# Is $a^0 = 1$ for a nonzero, real number $a$, a theorem or an axiom?

For the students of grade 9: Is $a^0 = 1$ for a non zero real a, considered a theorem or an axiom?

• Well I always trying to avoid axioms since nothing for the mind to think .. so in fact I used to prove it as a theorm depending on the defintion of a ^ n – Abdallah Abusharekh Nov 3 '14 at 16:07
• Of course, the best way to define exponentiation of positive integers is by defining $n^m = \{ f: \{1,\ldots,m\} \to \{ 1,\ldots, n\} \}$ for which the fact that $a^0 = 1$ just follows from the fact that the empty set is the initial object in the category of sets. I'm sure the grade 9 students will love this :P – Tyler Holden Nov 6 '14 at 20:37
• From the standpoint of mathematics education: Students in grade nine are not at the level of rigor to have developed an understanding of what constitutes a "real number $a$." Moreover, the function $a^{x}$ for fixed nonzero $a \in \mathbb{R}$ cannot really be explained meaningfully and formally to (most) ninth graders; however, if we limit ourselves to the function with $x \in \mathbb{Z}$, then it can be presented in the standard way. In this last case, "$a^0 = 1$" is a definition, but it is also the only sensible one given the meaning of $a^1, a^{-1}, a^2, a^{-2}, a^3, a^{-3},$ etc. – Benjamin Dickman Nov 8 '14 at 5:23
• Is it a theorem or an axiom? I would say: neither, it is a definition. – Gerald Edgar Dec 14 '14 at 14:15
• One of the great things about math is that you can tell the story in whatever way you choose. It's very common for some important theorem to be later taken as the definition of a thing. (Or similarly, to be held as an axiom). For this particular case, I think it makes most sense to explain that x^n is the n-fold multiplication of x. What this means is obvious for positive natural numbers n. It can be extended to 0 by saying x^0 = 1 since x * x * x * x is really the same as 1 * x * x * x * x, and so x^n = 1 * x * ... * x, and then finally, at n = 0, you get x^0 = 1. – Tac-Tics Jun 25 '15 at 18:16

It is more a 'corollary' or a 'property' derived from the basic power rules. Once you have shown how products and divisions of powers with the same base work, $a^0=1$ is just a consequence.

$$\frac{a^m}{a^n}=a^{m-n}$$ Then if we set $m=n$, we obtain $$\frac{a^m}{a^m}=a^{m-m}=a^0$$ and as $\frac{a^m}{a^m}=1$, we can state that $a^0=1$.

• I agree with your reasoning for why it is defined as it is, but technically it is still an axiom. You still have to declare that it's ok to write $a^0$ at all, which you can only really do by defining what it is. You can get around it by assuming it will have a definition eventually, but at the end I still think you need to say something like, "and this is why we need to define $a^0 =1$". – DavidButlerUofA Nov 3 '14 at 6:55
• I would add that this explanation (and Rory's) also explains why this doesn't work when a = 0. – Ben Hocking Nov 3 '14 at 7:34
• I agree with @DavidButlerUofA, it's still an axiom. And at grade 9, a formal introduction to what a power is, is not required. They haven't the necessary background yet. The thing is that stablishing $a^0=1$ as an axiom at that age could have a negative impact on their believes about maths. Axioms are interpreted by the students as because the teacher say that, or because it is so, and they must be avoided when possible. At these ages, powers are just an abbreviation for a lot of equal products. – Pablo B. Nov 3 '14 at 8:16
• I'm not convinced it would have a negative impact as you describe. – DavidButlerUofA Nov 3 '14 at 8:44
• I think giving students a circular proof has a greater negative impact. – Matthew Leingang Nov 3 '14 at 15:51

This is what really happens during education (not in the mind of a mathematician): The teacher introduces $4^9$ as just a way to abbreviate $4\times 4\times 4\times 4\times 4\times 4\times 4\times 4\times 4$. So, the $a^n \times a^m=a^{m+n}$ and othere rules can be shown to be true, simply via thinking to the meaning of this abbreviation.

