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I am a programmer, but grew up with a pretty weak math education. While I can cobble together enough understanding to build something like these charts solo from raw data, my level of true math literacy has always felt very low in my own opinion.

I recently watched a video that was seductively titled, "What is 0 to the power of 0?" on youtube.

In it, an educator demonstrates that going from $0.9^{0.9}, 0.8^{0.8}, 0.7^{0.7} ...0.1^{0.1}$ shows an odd pattern--somewhere between $0.3$ and $0.4$ the numbers stop going down, and start going up, and if you chase this pattern ($0.000001^{0.000001}$, etc.), you continue approaching $1$.

Now, aside from the fact that I wrote a little sieve to go figure out what's the most precise number javascript can show me as to what that 'splitting' number is, where it starts going up instead of down, the weirdness of that behavior made me realize I don't deeply really seem to understand exponents.

As I've always understood them, the exponent was just a counter for how many times to multiply the base within an operation.

Turns out, as I discovered in my attempts to self-educate, $0^0$ looks to be 1 (though some seem to disagree...?), anything to the 0 power is $1$ (that's irritating, and the kind of thing that makes me hate math--where the hell is that coming from?), and fraction exponents...

...what I'm really getting to is, what the hell is a fractional exponent? What does it really mean? When I picture multiplying a whole number like $7$ by a decimal, like $.5$, I picture the $7$ getting counter $.5$ times, or a jar of quantity $7$ filled up halfway.

What am I supposed to imagine for $7^.5$?

I tried working out my thoughts by writing a little javascript function that calculates exponents. It handles 0,1,negative numbers, and regular numbers all reasonably cleanly, but I really couldn't see how to possibly put a decimal into this framework.

My girlfriend is a mathematician, so I asked her. The best answer she was able to give me (it's hard for us to talk about math--it's like asking a native speaker a question about their language, it's just so hard to understand what it's like to experience not being a native) was (as I understand it), a way around the problem:

Just multiply the exponent by a high enough number that the decimal goes away, and then get the corresponding root of the number after you finish.

Basically, recursion.

I implemented that into my code, and it works. I get something that returns the same as the built-in Math.pow() function in javascript 89% of the time, and disagrees only at extreme precision levels the other 11% of the time.

But I don't feel like I deeply understand what the dance of numbers is that leads $2^{.9}$ to be $1.866[...]$, or, to my original point, what the hell is going on when we move from $0.4^{0.4}$ to $0.3^{0.3}$ (or, more specifically, from $0.367^{0.367}$ to $0.366^{0.366}$, where the downward trend turns upwards mysteriously).

What am I missing about exponents, and why is it so hard to find an explanation in these terms? Is my question just somehow deeply flawed?

Bonus, if it helps, here's my internal thought process for reading an exponent turned into code:


var powerify = function(base, exponent) {
  let i = exponent;

  let answer = 1;

  if (exponent % 1 !== 0) {
    answer = powerify(base, exponent*10) ** .1
  }
  else if (i > 0) {
    while (i > 0) {
        answer = answer * base;
        i = i - 1;
    }
  }
  else if (i < 0) {
    while (i < 0) {
        answer = answer / base;
        i = i + 1;
    }      
  }

  return answer;
}
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    $\begingroup$ Related. $\endgroup$
    – algae
    May 16, 2020 at 3:14
  • $\begingroup$ Thanks for that link. I found math.stackexchange.com/a/1475935/788921 <--this answer, which is great, but doesn't dive into fractional exponents (or negative ones), unfortunately. Something along that type of thinking is exactly what I crave, though. $\endgroup$
    – Kyle Baker
    May 16, 2020 at 3:22

5 Answers 5

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As long as the exponent is rational and you are dealing with positive numbers, you can rely on the law of exponents $\left(a^b\right)^c=a^{bc}$. If you are interested in $7^{0.5}=7^{1/2}$ you know that $\left(7^{1/2}\right)^2=7$ so $7^{1/2}=\sqrt 7$. This establishes the relationship between fractional exponents and roots. This process can handle any rational exponent.

