule)

Answer 1

By your definition, for $n\in\mathbb N,\;x\neq0$, $x^{-n}=\frac1{x^n}$. From this definition, we know that $x^n\cdot x^{-n}=1$ for all $n\in\mathbb N$. So, given $p,q\in\mathbb Z$, we have a few cases:

Case 1: $p,q<0$. Then, $$x^{p}\cdot x^{q}=\frac1{x^{-p}\cdot x^{-q}}=\frac1{x^{-p-q}}=x^{p+q}$$

Case 2: $p\geq0, q<0,\;p+q\geq0$. We know that $x^{p+q}\cdot x^{-q}=x^p$ since $p+q,-q\geq0$, so $$x^{p+q}\cdot x^{-q}\cdot x^q=x^p\cdot x^q$$

Case 3: $p\geq0,q<0,\;p+q<0$. We know that $x^{-p-q}\cdot x^p=x^{-q}$, since $-p-q,p\geq0$. But that means $$\frac1{x^{p+q}}=\frac1{x^p\cdot x^q}$$

If $q\geq0,\;p<0$, just switch the variable names. And if both are non-negative, we already know the rule holds.

Answer 2

Just because $-n$ has a negative sign doesn't mean it is negative. In fact if $n < 0$ then we know $-n = |n|$ is positive.

If $m> 0$

So $x^mx^n = \frac {x^m}{x^{-n}} = \frac {x^m}{x^{|n|}}$. So if $m \ge |n|$ we have $x^mx^n = \frac {x^m}{x^{|n|}}=x^{m-|n|}$ And if we have $m < |n|$ then we have $x^mx^n = \frac {x^m}{x^{|n|}} = \frac 1{x^{|n| - m}} = x^{-(|n|-m)}$.

In either case $m-|n| = m+ n$ and $-(|n|-m)=m+n$ and $x^mx^n = x^{m+n}$.

And if $m < 0$ then ... well... nothing to it:

$x^mx^n = \frac 1{x^{-n}}\frac 1{x^{-m}} = \frac 1{x^{-n}y^{-m}}$ and as $-n$ and $-m$ are both positive... $= \frac 1{x^{-n-m}} = \frac 1{x^{-(n+m)}} = x^{n+m}$.