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From high school to introduction courses in university, the expression $0^0$ is some (psychological) problems. High school students just apply it to their calculator and either the result is $1$ or undefined.

In calculus, you may define $0^0$ as a limit of $x_n^{y_n}$ for $x_n,y_n\to 0$, but the limit depends on the given sequences.

Can you give arguments which explain/legitimate the "definition" of $0^0:=1$? The arguments should be accessible to the students at their point of knowledge, not contain anything like "If you want to proof some theorem in the next two years, then you don't have to specify on that particular case that the base is $0$".

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  • $\begingroup$ Related: math.stackexchange.com/questions/11150/… $\endgroup$ – Incnis Mrsi Nov 9 '14 at 13:30
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    $\begingroup$ It seems to me that your one-sentence explanation using the $x_n$ and $y_n$ is exactly the right explanation. Just point out that in the most important special case, where $x_n=y_n$, you do get 1. $\endgroup$ – Ben Crowell Dec 27 '14 at 1:09

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I agree with vonbrand that it is important to stress that this is a convention that is used sometimes but not others. But I would add the emphasis that all conventions are local. There are places where it is helpful to adopt this convention but other settings in which it would be a disaster. My preference would be to make sure students understand that the expression $0^0$ places two properties in direct conflict with one another. First, there is the fact that $x^0=1$ for $x>0$. Second, there is the fact that $0^x=0$ for $x>0$. There is no way to define $0^0$ so that both of these are preserved and extended, so any definition you adopt is going to lead to some kind of undesirable discontinuity. Understanding that is, in my opinion, much more important than understanding why in some particular contexts one convention is more useful than the other.

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    $\begingroup$ Notably, your discontinuity argument, although valid, relies on the use of real numbers both in the base and exponent (see also supercat’s posting) or, at least, on the use of “standard” topology on ℚ. It is only one of possible topologies on rational numbers. The 0⁰ = 1 has arithmetical arguments behind it. They are valid for the base in any field (or unital ring) and any integer exponent; topology and metric are irrelevant. It is not a “local” convention. It is a universal arithmetico-algebraic convention, whereas “$0^x=0$ for $x>0$” is a specific feature of exponents from ordered rings. $\endgroup$ – Incnis Mrsi Oct 28 '14 at 20:55
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The most intuitive reason I know for $0^0 = 1$ comes from interpretation in terms of functions, namely

$$\text{There are } |B|^{|A|} \text{ functions } A \to B \text{ for any finite $A$ and $B$}.$$

Now, there are no functions $\{\spadesuit\} \to \varnothing$, so $0^1 = 0$, but there exists exactly one function $\varnothing \to \varnothing$ (its set of input-output pairs is empty), so $0^0 = 1$.

I hope this helps $\ddot\smile$

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In the context of polynomials, $0^0 = 1$ is a convenient convention to be able to write $p(x) = \sum_{0 \le k \le n} a_k x^k$ and have the formula valid for $x = 0$ (perhaps explain that here the $0$ is fixed, while $x$ is variable; so the "indeterminate unless the exact path followed in $x^y$" just doesn't come up).

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Since there is much choice in Terms of limits as you said, a definition should be reasonable in terms of convenience. If you define the exponential function via the series $e^x:=\sum_{n=0}^{\infty} \frac{x^n}{n!}$, then in $e^0$ you need $0^0=1$. The same is true for the binomial theorem.

Edit: Also have a look at http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_power_of_zero resp. http://de.wikipedia.org/wiki/Potenz_%28Mathematik%29#Null_hoch_Null

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If it is feasible, you could start by introducing the big product operator ($\prod$) and use it to define (or just write) such things as the factorial, the exponentiation of natural numbers or binomial coefficients:

$$n! := \prod\limits_{i=1}^n i;\quad a^n := \prod\limits_{i=1}^n a;\quad \binom{n}{k} := \prod\limits_{i=1}^k \frac{n+1-i}{i}$$

Now, it is convenient¹ to define the empty product as 1, since it allows us to get the following results from the above definitions without needing to define an exception²: $$0! = \prod\limits_{i=1}^0 i = 1;\quad 2^0 = \prod\limits_{i=1}^0 2 = 1;\quad \binom{n}{0} = \prod\limits_{i=1}^0 \frac{n+1-i}{i}=1$$

In particular for the binomial coefficients, this makes some sense in “application”, as a set with $n$ elements has exactly one subset with 0 elements, namely $\emptyset$.

