# Why is 10^0 = 1 ???

A common question I ask my students – their answers make me question the quality of their math education…

I often find that students do not know why 10 raised to the zero power equals one, i.e., 10^0=1. Frequently they are either completely at a loss or tell me that a teacher told them that it is just defined that way.

After all, we all know that ten squared means multiply 10 by itself two times to get 100, 10^3 means multiply ten by itself three times to get 1000, etc.

How can one multiply 10 by itself zero times and get 1 ???  Think about it!  The answer follows below!

Why is 10^0 = 1?

As stated above, the meaning of ten to a positive power is easy to understand, e.g., 10^5 = 10x10x10x10x10 = 100,000. But how can one multiply ten by itself zero times and get 1 ?? We will get to the answer to this question by reversing the process and looking at division!

If I divide 100,000 by 1,000 I get 100 or, using exponents, 10^5 / 10^3 = 10^2 = 10^(5-3). Note that when we divide we subtract the exponents. This is because we have 5 tens in the numerator and three in the denominator, and the three in the denominator “cancel” (i.e., divide to a result of 1) three of the 5 tens in the numerator, leaving only 2 tens (or 10^2) in the numerator.

What happens if we divide 1000 by 1000? We clearly get 1 for the answer, but if we do this using exponents we get 10^3 / 10^3 = 10^(3-3) = 10^0 = 1! In effect, there are zero tens present in the answer, but a one remains after the division, not a zero!

In the following table the pattern is clear, and not only explains why 10^0 is 1, but also explains the meaning of negative exponents.  Note from each row to the next we divide by 10 or 10^1 on both sides of the equation.  On the left side when dividing by 10^1 we subtract the exponents, e.g. 10^4/10^1=10^(4-1)=10^3:

10^4 = 10000

10^3 = 1000

10^2 = 100

10^1 = 10

10^0 = 1

10^(-1) = 1/10 = 1/(10^1)

10^(-2) = 1/100 = 1/(10^2)

10^(-3) = 1/1000 = 1/(10^3)

etc.

In general 10^(-n) = 1/(10^n)

Finally, as long as we are on the mysteries of exponents, what is the meaning of a fractional exponent, e.g. 4^(1/2)?

Note that 4^(1/2) times 4^(1/2) = 4^((1/2)+(1/2)), because we add exponents when multiplying the same base (4 in this case). Clearly 4^(1/2) times 4^(1/2) = 4^1 = 4.

This means that if we multiply 4^(1/2) by itself we get 4. That is the meaning of a square root!  2 is the square root of 4 because if you multiply 2 times itself you get 4. Similarly 9^(1/2) * 9^(1/2) = 9^1 = 9 so 9^(1/2) is the square root of 9 or 3.

By similar logic 8^(1/3) * 8^(1/3) * 8^(1/3) = 8^(1/3 + 1/3 + 1/3) = 8 ^ 1 = 8. If you multiply 8^(1/3) by itself three times you get 8, so 8^(1/3) is the cube (third) root of 8 which is 2. 2*2*2=8.

By similar logic 10000^(1/4) is the fourth root of ten thousand which is 10.

Sadly when I talk about this with many high school students in precalculus and sometimes higher math, I often come away with the impression that the above explanation is completely new to them…

PS – this leaves us with another tantalizing puzzle.

Is it possible to have an irrational exponent like the number pi??

Does 2 ^ pi mean anything??

If pi was equal to 22/7, which I believe was decreed by law in the state of Indiana at one time, then 2^pi = 2^(22/7) = seventh root of 2^22.

This is an operation we could carry out on a calculator, but pi is a never ending decimal with no pattern and cannot be expressed as a fraction.

“Irrational” numbers are called irrational, not because they are crazy, but because they can not be expressed as ratios, i.e., fractions!

Is it even possible to compute 2 ^ pi ???

I’ll let you look that one up if you are interested!

## Author: David Kristofferson

Retired scientist, teacher, bioinformatician, IT director, software product manager, AAAS Fellow, avid cyclist (7690 miles and 724,300 feet of climbing in 2015), backpacker, you name it! Current avocation is tutoring high school students near San Mateo, CA in mathematics, physics and chemistry. Please see the Bio link in the right sidebar for my detailed background information.

## 7 thoughts on “Why is 10^0 = 1 ???”

1. John S says:

Aye, non-positive integer exponents.

As a math major, I actually have slightly different answer to why a^0=1. I think the students (and some teachers) are RIGHT in claiming that we DEFINE a^0 to be 1 (as long as a \neq 0), BUT, the twist is that it’s one and the ONLY way to extend the exponentiation to 0 and integers while keeping the properties (in this case, is a^(m+n)=a^n \cdot a^m). Think about it: we defined raising a to nth power (for positive integers n) to be a multiplied by itself n times – but this reasoning of “a multiplied by itself n times” simply doesn’t make sense when n is not positive (like what does it mean to multiply a itself by -3 times, or 2.33 times?). So we need to acknowledge that 1) we’re EXTENDING the definition of exponentiation (and so we need to be explicit on how we define exponentiation for natural numbers and rationals), and 2) we’re doing it while**keeping the properties true** (and that’s what you were “proving” when you said that a^{1/2}=\sqrt{a} by claiming that (a^{1/2})^2=a^{1/2+1/2}=a).

As for extending exponentiation to arbitrary real numbers, hint hint, continuity. I actually have the details worked out with basic tools from real analysis, but I don’t want to spoil the fun here. Perhaps, if anybody asks me, I’ll give out the answer.

