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I've understood how to apply this operation for many years but I recently was reminded that:

$n^0 = 1$

Which in high school I accepted and just "solved for", but I'm now curious. Why is that true? So my own cursory research on the internet led me to the most common proof (using recurrence):

$b^1 = b$

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

$b^0 = \frac{b^1}{b} = 1$

The most common proof for it relies on the fact that $\frac{n}{n} = 1$ so I started wondering why is that so? There's the rather obvious examples given by teachers using a pie but it feels like there's something else there. Maybe that something else involves Mathematics that is out of my league at the moment (the highest amount I've taught myself is through Algebra).

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There certainly is something else there, in fact what is there is what's important, unless you're a physicist and then you just don't care. – Git Gud Jun 12 '13 at 21:33
Division is normally viewed as the inverse of multiplication, e.g. introduction at if you believe 1*n=n it follows that n/n=1 – Ronald Jun 12 '13 at 22:24
Note that $n^k$ can be defined, at least for integral $k \geq 0$ as $1\times n \times ... \times n$, $n$ repeated $k$ times. If $n$ is repeated zero times ($n^0$), then it reduces to $1$. – Kaz Jun 12 '13 at 23:49
The proof you sketch doesn't hold up when $b$ is 0 ($0^1 = 1$). I think that edge case is usually taken as a definition rather than proven, though. – Dan Jun 13 '13 at 1:39
Dan, $0^1 = 0$. What are you talking about? The weird edge case is $0^0$. – dfeuer Jun 13 '13 at 5:59

16 Answers 16

up vote 47 down vote accepted

$\frac ab$ is, by definition, the solution of the equation $bx=a$. Thus $\frac nn$ is the solution to the equation $nx=n$. Assuming $n\neq0$, this equation has the unique solution $1$.

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"$\frac{a}{b}$ is the solution of the equation $bx = a$" - why is that? – omer Jun 12 '13 at 21:41
@OmerPT: it is the definition of division – Carl Mummert Jun 12 '13 at 21:44
Okay then I find no flaws in your answer. – omer Jun 12 '13 at 21:57
@OmerPT if you substitute x with (a/b), you get a = a. Also, divide both sides of bx = a by b. – NicolasMiari Jun 13 '13 at 2:02
This makes a lot of sense. I was exaggerating the situation in my head by a lot; the "definition of division" makes the whole thing very simple and straight forward. I also assumed there was something direct there when in fact division is just an abstraction on top of an intrinsic property and application of multiplication (in the same sense that multiplication is of addition). I knew that (how to "check" your multiplication work), but I suppose I didn't consider that was the way the identity of division is defined. – Ixmatus Jun 13 '13 at 2:50

In higher math, you usually take the equation $n/n=1$ as the definition of division by $n$. Specifically, $1/n$ is defined to be the unique number such that $1/n$ times $n$ is 1. Then you go on from there to define $2/n$ and other numbers using ideas called equivalence classes.

So it's that way because mathematicians feel like it. Like Bill Thurston said, math isn't real, it's just a way of organizing human thought.

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I like it too, although I feel bad for paraphrasing. Here's more of it and a reference: – Brian Rushton Jun 12 '13 at 21:52
It is easy to forget that mathematics is primarily a tool for human thought. … The most important thing about mathematics is how it resides in the human brain. … When mathematics loses its connection to our minds, it dissolves into a haze. … In mathematics, what is intriguing, puzzling, interesting, surprising, boring, tedious, exciting is crucial; they are not incidental, they shape how we think. (Foreword to Teichmuller theory and applications to geometry, topology, and dynamics; Volume I: Teichmuller theory, by John Hubbard) – Brian Rushton Jun 12 '13 at 21:52
Von Neumann said "One does not understand concepts in mathematics - one simply gets used to them." – Zen Jun 12 '13 at 21:55
@Ronald It is implicitly assumed that $n \neq 0$. You can derive that the system of numbers that you work in is trivial if you have a reasonable division by $0$. – Gil Jun 12 '13 at 23:14

You don't need division. $\ b^{n+1} = b\, b^n\stackrel{\large n=0}\Rightarrow b = b\, b^0\,\Rightarrow\,b(b^0\!-1) = 0\,\Rightarrow\, b^0 = 1\,$ if $\,b\ne 0.$

