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I think a zero exponent can be defined logically from various premises;

  1. We can define an exponent as the number of times the base "appears" in a multiplication process. Thus: $2^3 = 2*2*2 $, $2^1 = 2 $, $2^0 = ( )$ doesn't appear at all.

Here's the thing, I know that we can literally define $2^{(something)}$ to be anything and as long as we are consistent with our definition, we can show that any result follows directly from our rules and definitions.

  1. But mathematicians observed (and defined) a very nice property of exponents, namely; $b^n * b^m = b^{n+m}$ Thus, if we we want to be consistent with this definition AND WE WANTED TO INCLUDE THE NUMBER ZERO IN THIS, and we put n=0, then we have that $b^0 * b^m = b^m$ And since we have already defined this process $Anything * 1 = anything .....(*)$ Then it should follow that in order to be consistent with our rules, bexp 0 has to be defined as 1. (ofcourse if (*) was defined such as $anything * 5 = anything$, then we'd have to say that $b^0 = 5$).

My question is, however, why do we need to check the case when $n=0$? Why is it important/helpful? What problems will we face if we simply ignored $0$? I mean, the number $0$ wasn't even defined before? ofcourse the number $0$ alone is very useful in other contexts, but why here?

Lastly, fractional exponents are defined in such a way that they "act" like roots; thus making writing and dealing with roots easier, negative exponents make quotients easier and so on.. Is there some situation where we would for example get (something raised to the zero) as a result and therefore we would have to KNOW what that must mean?

I'm writing from a mobile version of the website so I can't use the Math symbols. Sorry about that. Thank you.

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    $\begingroup$ To make negative exponents meaningful. $\endgroup$ – Bernard Feb 14 '17 at 18:53
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    $\begingroup$ It's the obvious choice if you want $\exp$ to be continuous everywhere on $\Bbb R$ (or even just $\Bbb R_{\ge 0}$). $\endgroup$ – Bobbie D Feb 14 '17 at 18:57
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    $\begingroup$ "I'm writing from a mobile version of the website so I can't use the Math symbols" I'm pretty sure you can notwithstanding. $\endgroup$ – MathematicsStudent1122 Feb 14 '17 at 20:15
  • $\begingroup$ To make continual exponents over the reals meaningful, which is to make continuous growth functions calculable. $\endgroup$ – fleablood Feb 14 '17 at 22:21
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You're absolutely right that it is not compulsory to define extended versions of $a^b$. However, it is very often convenient to extend notational devices to situations where the original sense does not directly apply. And, naturally, we'd like the extensions to behave as much like the originals as possible, so that our acquired intuition and reflexes for manipulation of the notation/idea can likewise extend to a larger class of situations. In particular, as you observe, the goal is not just to define things, but to set notational conventions compatible with previous more limited use. (Otherwise we're pranking ourselves by setting ourselves up for silly errors.)

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First I'd say that $b^0=1$ is a natural generalization of $b^1=b,$ $b^2=b*b,$ etc. To see this, note we could write $b^1 = 1*b,$ $b^2 = 1*b*b,$ etc. Since $1$ is the multiplicative identity, $1$ is the natural answer to the seemingly-nonsensical question "what if I multiply $b$ zero times".

Secondly, as you correctly note in point $2$, the left hand side of the equation $ b^{m+n} = b^m b^n$ makes sense when $m$ or $n$ is zero (or, for that matter when one of them is negative but still $m+n>0$) and the definitions $b^{0} = 1,$ $b^{-1}=1/b,$ $b^{-2}=1/b^2$ make the right hand side equal the left hand side in these cases.

Being that we have a nice natural extension of exponentiation to the integers, why not define it that way? You ask whether you ever need to take something to the zero-th power. Well, in a literal sense that's asking if I ever need to take an arbitrary number and turn it into $1,$ so yeah that's not that useful. However since the definition's natural it often happens that we can and want to interpret the expression $b^0$ when it comes up. Here's a couple of examples:

  1. We put $\$100$ in a bank account at $5\%$ (I wish) annual interest, continuously compounded. The amount of money we have after time $t$ is given by $$ M = 100e^{0.05t}.$$ Now if we plug $t=0$ into the expression that gives $M=100.$ Is that right? Well, yes. $t=0$ is right now, and we have $\$100$ right now before receiving any interest. The equation also makes sense for $t=-1.$ $M=100e^{-0.05}$ is the amount of money we would have had to had in the account a year ago in order to have $\$100$ today. (Note this formula also highlights the usefulness of considering powers that are real numbers, not just integers. To that end we note that the exponential function $\exp(x) = e^x$ for real- (or even complex-) valued $x$ is ubiquitous and one of the most well-studied and important functions. Does it have $\exp(0)=1$? You bet. )

  2. A power series is an expression $f(x) = a_0 + a_1x + a_2 x^2\ldots.$ So there is a term proportional to $x,$ a term proportional to $x^2$, etc. There is also a constant term that does not depend on $x,$ but it can be convenient to think of "not depending on $x$" as "being proportional to $x^0$." For instance, we often write $$ f(x) = \sum_{n=0}^\infty a_n x^n$$ and the $n=0$ term naturally gives us our constant with no notational need for a special case. It is sometimes even useful to study power series with negative exponents, e.g. $f(x) = \ldots 1/x^2 + 1/x + 1 + x\ldots$ (these are called Laurent series)

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One part of the answer is "calculus". You want to have some theory of continuous functions, limits, derivatives, integration, ODE and so on; in 18. and 19. century, it has been remarkably successful to describe phenomena from physics. Exponential functions are one of the cornerstone of all of this; even the simplest equations such as $x'=x$, $x'=-x$ or $x''=-x$ have solutions that are various exponential functions. But for any of this, defining $b^n$ for integral $n$ is not good enough.

