Prove the laws of exponents by induction

Question

We inductively define $a^1=a, a^{n+1}=a^n a$. I want to show that $a^{n+m}=a^n a^m$.

By definition, this is true if $m=1$. Now for $m=2$, we have $$ \begin{align} a^{n+2} =& a^{(n+1)+1}\\ =& a^{n+1}a \\ =& \left(a^{n}a\right)a \\ =& a^{n}\left(aa\right) \\ =& a^{n}a^2 \end{align} $$ How do I finish this proof?

Answer 1

Then assume $a^na^m = a^{n+m}$ and prove $a^na^{m+1} = a^{n+m+1}$. I would say this is very similar to your $m=2$ proof above.

Answer 2

Step one: show that $a^n\cdot a=a^{n+1}$ (given).

Step two: assume that $a^n\cdot a^k=a^{n+k}$.

Then $a^n\cdot a^{k+1}=a^n\cdot a^k\cdot a=a^{n+k}\cdot a=a^{n+k+1}$

then add a bunch of useless words and you're done!

Answer 3

assume $m=k\Rightarrow a^{n+k}=a^na^k$
then we should prove that $m=k+1\Rightarrow a^{n+(k+1)}=a^na^{k+1}$
$$a^{n+(k+1)}=a^{(n+k)+1}=a^{n+k}a\qquad\qquad because\;a^{n+1}=a^na$$ $$\Rightarrow a^{n+(k+1)}=a^{n+k}a=a^na^ka=a^na^{k+1}\qquad\qquad because\;a^{n+1}=a^na$$

Answer 4

Fix any $a\in\mathbb R$ and any $m\in\mathbb N$ and let $P(n)$ be the statement $a^ma^n=a^{m+n}$ and show that for all $n\in\mathbb N$, $P(n)$ is true.

For $n=1$, $P(n)$ is true since by definition, $a^ma^1=a^ma=a^{m+1}$.

Now suppose that for some $n=k$, $P(n)$ is true, i.e., $a^ma^k=a^{m+k}$. Then $a^ma^{k+1}=a^m(a^ka)=(a^ma^k)a=a^{m+k}a=a^{(m+k)+1}=a^{m+(k+1)}$.

Thus, $P(n)\implies P(n+1)$, and since $P(1)$ is true, $P(n)$ is true for all $n\in\mathbb N$.