My daughter is stuck on the concept that $$2^0 = 1,$$ having the intuitive expectation that it be equal to zero. I have tried explaining it, but I guess not well enough.

How would you explain the concept to a child, other than the teachers "that is just the rule" approach?

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    $\begingroup$ @Djaian, but the multiplication rule can be verified by a kid: just add sufficiently many times: you shoul dget the same result as in the table. In fact, they should do this, so that they realize that the table is not random. $\endgroup$ – Mariano Suárez-Álvarez Nov 10 '10 at 14:43
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    $\begingroup$ Duplicate? math.stackexchange.com/questions/6832/… $\endgroup$ – Qiaochu Yuan Nov 10 '10 at 14:44
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    $\begingroup$ @Djaian: but it is just like an empty sum. The empty sum is the identity for addition, which is 0. The empty product is the identity for multiplication, which is 1. One has to distinguish between what the empty set is doing and what 0 is doing because they are not the same thing. $\endgroup$ – Qiaochu Yuan Nov 10 '10 at 16:12
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    $\begingroup$ How old/advanced is the daughter? $\endgroup$ – T.. Nov 10 '10 at 19:17
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    $\begingroup$ Apart from understanding 2^0 there is understanding 2^n where n is explicitly a variable. You might need the second, at least to a considerable extent, to clear up confusion about the first. So it might be better to first get her used to variable exponent (or variables of any kind, really). $\endgroup$ – T.. Nov 11 '10 at 0:38

19 Answers 19


I want to extend the answer by @Qiaochu Yuan.

I assume the kid accepts $2\times 0 = 0$. In other terms:

"Some number times $0$ yields the no-changer of plus."


"Some number to the power $0$ yields the no-changer of times."

By no-changer I refer, of course, to the unit element. That this can be added/multiplied to anything without resulting in a change should be accepted. I am unsure wether this approach helps understanding the hierarchy of arithmetic operators or wether you need the hierarchy for understanding the approach.

  • $\begingroup$ Thanks, this is actually the approach that worked best (at least she moved on to another question after this explination, so I think it got somewhere). $\endgroup$ – Yishai Nov 11 '10 at 14:18
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    $\begingroup$ Great! One down, many to go ;) $\endgroup$ – Raphael Nov 11 '10 at 14:58

I will give a different answer than the answer I gave in the other thread which tries to appeal to intuition. I am sure your daughter has no problem accepting that $2\times 0 = 0$. Intuitively this is because if you add $2$ to itself zero times, you get zero. Or, to be concrete, if someone gives you two apples zero times, you have zero apples.

For repeatedly adding $2$, talking about collections of apples is a good model. But for repeatedly multiplying by $2$, it isn't necessarily, since you can't multiply apples and apples (at least, not in a way that makes sense to a child). But you can multiply apples by numbers; that is, you can start with $1$ apple, then double the number of apples you have to get $2$ apples, then double the number of apples you have to get $4$ apples, and so forth. In general if you double your apples $n$ times, you have $2^n$ apples.

What happens if you double your apples zero times? Well, that means you haven't started doubling them yet, so you still have $1$ apple. If you want your notation to be consistent, then you should say $2^0 = 1$.

This is a subtly different argument from the argument I gave before. It's intuitive what it means to add different amounts of apples, and it's intuitive what it means to have zero apples. But the twos I am now working with aren't numbers of apples, they're just abstract numbers; in other words, they're unitless, so it's harder to get a grip on them. What $2^n$ really represents above is an endomorphism of the free commutative monoid on an apple, which is much less concrete than an apple.

There is a way to gain intuition here which sort of involves units, but I don't know if you can convince your daughter that it makes sense. One way to interpret $2^n$ is that it is the "size" of an $n$-cube of side length $2$ in dimension $n$. For example, the length of a segment of side length $2$ is $2$, the area of a square of side length $2$ is $4$, and so forth. One way to say this is that $2^n$ is the number of $n$-cubes of side length $1$ that fit into an $n$-cube of side length $2$.