Then, someday, $a^0$ should be introduced; a total nonsense regarding to the abbreviation mentioned above! An approach (I thikng the best one) in this point is as follows:

1. Giving motivation: Explaining that accepting $n$ to be $0$ in $a^n$ helps us in our mathematical practice. We can say it's nonsense or give it a meaning. Which one is better?
2. Recognizing different ways to define $a^0$ It's not anyone outside the classroom to tell us what to define $a^0$! We can define it anything we want, but we will pay the price for bad choices!
3. Good definition for $a^0$ What is the value for $a^0$ which won't make us to add new rules to our rulebox? What is the value which is consistent with what we know about exponents and makes the least trouble? It's $a^0=1$!
• Best answer by far. Makes them think instead of blindly applying rules that they don't understand. They should also easily figure out what $a^{-1}$ is, or $a^{1/2}$. – gnasher729 Nov 3 '14 at 17:26
• Well...I guess you can remember that $4 \times 4 \times 4 ...$ is the same as $1 \times 4 \times 4 \times 4$ So if $4^3$ is 3 fours written as $1 \times 4 \times 4 \times 4$, $4^0$ is 0 fours writen as $1$. – azdahak Dec 16 '14 at 4:54

In grade 9, you should teach that as a result of the usual properties, as Pablo B. wrote.

$$a^0=a^{n-n}=a^n/a^n=1$$

(whatever $a^n$ is: it just needs to be non-zero).

In higher grades, once the students have learned mathematical induction, you would make it an axiom, together with

$$a^{n+1}=a \cdot a^n$$

Those two axioms define $a^n$ for all non-negative integers $n$.

So, for your students, your equation starts as a theorem then later becomes an axiom. Of course, many or most secondary students never get that far.

The correct answer is that it's a definition. This is clearly labelled as such in any textbook I've ever seen (e.g., Martin-Gay Prealgebra & Introductory Algebra, Bittinger Intermediate Algebra, Sullivan Algebra & Trigonometry, Ratti & Mcwatters Precalculus, Stein & Barcellos Calculus and Analytic Geometry, etc., etc.)

Moreover, consider the following article by Esther Levenson (2012). In it, Levenson uses specifically this question in interviews with experienced junior-high teachers (in this case, Israeli) to assess whether they understand the status of different kinds of mathematical statements.

Abstract

This paper focuses on three junior high school mathematics teachers and their knowledge of the nature of definitions. The mathematical context of exponentiation is used as a springboard for discussing two aspects of definitions: their corresponding domains and the distinction and relationships between definitions, proofs, and theorems. Through interviews it was shown that some teachers are not aware that definitions and domains are intrinsically connected and some teachers believe that definitions may be proved. Findings also indicate that knowledge of the nature of definitions may be dependent on the context.

Highlights

• Some junior high school teachers are not aware that $a^0 = 1$ is a definition. Instead, they believe that they may prove this equality.
• Some junior high school teachers are not sensitive to the issue of the domain of a definition.
• Some junior high school teachers believe that some definitions may be proved.
• Junior high school teachers’ knowledge of definitions and proofs is more stable within the context of geometry than it is within the context of exponentiation.

Levenson, Esther. "Teachers’ knowledge of the nature of definitions: The case of the zero exponent." The Journal of Mathematical Behavior 31.2 (2012): 209-219. (Link)

• Seems like these sort of problems could be overcome by making a greater emphasis on mathematics education of teachers. Or, perhaps, by a continuing education requirement. Interesting paper. – James S. Cook Jul 6 '16 at 2:51

Neither, exponentiation is defined as: $$a^0:=1 \quad (a\not= 0)$$ $$a^{n+1}:=a \cdot a^n$$

So, it holds by its Definition.

To clarify things here:

• A Definition is a precise description of the meaning of a mathematical term. In this case it is not much more than an abbreviation.

• An Axiom is a mathematical statement that is assumed to be true without proof – often it is not possible to prove it.

• A Theorem is a mathematical statement that is proved.

• A Corollary of a Theorem is a statement which can be easily proved by the theorem.