All of this works perfectly well as long as you don't ask about $0^0$. The problem with $0^0$ is that we would like $x^y$ to be a continuous function in both $x$ and $y$. Unfortunately, if you first set $x$ to $0$ and vary $y$ then $x^y=0$. If you first set $y$ to $0$ and vary $x$ then $x^y=1$. There is no way to reconcile that, which we can demonstrate even more clearly using complex variables. Most mathematics just says $0^0$ is undefined. Some parts of mathematics find it more convenient to define it, sometimes as $1$ and sometimes as $0$.

When you get to real exponents we define $x^y=e^{y \log x}$. There are many routes to get to this definition, but fundamentally they are all motivated by the need to get to it. This echos the problem with $0^0$ but otherwise works perfectly.

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  • $\begingroup$ Just out of curiosity, are there actually any branches of mathematics where $0^0 = 0$ by established convention, and if so, what are they? From what I've seen, pretty much anywhere that $0^0$ is defined at all, it's defined as the empty product (and thus equal to $1$). $\endgroup$ May 16, 2020 at 12:30
  • $\begingroup$ I'm pretty sure I have seen it, but I don't remember where. $\endgroup$ May 16, 2020 at 13:49
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Answer for non-mathy people:

The answers given may be perfectly good for other more mathematically inclined people, but they are, unfortunately, over my head due to being too 'mathy'. :/

(After thinking about this for a couple days, and then writing this out over a couple days, I can finally make sense of half of one of the other answers.)

However, diving into deeper research driven by the 'related questions' now popping up (I did try to search before!), in the 8 or so more tabs on this stack exchange, I finally found one that made it click for me. That answer is here, but to avoid being a 'link only' answer, I'll try my best to highlight it in the words of someone as illiterate in math as I am, in the hope that I may speak to someone someday yet further down the totem-pole.

I think the short form of my personal epiphany would be expressed like this:


Multiplication is an order higher than addition

$8*1/4$ = the number that, when you add it four times, you get 8

so, $8*1/4 = 2$

and

$2+2+2+2 = 8$


Exponentiation is an order higher than multiplication

$8^{1/4}$ = the number that, when you multiply it four times, you get 8

so, $8^{1/4}=1.6817[...]$

and

$1.6817[...]*1.6817*1.6817*1.6817=8$ (if the decimals were truly precise)

(for a cleaner example, note that $8^{1/3}=2$ and $2*2*2=8$.)


Just as division is the opposite of multiplication, roots are the opposite of exponents

$4\overline{)8}$ = the number that, when you subtract it 4 times, you get 0

so, $8-2-2-2-2$ = 0.

$\sqrt[4]{8}$ = the number that, when you divide 8 by it four times, you get 1

so, $8/1.6817/1.6817/1.6817/1.6817=1$.

(why 0 vs 1? Don't ask me, but I assume it has to do with deep truths about why 1 behaves uniquely in multiplication and is just another number in addition--e.g., $8+1=9$, an increase of 1, but $8*1=8$, an increase of 0...)

(edit: my mathematician girlfriend informs me that 1 is the "neutral element" of multiplication, and 0 is the "neutral element" of addition--revealed by $8+0=8$ and $8*1=8$)


Now, to try in the words that connect to me:

If I multiply $x$ by 8, just one time, that is framed in this idea as multiplying $x$ by $8^{1}$... so: $8^{1} = 8$

Cool. So what's up with $8^{.5}$? That number is $2.8284271247461903$, not $4$.

My girlfriend attempted to communicate this truth in the language of square roots, and I see why now, though it made no 'deep' sense to me as someone who is unfamiliar with the idea of square (and other) roots at a meaningful level barely beyond pure abstract rote definition.

Much like a square root represents an idea, in that 'the square root of 5' means 'the number that, if you multiplied it by itself, you'd get 5', the number 8$^{.5}$ is also an idea, and that idea is, 'the number that, if you multiplied by it twice*, would be the same as multiplying by 8'.

