We’ve been looking at oddities of zero. Because “nothing” behaves differently than “something”, operations with it can be surprising. Although students learn that \(x^0=1\) for any non-zero number *x*, they often wonder, *why*?? I’ve selected a few out of at least a dozen such questions in our archive.

## Using the quotient rule for exponents

We’ll start with a question from 1996:

0 PowerDear Dr. Math,Why is any number to the 0 power equal to one?Thank you!Levi Goins

Doctor Anthony answered:

The rules of algebra are designed to beconsistentas far as this is possible.Why is anything to power 0 equal to 1?Consider first a^5/a^3 . As you know this is the same as (a*a*a*a*a)/(a*a*a) = a^2So to get the result wesubtracted the powersto give 5-3 = 2 .What about (a*a*a)/(a*a*a) = a^(3-3) = a^0 ?But we know that a^3/a^3 = 1, and so a^0 = 1 .This does not depend on a, and is true in the general case.

This **quotient rule** can be easily seen by “canceling”: $$\frac{a^5}{a^3}=\frac{a\cdot a\cdot a\cdot a\cdot a}{a\cdot a\cdot a}=\frac{a\cdot a}{1}\cdot\frac{a\cdot a\cdot a}{a\cdot a\cdot a}=\frac{a\cdot a}{1}\cdot1=a^2$$ The general rule is $$\frac{a^m}{a^n}=a^{m-n}$$

But the same rule applied to \(\frac{a^3}{a^3}=a^{3-3}\) tells us that \(a^0=1\). We’ll be looking at this same idea from a higher perspective later.

## Extending a pattern

Jenna wrote on behalf of her class in 1997:

Anything to the 0 PowerDear Dr. Math,Everyone in my 7th grade class wants to know the answer to this question:Why is anything to the 0 power 1?No one in my class knows the answer, not even my teacher. She asked me to e-mail Dr. Math and we are eagerly waiting to hear the answer. Thanks for your time,Jenna

It was common then for classes to write to us like this; teachers might even pretend not to know the answer so their class could have this experience! Doctor Steven answered:

Look at the powers of any number, say 8: 8^1 = 8 8^2 = 64 8^3 = 512 ...To get from 8^3 to 8^2 we have to divide by 8. And to get from 8^2 to 8^1 we have to divide by 8 again. The logical idea is tocontinue doing thisso we get: to go from 8^1 to 8^0 we should divide by 8.So 8^0 = 8^1/8 = 8/8 = 1. This works for any nonzero number. 0^0 is undefined.

Going forward in the list, we multiply by the base, 8; going backward, we divide by the base. So we just keep going backward past the start.

The same reasoning can be stated algebraically: Since \(a^{n-1}=a^n\div a\), when \(a=1\), \(a^0=a^{1-1}=a^1\div a=1\).

But note the comment: All of this assumes the base is not zero, so it doesn’t prove anything about \(0^0\). We’ll get to that even more special case (*two* zeros!) next week.

We can even continue this process to createnegative exponents: to go from 8^0 to 8^(-1) we divide by 8So 8^(-1) = 8^0/8 = 1/8.Now we have this: ... 8^3 = 512 8^2 = 64 8^1 = 8 8^0 = 1 8^(-1) = 1/8 8^(-2) = 1/64 8^(-3) = 1/512 ...Hope this helps.

## Extending the definition to retain the product rule

A father, Rick, in 1996, adapted that answer, or one like it:

Zero as an ExponentMy 7th grade son has a 5^0 question on a study guide and didn't know the answer. I told him thatI thought that the answer was 1based on my math training in years gone by, butI didn't know why. I searched the net and found your page which explained it well enough for me to understand.I went on to explain it to him in this way: 5^1=(5*1)/1, 5^2 = (5*5*1)/1, 5^(-1)=1/(5*1), 5^(-2)=1/(5*5*1). Following this flow, 5^0 would be viewed as 1/1 with no 5's. Then of course 1/1 = 1 or 5^0 = 1. Does this make sense? Have I got it figured correctly? - Rick Humphreys

His attempt assumes that negative exponents are already understood, but is otherwise similar to the **pattern** in the last answer, and to a “redefinition” we’ll see later: $$5^2=\frac{5\cdot5\cdot1}{1},5^1=\frac{5\cdot1}{1},{\color{Red}5^0=?},5^{-1}=\frac{1}{5\cdot1},5^{-2}=\frac{1}{5\cdot5\cdot1}$$

Doctor Ceeks suggested an alternative inspired by Rick’s suggestion, starting with the **definition of a power**, and the **product rule** derived from it:

I do not think your answer is the best answer because it doesn't arise out of any natural sequence of ideas. I think this is more natural:First, the exponential was defined as a notational method to represent the process of multiplying a given number over and over. Thus,5^n = 5 times 5 times 5 times 5, n times, where n is a positive integer. It then follows that 5^(a+b) = 5^a 5^b.

