Why is $n$ divided by $n$ equal to $1$?

Question

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).

Answer 1

$\frac ab$ 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 $1$.

Answer 2

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.

Answer 3

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 $a^{0} = 1$.

Answer 4

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:

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

Answer 5

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 $[n/n]=[1/1]=1\in\mathbb{Z}$.

Answer 6

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 $\,b\ne 0.$

Answer 7

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.

Answer 8

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.

Answer 9

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 $\Pi(\varnothing)$.

Answer 10

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.

Answer 11

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, $\textbf{n/n = 1}$.

Answer 12

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.

Answer 13

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.

Answer 14

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 $\lim\limits_{x\to0}x^x$.

Answer 15

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 $b=0$.

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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.

Answer 16

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: $$\dfrac rr=1.$$ $\blacksquare$

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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 $\frac{1}{\frac{1}{X}}=X$?