Seeking elegant proof why 0 divided by 0 does not equal 1 Several years ago I was bored and so for amusement I wrote out a proof that $\dfrac00$ does not equal $1$. I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$. As this is clearly false and if all the steps in my proof were logically valid, the conclusion then is that my only assumption (that $\dfrac00=1$) must be false. Unfortunately, I can no longer recall the steps I used to arrive at the contradiction. If anyone could help me out I would appreciate it.
 A: In lay terms, evaluating 0/0 is asking "what number, when multiplied by zero, gives zero".  Since the answer to this is "any number", it cannot be defined as a specific value.
A: The accepted definition of division on the natural numbers is something like:

For all natural numbers $x, y, z$ where $y\ne 0$, we have  $x/y = z$ iff $x=y\times z$. (Also works for the integers, rational numbers and reals.)

Using this definition, you can neither prove nor disprove that $0/0=1$. You wouldn't be able to draw any inferences from your assumption that $0/0=1$. If $y$ (the divisor) is $0$, this definition tells you nothing. 

Suppose we did not have the restriction $y\ne 0$ and that, instead, we simply defined $x/y = z$ iff $x=y\times z$ for any natural numbers $x, y$ and $z$. 
Then, consider two cases: $x=0$ and $x\ne 0$.
If $x=0$, then the definition would be inconsistent with our definition of the natural numbers. 
$0/0$ could be $0$ because $0\times 0 =0$
$0/0$ could be $1$ because $0\times 1 = 0$
$0/0$ could be $256$ because $0\times 256 = 0$
All natural numbers would have to be equal (a contradiction). This alone would be enough to reject our restriction-free alternative definition. It is inconsistent.
If $x\ne 0$, then no natural number would work for $x/0$. For any natural number $z$, we could not have $x=0\times z$. Zero times any number is always zero.
Either way, the alternative definition simply doesn't work.
A: Let's view the problem within ring theory, i.e., an algebraic structure with addition and multiplication following familiar axioms. Usually when we write $0$ in this context, it means an additive neutral element, i.e. $x+0=0+x = x,\ \forall x$. In any ring it follows from distributive laws that for any $x$ we have: $$x\cdot 0 = x\cdot(0+0) = x\cdot 0 + x\cdot 0\implies x\cdot 0 = 0$$ Now, when we write $\frac xy = z$, in this context we mean that there is unique $z$ such that $x = zy$. If there are more than one $z$ meeting this condition, $\frac xy$ wouldn't be well defined. 
Let us assume that we have a ring $R$ with additive neutral element $0$, such that division by $0$ is well defined for some $x$, that is, there is unique $y\in R$ such that $y\cdot 0 = x$. First we note that $x = y\cdot 0 = 0$, so only $\frac 00$ might be defined. Now, assuming $\frac 00 = y$, as we said before, it means that $y$ is the unique element in $R$ such that $y\cdot 0 = 0$. But since for any $y\in R$ we have that condition, we conclude that $R$ has exactly one element, namely $R=\{0\}$. We call this ring zero ring.
TLDR: Division by $0$ is possible, but only in a trivial (zero) ring, and nowhere else.
A: If $0/0$ were equal to $1$, then $1=\frac{0}{0}=\frac{0+0}{0}=\frac{0}{0}+\frac{0}{0}=1+1=2$.
A: Yes. There is a algebraic reason. In a field there is no reasonable way we can divide by zero, because one cannot have  both the identities $(a/b)\times b =a$ and $c\times0=0$ hold simultaneously if $b$ is allowed to be zero.
Note that the cancellation law depends of non-zero divisors:
Proposition (Integers have no zero divisors). Let $a$ and $b$ be integers such that $ab=0$. Then either $a=0$ or $b=0$ (or both).
Corollary (Cancellation law for integers). If $a, b, c$ are integers such that $ac=bc$ and $c$ is non-zero, then $a=b$.

EDIT. By other hand, is possible to construct a algebraic structure with $0/0=1$ (similar a ring, but adding another axioms, maybe as $0/0=1$). But in that case, we must consider that we are no working with the rationals $\Bbb Q$ or the reals $\Bbb R$, since they are fields, so their theorems couldn't be true.
A: Assume $\frac{0}{0} = 1$ and define $x := \frac{0}{0^2}$. Then
$$
0 = 0\cdot x = 0 \cdot \frac{0}{0^2} = \frac{0}{0} = 1.
$$
A: Here is one of the famous "fake" proof that $0=1$.
Let $a=b=1$ (But here you can choose another number is you want). 
$a²=ab$
$a²-b²=b(a-b)$,
$(a-b)(a+b)=b(a-b)$
Then you divide each term by $(a-b)$ (here is the clincher, of course...) and you get
$a+b=b$, hence $a=0$ (But remember that $a=1$...).
Of course in order to "get" this result, you have to divide by $0$, which as NO meaning whatsoever... And you can get whatever result you want, which will be meaningless, since you did something meaningless. 
A: Here are two algebraic proofs(plus some calculus).
\begin{align*}
\frac00={}&0^0 \\
\log\left(\frac00\right)={}&\log\left(0^0\right) \\
\log\left(\frac00\right)={}&\log(0)\cdot0 \\
\end{align*}
In calculus, the limit of log(x) is equal to the limit of -1/x as they approach 0+
\begin{align*}
\log\left(\frac00\right)={}&-\left|\frac10\right|\cdot0 \\
\log\left(\frac00\right)={}&\frac00
\end{align*}
For $\log\left(\frac00\right)={}\frac00$, if $\frac00=1$, then 0=1.
\begin{align*}
\frac00={}&x \\
0\cdot\frac00={}&x\cdot0 \\
\frac{0\cdot0}{0}={}&x\cdot0 \\
\frac00={}&\frac00\cdot0.
\end{align*}
If $\frac00=1$, then 1=0.
A: The other answers are quite helpful.I only want to add that $\frac00$ is just not defined any value.It is one of the several indeterminate forms of mathematics-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
A: Here is my theory: if anything times zero is zero, than zero divided by anything is zero, and zero divided by zero is anything. When I say "anything" I mean, any number that exists, even an imaginary number. 0*anything=0, so that should mean 0/anything=0, and 0/0=anything. So if I say 0/0=0, that would be true, and if I say 0/0=1, that would be true as well. 0/0=infinity would also be true, 0/0=(-infinity) would be true. Think of it like this: nine divided by three is three, because three goes into nine three times. But how many times must zero be added to get to zero? If zero is added once, it'll be zero. If zero gets added twice, it'll be zero. If it's added three times, four, five, six, seven, as much as infinity, a negative number, a decimal/fraction, a whole number, an integer, a rational number, an irrational number, a real number, an imaginary number, ALL NUMBERS, it'll still be zero. So basically zero divided by zero can be anything, in my theory.
A: Here is an argument for the kindergarten notion of division, i.e., to compute $a/b$ for natural numbers $a,b$, pretend you take $a$ cookies and divide them among $b$ kids, and ask yourself how many cookies will each child get. Theorem: $a/0$ does not have a unique value, and therefore it does not equal anything. Proof: Take $a$ cookies and divide them among $0$ children. It is vacuously true that each child gets $5$ cookies. It is also vacuously true that each child gets $m$ cookies, for any value of $m$. QED. 
A: We know that
$$ 0 \times 1 = 0 \times 2 $$
If we assume $0/0 = 1$, then we can divide both side by zero, and get
$$\begin{align*} \frac{0}{0} \times 1 &= \frac{0}{0} \times 2  \\
                       1 &= 2\end{align*} $$
We know that $1 \ne 2$, and so $\frac{0}{0} \ne 1$.