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You divide by b... – RobAu Jun 13 '13 at 11:18
@RobAu: There is no division by b. The first step is a power identity. The second step substitutes 0 for n. The third step subtracts b from both sides and factors the polynomial; this is a polynomial identity; you don't need division. The fourth step observes that for a product to be zero, one of the multiplicands must be zero. Where's the division? – Eric Lippert Jun 13 '13 at 16:29

The way that I see $a^{0} = 1$ is that people want to have the following rule to hold in any number system with a multiplication: $a^{m + n} = a^{m}a^{n}$.

In order for this rule to hold, it is inevitable to define that $a^{0} = 1$.

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I like to describe this as an example of the "the tail wagging the dog" phenomenon in mathematics. We build a set of rules ---observations, really--- about things that make absolute sense, such as that $a^{m+n} = a^m \; a^n$ for nice, non-controversial, positive integers $m$ and $n$. Then, when we take it upon ourselves to consider what $a^0$ might mean, the rules take over and direct our thinking; as you say: it is inevitable to define that $a^0 = 1$ (or, say, that $a^{1/2} = \sqrt{a}$). The rules start making demands of us, instead of the other way around. – Blue Jun 13 '13 at 6:00
For me, I see it as a group map that maps integers to $\{a^{n}\}_{n \in \mathbb{Z}}$, whatever this symbol means depending on the given context. It feels more "obvious" to me that $a^{0} = 1$, since it simply says identity maps to the identity. – Gil Jun 13 '13 at 6:24
@Blue I assume you don't see the "the tail wagging the dog" phenomenon a bad thing in this case, it is just a logical extension to what we do see as non-controversial. – Mark Hurd Jun 13 '13 at 6:54
@MarkHurd: It's not a bad thing at all, just a dynamic worth pointing out explicitly. It's one of the touchstones I use when describing the mathematical experience to students. My overly-wordy answer here talks through a dog-wagging way of understanding how trig functions "want" to behave outside of the First Quadrant. It's neat when our rules provide guidance beyond their original intent; it's like mathematics is trying to tell us something. – Blue Jun 13 '13 at 7:27
@badp: Indeed. (The "demand" aspect is just for dramatic effect. :) Your point is sorta what I'm getting at by saying that "it's like mathematics is trying to tell us something" ... specifically, that the patterns might "easily" observe often extend naturally into places we hadn't anticipated, which helps make mathematics an on-going adventure. – Blue Jun 13 '13 at 12:19

For another neat definition of $a^b$ when $a,b$ are nonnegative integers: It is the number of maps from a set $B$ of $b$ elements to a set $A$ of $a$ elements. If $B$ is empty, there is exactly one such map, no matter what $A$ is. Why? To describe such a map we need to specify for each element of $B$, which element of $A$ it is mapped to. So, let's try to do this explicitly when $B$ is empty. Ready? Here we go. Starting ... done.

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A combinatorial argument, cool! – Frudrururu Jun 13 '13 at 3:49
Unlike division, this also defines unambiguously $0^0=1$: There's certainly exactly one way to map nothing to nothing. – celtschk Jun 13 '13 at 6:10

Here is how an algebraist might answer the question.

Take the rationals to be the fraction field of the integers, and embed $\mathbb{Z}\hookrightarrow \mathbb{Q}$ by associating $a$ to the class $[a/1]$. In $\mathbb{Q}$ we have the relation $[a/b]=[c/d]$ if and only if $ad=bc$. Apply this to show that $[n/n]=[1/1]=1\in\mathbb{Z}$.