• As I remarked under Philip Gibbs' answer, that is the only definition of $a^0$ for $a\neq0$ that is compatible with other properties of powers. Outside universities I would prefer to consider that as a consequence of the other power rules, although it may technically be a definition. – Joonas Ilmavirta Nov 3 '14 at 10:15
• The properties of powers are consequences of its definition, not the other way around. The meaning of $a^{n-m}$ for example can't be a good motivation of the definition. The motivation of exponentiation is simple: $a^5$ is an abbreviation of $a \cdot a\cdot a\cdot a\cdot a$ – schnittstabil Nov 3 '14 at 10:25
• The definition that $a^n=a\times\cdots\times a$ (with $n$ copies of $a$) is natural for students if $n\geq1$. The empty product does not make that much sense for students that learn these things for the first time. They can observe that $a^{n+m}=a^na^m$, and this provides a good justification for letting $a^0=1$. In schools students don't necessarily start from definitions when they think. Universities are different. – Joonas Ilmavirta Nov 3 '14 at 10:29
• You're right, one may start with a definition $n>=1$ and $a^{n+m}$ motivates $a^0:=1$, but at this moment you may want to extend the definition! It's neither a theorem, nor an axiom that $a^0=1$ holds. – schnittstabil Nov 3 '14 at 10:34

I will say something a bit different than other people.

First, there is the question about what the correct mathematical definition is. How do you define $a^x$ for a real number $x$ (and non-negative $a$). To do this one could define $a^x$ first for natural numbers. Then we can extend to the non-zero integers in the obvious way. To define the expression at a real number you can then take limits in the usual way. If you do all this, then $a^0 =1$ for all $a > 0$.

Second, there is the question on how you should teach this in your 9th grade class. Often we have to teach students have to do stuff that we can't really define rigorously. My favourite example of this is sets. We learn about sets early on, but we never really define what a set is until (most often) we reach graduate school and learn about set theory. So I think you have the same problem with the example of $a^x$. So then you will hear different opinions on how to convince the students that $a^0 = 1$ makes sense. My preferred way would simply be to first define $a^n$ for a natural number the obvious way and then just say that we define $a^0 =1$. You can then point out that with this definition the usual nice rules (for example $a^ma^n = a^{m+n}$) hold. But I be careful what you say and how you say it.

Third, what ever you do don't lie to the students. Don't tell them that it is an axiom. This will just confuse them when they later on possibly have to do real work with axioms.

• +1: You probably want to extend from $\mathbb Z$ to $\mathbb Q$ first though, then use completions and continuity. This answer probably best encapsulates what is happening. There are many ideas which start on the naturals, are extended next to the integers, then the rationals, then the reals (for example, multiplication, or say the power rule from calculus). – Tyler Holden Nov 6 '14 at 20:32

To me it is clearly a definition. Maybe, in addition to Behzad's post have one more idea: What if $a=0$? Then $a^n=0$ for positive values of $n$. So $3^0, 2^0, 1^0$ all equal $1$ wheras $0^3, 0^2, 0^1$ equal $0$. Which value should $0^0$ have? Both choices, $0$ and $1$ have good reasons. I think this is a good starting point to learn, that mathematics is made by humans (by us!) and we may decide (define) some things.

For students in the ninth grade, evaluate $3^1$, $3^2$, $3^3$ and ask them how they would get the next value, $3^4$ (multiply). Then if we traverse this table in the opposite way, how do we get 'next' values? If they can tell you how to get to 9 from 27 and how to get to 3 from 9, then perform that operation on $3^1$ to evaluate $3^0$, and similarly, $3^{-1}$.

It is more of a definition than an axiom. I.e. it is part of the definition of the power function which would proceed by defining the function first for whole numbers inductively starting from $a^0 = 1, a \ne 0$ and $a^{n+1} = a \cdot a^n$ , then you would define it for negative integers, rational numbers, reals, complex in the well known ways. There are alternatives such as defining $e^x$ as a power series and then using $a^x = e^{(\ln{a})x}$

It is not usual to define functions like this axiomatically but you could do it if you wanted to teach the axiomatic method. In that case it would be better to start with axioms $a^1 = a$ and $a^x a^y = a^{x+y}$ This would be sufficient to derive the power function rules for rational numbers. You would need to assume an axiom of continuity to extend to the reals. If you did this you would have to justify the assertion that the axioms are consistent. This is why it is better to use a definition rather than axioms. Axioms should be kept to a minimum and used only for the essential elements of the mathematical system you are working in.

