Probably a simple question and possibly not asked very well. What I want to know is..

In binary, a decimal value of 1 is also 1. It can be expressed as $x = 1 \times 2^0$

Question: Why is two to the power of zero equal to one? I get that two to the power of one is equal to two, or binary 10, but why is to the power of zero equal to one, is this a math convention? is there a link I could read?

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    $\begingroup$ Although it isn't explicitly stated, the answers below are saying yes, it is a 'convention', but the reasons for adopting it are so strong that it would be unnatural to define it any other way (or to not define it). $\endgroup$
    – yasmar
    Commented Oct 15, 2010 at 4:18
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    $\begingroup$ Well, I am saying something a little more. I am saying that exponentiation has a better definition than you are taught in school and that the result follows directly from this definition. $\endgroup$ Commented Oct 15, 2010 at 23:48
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    $\begingroup$ $2^n$ in binary is a 1 followed by $n$ zeros. Viola. $\endgroup$ Commented Nov 15, 2010 at 3:11

5 Answers 5


Because we want $2^{m+n} = 2^m \cdot 2^n$, and if $n = 0$ this requires that $2^0 = 1$. More combinatorially, $a^b$ is the number of functions from a set with $b$ elements to a set with $a$ elements, and there is exactly one function from the empty set to any other set (the empty function). This is the same reason that $0! = 1$.

  • $\begingroup$ marking Qiaochu's answer as accepted due to additional explanation given of exponentiation. However most of it went over my head :) $\endgroup$ Commented Oct 17, 2010 at 23:18

I think it's helpful to think about sums for a bit, and then come back to products.

With sums… we’re all happy with the idea that you can add up a list of 2 things, or 3 things, or more (so e.g. the sum of the list $(x,y,z)$ is $x+y+z$). And after a moment’s thought, we’re happy with the idea that you can add up a list of 1 thing, eg $(x)$, and get just $x$ itself. And it takes another moment’s thought, but it's still pretty intuitive, to decide that you can add up the empty list $()$, and that its sum is $0$.

(“I went shopping today, with nothing on my shopping list. How much did it cost in total?”)

Now, play exactly the same game with products! It's just as easy until you get to the empty list. For some reason, it's much less intuitive at first (at least it was for me, and I think it is for most people) that one can make sense of the product of the empty list, and that it should be $1$. But every argument I know to justify why the sum of the empty list should be $0$ also works as an argument that its product should be $1$! (See the arguments about making exponents work nicely, etc, in the earlier comments and answers.) And, of course, once the lightbulb flashes, it suddenly becomes completely clear and natural that the product is 1, and that that's not just a convention, and how could anyone possibly think otherwise!? :-P

…unfortunately I don't know an example for products that’s quite as intuitively convincing as the shopping list example is for sums. The best I can do: at the checkout, they apply a bunch of taxes and/or discounts. Each one of those is a multiplier: e.g. 7% sales tax multiplies my total by 1.07, a 20% discount is a multiplier of 0.8, and so on. To get the overall multiplier from all the taxes/discounts currently in effect, you just multiply the list of them together (of course)! But if there are no taxes or discounts at the store today, what's the overall multiplier on my shopping?

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    $\begingroup$ good point, i like your lay example its certainly much easier to follow. $\endgroup$ Commented Oct 17, 2010 at 23:20
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    $\begingroup$ Product example analogous to the shopping-list example: compounding at a rate of $r$ $n$ times multiplies the principal by $(1 + r)^n$. Compounding $0$ times multiplies the principal by $(1 + r)^0 = 1$. $\endgroup$
    – LSpice
    Commented Dec 20, 2020 at 23:06
  • $\begingroup$ Addition is a binary operation, making the sum of one thing not as intuitive as you're making it seem. When a person says sum from i = 1 to n of N_i, there is an implicit 0 + N_1 + N_2 + ... + N_n. This is similar for multiplication except you use 1. $\endgroup$
    – user974006
    Commented Nov 5, 2023 at 20:52
  • $\begingroup$ @user974006: When a binary operation is associative and has an identity element, you can view them together as an operation on lists — this is natural and fruitful from a high-level algebraic viewpoint (operads, and other modern theories of algebraic structures) but also ubiquitous in real world usage (the examples in my answer). So the binary version is certainly important as the simplest non-degenerate case, but it’s not obviously more primitive than the operation on lists — if it appears so, that’s arguably an accident of current maths teaching. $\endgroup$ Commented Nov 5, 2023 at 21:37

Any number (except zero) to the zero power equals one. There are many ways to justify it. Easiest (to me) is to make the laws of exponents work. We know that $a^b/a^c=a^{b-c}$. Now put b=c

  • $\begingroup$ ah right i get it. $\endgroup$ Commented Oct 17, 2010 at 23:17

The easy explanation is:

  • $2^0 = 1_2$
  • $2^1 = 10_2$
  • $2^2 = 100_2$
  • $2^3 = 1000_2$
  • $2^4 = 10000_2$

If we multiply $2^1$ by $2$, we get $2^2$; naturally if we divide $2^1$ by $2$, we should get $2^0$. And it happens that ${2^1 \over 2} = {2 \over 2} = 1$.

  • $\begingroup$ I really like this answer. It is very concise and to the point. $\endgroup$ Commented Nov 15, 2010 at 4:58

The definition $\ 2^0 = 1\ $ is "natural" since it makes the arithmetic of exponents have the same structure as $\mathbb N$ (or $\mathbb Z\:$ if you extend to negative exponents). In more algebraic language: the definition is the canonical extension of the powering homomorphism from $\rm\ \mathbb N_+\: $ to $\rm \mathbb N\ $ (or $\rm\: \mathbb Z\:$),$\ $ viz. $\rm\ 2^n\ =\ 2^{n+0}\ =\ 2^n\ 2^0\ $ $\rm\Rightarrow\ 2^0 = 1\:$. It's just a special case of the fact that the identity element must be preserved by structure preserving maps of certain multiplicative structures (e.g. commutative cancellative monoids).

It may be viewed as a special case of adjoining an identity element to a commutative semigroup. And it proves very convenient to do so, for the same reason it proves convenient to adjoin the identity element 0 to the positive natural numbers, e.g. it allows every element to be viewed as a sum, so one can write general formulas for sums that work even in extremal cases where an element is indecomposable (e.g. by writing $ 1 = 1 + 0 $ vs. having to separate a special case for the sum-indecomposable element $1$ or, in $\,2^{\Bbb N},\,$ for $\, 2 = 2\cdot 1 $). Empty sums and products prove quite handy for naturally founding inductions and terminating recursive definitions.

  • $\begingroup$ See here for further remarks. $\endgroup$ Commented Nov 25, 2019 at 2:16

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