definition of exponents

Question

if you wanted to define $a^x$ for all integers would this be ok:

For non negative integers:

$a^0=1$

$a^{(n+1)}=a^n\cdot a$

For negative integers:

$a^{-n} = \frac{1}{a^n}$ an identity which holds for $n$ a positive integer, by definition.

So what my question really is: is the purpose of this definition to give $a^x$ meaning and so to say i wanted to prove this definition would just not be valid? But Then would it be valid to prove the identity $a^n*a^m=a^{(n+m)}$ From this definition for all integers? And lastly could you prove that the identity $a^{-m}=\frac{1}{a^m}$ (which is i think by definition true for positive m) is true for all integers by the defintion given above.

Sorry if I'm way off the mark here, thanks.

Can you clarify what your question is? Are you trying to understand the notation, derive formulas for $a^x$ from the nonnegative integer formulation, or understand someone else's construction of powers? — Michael Burr, May 19 '18 at 11:35
is a^-n = 1/a^n true for positive n by definition? how do you show its true for all integer n? — James Doherty, May 19 '18 at 11:49
If you agree with the basic definition and the proof of the property that $a^{n+m}=a^na^m$ we have that $a^{n}a^{-n}=a^{n-n}=a^0=1$.Thus, form $a^{n}a^{-n}=1$ we have that $a^{-n}= \dfrac {1}{a^{n}}$. — Mauro ALLEGRANZA, May 19 '18 at 18:35
Possible duplicate of what does define $a^x$ for any $a \in \Bbb R$ and $x \in \Bbb Z$ mean? — Mauro ALLEGRANZA, May 19 '18 at 18:55

score 0 · Answer 1 · answered May 19 '18 at 11:37

By definition

$$\overbrace{a\cdot \dots \cdot a}^{\text{n terms}} =a^n$$

thus

$$\overbrace{a\cdot \dots \cdot a}^{\text{n terms}} \cdot \overbrace{a\cdot \dots \cdot a}^{\text{m terms}} =a^{n+m}$$

and you can check that the definition is consistent also with the other rules.

score 0 · Accepted Answer · answered May 19 '18 at 22:20

You propose the following as a definition of $a^n$ for $n\in\mathbb Z$:

$a^0=1$
$a^{n+1}=aa^n, n>0$
$a^{n}=\frac1{a^{-n}},n<0$

This is indeed a valid definition. And you're correct that it's never meaningful in mathematics to talk about "proving" a definition. You can't prove a definition, a definition just is.

You have two mathematical questions:

Can you prove $a^{n+m}=a^na^m$ for $n,m\in\mathbb Z$ from this definition?
Can you prove $a^{-n}=\frac{1}{a^n}$ for $n\in\mathbb Z$?

The answer to both questions is yes, although the details are kind of annoying and you have to break the proof into lots of cases. For the first property, it's probably simplest to fix $n$ and prove the result by induction for $m$. For the second, you're correct that for $n>0$, this is true by definition. For $n<0$, $-n$ is positive, so by point (3) of your definition, $a^{-n}=\frac1{a^n}$, which is equivalent to $a^n=\frac1{a^n}$. For $n=0$ the property is obvious since by definition $a^0=1$.

This is what i was asking thanks. For the second part of the definition do you have to define that a^1=a? And lastly why is it that you can prove parts of the definition from other parts for example 2 can be rearranged then if you sub 0 you get one? was it that they started with 2 defined a^0=1 so that it fits? what was the process in creating this definition? Thanks — James Doherty, May 20 '18 at 09:09
@JamesDoherty $a^1=a$ follows from points (2) and (1) in the definition. You're also correct that point (2) is really the most fundamental. If we want to have $a^{n+1}=aa^n$ for all $n$ (including negative $n$), then there's only one possible way to extend the definition "backwards" into the negative values of $n$. — Jack M, May 20 '18 at 09:16

definition of exponents

2 Answers2