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.

Solution 1:

If $0/0$ were equal to $1$, then $1=\frac{0}{0}=\frac{0+0}{0}=\frac{0}{0}+\frac{0}{0}=1+1=2$.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.

Solution 5:

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.