Keep in mind that *n* is the *number of 5’s*, not the *number of multiplications*. If we multiply together \(a\) 5’s, and then \(b\) 5’s, then we are multiplying \((a+b)\) 5’s in all: \(5^a\cdot5^b=5^{a+b}\).

In mathematics, it often happens that one would like toextend the definitionof something. How can we extend the definition of the exponent to all the integers? What property of the exponential can guide us beyond the positive integers? We have the beautiful law that 5^(a+b) = 5^a 5^b. Is it possible to extend the definition so as toretain the fundamental propertythat 5^(a+b) = 5^a 5^b? The answer is yes...it can be extended, and it can be extended in only one way. First, 5^(-1) must be 1/5, because we demand that 5^(n-1)=5^n 5^(-1). But then we see that 5^0 = 5^(1-1) = 5^1 5^(-1) = 5 * 1/5 = 1.

Here he first defined a negative power, as Rick did; for example, since \(5^3=5^{4-1}=5^4\cdot5^{-1}\), but we get from \(5^4\) to \(5^3\) by dividing by 5, we conclude that \(5^{-1}=\frac{1}{5}\).

But then $$5^0=5^{1+(-1)}=5^1\cdot5^{-1}=5\cdot\frac{1}{5}=1$$

## Extending the definition by dividing

Fifth grader David wrote in 1998:

The Zero Power of TwoDear Dr. Math,In fifth grade we've learned that 2 to the third power = 8, two squared = 4, 2 to the first power = 2, and 2 to the zero power = 1. Could you please explain how 2 the zero power = 1 because I'm having trouble understanding this. For example, 2 cubed means that you multiply 2 by itself 3 times.How do you multiply 2 by itself 0 timesin 2 to the zero power?I understand the pattern of 2 cubed, squared, to the first power, and to the zero power (8, 4, 2, 1), but I'm still having trouble with this idea.Could you help? I looked through your elementary archives and found nothing on this subject.David Burns

Doctor Rick answered with an extended version of our first answer:

Hi, David. Good question! Actually we do have material on why a number to the zero power is 1, but I'm not surprised thatit isn't in the Elementary Archives. Questions about why numbers behave as they do are best answered when you get to study algebra.

Most of the answers we’ll see are listed under Middle School. Can we explain this at a pre-algebra level?

You know, there was a time when the only numbers people knew were the counting numbers 1, 2, 3, ... .Zero hadn't been invented yet, so nobody could ask your question. Then zero and negative numbers were invented, and fractions and decimals, and even more that you probably haven't heard of yet.Each time new numbers were invented, mathematicians had to figure out how those numbers behave. You don't want to have a whole new set of rules for the new numbers - you want them tofollow the same old rules, but to take themwhere no number has gone before. This is what happened with powers. When zero is added to the counting numbers, you need to figure out what 2^0 (2 to the 0 power) is. The old definition doesn't help you, because as you say,multiplying zero 2's together doesn't make sense. But you want powers to keep working the same way they always did, and one rule is this: if you divide a number to a power by the same number to a different power, the answer is the same number raised to the difference of the first two powers.

This is our quotient rule, \(\frac{a^m}{a^n}=a^{m-n}\).

For example, 3 2 (3-2) 1 ---- = 2 = 2 2 2What happens when the powers in the numerator and denominator are the same? 3 2 (3-3) 0 ---- = 2 = 2 3 2But you know that 8/8 = 1. So 2^0 must equal 1.

We didn’t have a meaning for \(2^0\) before, but if this rule is to keep working, it has to be 1.

You can do the same sort of thing to figure out what 2^(-1) should be, or what 2^(1/2) should be.I hope this helps you. Keep asking those "why" questions, and you will be all set for algebra, and more!

