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This question already has an answer here:

I understand that $$3^2=9 \text{ because } 3\times3=9$$

But is it possible to explain in same simple terms how $3^0=1$?

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marked as duplicate by Xander Henderson, Holo, Don Thousand, Saad, Lord Shark the Unknown Oct 8 '18 at 3:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0". $\endgroup$ – Mauro ALLEGRANZA Oct 4 '18 at 14:30
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    $\begingroup$ 3^0 =3^(1-1)=3/3=1 $\endgroup$ – Avinash N Oct 4 '18 at 17:46
  • $\begingroup$ k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1 $\endgroup$ – asmgx Oct 5 '18 at 2:23
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    $\begingroup$ Also of: math.stackexchange.com/questions/454670/… $\endgroup$ – Holo Oct 7 '18 at 23:37
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    $\begingroup$ See also Why is $\ \large 2^0 = 1\ $? $\endgroup$ – Bill Dubuque Oct 7 '18 at 23:47

15 Answers 15

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$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.

We start by defining $x^k$ as $\underbrace{x\cdot x \cdots x}_{k\text{ times}}$ because it's a useful way of writing the product.

Using this definition, we can notice the following interesting property:

For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $x\in \mathbb R$ we have $$\frac{x^{k_1}}{x^{k_2}} = x^{k_1-k_2}$$

Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.

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    $\begingroup$ If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3. $\endgroup$ – Monty Harder Oct 4 '18 at 15:38
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    $\begingroup$ Important: $x$ must not be zero $\endgroup$ – Barranka Oct 5 '18 at 1:04
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    $\begingroup$ @HRSE I agree with $0^0=1(=\exp(0))$ but not on base of what you mention. The middle expression is not defined because $\ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication. $\endgroup$ – Vera Oct 5 '18 at 7:13
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    $\begingroup$ For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^{k_1} \cdot x^{k_2} = x^{k_1+k_2}$, so if we extend it to $0$, $x^{k} = x^{k+0} = x^{k} \cdot x^{0}$ forces $1 = x^{0}$". $\endgroup$ – Pablo H Oct 5 '18 at 12:21
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    $\begingroup$ This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/… $\endgroup$ – user202729 Oct 5 '18 at 12:57
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$$3=3^1=3^{1+0}=3^1\times3^0=3\times3^0$$implying that $3^0=1$.

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    $\begingroup$ “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^{\tfrac12}\times(-1)^{\tfrac12} =$ product of two square roots, square roots are always positive $\Longrightarrow$ $-1$ is positive”? $\endgroup$ – leftaroundabout Oct 4 '18 at 23:42
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    $\begingroup$ It's nonsense because $x^{1/2}$ is not the same thing as $\sqrt{x}$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all. $\endgroup$ – Nij Oct 4 '18 at 23:56
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    $\begingroup$ How is $x^{1/2}$ not the same as $\sqrt{x}$? $\endgroup$ – GraphicsMuncher Oct 6 '18 at 4:51
  • $\begingroup$ @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots. $\endgroup$ – Daniel Wagner Oct 6 '18 at 12:04
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    $\begingroup$ @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately. $\endgroup$ – leftaroundabout Oct 6 '18 at 13:03
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5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$

We have $$x^k = \underbrace{x\times x \cdots x}_{k\text{ factors of }x} = 1 \underbrace{ \times x\times x \cdots \times x}_{k\text{ multiplications by }x}$$

Therefore $$x^0 = 1 \underbrace{ \times x\times x \cdots \times x}_{0\text{ multiplications by }x} = 1 $$

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    $\begingroup$ This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x. $\endgroup$ – Monty Harder Oct 4 '18 at 15:34
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    $\begingroup$ I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post. $\endgroup$ – Paul Sinclair Oct 4 '18 at 16:30
  • $\begingroup$ In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 \cdot x \cdot x...$ [$n$ times], and $x^{-n} = 1 \div x \div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder). $\endgroup$ – Daniel R. Collins Oct 4 '18 at 18:08
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    $\begingroup$ Couldn't you just as easily say that $x^k = x \times x \cdots x = 0 + x \times x \cdots x$, so $x^0 = 0$? $\endgroup$ – Arcanist Lupus Oct 5 '18 at 1:00
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    $\begingroup$ @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out! $\endgroup$ – F.Carette Oct 5 '18 at 7:39
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The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $${\displaystyle \prod_{a \in A} a}$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $A\cup B$ will be $${\displaystyle \prod_{x \in A \cup B} x}= \left( {\displaystyle \prod_{a \in A} a} \right) \left( {\displaystyle \prod_{b \in B} b} \right).$$ For example let's take the two sets $M:=\{x,y\}$ and $N:=\{z\}$. Then $M\cup N = \{x,y,z\}$, and by the associative law $${\displaystyle \prod_{p \in M \cup N}p } = x*y*z=(x*y)*(z)= \left( {\displaystyle \prod_{m \in M} m} \right) \left( {\displaystyle \prod_{n \in N} n} \right).$$ Now, let's try to extend this notation to the empty set $\{\}$. What should ${\displaystyle \prod_{x \in \{\}} x}$ be? If we want to keep the rule $${\displaystyle \prod_{x \in A \cup B} x}= \left( {\displaystyle \prod_{a \in A} a} \right) \left( {\displaystyle \prod_{b \in B} b} \right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M \cup \{ \}$, it follows that $${\displaystyle \prod_{m \in M} m} \ = {\displaystyle \prod_{x \in M \cup \{\}} x}\ = \left( {\displaystyle \prod_{m \in M} m} \right) \left( {\displaystyle \prod_{x \in \{\}} x} \right).$$ For this to hold the empty product ${\displaystyle \prod_{x \in \{\}} x}$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.