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Nice, but that relation is derived by multiplication of both sides of the equation by B and by D, creating B/B and D/D on their respective original sides of the equation, which are understood to equal 1 and thus cancel. The fact that n/n=1 is tautologous; it is its own proof. Not quite a postulate, but close. – KeithS Jun 13 '13 at 0:37
Except that fields, rings, integral domains, and abstract algebra in general are defined in terms of the operations of elementary arithmetic, and require the acceptance of those postulates as true because the properties of the abstract operations are the same as the elementary ones by definition. Division in abstract algebra is still the inverse of multiplication. You're doing nothing more than using the abstraction of the operations via AA to claim that their definitions are somehow separate from the elementary ones at the heart of the question, when in fact they're congruent by design. – KeithS Jun 13 '13 at 1:56
Case in point; you're defining Q as the fraction field of Z, ignoring the fact that the fraction field is by definition the field in which any element of the domain can be produced by multiplication of some two elements of the field. The fraction field is defined by the operation of division as the inverse of the multiplication operation defined for the fraction field (which for the rationals is elementary multiplication). Abstract multiplication, in turn, is defined by the properties of elementary multiplication, including n*1=n and n*0=0, with the requisite definition of both 1 and 0. – KeithS Jun 13 '13 at 2:06
@KeithS I have no idea what you are talking about. When we define the field of fractions given an integral domain $D$, we define an equivalence relation $\langle a,b\rangle\sim\langle c,d\rangle\iff ad=bc$, and from this you can prove that $\langle n,n\rangle\sim\langle 1,1\rangle$, because $n\cdot 1=1\cdot n$. At no point is anything about $n/n$ assumed to be true. It may well be that the original choice of definition $ad=bc$ was motivated by experiences with "$n/n=1$ ought to be true", but that is irrelevant to the definition. – Mario Carneiro Jun 13 '13 at 5:02
@KeithS In particular, when we start from an integral domain $D$ and build its field of fractions $F$, we do not assume any division properties in $D$, only addition and multiplication. By the choice of definition $ad=bc$, we are able to prove that $F$ is a field without any extra postulates. – Mario Carneiro Jun 13 '13 at 5:09

Most of the math systems you're familar with (the standard definitions of addition and multiplication over real numbers, rational numbers, or complex numbers) are fields, so you need to look at the field axioms. The most relevant here are:

  • There exists an additive identity 0 such that for all a, a + 0 = a. For example, in all of the familiar fields I mentioned above, the additive identity is the number 0.

  • There exists a multiplicative identity 1 such that for all a, a · 1 = a. For example, in all of the familiar fields I mentioned above, the multiplicative identity is the number 1.

  • For all a except 0, there exists a multiplicative inverse (denoted as a-1) such that a · a-1 = 1.

n/n is also written as n · n-1 --- in other words, it is n multiplied by its multiplicative inverse. By the third axiom, this is equal to the multiplicative identity, which is 1.

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Yes, in someone else's answer they used sets and it was clear the use of "unit" was necessary for establishing the identity of an empty set. – Ixmatus Jun 13 '13 at 12:31

There are many different interpretations of division in elementary mathematics:

The expression $\frac{a}{x}$, or $a\div x$, can be interpreted as:

  1. The number of times that $x$ goes into $a$. For example, $\frac{10}{5}=2$ because $5$ goes into $10$ twice. The symbolic way of putting this is $\frac{a}{x}=q \equiv xq=a$ (here the symbol "$\equiv$" denotes equivalence). With this interpretation, $\frac{x}{x}=1$ because $x$ goes into $x$ once, or because $x1=x$.
  2. The number that goes into $a$ $x$ times. For example, $\frac{10}{5}=2$ because $2$ goes into $10$ $5$ times. The symbolic way of putting this is $\frac{a}{x}=q \equiv qx=a$. With this interpretation, $\frac{x}{x}=1$ because $1$ goes into $x$ $x$ times, or because $1x=x$. This interpretation is the same as interpretation 1 because $qx=xq$, the commutativity of multiplication.
  3. The ratio of $a$ and $x$, i.e. $a$ per $x$. This can be thought of best in terms of rates. Think of two objects, $A$ and $X$, traveling along the number line at constant velocity. $A$ travels from $0$ to $a$ in the same time interval in which $X$ travels from $0$ to $x$. Then $\frac{a}{x}$ is the distance that $A$ travels for each unit distance that $X$ travels, or the distance that $A$ travels in a certain time interval divided by the distance that $X$ travels in that same time interval. For example, $\frac{10}{5}=2$ because in this case $A$ travels a distance of $2$ every time $X$ travels a unit distance, and in any time interval, $A$ will travel twice as far as $X$. With this interpretation, $\frac{x}{x}=1$ because $X$ will travel a distance of $1$ for each unit distance that an object of identical trajectory travels.
  4. $a$ $x$ths. For example, $\frac{10}{5}$ is $10$ fifths, which is $2$. The symbolic way of putting this is $\frac{a}{x}=a\times\frac{1}{x}$. With this interpretation, $\frac{x}{x}=1$ because $x$ $x$ths of something is $1$.