(*quick clarification: twice, because 0.5 * 2 = 1, which is a lot more obvious if you use fractions instead, so, $8^{1/2}$)

In other words,

$8^{1/2}$ is the square root of 8. and, $8^{1/3}$ is the cube root of 8.

and while $8^{1/4} = 1.6817928305074292$, which looks like a crazy number, it's the one number that, if you do the following:

$2 * 8^{1/4} * 8^{1/4} * 8^{1/4} * 8^{1/4}$

is the same as

$2 * 8$.

Looking at it the reverse way, if you take that number (let's call it $a$, so $a = 1.6817928305074292 = 8^{1/4}$), it's the one number that if you divide 8 by it four times, you'd get $1$, so:

$a = 8^{1/4}$

$1 * a * a * a * a = 8$

$8 / a / a / a / a = 1$

Or, finally, what my girlfriend tried to show me in the first place,

$\sqrt[4]{8^{1}} = a = 8^{1/4}$

(Hope I got all that right--if not, comment/edit away, I'm not a real mathematician)

Now, that paragraph in front of it is what I had to see to understand it. But now that I've seen that, I can understand the connection.

Likewise, working from there, the fractional exponent of a fraction should also be pretty clear, in theory.

$.1^{.1}$ is the number that, when you've multiplied by it ten times, you've done the equivalent of multiplying by $.1$.

So,

$(1/3^{1/3})^{3} = 1/3$

$(.25^{.25})^{4} = 0.25$

Here we take the base 1/3 and make it three orders "smaller", and then three orders "bigger", to arrive right back at itself. (At least, that's how I think of it--keep in mind, it doesn't actually correlate with size when I say order, that direct correlation with size breaks down when dealing with decimals, just like multiplying by a decimal results in a smaller number, not a bigger one.)

Now with that understanding, I can begin approaching my original question, about the behavior of exponents approaching 0.

Looking at $0^{0}$ & $x^{0}$

I notice now that every number looks reeeaaalllly close to 1 when you use really small decimal exponents ($100^{0.000000001}$ & $0.0001^{0.00000001}$ are both basically 1), showing why anything to the exponent of $^{0}$ is considered 1. That second number there ($almost0^{almost0}$) points to why there is a tendency to include $0^{0}$ in that category as well.

Looking at negative exponents

The numbers that come out of negative exponents are somehow extremely familiar, and easy to make rules about, but it takes a bit to thing about how this language above applies. $10^{-1}$ is $1/10$. $10^{-2}$ is $1/100$. $2^{-1}$ is $1/2$. So, the pattern is clear already.

As far as I can understand, I'd say that the english language version idea of a negative exponent is this:

Negative is inverse positive, division is inverse multiplication

So, where $8^{2}$ is the result of multiplying 8 twice ($1*8*8$), $8^{-2}$ is the result of dividing by 8 twice ($1/8/8$) (so, 1/8th of 1/8th).

And negative fraction exponents, then, should be the obvious extension of that, right? Let's see. I'll start with English, make a prediction, and then check it.

"$8^{-1/4}$ should be the number that, if you divide by it 4 times, should be the equivalent of dividing by 8."

Is it right? I just checked. It's close, and the answer is only one word switch around.

$8^{-1/4}=0.5946035575013605=b$

We try, and find:

$8/b/b/b/b=64$

So, to divide by this number is the equivalent of multiplying by 8... so on a quick hunch, we instead find:

$8*b*b*b*b=1$

So, the correct sentence is this:

"$8^{-1/4}$ should be the number that, if you multiply by it 4 times, should be the equivalent of dividing by 8."


...now, if I can just figure out what's special about $0.367879^{0.367879}$, I'll let you know.


Bibliography:

Here some other resources on exponents I found while trying to figure this out, most of which I couldn't understand but may help you if you're at all more math-literate than me (which is a low bar):

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I'm not sure if this is exactly what you are looking for (and there are certainly much more technical explanations that can be given), but perhaps a good "geometric" picture of fractional exponents, of the form $a^{p/q}$ can be given as follows.

