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My question is two-fold. First: is $q^{0}=0$ a result or a definition? If it is a definition, is it a vacuous definition?

I'm producing a collection of "review notes" to shore up the countless holes in my mathematical skills and understanding. One of the points I'm unsure of is how, exactly, the zeroth power of a non-zero rational number is established.

Let's assume we have already established the following for $n,m\in\mathbb{N}-0$ and $q,r\in\mathbb{Q}-0:$

$$\frac{q}{q}=1,$$

$$1^{n}=1,$$

$$q^{-n}\equiv\left(\frac{1}{q}\right)^{n},$$

$$q^{n}r^{n}=\left(qr\right)^{n}\text{ and},$$

$$q^{n}q^{m}=q^{n+m}.$$

We arrive at the problem of what to do with:

$$q^{n}q^{-n}=q^{0}=?$$

Since $q^{-n}\equiv\left(\frac{1}{q}\right)^{n}$, we have

$$q^{n}q^{-n}=q^{n}\left(\frac{1}{q}\right)^{n}=\left(\frac{q}{q}\right)^{n}=1.$$

That doesn't look like a definition. It looks like a result, to me.

Until recently, I was unfamiliar with the notion of vacuous definition. I take it to mean that, if there is some statement or condition which has no current definition, we are free to assign it any arbitrary definition, providing it doesn't result in a contradiction.

In the case of $q^{0}$, the existing definition of $q^{n}$ with $n>0$ as a rational number multiplied by itself $n$ times leads to $q$ multiplied to itself $0$ times. Which has no immediate meaning prior to the above development.

But now, simply by following our established operation, we arrive at the assertion of equivalence that for all $n\in\mathbb{N}-0$ and $q\in\mathbb{Q}-0:$

$$q^{n}q^{-n}=q^{n-n}=q^{0}=1.$$

Another motivation for liking the assumption that $$q^{0}=1,$$ is that it allows us to begin exponentiation from zero. that is $q^{0}=1;$ $q^{1}=1\times q;$ $q^{2}=1\times q\times q;$ etc.

Steven Thomas Hatton
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  • For natural numbers $A$ and $B$, $A^B$ is the number of functions from a set with $B$ elements to a set with $A$ elements. If $B=0$, this is the number of of functions from an empty set to a set with $A$ elements. There is exactly one such function, namely the empty function. So $A^0=1$ for natural numbers, and one extends to rational numbers via the rules for exponents. – WillO Jul 19 '18 at 21:27

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Following the definition of exponents that most people first encounter, $q^0$ doesn't make sense (yes, there's the empty product, but that's still too abstract art that age), and neither does $q^{-1}$. Working with that original definition, we see how exponentiation behaves. Most notably $q^mq^n=q^{m+n}$ and $(q^m)^n=q^{mn}$.

If we now wish to expand our domain of exponents to cover all the integers, then there are infinitely many ways to do so. If we want to do it in a way which is useful, then we need to specify what we want from the generalisation. Collectively we have decided that keeping the two properties $q^mq^n=q^{m+n}$ and $(q^m)^n=q^{mn}$ is useful. If you want to keep these two properties, there really is only one way to expand the notion of exponents to include integer exponents (and even rational exponents, as long as $q>0$).

That's all there is to it, really. Start out with a useful, but limited concept, try to expand said concept while retaining usefulness, and then see what you end up with. $q^0=1$ is a result of this process.

So to answer your first question, it can be both. We have the result "If we want to keep exponents as defined on the naturals useful when we expand to the integers, then $q^0$ must be $1$", then there is the definition "$q^n$ for integer $n$ is given by $q^n$ if $n$ is positive, $1/q^{-n}$ if $n$ is negative and $1$ if $n=0$".

Arthur
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  • So is $q^0=1$ the necessary consequence of prior definitions, or is it a new definition? Particularly, in the way I arrived at it. – Steven Thomas Hatton Jun 14 '18 at 14:55
  • Since Arthur’s answer didn’t respond to your question of whether taking $q^0=1$ for $q\neq0$ is a vacuous definition, I’ll add that I don’t think mathematicians use the phrase “vacuous definition.” Mathematicians use vacuous to describe propositions that are true because there is nothing to verify. For example, it is “vacuously true” that the empty set is a set of prime numbers, because there are no elements in the set to verify the primality of. – Steve Kass Jun 14 '18 at 14:56
  • @StevenHatton I added a paragraph explaining how $q^0=1$ can be both a result and a definition, depending on how you look at it. – Arthur Jun 14 '18 at 14:59
  • @SteveKass Touché. I should have asked if it is vacuously true. https://en.wikipedia.org/wiki/Vacuous_truth – Steven Thomas Hatton Jun 14 '18 at 15:04
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    @StevenHatton It's definitely not vacuously true. If you try to state, for instance, "for any number $q$, we have $q^0=1$" then that may be true, but it's not because there are no numbers $q$ to check in the first place. – Arthur Jun 14 '18 at 15:08