The above list is not exhaustive; there are other ways of interpreting division.

However, you don't need division to show that $x^{0}=1$. Here is an alternative way.

Let $S$ be a multiset of numbers (a multiset is a set in which elements can appear more than once), for example $\left\{2,3,3,6\right\}$. Let $\Pi(X)$ denote the product of all the numbers in the set $X$, and let $\Sigma(X)$ denote the sum of all the numbers in the set $X$. When adding and multiplying, the order and grouping of the terms or factors has no effect on the result (this follows from an inductive argument on the number of terms/factors that uses the associative and commutative properties). Thus if $S$ is separated into two disjoint sets $S_{1}$ and $S_{2}$ (for the above example, these could be $S_{1}=\left\{2,3\right\}$ and $S_{2}=\left\{3,6\right\}$), it is the case that $\Pi(S)=\Pi(S_{1})\times\Pi(S_{2})$ and $\Sigma(S)=\Sigma(S_{1})+\Sigma(S_{2})$. The result about the order and grouping of terms/factors only entails this if $S_{1}$ and $S_{2}$ are nonempty, but it would make sense to define $\Pi(X)$ and $\Sigma(X)$ such that it works for all sets $S_{1}$ and $S_{2}$, even empty ones. Let $S_{1}$ be the empty set, denoted $\varnothing$. Then $S_{2}=S$. We want to have $\Pi(S)=\Pi(\varnothing)\times\Pi(S)$ and $\Sigma(S)=\Sigma(\varnothing)+\Sigma(S)$, so we must define $\Pi(\varnothing)=1$ and $\Sigma(\varnothing)=0$.

What we have shown is that the sum of all the elements in the empty set is $0$ and the product of all the elements in the empty set is $1$. When we write $x^{n}$, we are taking the product of all the elements in the multiset that has $x$ in it $n$ times and nothing else. So $x^{0}$ is the product of all the elements in the empty set, or $\Pi(\varnothing)$. Therefore $x^{0}=1$. This argument can also be used to show that $0!=1$, as this is also $\Pi(\varnothing)$.

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I like the use of sets; I don't know much about set theory so bear with me if this is a naive question: the definition of the product of an empty set is relying on the concept of "unit" - which is also represented as "one", correct? – Ixmatus Jun 13 '13 at 12:12
Well it is relying on the fact that $1$ is the multiplicative identity, i.e. $1$ is the unique $y$ such that $x\times y=x$. Similarly, $0$ is the additive identity, i.e. $0$ is the unique $y$ such that $x+y=x$. – Joshua Meyers Jun 13 '13 at 13:56
I'm not sure if that is what you are asking, though. – Joshua Meyers Jun 13 '13 at 13:58
Yes. I was clarifying that the establishment of identity is crucial. – Ixmatus Jun 13 '13 at 16:59

Given $n \neq 0$, lets assume $n/n = x$, where $x \neq 1$. Now,

$n/n = x\\n = x \times n\\n - x\times n = 0\\n\times (1-x) = 0$

Now, as $n \neq 0$, so $\textbf{(1-x) = 0}$ must be true. Otherwise the last line can not be true.

$(1-x) = 0\\x = 1$

We come to a contradiction. So, $n/n = x$, where $x \neq 1$ is false. So, $\textbf{n/n = 1}$.

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OK, but to get from $n - x\times n$ to $n\times (1-x)$, you have to assume that $n/n = 1$, don't you? – LarsH Jun 13 '13 at 13:59
@LarsH: You're saying that you need division to show that 2 x (3 + 4) is equal to 2 x 3 + 2 x 4? – Eric Lippert Jun 13 '13 at 16:32

Here is another proof of $b^0=1$ for every non-zero $b$ in $\Bbb Z$, which doesn't require division.