In any case this would be more than you wanted to explain to most grade 9 students. It is better to just call them definitions. You can motivate them with statements like $a^0 = a^n/a^n$ to help memory and understanding, but you should not confuse definitions with axioms.

• It is a definition, yes, but it is the only definition that works well with the other properties of powers. In that sense one could see $a^0=1$ as a consequence or the natural extension of the other properties. – Joonas Ilmavirta Nov 3 '14 at 10:10
• Joonas I agree with your comment and feel I covered the point in my last paragraph. It is important for a mathematician to appreciate the difference between the motivational reasoning that leads to certain definitions and the rigourous process of axioms, definitions and theorems that builds up our solid mathematical knowledge. In the past they used Euclidean geometry in schools to teach this, but now it is left for higher education. Nevertheless it is important to not confuse axioms, definitions and theorems at any level. – Philip Gibbs Nov 3 '14 at 12:38

$a^b$ is a notation for the multiplicative operand equivalent to $b$ repeated multiplications by $a$, the same way that $a*b$ is a notation for the additive operand equivalent to $b$ repeated additions of $a$.

$a*0 = 0$ for all $a$, since adding no $a$s, ie, doing no addition, is the same as adding zero.

$a^0=1$ for all $a$ (including $0$), since multiplying by no $a$s, ie, doing no multiplication, is the same as multiplying by one.

I find it best to think of exponents as repeated ratios in a geometric progression. When defined in this way, it automatically solves the edge cases via definition rather than requiring an axiom or theorem. It also builds a good foundation for understanding how powers of different bases relate to each other, logarithms, binomial powers etc.

As such I think exponent should always be thought of as $$x^n = 1\times x \times x \times x ...$$

Year 9 students should be able to recognise an easy geometric progression: $3, 6, 12,24,48$ $$24 = 3\times 2 \times 2\times 2= 3\times 2^2$$ $$12 = 3\times 2 \times 2= 3\times 2^1$$ $$6 = 3\times 2 = 3\times 2^1$$ So what was the first in the series? $$3=3\times 2^0$$

This leads to a general formula for the $n^{th}$ term in a geometric progression, starting from $a$ increasing with a ratio $b$ : $a\times b^n$. $$a\times b^2 = a\times b\times b$$ $$a\times b^1 = a\times b$$ $$a\times b^0 = a$$

So now we understand geometric progressions.

The only remaining question is, when $a$ is unwritten, what do we place in front of the repeated ratios? $1$ is the only choice: $$b^2 = 1\times b^2 = 1\times b\times b$$ $$b^1 = 1\times b^1 = 1\times b$$ $$b^0 = 1\times b^0 = 1$$

To further demonstrate that the above is the standard way of understanding exponents, mathematicians now generally agree the following (though in the past there was more disagreement, so probably not for discussion in a 9th grade class!) $$0^2=1\times 0\times 0$$ $$0^1=1\times 0$$ $$0^0=1$$

If $0^0$ is not treated this way, all sorts of mathematical formula break, for instance the binomial theorem (the leading coefficients are the binomial coefficients from Pascal's triangle) $$(1+2)^2=1\times 2^0 + 2\times 2^1 + 1\times 2^2=9$$ $$(1+2)^1=1\times 2^0 + 1\times 2^1=3$$ $$(1+2)^0=1\times 2^0 =1$$ $$(1+0)^2=1\times 0^0 + 2\times 0^1 + 1\times 0^2=1$$ $$(1+0)^1=1\times 0^0 + 1\times 0^1=1$$ $$(1+0)^0=1\times 0^0 =1$$

In conclusion, the leading "$1\times$" in the style of a geometric progression does not need to be used all the time for exponents, but defining exponents in this way does clarify the edge cases nicely.

As for your original question, I would say that $x^0 = 1$ is simply part of the definition of exponents. However, the fact that their has been significant disagreement about another edge case ($0^0$), with people sometimes arguing about "what it is" rather than "what it should be defined as" indicates that in practice the line between definition and axiom can be blurry.