## Reformulating the definition

Aron, in 2008, knew how to get the answer from the product rule, but it still felt wrong:

How Can a Number Raised to the Zero Power Be One?I understand that 3 raised to the 7th multiplied by 3 raised to the 0 is equal to 3 raised to the 7th, because the rule is that you would add the exponents which would give you 3 to the 7th whichclearly alludes to the fact that 3 to the zero equals oneso that 3 to the 7th is equal to itself........but WHY?How, mathematically, is 3 to the zero equal to one????

His argument is simple: \(3^7\cdot3^0=3^{7+0}=3^7\), so \(3^0\) must equal 1. It’s logically true … but doesn’t fit our understanding of what a power *is*.

I answered:

Hi, Aron.Here is another way to make sense of it:Rather than think of x^n (that is, x raised to the nth power) as x^n = x * x * ... * x \_____________/ n copies of xwe can think of it asstarting with 1and then doing n multiplications--literally "multiplying by x, n times" x^n = 1 * x * x * ... * x \_______________/ n multiplications by xTaking it this way, we have: x^2 = 1 * x * x \_____/ 2 multiplications x^1 = 1 * x \_/ 1 multiplication x^0 = 1 \/ 0 multiplicationsSo x^0 is 1, for any x.

This reformulation of exponentiation for positive exponents naturally extends to zero exponents. It doesn’t help with negative exponents, but makes this “nothing” step feel more reasonable.

## Making sense of multiplying nothing

Connor, in 2007, had a similar concern:

More on Why n^0 = 1Why does n^0 = 1? At first my question may seem the same as the other question in your archive, but read the whole thing please, because it's different. I saw your other question on n^0, and you gave the same response as the one I got from my math teacher. My question, which she couldn't answer, is if 2^3=2*2*2, 2^2=2*2, and 2^1=2,shouldn't 2^0= , meaning that 2^0=0?I already know that 2^5=2^6/2 and so on. I've tried 2^3=1*2*2*2, but it didn't seem to make sense to have to add that 1. (In that case 2^2=1*2*2, 2^1=1*2, and 2^0=1.)

He evidently knows the pattern, that reducing the exponent by 1 divides the value by 2, which leads to the answer of 1; and even seems to have seen what I showed in the last answer (starting with 1), but again …

If you’re multiplying **nothing**, shouldn’t you get **nothing**? This is probably what lies behind many other students’ doubt!

I explained a little more fully:

Hi, Connor.We get LOTS of questions about this, and there is more than one answer to be found in our archive. I'll suppose that what you found is the FAQ: N to 0 power http://mathforum.org/dr.math/faq/faq.number.to.0power.html My way of explaining it fits the form of your question nicely. It starts with the observation that talking about a power a^n as "multiplying a number by itself n times" is very awkward, sincereally there are n-1 multiplications! You have to say something like "multiply together n copies of the base". But there's a way to say it thatdoes involve n multiplications: start with 1, and multiply BY the base n times: a^n = 1 * a * a * ... * a \_______________/ n timesThis is the definition you mention; I don't see why you say it doesn't make sense. It clearly gives the same result for positive integer exponents, since multiplying anything by 1 doesn't change it; and it provides a neater way to state the definition. Does what I said to introduce it help it make more sense to you?

I prefer this formulation not just because it gives us \(a^0\), but because it makes more sense of how we *say* it.

Now, when you see it this way the answer becomes obvious: 2^3 = 1*2*2*2 = 8 2^2 = 1*2*2 = 4 2^1 = 1*2 = 2 2^0 = 1 = 1The answer falls right out of the definition this way.

This definition can also be extended a little to handle **negative exponents**: Just as positive exponents mean starting with 1 and repeatedly **multiplying** by the base, negative exponents mean starting with 1 and repeatedly **dividing** by the base (that is, “un-multiplying”):

2^-1 = 1/2 = 1/2 2^-2 = 1/2/2 = 1/4 2^-3 = 1/2/2/2 = 1/8

But there’s more:

Note that when you are doing addition,a blank ("adding nothing") means zero. But when you are doing multiplication,a blank ("multiplying nothing") means 1! The same thing happens when you cancel in a fraction: / / 2*3 ? ----- = --- 2*3*5 5 / /What's left when you'vecanceled everything? Not 0, but 1, since it's really / / 1*2*3 1 1 ------- = --- = --- 1*2*3*5 1*5 5 / /So the role of 1 here is common throughout algebra.Now, do you know about negative exponents? How could you extend this definition to cover that case?