From this, you also get: $$0^0 = \prod\limits_{i=1}^0 0 = 1$$


¹ And I would make clear that this is nothing more.
² Also, it is somewhat consistent to have all the big operators return the neutral element of their respective operation, when empty: $$\sum\limits_{i=1}^{0}a_i = 0;\quad \prod\limits_{i=1}^{0}a_i = 1;\quad \bigcup\limits_{i=1}^{0}a_i = \emptyset $$

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Others have already given good reasons for the usual convention that $0^0=1$. A while ago, I explained at https://math.stackexchange.com/questions/475337/ why I consider the most common alternative, namely to declare $0^0$ undefined, to be inadequately motivated. Specifically, it serves only to protect calculus students from certain sloppy mistakes, and alternative protection is available, namely not to assume continuity at points of discontinuity.

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Answers of @dtldarek and @vonbrand are excellent, but Ī’m ready to add my few cents.

Consider k random independent events of zero probability each (for example, Ī shoot for the Moon). Which probability has “all successes” event? Of course, 0k.

We have no problems to realize that for k = 1, 2, … “all successes” is a zero-probability event. But what for k = 0? Ī didn’t attempted any shoots. Ī didn’t fail any time. Hence, Ī won with probability 1.

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Raising a number to a natural-number power, and raising a number to a real-number power, are different operations whose results mostly coincide. In any abstract algebraic ring, raising any element $x$ to a natural number power $n$ is equivalent to multiplying the multiplicative identity of that ring by $x$, $n$ times. If $n$ is zero, the result in any ring will always the multiplicative identity, regardless of $x$.

Note that raising the additive identity ("zero" for most kinds of numbers) to any natural-number power other than zero will always yield the additive identity ("zero") and raising it to the zero power will yield the multiplicative identity ("one"), this doesn't represent a "discontinuity" since the whole numbers represent a set of discrete values rather than a continuum.

Raising things to powers that are not natural numbers makes things more complicated. The function $x^y$ is continuous for all $x$ and $y$ when $x>=0$, except when $x$ and $y$ are both zero; the discontinuity at (0,0) means that many rules which are true for any other values $x$ and $y$ in the domain cannot hold true there. When using real-number exponents, it is simpler and more practical to say that the rules applicable to such exponents apply in all cases where a value is defined, but say that the value of $0^0$ is undefined, than it would be to define $0^0$ as equal to 1 but say that many of the rules of exponentiation don't hold there.

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  • $\begingroup$ Could you specify explicitly which arithmetic rule does the “0 in the real power 0 yields 1” hypothesis break? Do not restate all the continuity stuff, please. $\endgroup$ – Incnis Mrsi Oct 29 '14 at 12:31
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    $\begingroup$ @IncnisMrsi: I think a lot depends upon where values "come from". If one has a computation like like $f(x)^{g(x)}$, and there exist values of $x$ where both $f(x)$ and $g(x)$ are zero, I would expect that the behavior of that computation around such points may or may not be consistent with a assigning a value of 1 at those points. $\endgroup$ – supercat Oct 31 '14 at 17:42
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Exponential notation is a shorthand for repeated multiplication in the same way that multiplication is a shorthand for repeated addition. Where multiplication returns an addend, exponentiation returns a multiplicand.

In multiplication, $0x = 0$ since adding no '$x$'s is the same as doing no addition, which is the same as adding zero.

In exponentiation, $x^0 = 1$ since multiplying by no '$x$'s is the same as doing no multiplication, which is the same as multiplying by one.

There is no need for special consideration in the case that $x=0$. Not multiplying by zero is the same as not multiplying by any other number.

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    $\begingroup$ Yes, your material conditional is true: “multiplication is (only) a shorthand for repeated addition” ⇒ “exponentiation is (only) a shorthand for repeated multiplication”. Both antecedent and consequent are false because of possible non-integer arguments. $\endgroup$ – Incnis Mrsi Aug 17 '15 at 12:53
  • $\begingroup$ You can't count three and a half of something? $\endgroup$ – NiloCK Aug 17 '15 at 14:35
  • $\begingroup$ “Half” is not expressible with repeated addition. It requires solving an equation on addition. $\endgroup$ – Incnis Mrsi Aug 17 '15 at 17:07
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If working with high school students, I would suggest mentioning that all of the usual exponent rules hold. It could be a fun exercise to ask your students to interpret the numerical value of $0^0$ and to write a brief paragraph explaining their reasoning.

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