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

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?

• This is double-induction. First show that for $m=1$ , the claim is true by usual induction. Then, assume that the claim is true for m and all n, and prove it for m+1 and all n. – Peter Apr 18 '18 at 7:51
• You can indeed do induction over multiple variables and you can do it in several different ways. One common way would be proving it it for $(n,m)=(1,1)$, then proving that it being true for $(n,1)$ implies it is true for $(n+1,1)$, then proving that it being true for $(n,m)$ implies it is true for $(n,m+1)$. Another option is to show that given it is true for $(1,1)$ then assuming it is true for $(n,m)$ show that it is simultaneously true for $(n+1,m)$ and true for $(n,m+1)$. Further methods work too. The point being that we want to show that it is true for arbitrary $(n,m)$. – JMoravitz Apr 18 '18 at 7:54
• Think of induction like knocking over dominos. So long as the initial domino or dominos get knocked down (our base cases) these cause other dominos to get knocked over too (our induction step). So long as all of the dominos get knocked down, it doesn't matter whether it was done in a single path, or whether the path branches off into multiple separate paths, but whatever path or paths the dominos fall in it must hit every domino that we hoped would be knocked down. – JMoravitz Apr 18 '18 at 7:56
• So, there are two steps need to be done for each one variable? – Shane Dizzy Sukardy Apr 18 '18 at 8:00

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