Suppose that $a\in \mathbb{R}$ and $a>1$ and we define in the recursive way the following operation: $$a^1:=a,$$ $$a^{n+1}:=a^n\cdot a \quad\text{for each} \quad n\in \mathbb{N}.$$

I want to prove that $a^{n+m}=a^na^m$ for any $n,m\in \mathbb{N}$.

Proof: Let $E=\{m\in \mathbb{N}: a^{n+m}=a^na^m \ \text{for any} \ n\in \mathbb{N}\}$

For any $n\in \mathbb{N}$ we have $a^{n+1}=a^na=a^na^1$ which shows us that $1\in E$.

Suppose that $m\in E$ and we'll try to show that $m+1\in E$.

Indeed, for any $n\in \mathbb{N}$ we have $a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{(n+m)+1}$.

As you see I am almost done but one probably stupid moment bothers me a lot. Probably I am too pedantic.

We know that $(n+m)+1=n+m+1$ by associativity of natural numbers. How it follows that $$a^{(n+m)+1}=a^{n+m+1} \ ?$$

I do not think the we can just substitute $(n+m)+1$ by $n+m+1$.

May be I cannot formulate my question quite explicitly but I'd be very thankful for your help!

  • $\begingroup$ You won't resolve your confusion if you write $n+m+1$. You surely mean $n+(m+1)$. $\endgroup$ – ancientmathematician Jan 24 at 12:18
  • $\begingroup$ @ancientmathematician, could you give more details please? I will appreciate your help! $\endgroup$ – ZFR Jan 24 at 12:20
  • $\begingroup$ That then reduces the question to proving that $+$ is associative on $\mathbb{N}$; which may depend on how you are defining $+$. $\endgroup$ – ancientmathematician Jan 24 at 12:20
  • $\begingroup$ You get to $a^{(n+m)+1}$ OK; then this equals $a^{n+(m+1)}$ by associativity of $+$ on $\mathbb{N}$, so $m+1\in E$. $\endgroup$ – ancientmathematician Jan 24 at 12:22
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    $\begingroup$ The definition you have given of $a^n$ is recursive: you give a base value then define $a^{n+1}$ in terms of $a^n$. It is true but non-trivial that there is one unique function that satisfies these conditions. And it is a function, that is each $n$ is related to precisely one $a^n$. [And no, this is not "injective" - quite the opposite in fact.] $\endgroup$ – ancientmathematician Jan 24 at 13:27

Yes, this last step is fine. If $x=y$ then $a^x=a^y$ and in this case you know that $(n+m)+1=n+m+1$.


I think you answered your own question. Because of associativity, $(n+m)+1=n+(m+1)$, so using your calculation, $a^na^{m+1}=\cdots=a^{(n+m)+1}=a^{n+(m+1)}$, so the inductive step is done.

Recall that associativity means $(a+b)+c=a+(b+c)$.

  • $\begingroup$ Sorry but I don't think that it addresses my question. $\endgroup$ – ZFR Jan 24 at 12:09
  • $\begingroup$ I think the question is "is using associativity valid?", not "how do I use associativity?" $\endgroup$ – No Name Jan 24 at 20:58

Another way of formulating it is that we have a function $exp: \mathbb R \times \mathbb N \rightarrow \mathbb R$ such that $exp(a,n+1) = n* exp(a,n)$ (in other words, I am rewriting $a^n$ as $exp(a,n)$). You are asking whether we can take $exp(a,(n+m)+1)$ to be the same as $exp(a,n+m+1)$. A fundamental property of a well-defined function is that if $x$ and $y$ are different expressions representing the same number, then $f(x)=f(y)$. So, for instance, $f_1(\frac a b) = \frac {2a}b$ is a well-defined function, because applying it to two different representation of the same fraction gives the same result (e.g. $f_1(\frac 1 2)=f_1(\frac 2 4)$). $f_2(\frac a b) = a+b$, on the other hand, is not a well-defined function, because $f_2(\frac 1 2)=3$, while $f_2(\frac 2 4)=6$.

You've said that you've defined $exp$ recursively, and implicit in saying that this is a definition is saying that the function is well-defined.


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