First of all $$ b^0=b^{0+0}=b^0b^0 $$ means that $b^0$ is an idempotent in $\Bbb Z$. Moreover, it is non-zero, since $$ 0\neq b=b^{1+0}=bb^0 $$ Now observe that the only non-zero idempotent in $\Bbb Z$ is $1$. Indeed, let $e\neq 1$ be another one. Then $$ e(e-1)=ee-e=e-e=0 $$ which is absurd, since by hypothesis $e\neq 1,0$ and there are no zero-divisors in $\Bbb Z$, i.e. each time $ab=0$ either $a=0$ or $b=0$.

Note: To prove that $a^{m+n}=a^na^m$ you don't need division. It is a straightforward consequence of the associative property of the product.

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Here's a way to think about it:

The following "proof" relies on the fact that multiplication by 1 doesn't change the value and that division is the inverse of multiplication:

We want to evaluate the fraction $\frac{n}{n}$:

The value will stay the same if we multiply it by 1, so: $\frac{n}{n} = 1 \cdot \frac{n}{n} $

This expression can also be written as:

$1 \cdot n \div n$

Meaning "1 multiplied by n and then the result divided by n".

Division is the inverse of multiplication, so multiplying by a number and then dividing by that same number has no effect, so basically:

$\frac{n}{n} = 1 \cdot \frac{n}{n} = 1 \cdot n \div n = 1$

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I have considered a more axiomatic approach to dividing a pie. Consider the real numbers $\mathbb{R}$. It is a field. This comes from properties of equivalence classes of all rational Cauchy-sequences. Then for any $n \in \mathbb{Z} \subset \mathbb{R}$, $n \neq 0$ there is a multiplicative inverse $n'\in \mathbb{R}$, that satisfies $nn'=1$. Denote $n'$ by $\frac{1}{n}$. The equation $nn'=1$ corresponds to taking $n$ pieces of a pie that is divided into $n$ pieces. The fraction $\frac{m}{n}$ means by definition $m \frac{1}{n}$ because it corresponds to taking $m$ pieces, that are of the size $\frac{1}{n}$. The definition of $\frac{m}{n}$ was constructed in an intuitive way but has also an axiomatic formalisation using the multiplicative inverse. Now we can calculate \begin{equation} \frac{n}{n} = n \frac{1}{n} = 1 \end{equation} whenever $n \neq 0$. Note that this isn't the definition in algebra studybooks, but because of the nature of the problem I tried to find an axiomatic formulation to the way, I have learned using fractions in the comprehensive school.

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Basically x/y can be defined as how many times you can take y from x till y becomes 0. So in this respect, 1 can be taken from 1 only 1 times and we are left with 0 and therefore 1/1 is 1. I hope this is a simple answer without much complications.

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It is simple and I appreciate it; it isn't something I already knew though - I already knew that if you have five apples and five boys and divide those five apples by five boys you get one apple for each boy. I intuitively understand it but I wanted a more formal answer as to why it is that way (which I got, in many different ways!). – Ixmatus Jun 13 '13 at 12:23

Definition: In mathematics, especially in elementary arithmetic, division ($\div$) is an arithmetic operation. Specifically, if $b$ times $c$ equals $a$, written:

$\color{red}a = \color{blue}b \times \color{green}c$

where $b$ is not zero, then $a$ divided by $b$ equals $c$, written:

$\color{red}a\div \color{blue}b = \color{green}c$

Using the definition and by letting $\color{green}c=\color{green}1$, $\color{blue}b=\color{blue}{1/n}$ where $n$ is a non zero real number, $\color{blue}b$ times $\color{green}c$ is equal to a number $\color{red}a$: $$ \color{red}a=\color{blue}{\dfrac1n}\times\color{green}1=\color{red}{\dfrac1n} $$ Therefore, from the definition: $$ \color{red}a\div\color{blue}b=\color{green}c \\\Downarrow\\ \begin{align} \color{red}{\dfrac1n}\div\color{blue}{\dfrac1n}&=\color{green}c \tag{1}\\\,\\ \color{red}{\dfrac1n}\cdot\dfrac n1 &=\color{green}c \tag{2} \\\,\\ \color{black}{\dfrac nn} &=\color{green}c\\\,\\&=\color{green}1 \end{align}$$ Thus for any non zero real number $r$ we have that: $$\dfrac rr=1.$$ $\blacksquare$

If you search for why the procedure I applied when going from step $\text{(1)}$ to step $\text{(2)}$ is correct, check this answer: Why is $\frac{1}{\frac{1}{X}}=X$?

