What is wrong with the following argument (if you don’t involve ring theory)?

Proposition 1: $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}$ is not equal to $0$$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction

Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$

Therefore, $\frac{0}{0} = 0$.

Q.E.D.

Update (2015-12-01) after your answers:

Proposition 2: $\frac{0}{0}$ is not a real number

Proof[Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:Suppose that $\frac{0}{0}= x$, where $x$ is a real number.

Then, either $x = 0$ or $x$ is not equal to $0$.

1) Suppose $x = 0$, that is $\frac{0}{0} = 0$

Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $

Contradiction

Therefore, it is not the case that $x = 0$.

2) Suppose that $x$ is not equal to $0$.

$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction

Therefore, it is not the case that $x$ is a real number that is not equal to $0$.

Therefore, $\frac{0}{0}$ is not a real number.

Q.E.D.

Update (2015-12-02)If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.

Proposition 3: $\frac{0}{0}$ is not a real number

Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$

$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$

Q.E.D.

Update (2015-12-07):How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?

Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.

Then, $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$

Q.E.D.

Suggested

definitionof division of real numbers:If $b \ne 0$, then

$\frac{a}{b}=c$ iff $a=bc$

If $a=0$ and $b=0$, then

$\frac{a}{b}=0$

If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.

A somewhat more minimalistic version:

Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.

Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$

Q.E.D.

**Answer**

The error is in the very first implication. If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$. There is no reason to believe that we can do this division and get a number.

Therefore you have a proof that $0/0$ cannot equal any nonzero $x$. Combine this with a proof that $0/0$ cannot equal zero, and you have proved that $0/0$ is not a number.

**Attribution***Source : Link , Question Author : Lennart , Answer Author : vadim123*