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1. As I mentioned in my response to your Comment on another article here about CPM math, sorry but I am swamped at the moment and will respond in more detail this weekend. I would be happy for you to post your explanation re arbitrary real number exponents in the mean time. When this post was first published some time back, I had a few people react to this question on our local Nextdoor newsfeed but no one commented here. I can’t guarantee you at this late date that a lot of readers will see your response unfortunately.

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2. John, your clarification is correct and demonstrates a higher level of detail expected from a math major, particularly a Ph.D. candidate. Best of luck with your studies!

I studied physics and physical chemistry in undergraduate school (UCSD), and, besides the standard lower division two years worth of Calculus and Linear Algebra, also took a year of upper division math covering topics such as Complex Variables and Fourier Series and Boundary Value problems. I also had an independent study course in symbolic logic as part of my philosophy minor which included Gödel’s Theorem. However, my general interest has always been in the application of mathematics to physical applications. I retain enough interest though to still read popular works and histories of mathematics, most recently Morris Kline’s book on the problems in the foundations of mathematics entitled “Mathematics: The Loss of Certainty.”

My probably too loose use of the word “defined” in the article above, simply means that most students that I run into think that “definition” means it is *completely arbitrary*, as no one has ever shown them the example that I used in the article. In fact I don’t recall ever seeing that example in my high school classes many years ago and only came upon it later in my personal reading about mathematics, I think in an article by Steven Strogatz from Cornell.

Regarding 2^pi, this was discussed locally on Nextdoor when the article was first published, and the use of limits like you describe in a later comment was also noted, but thanks very much for putting it here in much greater detail for other readers to see!

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2. John S says:

With your consent (and remember folks, consent is really important!), I’ll explain what it means to raise a positive number a to a real number power.

As I’ve mentioned, integer and rational exponentiation is really the unique extension of positive integer exponentiation so that the property a^{n+m}=a^n \cdot a^m holds true. Moreover, it’s not hard to verify that other properties (such as (a^m)^n=a^{mn}, (ab)^n=a^n \cdot b^n) holds true. This raises the question: is there a unique way to extend exponentiation to reals while keeping the algebraic properties true (e.g. a^{n+m}=a^n \cdot a^m)? For the property a^{m+n}=a^n \cdot a^m alone, the answer is an **emphatic no**. There are infinitely many such as extensions, and in fact, as many as functions from R to R, which is A LOT (and this takes a bit of work to show, with a solid understanding of vector spaces over arbitrary fields and the Axiom of Choice; see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation). For property a^{nm}=(a^n)^m together with the original, I’m not sure, but I suspect that the extension is NOT unique. So, we can’t extend exponentiation to reals with *just* algebraic properties.

With a quick series of exercises, one can indeed, prove that for a>0, the function Q->R, q ->a^q is indeed, continuous. So for an arbitrary real number r, we can DEFINE a**r (denoted with a**r instead of a^r to indicate a different definition) to be the lim q ->r a^q (with q being only allowed to take rational values as q approaches r).

I.e. a**r=lim q ->r a^q.

e.g. 2^pi=lim q ->pi 2^q, 10^\sqrt{2}=lim q ->\sqrt{2} 10^q

Then, with a bit of real analysis, one can indeed, show that 1) the new definition agrees with the old one when r is a rational number (i.e. a**q=a^q), which is an immediate consequence of continuity of exponentiation in Q, 2) the new exponentiation function R ->R r -> a**r=a^r (and we’ll denote it as a^r from now on, unless there’s a confusion) is also continuous and monotone, and 3) the old algebraic properties hold true. Moreover, this is the UNIQUE extension of exponentiation while keeping continuity, so it’s a very natural extension.

Interestingly enough, extending exponentiation to integers, rationals, and then, to reals, forms a strong parallel to the construction of integers, rationals, and reals. Integers and rationals were conceived in order to allow subtraction and division between (and hence, additive and multiplicative inverses of) arbitrary numbers, and real numbers are understood as “completion” of rationals (see https://en.wikipedia.org/wiki/Real_number for more details), involving some sort of “analytic,” “limiting” or a “continuous” process. And so it makes sense that extending exponentiation to integers and rationals involve keeping algebraic properties, whereas, extending to reals involve some appeal to continuity.

So, that’s my long-winded answer to what real number exponentiation is. Basically, a^r is the limit as q ->r of a^q, which is the natural definition as it keeps continuity AND as a bonus, algebraic properties of exponentiation.

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1. John S. says:

I should add that I wasn’t super careful with the continuity part.

1) The existence of lim q -> r a^q **does not** follow from continuity of a^q alone. For instance, the function Q -> R, x ->1/(x-\sqrt{2}) is a continuous function (since \sqrt{2} is not part of the domain), BUT we know that lim x ->\sqrt{2} 1/(x-\sqrt{2}) does not exist, even if we allow x to take real values. The existence of the limits DO follow from monotonicity and continuity together.

2) Speaking of limits, what does it mean lim q ->r a^q with q only being allowed to range over rationals? On some level, we know given any real number r, some rational numbers do get closer and closer to r, and it is this property that allows us to take the limits. Obviously, if we had a function, say f: [0, infty) ->R, we wouldn’t be able to take

lim x -> -2 f(x) (as x ranges over nonnegative real numbers)

simply because nonnegative numbers don’t get really close to -2. The exact details are a bit subtle, and I’ll let you and other ponder this a little bit. If you’re still interested, see https://en.wikipedia.org/wiki/Limit_point

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1. PS – As a curious aside, Kline, in his book that I cited above, also discusses the work of the mathematician Leopold Kronecker who questioned whether transcendental irrational numbers like pi even “existed” – see p. 279 of that book.

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