To get a meaningful interpretation of the above when $n = 0$ we need to decide what $0$-dimensional objects are. Well, if $2$-dimensional space is a plane and $1$-dimensional space is a line, then $0$-dimensional space must be... a point. In particular, a $0$-cube, of any side length, is a point, and so exactly one $0$-cube of side length $1$ fits into a $0$-cube of side length $2$. Hence $2^0 = 1$.

(I'm really curious what her response to this argument will be, actually. Could you report back on this?)

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    $\begingroup$ I like this a lot, and I've never thought of it precisely this way. Thinking of $2^n$ as how much we multiply by if we double $n$ times makes it clear that it should be $1$ if we don't double at all. $\endgroup$ – Jonas Meyer Nov 10 '10 at 17:24
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    $\begingroup$ +1 A bit long, yet the core "you haven't started doubling them yet" is brilliant :) $\endgroup$ – Asaf Nov 10 '10 at 20:15
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    $\begingroup$ I tried the point-line-square-cube argument on my daughter, and her reaction was that a "point" could be zero as well. I tried the doubling argument, and that went a bit better. $\endgroup$ – Yishai Nov 11 '10 at 14:16
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    $\begingroup$ @QiaochuYuan "Intuitively this is because if you add 2 to itself zero times, you get zero." Your answer contains a common mistake that even very smart mathematicians seem to make quite often. 2 * x does not mean adding 2 to itself x times. That would indicate there are x + 1 copies of 2 total. 2 * x means you are adding up x copies of 2. So, 2 * 0 indicates you are adding up 0 copies of 2, not that you are adding 2 to itself 0 times. If you add 0 copies of 2 to 2, you get 2, not 0. $\endgroup$ – Graphth Dec 5 '11 at 15:00
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    $\begingroup$ I think I will remember "an endomorphism of the free commutative monoid on an apple". What a phrase. +1 $\endgroup$ – Newb Jan 31 '14 at 22:07

How about this: There's always an implicit 1 in the expansion:

$$2^{3} = 2 \cdot 2 \cdot 2 \cdot 1 = 8$$

$$2^{2} = 2 \cdot 2 \cdot 1 = 4$$

$$2^{1} = 2 \cdot 1 = 2 $$

$$2^{0} = 1 = 1 $$

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    $\begingroup$ This is always how I thought of it as a child. The exponent counts how many additional factors of $2$ you throw onto the implicit $1$ (via multiplication). If the exponent is negative, you are throwing the factors off of the implicit 1 (i.e, removing them via division). $\endgroup$ – MPW Jan 17 '14 at 4:55

I'd demonstrate this using a pattern.

$2^3 = 8$

$2^2 = 4$

$2^1 = 2$

$2^0 = 1$

$2^{-1} = 1/2$

$2^{-2} = 1/4$

When you decrease the exponent, you divide by 2. So, when you go from 21 to 20, of course you divide by 2, which gives you 1.

From there, you can segue into negative exponents, if you'd like. Just keep dividing by 2.

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    $\begingroup$ This is the one which made sense to me when I was learning it. But I was an odd kid who enjoyed formal logic and whose idea of fun was teaching my little sister simultaneous equations. (She's now doing a maths PhD.) $\endgroup$ – TRiG Dec 14 '10 at 21:20
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    $\begingroup$ +1 As a classroom teacher I have found this argument to be effective with low- and average-skilled middle and high school students. $\endgroup$ – Ben Blum-Smith Jan 9 '12 at 1:30

I would try to explain it in terms of exponents

$2=2^{1}=2^{1+0}=2^{1}\times 2^{0}=2\times 2^{0}$

and by a division

$\dfrac{2}{2}=\dfrac{2}{2}\times 2^{0}$

$1=1\times 2^{0}=2^{0}$.

Note: Applying the same argument to any number different from $0$ gives the same result.


When I was a kid, I had to teach this concept (and fractions) to some kids in a lower grade. The best method is to work backward with division.

2^3 = 8

2^2 = (2^3)/2 = 8/2 = 4

2^1 = (2^2)/2 = 4/2 = 2

2^0 = (2^1)/2 = 2/2 = 1

And so on. This is also the only way to wrap your head around negative powers.

