This is my proof:
So, $0\cdot0=0$
And we know that
$a-a=0$
By substitution,
We have $(a-a)(a-a)=0$
Then by simplifying, $a^2-a^2+a^2-a^2=0$
and the we have $0-0=0$,
Therefore, $0=0$.
I am not sure about my answer. Will you please show me another way of proving it or some way to improve my answer? Thank you!
$0 \cdot 0 = 0$ because $a \cdot 0 = 0$, which is an axiom of multiplication in Peano arithmetic.
The way I read your proof, you use the $=$ sign to denote what you intend to proove, not what you already have established. You might make that clearer by writing $\overset?=$ instead. But I'll leave that notational aspect to other answers, and concentrate on what I believe you are trying to say.
You could improve your derivation by stating clearly what set of axioms you assume, what theorems you already derived from these axioms (as far as they are or at least might be relevant), and which of these axioms or theorems you use at each step.
If you look at e.g. the definition of a ring in Wikipedia, and compare that with your proof, you could end up with something like this:
At this point you can see that your proof is flawed, and can either look for ways to fix it, or start in a different direction. And I hope you will notice how shorthand notation like the use of minus as a binary operator can lead to oversights when dealing with axiomatic systems at such a low level.
If you apply the same notation to a variant of the proof Mirko suggested, you get
Taking everything together you conclude that $\forall a:0\cdot a=0$. Using essentially the same steps, you can show that $\forall a:a\cdot 0=0$. Either of these will include $0\cdot0=0$ as a special case.
If you don't like the use of $b$ as an abbreviation for $0\cdot a$, or if you prefer dealing with terms instead of equations, you can also write this whole thing as a sequence of such term transformations:
\begin{align*} 0\cdot a &= 0\cdot a + 0 &&\textbf{additive identity} \\&= 0\cdot a + \Bigl(0\cdot a + \bigl(-(0\cdot a)\bigr)\Bigr) &&\textbf{additive inverse} \\&= (0\cdot a + 0\cdot a) + \bigl(-(0\cdot a)\bigr) &&\textbf{associativity} \\&= (0 + 0)\cdot a + \bigl(-(0\cdot a)\bigr) &&\textbf{right distributivity} \\&= 0\cdot a + \bigl(-(0\cdot a)\bigr) &&\textbf{additive identity} \\&= 0 &&\textbf{additive inverse} \end{align*}
If you are not talking about rings, then what else are you talking about?
You have your "if"s and "then"s in the wrong order.
You say "we have $(a-a)(a-a)=0$." But you don't have that unless you know $0\cdot0=0$. Then you go on to say that by simplifying, you get $a^2-a^2+a^2-a^2=0$. But the fact that $a^2-a^2+a^2-a^2=0$ is something you know before you're done proving that $0\cdot0=0$. So you should say first $a^2-a^2+a^2-a^2=0$ and then from that deduce (by factoring) that $(a-a)(a-a)=0$. Which one you deduce from which is what you've got backwards.
At the end you say "Therefore $0=0$", but again that's something you know before you're done proving that $0\cdot0=0$, so again you've got your "if" and your "then" interchanged.
Say $0\cdot0=p$. We need to prove that $p=0$. Use that $0+0=0$.
Then $p=0\cdot0=0\cdot(0+0)=0\cdot0+0\cdot0=p+p$.
Hence $p=p+p$, and $0=p-p=p+p-p=p$, which completes the proof.
I asked in the comments what is $0$ and what is multiplication.
Your answer was "Its just a simple multiplication of zero".
So then, what is "simple multiplication of zero"? And, what is zero?
The above is one possible proof, depending on what assumptions you start with. The same way you could prove that $r\cdot0=0$, for each $r$. Indeed, let $r\cdot0=q$. Then $q=r\cdot0=r\cdot(0+0)=r\cdot0+r\cdot0=q+q$.
Hence $q=q+q$, thus $0=q-q=q+q-q=q$.
To make the proof look like yours, start with $0\cdot0$ and transform this to $0$.
Indeed $0\cdot0=(a-a)(a-a)=a^2-a^2+a^2-a^2=0+0=0$.
The set-theoretical proof: $$ 0 = |\emptyset|,\quad |A|\cdot|B| = |A\times B|\implies 0\cdot 0 = |\emptyset|\cdot|\emptyset| = |\emptyset\times\emptyset| = |\emptyset| = 0. $$
If we take real numbers:
$0\cdot0 = (0 + 0) \cdot 0\qquad$ ($0 = 0+0$ because $0$ is the additive zero element)
$\qquad= 0\cdot0 + 0\cdot0\qquad$ (distributive law, $(a+b)\cdot c = a\cdot c + b\cdot c$)
Since $x = x+a$ with $x = 0\cdot0$ and $a = 0\cdot0$, this makes $0\cdot0$ the additive zero.
i will try the proof with "Reductio ad absurdum": 0x0<>0
(a) If 0x0<>0 then (a.1) 0x0>0 or (a.2) 0x0<0.
For example - a.1:
0x0>0 => multiply by 1 both sides, we have the following equation
(3) 1*0*0>1*0
And for the same logic to a.2, we have the following:
(4) 1*0*0<1*0
For (3) and (4), we have that it's NOT possible that the same numbers:
1*0*0 and 1*0 are less and greater simultaneously. Therefore, The only way is that the numbers are equal. For that, We Can conclude that 0*0=0.
I don't pretty sure for the details of applying "Reductio ad absurdum" but I believe that could be a possible way to prove it.