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Any algebraic proof that division of a number by itself equals 1 is a tautology; the proof invariably accepts that it is true, therefore it's true because it's true.

For instance, the top-voted answer states that $a/b$ is the solution to the equation $bx=a$ because it is "the definition of division" (which is true). However, to algebraically show this in small logical steps, we divide both sides of this equation by $b$, producing the intermediate $bx/b = a/b$, and then by the associative property of both multiplication and division, $(b/b)x = a/b$. Now, to take the last step, we must assume two things to be true: that $b/b=1$ so that $1*x = a/b$, and that $x*1=x$ so that $x=a/b$.

Therefore, we cannot prove $n/n=1$ except in terms of itself (not even by generalizing using abstract algebra and its definition of multiplication). It is an axiom, a postulate, as is $n*1=n$ (the multiplicative identity, which is also axiomatic in the abstract realm); something self-evident that forms the basis of more complex reasoning. You must accept it as true for our system of mathematics to function.

If $n/n \neq 1$, then multiplying both sides by $n$ produces $n \neq 1*n$, and thus if you still accept the multiplicative identity axiom, $n \neq n$ violating the reflexive property of equality. To reject that $n/n=1$ thus means you must either also reject that 1 is the multiplicative identity (and thus it changes $n$ into "not $n$"), or you must reject that equality is a reflexive comparison operation (that something always equals itself). If you do either of those, you're no longer speaking our mathematical language.

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This answer does not make any sense to me. By definition, $a/b$ is that number $x$ that satisfies $bx=a$, if one exists. In a field $F$, such elements are postulated to exist for all $a,b\in F$ such that $b$ is not the additive identity. Given an equation $bx=a$, we know there is a solution and we denote it by $x=a/b$, where $/$ is a new symbol invented to represent this concept. Note the lack of any tautological manipulations. Since we know that $n\cdot 1=n$ by the definition of $1$, we know there exists a solution to $nx=n$, and it is unique iff $n\ne0$. In this case, we can say $1=n/n$. – Mario Carneiro Jun 13 '13 at 5:26
As I discussed with Jared, the abstract operations are defined in general terms for any mathematical object based on the properties we observe axiomatically with numbers. For instance, we define how addition and multiplication work within rings in general based on how they work with integers. The fraction field of an integral domain is defined based on the equality we can show in fractions of integer terms, and so the idea that the slash symbol, representing the inverse of multiplication, is by any means "new" in this meaning is laughable; it is the abstraction of division and its axioms. – KeithS Jun 13 '13 at 5:55
I think you are conflating the history and development of math with the definitions involved. The division symbol is "new" in the sense that we are only given the field axioms to start with, and the division operation is defined given the inverse existence axiom. Obviously we've all seen division before in grade school or other contexts, but you can't use any previous knowledge like that to prove anything, unless you have the basics down pat. And since this is a question about the basics, you definitely can't appeal to your common notion of what division is to make a proof. – Mario Carneiro Jun 13 '13 at 6:02
You are correct that most (all?) of abstract algebra is motivated by the desire to generalize more common structures that are seen in basic arithmetic, but once the definitions are made, they exist independent of their "ancestry", so to speak. But elementary experiences with numbers are poorly axiomatized, so it is hard to prove anything without appealing to some "reader's intuition", which is exactly what a rigorous proof should avoid. Even if you want to prove something like $n/n=1$ for rational numbers, the easiest approach is just to show the rationals are a field and then use the axioms. – Mario Carneiro Jun 13 '13 at 6:13

Because $1$ is the multiplicative identity element and the multiplication is injective in $\mathbb{R}\setminus\{0\}$. And that's why $\frac00$ isn't defined, the multipication isn't injective in $\mathbb{R}$. And $0^0$ is defined just as $\lim\limits_{x\to0}x^x$.

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