Incidentally, the best way to teach fractions is by having pizza for dinner.

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    $\begingroup$ Unfortunately, I understood it too early, and never got any pizza. :( $\endgroup$ – Mateen Ulhaq Nov 11 '10 at 4:15
  • $\begingroup$ This is very clear and breaks down into steps comparable with what the child already understands. $\endgroup$ – Simon Hoare May 8 '18 at 5:15
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    $\begingroup$ +1 for that last line. When this occurred to me as an adult, I couldn't help but wonder if students who frequently ate pizza (specifically, the round pie, cut into 8 sections thru the center) developed better number sense. $\endgroup$ – JTP - Apologise to Monica Jul 4 '19 at 13:11

I think that it is worth pointing out that, strictly speaking, explaining “$2^0$ equals $1$” is not really what people do in this situation. The best we can do is to convince a child of the following facts:

1. If $2^0$ is any number, it makes more sense to consider that $2^0=1$ than considering $2^0$ as any other numbers (such as $0$).
2. It is more interesting to consider $2^0$ to be $1$ than giving up.

Some of the other answers provide good ways to convince a child of these facts.

However, the reason that $2^0$ equals 1 rather than $2^0$ is undefined is really conventions and experiences: it is much more convenient to define $2^0=1$ than leaving $2^0$ undefined. I do not think that it is possible to convince a child of this fact.

Compare this to the following. Let $f(x) = \frac {\sin x} x$. What is $f(0)$?

1. If $f(0)$ were any number, it would make more sense to consider that $f(0)=1$ than considering $f(0)$ as any other values.
2. It would be more interesting to consider $f(0)$ to be 1 than giving up.

However, these “evidences” do not make $f(0)=1$ under the usual way of mathematical writing. I think that the only things that differentiate these two cases are conventions and experiences.

  • $\begingroup$ I believe it's easy to convince a kid that 2^0 = 1 by first explaining to them how 2^(a+b) = 2^a * 2^b for a,b>0. Just write it out for some small values of a,b. From there, trivially 2^5 = 2^5 * 2^0. $\endgroup$ – MSalters Nov 9 '15 at 11:01
  • $\begingroup$ @MSalters: Sorry for asking this, but did you read my post before commenting on it? $\endgroup$ – Tsuyoshi Ito Nov 9 '15 at 13:59
  • $\begingroup$ Obviously. I didn't come up with "convince a kid"; that's in direct response to " I do not think that it is possible to convince a child ". This can all be done using positive integers, it doesn't even need fractions or negative numbers. $\endgroup$ – MSalters Nov 9 '15 at 14:14
  • $\begingroup$ @MSalters: Please take your time to read the rest of my post. What you wrote only convinces that “If 2^0 is any number, it makes more sense to consider that 2^0 = 1 than considering 2^0 as any other numbers” (item 1 in my post), and the point of my post is that this fact is not sufficient to establish that 2^0 = 1. $\endgroup$ – Tsuyoshi Ito Nov 9 '15 at 14:46
  • $\begingroup$ You'd have to introduce a fairly arbitrary exception to the 2^(n+1) = 2^n * 2; 2^(n-1) = 2^n / 2 rule to make 2^0 not-a-number. 2^0 is defined by default. Your sin(x)/x example isn't comparable : even children know you can't divide by zero. It's undefined by default. $\endgroup$ – MSalters Nov 10 '15 at 9:39

I would appeal to the laws of exponents. Show for $a,b>0$ that $2^a* 2^b=2^{a+b}$, then extend it to negative $b$, then set $a=b$.