So the answer is not about one particular example where "writing $b^{-1}$ is provably better than writing $1/b$". There is no such example. But without real functions such as $b^x$ and $e^x$, you (arguably) couldn't even build up calculus in the form in which it is used and applied in recent centuries.

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In a world where we define exponentiation (of positive) by zero:

Compute $$ \int_a^b e^x \, \mathrm{d}x $$


An antiderivative of $e^x$ is $e^x$, so $\int_a^b e^x \, \mathrm{d}x = e^b - e^a$

In a world where we don't

Compute $$ \int_a^b e^x \, \mathrm{d}x $$


We prove by cases that the answer is

$$\int_a^b e^x \, \mathrm{d}x = \begin{cases} e^b - e^a & a \neq 0, b \neq 0 \\ 1 - e^a & b = 0, a \neq 0 \\ e^b - 1 & b \neq 0, a = 0 \end{cases} $$

Case 1: $0 < a \leq b$ or $a \leq b < 0$

An antiderivative of $e^x$ is $e^x$, so $\int_a^b e^x \, \mathrm{d}x = e^b - e^a$

Case 2: $0 = a < b$

This is an improper integral which we compute as

$$ \int_a^b e^x \, \mathrm{d}x = \lim_{t \to 0^{+}} \int_t^b e^x \, \mathrm{d}x = \lim_{t \to 0^{+}} (e^b - e^t) = e^b - 1 $$

Case 3: $a < b = 0$

This is an improper integral which we compute as

$$ \int_a^b e^x \, \mathrm{d}x = \lim_{t \to 0^{-}} \int_a^t e^x \, \mathrm{d}x = \lim_{t \to 0^{-}} (e^t - e^a) = 1 - e^a $$

Case 4: $a < 0 < b$

$$ \int_a^b e^x \, \mathrm{d}x = \int_a^0 e^x \, \mathrm{d}x + \int_0^b e^x \, \mathrm{d}x = (1 - e^a) + (e^b - 1) = e^b - e^a $$

Case 5: $a = b = 0$

$$\int_a^b e^x \, \mathrm{d}x = \int_0^0 e^x \, \mathrm{d}x = 0 $$

Case 6: $b < a$

$$\int_a^b e^x \, \mathrm{d}x = - \int_b^a e^x \, \mathrm{d}x $$

and look up the right hand side by appealing to the appropriate case to determine

$$\int_a^b e^x \, \mathrm{d}x = \begin{cases} e^b - e^a & a \neq 0, b \neq 0 \\ 1 - e^a & b = 0, a \neq 0 \\ e^b - 1 & b \neq 0, a = 0 \end{cases} $$

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  • $\begingroup$ This is a bit confusing as $e^x$ doesn't make sense without the existence of the real function $e^x$ and there is nothing to integrate $\endgroup$ – Peter Franek Feb 14 '17 at 23:08
  • $\begingroup$ That showed convincingly that without $0$, math would be extremely cumbersome! Another example: if we didn't let $x^0$ be $1$ for every $x$ (yes, including $0$), then every instance of $x^m y^n$ in math texts would have to be replaced by $4$ cases; depending on if $m=0$, or $n=0$, or both, or neither. $\endgroup$ – Mark Feb 16 '17 at 15:53
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Regarding your question: What problems will we face if we simply ignored $0$? In this case lots of formulas/theorems/programs/tables/etc. become longer than they need to be.

Suppose for example a store makes a table, the amount sold each day. If you don't allow the number $0$, you'd need two tables "were there any sales that day, yes/no", and "if so, how much". The invention of the number $0$ allows you to do this with just $1$ table.

Likewise, disallowing $0$ in any context (addition, multiplication, exponentiation, etc.) means that you have to distinguish $2$ cases (where $1$ case would otherwise have been sufficient). It would make many formulas longer than they need to be.

In ancient Greece, not only was $0$ not considered a number, but neither was $1$ (so for them, you would not have a number of apples if you have only $1$ apple). Imagine how cumbersome it would be to run computations, make tables, write programs, formulas, theorems, etc., without allowing the numbers $0$ or $1$. The number $0$ is the most important invention in mathematics of all time, not allowing this number is highly inconvenient.

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Continuous growth or decay. We must be able to calculate growth/decay as though it is over a continuum. If it grows at such and such rate over an hour how mean does grow after 20 minutes. (The third root of the rate? What a coincidence!) At what rate is it growing at exactly $12:33$. That'd be $\lim_{t\rightarrow 0}\frac{I*(1+r)^{\frac {33 + t}{60}} - I*(1+r)^{\frac {33}{60}}}t= \lim_{t\rightarrow 0}\frac{I*(1+r)^{\frac {33}{60}}((1+r)^{\frac {33 + t}{60} -\frac{33}{60}} - (1+r)^{\frac{33}{60}-\frac{33}{60}})}t=\lim_{t\rightarrow 0}\frac{I*(1+r)^{\frac {33}{60}}((1+r)^{\frac t{60}} - (1+r)^{0})}t$

...hmmm? Does $(1+r)^0 =1$?

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