Prove by induction that $a^{m+n} = a^m \times a^n$ and $(a^m)^n = a^{mn}$, for $a \in \mathbb{R}$ and $n, m \in \mathbb{N}$.

Question

Prove by induction that $a^{m+n} = a^m \times a^n$ and $(a^m)^n = a^{mn}$, for $a \in \mathbb{R}$ and $n, m \in \mathbb{N}$.

Using mathematical induction, we should check that for $m, n = 1$, the conclusion is true. Now, what about $k + 1$ and $j + 1$? Is it true that mathematical induction can be generalised to 2 variables like this?

Answer 1

For this one, you don't need to do an induction on the two variables at once. First, fix $m \in \mathbb N$. Then prove by induction on $n$ regardless of $m$. Hence what you have to prove is:

$$a^{m+0} = a^m \cdot a^0,$$ $$a^{m+(n+1)} = a^m \cdot a^{n+1} \text{ assuming } a^{m+n} = a^m \cdot a^{n}.$$