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    $\begingroup$ No need to do negatives just yet; it suffices to only show that $\frac{2^m}{2^n}=2^{m-n}$, and then consider what happens when you divide a number by itself... $\endgroup$ – J. M. is a poor mathematician Nov 10 '10 at 15:15

I think it's quiet simple to understand:

The base point is 1 (I'm not talking about the base... read on...).

from this initial point of 1, you are applying the base to the result n times:

2^3 = 1 * (2 *2 *2) = 8

2^-3 = ((( 1 / 2) /2) / 2) = 1/8

The base point is always 1 so when you are in the 0 point (n^0) you don't have to multiply the very basic 1 so you have "1"

the starting point is the "1" not the base.

edit: I've just noticed that Zarkonnen already gave this answer, yet I think that the way I presented the answer is easier for a child to understand- So Zarkonnen, I hope you are ok with this post ;)


Here is a very prosaic answer. If you go into a grocery store with an order, you expect the cash register to read 0 before any items are rung up. This is because $0 + x = x$ for any number $x$. It is the fact that zero is the neutral element of addition that makes it a valid start for the register; starting there will not cheat you on your order or give you any unfair advantage. To wit: an empty sum is 0.

Likewise, suppose we are multiplying numbers in a multiplying machine. What is neutral there? We know $1*x = x$ for any number $x$. Since an empty product, like an empty sum should be neutral and not affect the product of the numbers coming in after it, the multiplying machine should be set at 1 before it starts work.

Now $2^0$ is an empty product so $2^0 = 1$. Said identity should be true for any base.


$$\begin{align} 2^0=&2^{7-7}\\ =&2^{7}\cdot2^{-7}\\ =& \frac{2^{7}}{2^{7}}\\ =&1 \end{align}$$

  • $\begingroup$ See here. $\endgroup$ – Yishai Nov 12 '13 at 19:08
  • $\begingroup$ @Yishai Why that argument is not convincing? $\endgroup$ – user93957 Nov 12 '13 at 19:10
  • $\begingroup$ kids sometimes just don't get it. It isn't intuitive to their world view, and abstractions are at times challenging for people, especially kids. Anyway, my daughter is now 14 and I think gets it. $\endgroup$ – Yishai Nov 12 '13 at 19:21
  • $\begingroup$ @Yishai Well, if you go slowly through the proof showing them step-by-step, then there is a huge chance that they get it. $\endgroup$ – user93957 Nov 12 '13 at 19:24
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    $\begingroup$ True. I up-voted your answer, it is a valid approach, and more clearly stated than the comment, although I might think about reordering it, depending which part of those steps they intuitively find agreement with. $\endgroup$ – Yishai Nov 12 '13 at 19:31

Why not just take out a calculator, let her punch in any number she likes, and then let her hit square root over and over. The limit is 1, and to show $2^0=1$ will take some explaining, but i think it's the best way to think about this in the long term. Actually this sequence will allow you to later on introduce $log$ and $e$ in a simple way.


\begin{align*}2^0 &= 2^{1-1}\\ &= 2^1\cdot 2^{-1}\\ &= 2\cdot \frac{1}{2}\\ &= \frac{2}{2}\\ &= 1 \end{align*}

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    $\begingroup$ This assumes the child is comfortable with negative exponentiation. A better argument is perhaps just that $2 = 2^{0+1} = 2^0 \times 2^1$, then cancel. $\endgroup$ – JHance Oct 13 '14 at 15:00
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    $\begingroup$ This is the exact same argument as @user93957 uses above.. $\endgroup$ – DanZimm Oct 13 '14 at 15:09

Sounds like you are looking for an intuitive understanding of exponents where exponent $=0$. Don't give up - it's there. Take a look at this url - it presents a more thorough understanding of exponents; it goes beyond the conventional explanation (i.e. the easy-to-remember-rules and the mathematical proofs) and reveals the underlying "sense" to it.

In short, an exponent 'transforms' the number $1$, so $3$ (or $3^1$). The exponent $1$ 'gives the number $1$ the power to transform into $3$. The exponent $0$ provides $0$ power (i.e. gives no power of transformation), so $3^0$ gives no power of transformation to the number $1$, so $3^0=1$. Once you have the intuitive understanding, you can use the simple rules with confidence.


As a more simple approach for someone like me who was trying to get to the depths of this question as I'm trying to learn binary and programming. Differentiating the process by which we calculate something as to that which a process or calculation actually means might be important.