As a first example, take first $a=2$ and $p/q=1$. Picture this as a line segment with length $2$ - a segment of the number line in $\mathbb{R}$. Now take the case $p/q=1/2$. Clearly, this is a number that squares to be $2$, by the property $(a^b)^c=a^{bc}$. The term "squares" is very suggestive, however - picture now a square in $\mathbb{R}^2$ with area exactly $2$. The side length of such a square is exactly $2^{1/2}$. Repeat for $2^{1/3}$ - picture a cube in $\mathbb{R}^3$ with volume 2. The side length of such a cube must have length $2^{1/3}$. In factor for any fraction $1/q$, you can picture $2^{1/q}$ as being the side length of a $q$-dimensional hypercube with hypervolume $2$!

Now pick your favorite nonnegative real number $a$, and a positive rational number $p/q$. Write $a^{p/q}=(a^p)^{1/q}$. Now by the same logic used for $2^{1/q}$, we can construct $(a^p)^{1/q}$ as the side length of the $q$-dimensional hypercube with hypervolume exactly $a^p$.

This construction obviously doesn't work for real exponents or negative $a$ (although those problems are more subtle), but what about for $0^0$? Here's a possible explanation (that I do not claim to be $100\%$ rigorous, but it paints a fair picture of the issue) for why we cannot define it so easily, using this visualization.

Consider $0^{1/q}$ for any positive integer $q$. This is the side length of a $q$-dimensional hypercube with hypervolume $0$ - clearly, this side length has to be zero for all $q$. We may choose $q$ arbitrarily large so that $1/q$ becomes arbitrarily close to $0$ - this would suggest that we should define $0^0=0$ (formally, in order to ensure continuity of the function $0^x$). But this does not agree with what you have noticed when considering the function $x^x$ - as $x\to0$, $x^x\to1$, not $0$, and so this suggests that we should actually define $0^0=1$ - so we cannot define $0^0$ in a way that makes the function $a^x$ continuous! For this reason we typically choose not to define it at all (except in special contexts).

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  • $\begingroup$ (As an aside, if anyone has a good way to visualize why $a^0=1$ using this method, I would welcome any feedback - it is late and I cannot think of one at the moment.) $\endgroup$
    – csch2
    May 16, 2020 at 4:29
  • $\begingroup$ Since $0=-0$, $a^0=\frac{1}{a^0}$. The only time when $n=\frac{1}{n}$ is when $n=1$. Therefore, $a^0$ must equal $1$. $\endgroup$ Mar 26 at 17:29
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Illustrating your described pattern; future readers, please feel free to add accompanying observations.

In it, an educator demonstrates that going from $0.9^{0.9}, 0.8^{0.8}, 0.7^{0.7} ...0.1^{0.1}$ shows an odd pattern

what the hell is going on when we move from $0.367^{0.367}$ to $0.366^{0.366}$, where the downward trend turns upwards mysteriously

enter image description here

$$\frac{\mathrm d}{\mathrm dx} x^{x}=\frac{\mathrm d}{\mathrm dx} \left(e^{x\ln x}\right)=\left(x\left(\frac1x\right)+\ln x\right)e^{x\ln x}=(1+\ln x)x^x;\\\frac{\mathrm d}{\mathrm dx} x^{x}=0\iff x=\frac1e.$$

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Here are a few steps to calculate an exponent expression with a non-integer index.

$1$. Express the index in fraction form: $$0.4^{0.4}=0.4^{\frac{2}{5}}$$ $2$. Use the law of exponents (which states that $(a^b)^c=a^{bc}$) to express it in the form $a^{\frac{1}{b}}$: $$0.4^{\frac{2}{5}}=\left(0.4^2\right)^\frac{1}{5}$$ $3$. Calculate the exponential inside the bracket: $$\left(0.4^2\right)^\frac{1}{5}=0.16^\frac15$$ $4$. Use the fact that $a^\frac1b=\sqrt[b]a$: $$0.16^\frac15=\sqrt[5]{0.16}$$ $5$. Calculate the root: $$\sqrt[5]{0.16}\approx0.693144843155146$$ And a proof that $0^0=1$: $$\text{Since $0=-0,a^0=a^{-0}$, so $a^0=\frac1{a^0}$. The only number $n$ where $n=\frac1n$ is 1, so $a^0$ must be equal to $1$ for any number $a$, and that includes 0.}$$

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