For positive integer $n$,

$$a^n:=a\cdot a\cdot a...a \ \ \ (n\ \text{times})$$ $$a^{-n}:=\frac{1}{a^n}$$ $$a^0:=1$$

I want to prove the properties, $$a^n\cdot a^m=a^{n+m}$$ $$\frac{a^n}{a^m}=a^{n-m}$$ $$(a^n)^m=a^{nm}$$ for all integer values of $n,m$. It's easy to prove them for positive integers using the 3 definitions. Is there any way I could extend the proof to any integer value (including 0) directly (perhaps by mathematical induction), or do I have consider each possible combination of positive, negative and 0 separately?

This might be a very basic question but I'm trying to improve by proof writing skills and would appreciate any help.


My text book says that 0 and negative exponents are defined the way they are as that is the only way to make the properties hold for all values of $n$ & $m$. Wouldn't that mean that I'll have to prove the 0 and negative cases separately, usingg 2 extra defintions?

  • $\begingroup$ This depends. Has addition been defined? Do we know that if you combine 7 apples with 5 apples that you will have 7+5 apples? Can you know that if you multiply $a$ by itself $n$ times and then continue with multiplying by itselft $m$ more times that you have multiplied it by itself $m+n$ times? I'd say it follows by definition. But, to play it safe, maybe you should prove it be induction (but then again how do you know proofs by induction works). To do this properly you have to know you basic definitions an axioms. $\endgroup$
    – fleablood
    Commented Aug 3, 2021 at 5:17
  • $\begingroup$ Addition and multiplication are assumed to be defined. I'm still in high school, so I don't think I would be able to grasp defining something as basic as addition and multiplication for non-natural numbers. $\endgroup$ Commented Aug 3, 2021 at 5:28
  • $\begingroup$ @fleablood: I presume defining the set of natural numbers $\Bbb N$ and then addition and multiplication on $\Bbb N$ (via Peano axioms) would be a prerequisite before even attempting to give a formal proof for this, otherwise what does $m+n$ even mean? Or what does $a\cdot a\cdots a$ ($n$ times) mean? $\endgroup$ Commented Aug 3, 2021 at 5:35
  • $\begingroup$ @PrasunBiswas But then what is there to prove? $a^n\cdot a^m =\underbrace{\underbrace{a\cdot a...a}_n\cdot\underbrace{a\cdot a...a}_m}_{n+m} = a^{n+m}$. What is there to prove? $\endgroup$
    – fleablood
    Commented Aug 3, 2021 at 5:41
  • $\begingroup$ @fleablood: I think it would be proving that the power laws that hold in $\Bbb N$ also hold in $\Bbb Z$ $\endgroup$ Commented Aug 3, 2021 at 5:46

1 Answer 1


Yes, a complete proof will cover all the cases. But you can simplify the work a lot: note that proving $a^{m+n}=a^m\cdot a^n~~~(\dagger)$ for integers $m,n$ imply the two other properties:

$$\frac{a^m}{a^n}=\begin{cases}a^m\cdot\dfrac 1{a^n}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m+(-n)}=a^{m-n}&,n\gt 0\\ \dfrac{a^m}{a^0}=a^m=a^m\cdot a^0\overset{\dagger}{=}a^{m-0}&,n=0\\ \dfrac{a^m}{a^{-(-n)}}=a^m\cdot\dfrac 1{a^{-(-n)}}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m-n}&,n\lt 0\end{cases}$$

$$(a^m)^0:=1=a^{0}=a^{m\cdot 0}$$

$$(a^m)^n=\begin{cases}\underbrace{a^m\cdot a^m\cdots a^m}_{n\text{ times}}\overset{\dagger}{=}a^{\underbrace{m+m+\cdots+m}_{n\text{ times}}}=a^{mn}&,n\gt 0\\ \dfrac 1{\underbrace{a^m\cdot a^m\cdots a^m}_{-n\text{ times}}}\overset{\dagger}{=}\dfrac 1{a^{\underbrace{m+m+\cdots+m}_{-n\text{ times}}}}=\dfrac 1{a^{m(-n)}}=\dfrac 1{a^{-(mn)}}=a^{mn}&,n\lt 0\end{cases}$$

So, all you need to do is to prove $(\dagger)$ for all integers $m,n$

Once you have proved it for positive integers $m,n$, you can wlog consider only one of $m,n$ to be negative and by symmetry, you prove the cases where exactly one of $m,n$ is negative. For the case of both $m,n$ negative, we note that $$a^{m+n}=a^{-(-m-n)}=\frac 1{a^{(-m)+(-n)}}\overset{\dagger}{=}\frac 1{a^{-m}\cdot a^{-n}}=\frac 1{a^{-m}}\cdot\frac 1{a^{-n}}=a^m\cdot a^n$$

where we apply $(\dagger)$ with positive integers $-m,-n$

which leaves us with the case of one or both of $m,n$ be $0$ which is easy to finish up. By symmetry, wlog prove with just one of $m,n$ being $0$ and for the case of $m=n=0$, we have $a^{0+0}=a^{0}=1=1\cdot 1=a^0\cdot a^0$. $_\square$

  • 1
    $\begingroup$ @OVERWOOTCH: I added cases in the $a^m/a^n$ part to complete (rigorify) the argument how $(\dagger)$ for integers $m,n$ imply that it is $a^{m-n}$ in general. You may want to check up on that and let me know if there's something missing. The point is that when $n\lt 0$, $-n\gt 0$ is a positive integer and we use the given axioms with $-n$ $\endgroup$ Commented Aug 3, 2021 at 6:27
  • 1
    $\begingroup$ In the case where exactly one of $m$ and $n$ is negative, do I have to separately prove the cases where the negative exponent has a greater/lower magnitude? I know choosing $m$/ $n$ to be negative is an arbitrary choice, but further choosing which exponent to be larger in magnitude is not arbitrary right? Since now $m$ and $n$ are no longer identical? $\endgroup$ Commented Aug 3, 2021 at 23:56
  • $\begingroup$ @OVERWOOTCH: Yes, wlog, considering $m$ to be the negative exponent and $n$ the positive one, you need to prove the two cases $0\lt |m|\le n$ and $0\lt n\le |m|$ separately ($|m|=-m$ since $m$ is negative). This has already been pointed out in the comments by fleablood. The first two axioms should be useful here: prove the lemma that $a^{u-v}=a^u\cdot a^{-v}$ for positive integers $u,v$ where $v\le u$ (so that $0\le u-v$ is a natural). The two cases can be now be proved using $u=n,v=|m|=-m$ and $u=|m|=-m,v=n$ in the proven lemma. $\endgroup$ Commented Aug 4, 2021 at 0:36
  • $\begingroup$ Alright, I think I got it. Thanks $\endgroup$ Commented Aug 4, 2021 at 2:48

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