$2$ to the power of $3 = 8 = 2 \times 2 \times 2$. If we were to look at $2$ to the power of zero and come to the conclusion that it is zero then we are taking an incorrect view of the system of calculation we are using as therefore $2$ to the power of $3$ would be $2 \times 3$ equaling $6$ which is obviously very wrong. But if we said $2$ to the power of zero is $2$ all on its own... then there is only one $2$. I think this sort of approach without going into too much depth is counter intuitive as it will where the mind is eager warrant further thought and possible investigation into mathematical system, structures and origins... which although I'm very new to I find quite fascinating, especially how even through something so simple a world of interconnections, possibilities and reasoning can be opened up.


Try thinking in terms other than math. Tell her to take any number of balls that are the same size. Take each one and stack them vertically on a table. Tell her to look down and tell you how many balls do you see? Tell her the top of the table equals the power of zero. Then tell her to lower her head to the level of the table. Then ask her to tell you how many balls do you see now? This will always stay in her mind whenever this is used in any equation in the future. Math by Association by Bill R. Association is a wonderful tool to use with any age. I hope this is helpful.


The sum of no number is $0$. Adding no number to a sum is the same as adding $0$.

Likewise, the product of no number is $1$. Multiplying a product by no number is the same as multiplying by $1$.

  • $\begingroup$ What you call "no number" is really the identity element of an operator. $\endgroup$ – MSalters Nov 10 '15 at 9:45
  • $\begingroup$ @MSalters: not really. You can say that combining to no number (i.e. not combining at all) has the same effect as combining with the identity. No number is no number. $\endgroup$ – Yves Daoust Nov 10 '15 at 9:52


so for example in every exponetation/multiplication there is a hidden 1

so 2^2 is 2x2=


but 2^3=2x2x2x1

x1 is always there but invisible its still part of the value hidden in every single calculation, its just not written, adding x1 makes no difference

so 2^1=2(x1)


so 0^1 is 0x1


but 0^0=1 and nothing

its ironic

  • $\begingroup$ This actually doesn't answer the question. Since there's a $1$ in all multiplication, as you said, $0 \cdot 1 = 1$, which is obviously not true $\endgroup$ – daOnlyBG Jan 27 '15 at 18:15

How about this:

Let's consider, their is 1bacteria in a dish. It doubles every hour (i.e x2)
So after 1 hr, no of bacteria = 2 = 1x2 = 2^1
After 2 hrs = 4 = 2x2 = 1x2x2 = 2^2
After 3 hrs = 8 = 4x2 = 1x2x2x2 = 2^3
After 4 hrs = 16 = 8x2 = 1x2x2x2x2 = 2^4

So 2^0 means the bacteria has not yet started multiplying. then the number of bacterial present is "1"

Lets take another example:

You have 1 dollar. Assume it triples every day (i.e x3).
So after 1 day, number of $ = 3 = 1x3 = 3^1
After 2days = 9 = 3x3 = 1x3x3 = 3^2
After 3days = 27 = 9x3 = 1x3x3x3 = 3^3
After 4days = 81 = 27x3 = 1x3x3x3x3 = 3^4

So 3^0 means the money has not yet started multiplying. then the number of $ = 1.

Remember, if their is no bacteria or no money,that means it cant multiply. so for there always exit a unit value. 'The Multiplicative Identity'.

If u still have doubt 'what will we do if we have 3dollar at the start itself?' Then, this 3dollar is actually 3 x 1 dollars (i.e 1+1+1) each of this 1 dollar triples by itself to become 9 after 1day(as stated above), So in "no" days, number of dollar = (3^0 + 3^0 + 3^0) = 3, which is the initial value.

So in general we can say that the physical meaning of m^0 is some unit value at 'no' time :)

  • $\begingroup$ Hello and welcome to MSE. I think you can improve the quality of your posts by being a bit more formal. Avoid using 'u' in place of 'you'. It is a good idea to start each sentence with a capital letter. Lastly, try to use TeX to render math. $\endgroup$ – caffeinemachine Jul 21 '15 at 9:03

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