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    $\begingroup$ This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness. $\endgroup$ – Joonas Ilmavirta Oct 5 '18 at 7:39
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    $\begingroup$ ...or disjoint union, $\sqcup$. $\endgroup$ – AccidentalFourierTransform Oct 5 '18 at 22:49
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To increase power by one, multiply by 3, for example $3^2\times3=3^3$. To decrease power by one, divide by 3, for example $3^{2}\div3=3^1$. If you repeat this, you get $3^{1}\div3=3^0$.

You can see it this way:

$$3^3=3\times3\times3$$

$$3^2=3\times3$$

$$3^1=3$$

$$3^0=3\div3$$

$$3^{-1}=3\div3\div3$$

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    $\begingroup$ +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer. $\endgroup$ – Wildcard Oct 5 '18 at 17:39
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Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).

Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where

  • for all $(x,y) \in M$, $x \in X$ and $y \in y$; and
  • for all $x \in X$, there is exactly one $y \in Y$ such that $(x,y) \in M$.

Now, if $B$ has zero elements (i.e., if it is the empty set $\emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = \emptyset$.

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Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.

If you have a list of numbers and you want to write a computer program to sum them you do something like

sum = 0
while there are unsummed numbers in the list
   sum = sum + next number

Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.

For multiplication you want

product = 1
while there are unmultiplied numbers in the list
   product = product * next number

That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list. This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.

answer = identity element for operation
while there are unused objects in the list
   answer= answer (operator) next object

For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.

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  • $\begingroup$ This way the definition for $x^k$ is also simpler. Compare $x^k = 1\cdot \underbrace{x\cdot\ldots\cdot x}_{\text{k times}}$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = \underbrace{x\cdot\ldots\cdot x}_{\text{k times}}$. Here we have effectively two base cases ($k = 0, 1$). $\endgroup$ – ComFreek Oct 5 '18 at 11:45
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A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.

Say you have an amount $b$ of fruit, each of a different kind.

  • For example an orange and a pear, which means $b = 2$.

Now say there are $a$ persons between whom you can distribute the fruit.

  • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

The power $a^b$ is the number of possibilities that you have for distributing the fruit between them. You could write up all the different possibilities, and find that there are $9$ of them:

  1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.
  2. Alice: orange. Bob: pear. Carol: Nothing.
  3. Alice: orange. Bob: Nothing. Carol: pear.
  4. Alice: pear. Bob: orange. Carol: Nothing.
  5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.
  6. Alice: Nothing. Bob: orange. Carol: pear.
  7. Alice: pear. Bob: Nothing. Carol: orange.
  8. Alice: Nothing. Bob: pear. Carol: orange.
  9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 \cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.


Now for the actual questions: You don't have any fruit, and there are three people that are ready to be given fruit.

Everybody doesn't get a fruit:

  1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

And this is the only one possibility that you have. Thus $3^0 = 1$.

While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.

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depends in who you need to explain this to, if beginners for abstract algebra then this may work

$x^k = 1 × k × ... × k $ (x times)

Therefore When the power = 0 the k will not be repeated at all

$ x^0 = 1 $

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I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 \times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.

$3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.

Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.

$0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.

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One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $\times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say $\times\text{{[2][3]}}$ is another way of saying 1 $\times$ 2 $\times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $\times\text{{[2][3]}}$ = 1 $\times$ 2 $\times$ 3 = 6 and the empty product is $\times\text{{}}$ = 1 = 1. It can also be show that for any product with at least one number expressed in that notation, the result is the same number as a standard multiplication string of the same numbers so $\times\text{{[2][3]}}$ also equals 2 $\times$ 3.

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Since $a^n=1\iff n\ln a=\ln 1=0$ so either $n=0$ or $\ln a = 0$. In this case, $\ln3>0$ so the result follows.

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The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property $$ x^{m+n}=x^{m}x^{n} $$ Note that also $x^1$ is problematic, because it is not a product to begin with.

Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.

Next we have $0+2=2$, so we need $$ x^2=x^0x^2 $$ and the only way to make it work is defining $x^0=1$.

At least when $x\ne0$, but then, why make differences?

Next we can check that the property $x^{m+n}=x^mx^n$ is indeed preserved with this definition.

Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely $$ x^0=1, \qquad x^{n+1}=x^nx\quad (n\ge0) $$ using recursion. This allows to prove the property $x^{m+n}=x^mx^n$ without “handwaving”, but with formal induction.

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If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.

This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.

You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.

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Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.

$a^{m+n} = a^m a^n,\ $ i.e. $p(m\!+\!n) = p(m)p(n)\,$ for $\,p(k) = a^k\in \Bbb R\backslash 0\,$ shows that powers $\,a^\Bbb {N_+}$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $\Bbb N_+$ under addition. If we enlarge $\Bbb N_+$ to a monoid $\Bbb N$ or group $\Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:

$$p(n) = p(0+n) = p(0)p(n) \,\Rightarrow\, p(0) = 1,\ \ {\rm i.e.}\ \ a^{0}= 1$$

$$1 = p(0) = p(-n+n) = p(-n)p(n)\,\Rightarrow\, p(-n) = p(n)^{-1},\ \ {\rm i.e.}\ \ a^{-n} = (a^n)^{-1}$$

The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^{\Bbb Z}.\,$ Because the power map on $\,\Bbb Z\,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.

For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $\,\gcd(m,n) = j m + k n\,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,\,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $\,a^m = b^m, a^n = b^n\,\Rightarrow\, a = b\,$ for integers $a,b,\,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)

This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).

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