There are other places where “empty products” appear, as well as “empty sums“. Whenever we are adding and find there is nothing to add, we default to 0; when we are multiplying and find nothing to multiply, we default to 1.

## Is this defined, or proved … or both?

We’ll close with this, from 2003:

n^0 Power = 1: Defined or Proved?I have been wondering if a^0 is DEFINED to be 1 or is PROVED to be 1. I think that we define a^0 = 1, but one of my friends said that a^0 is proved to be 1. I think thatwe have to define it first, because as we define a^n = a.a.a....a (n times) with n a whole number, this definition will not work for n=0. Therefore, we must define a^0 in another way, which isa^0 = 1(with a not equal to 0). As for the reasonwhy we define a^0 = 1but not any other number, it is justto keep the formulaa^m.a^n = a^(m+n), and (a^m)^n = a^(m.n)But as my friend said, we define a^n first in the way I used above, but n can be any number including 0. As we can't figure out what a^0 is from this definition, (in other words, it is meaningless to say a^0 is defined to be a.a...a [0 times]) we have to calculate it, andhe proved to methat 1 = a^n/a^n = a^(n-n) = a^0. Therefore he said a^0 is PROVED to be 1.I think before we talk about something, we must first define it, and the definition must show what that something is (and we don't need to go through some theorems to figure out or prove what we have already defined, as with my friend's argument). I think this is just normal mathematical logic thatwe have to define something clearly first before using it. Am I right?

I answered:

Hi, Tam.You are correct, but your friend is not too far off.Since we have no preexisting definition of a^0, we have to choose adefinitionfor it; but we canprovethat this particular definition is the only one that retains all the important properties of exponents. The only thing wrong with your friend's approach is that you can't make a definition that supposedly includes the case n=0, but then say that it can't be applied in that case. However, if he said that exponentiation was defined in generalby its properties, rather than as repeated multiplication, then he could prove that a^0 = 1 for all a other than 0 strictly from those properties (which would be taken asaxioms).

The friend did not prove \(a^0=1\) from his definition, but from an unstated assertion that anything that preserves the properties is correct. In fact, if his definition had been, say, that exponentiation is **defined by two axioms**, that \(a^1=a\) and that \(a^m\cdot a^n=a^{m+n}\), that would be sufficient. See below for more.

Adefinition with a reason given(as in your presentation) is not too far from aproof; and aproof based on a non-standard definition(as in my version of your friend's) is not far from adefinition.We discuss these ideas in the Dr. Math FAQ: N to 0 power http://mathforum.org/dr.math/faq/faq.number.to.0power.html But it is actually possible to start with a definition that includes 0 (but not negative exponents). Just say a^n = 1 * a * ... * a \___________/ n multiplicationsand it makes sense to say that a^0 is 1, not multiplied by anything. You still have to extend this definition to negative, and then rational, and then real, exponents.

In the FAQ, we state the argument from pattern and the argument from rule, including a statement about extending the definition by consistency.

So ultimately, which of you is rightdepends on what axiomatic system you start with, andhow you define exponents initially. I myself would say that we are making an extended definition, andproving it to be consistent.

Let’s take a look at my axiomatic suggestion. If we start only with \(a^1=a\) and the product rule, then we can immediately prove that \(a^0=1\) because \(a^0\cdot a=a^0\cdot a^1=a^{0+1}=a^1=a\), and dividing through by *a* (which is assumed not to be zero), we conclude that \(a^0=1\). But then for any positive integer n, $$a^n=a^{\overset{n\text{ times}}{\overbrace{1+1+\cdots+1}}}=\overset{n\text{ times}}{\overbrace{a^1\cdot a^1\cdot\cdots\cdot a^1}}=\overset{n\text{ times}}{\overbrace{a\cdot a\cdot\cdots\cdot a}}$$ giving our original definition. (This would be stated differently, likely by induction, for a proper proof, but this gives the idea clearly.)

From here, everything else about exponents can be derived.

In this way, we are explicitly stating the need for consistency, and getting everything else from that.