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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!

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15  
what is $0$? What is multiplication? Say $0\cdot0=p$. Since $0=0+0$, we have $p=0\cdot0=0\cdot(0+0)=0\cdot0+0\cdot0=p+p$. Thus, $p=p+p$, hence $p-p=p+p-p$, that is, $0=p$. – Mirko Feb 23 at 3:02
67  
You are trying to prove that $0*0=0$. But you have assumed at the start, without proof, that $0*0=0$. So there is no possible way that this can work. It is also impossible to answer your question without knowing what you are allowed to assume. (Group axioms? Field axioms? Something else?) – David Feb 23 at 3:02
21  
What qualifies it as "simple multiplication of zero"? For proofs of basic properties from the axioms, it is critically important that the axioms are specified. – T. Bongers Feb 23 at 3:06
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Before you "start the flow of [your] proof", you have to make it clear what the question is. See my previous comment. – David Feb 23 at 3:08
9  
As @David mentioned, you started by assuming what you want to prove, and this does not make a valid proof. But also notice that your conclusion is wrong: you write "Therefore $0=0$", but this isn't what you need to prove. So in fact, your proof is a bit upside down: you could start with $0=0$, since this follows from the definition of equality, and try to finish with $0\cdot0 = 0$. (Although in this case a better proof is given based on an axiom by @CommonerG.) – Théophile Feb 23 at 3:16
up vote 47 down vote accepted

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.

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2  
Basically, write your proof in reverse order and adjust the connectives – Stella Biderman Feb 23 at 15:13
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In my post I claim that at least when using the axioms for rings, the original proof is flawed in a more fundamental way, by essentially making use of the proposition while simplifying $(-a)a$ to $-(a^2)$. You didn't state what axioms you assumed, so what did you have in mind? Do you know of a commonly used set of axioms which would not have this problem, which would not be Peano arithmetic, and where “reversing the proof” would indeed make it valid? – MvG Feb 23 at 19:19
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@MvG : Actually I wasn't concerned with making the proof complete and valid, but only with pointing out a problem with it, which is a thing to be concerned about in virtually all proofs. The proposition you mention, $(-a)a = -(a^2)$, is true in rings in general, so at most this example shows the proof is incomplete. $\qquad$ – Michael Hardy Feb 23 at 19:26
    
Proving something you already understand is very hard to do objectively! – corsiKa Feb 24 at 23:55
    
@corsiKa : Perhaps, but certain things about writing proofs still need to be borne in mind, including the difference between "if A then B" and "if B then A". And besides, this may not be something you already know: you may know that it's true of the number $0$, but is it true of the $0$ element in every structure satisfying the ring axioms? That's a different question. – Michael Hardy Feb 24 at 23:57

$0 \cdot 0 = 0$ because $a \cdot 0 = 0$, which is an axiom of multiplication in Peano arithmetic.

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Huh, usually I have had to derive this. Although maybe it depends on what you take for granted... $a \cdot 0 = 0$ but $0 + 0 = 0$ therefore, $a \cdot 0 = a \cdot ( 0 + 0 ) = a \cdot 0 + a \cdot 0$. So $a \cdot 0 = a \cdot 0 + a \cdot 0 \implies 0 = a \cdot 0$ – Dair Feb 23 at 4:09
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@Dair: Are you sure you're thinking of Peano arithmetic? In every formulation I've seen of Peano arithmetic, $a \cdot 0=0$ is part of the definition of multiplication (and the distributive property is a theorem rather than an axiom). Your proof sounds more like field theory. – ruakh Feb 23 at 5:39
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@ruakh: Oh yes, that's probably it: I'm thinking of fields. – Dair Feb 23 at 5:41
18  
Dairy is thinking of field theory but he raises the point, that the question cannot be answered unless we are in agreement of what system we are defining and deriving the terms. I'm a field person myself. – fleablood Feb 23 at 9:14
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I have to downvote, because nowhere is it mentioned that we are working with natural numbers/rational numbers/real numbers... I take the question to be asking, how can we show that an additive identity element is a multiplicative annihilator. – Morgan Rodgers Feb 23 at 12:56

Making your proof more rigorous

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:

  1. Let $a$ be an arbitrary element of the ring. You could simplify the proof by specifically choosing $a=0$ or $a=1$, but I'll follow your approach. The important fact is that the underlying set cannot be empty, which is guaranteed by the existence of the additive and multiplicative identities (which might be the same).
  2. Let $-a$ be the additive inverse of $a$, so $a+(-a)=0$. Writing this as $a-a$ is a syntactic simplification which can obscure what exactly you may assume at any given point, so I'd not do this here.
  3. Substitute that into both occurrences of $0$ in $0\cdot 0$, to obtain $\bigl(a+(-a)\bigr)\bigl(a+(-a)\bigr)$. It might be enough to substitute one, since in fact $0\cdot a=0$ for any $a$, but again I follow your approach.
  4. Use left distributivity to split the right paren: $\bigl(a+(-a)\bigr)a + \bigl(a+(-a)\bigr)(-a)$.
  5. Use right distributivity twice to obtain $\bigl(a^2 + (-a)a\bigr) + \bigl(a(-a)+(-a)^2\bigr)$.
  6. Next you essentially simplify $(-a)a$ to $-(a^2)$. But how do you know that's valid? You might be tempted to derive that from $a^2+\bigl(-(a^2)\bigr)=0=0\cdot a=\bigl(a+(-a)\bigr)a=a^2+(-a)a$. But at the second $=$ you'd need to show that $0=0\cdot a$, which is a more general version of what you're about to prove.

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.

A working example of a pretty rigorous proof

If you apply the same notation to a variant of the proof Mirko suggested, you get

  1. Consider $0\cdot a=b$ which includes the special case of $a=0$.
  2. Since $0$ is the additive identity, you have $0+0=0$.
  3. Substitute 2. into 1. to get $(0+0)\cdot a=b$.
  4. Use right distributivity to obtain $0\cdot a+0\cdot a=b$.
  5. Substitute 1. into this to obtain $b+b=b$.
  6. Add $-b$, the additive inverse of $b$, to both sides: $(b+b)+(-b)=b+(-b)$.
  7. Apply associativity on the left hand side to obtain $b+\bigl(b+(-b)\bigr)=b+(-b)$.
  8. Use the fact that $b+(-b)=0$ since $-b$ was chosen to be the additive inverse. So you get $b+0=0$.
  9. And since $0$ is the additive identity this simplifies to $b=0$.

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*}

Alternate subject areas

If you are not talking about rings, then what else are you talking about?

  • If you are talking about natural numbers, there are several ways to define those – or rather the multiplication operation on these.
    • One could define them using Peano arithmetic, and as CommonerG pointed out, $a\cdot0=0$ is part of the definition of multiplication there.
    • One could define them as the cardinals of finite sets, with multiplication defined the way Martín-Blas Pérez Pinilla used it.
    • Other set-theoretic definitions represent the numbers themselves as sets. I'm no expert on how multiplication is defined in each of these formalisms, but I guess it will likely boil down to Peano arithmetic again.
  • You might be talking about integers, rationals, reals or complex numbers. Each of these is usually either defined axiomatically in a way that includes the ring axioms, or constructed (, , , ) in a way that eventually builds on natural numbers, so you'd prove it there and then use the details of the construction to propagate that fact.
  • Transfinite cardinal arithmetic is a generalization of this cardinal-based set-theoretic definition to infinite cardinals. Doesn't make a difference for $0$. Transfinite ordinal arithmetic has different definitions, so be precise which one you use.
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2  
+1 for making it clear that $a0 =0$ and the distributive law are interdependent at the foundation level. In rings the latter is an axiom. When defining arithmetic inductively in the natural numbers it's a consequence of the former. – Ethan Bolker Feb 23 at 14:47
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@MvG this is an excellent answer. Taking a learners own workings and extending. Should be the accepted answer in my opinion. – Karl Feb 23 at 19:31
    
I just realized that the thing to prove is a variant Fields - Proof that every multiple of zero equals zero. More special since it's only $0\cdot0$ not $0\cdot a$, but at the same time perhaps more general since it's not explicitely about fields, and therefore may be about rings. My proof as a sequence of term transformations closely resembles the one by Adriano over there. – MvG Feb 24 at 9:27

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 $0\cdot0=(a-a)(a-a)=a^2-a^2+a^2-a^2=0+0=0$.

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Thank you very much! :) – yourkeyfence Feb 23 at 3:38
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@Mirko Just a small remakr: In the third line it's pretty confusing that you write $0$ again instead of $p$ (in the last equation). – noctusraid Feb 23 at 7:34
    
@noctusraid Thank you, corrected $0=p-p=p+p-p=0$ to $0=p-p=p+p-p=p$, the latter was meant but I must have overlooked what I had typed. – Mirko Feb 27 at 6:23

The set-theoretical proof: $$ 0 = |\emptyset|,\quad |A|\cdot|B| = |A\times B|\implies 0\cdot 0 = |\emptyset|\cdot|\emptyset| = |\emptyset\times\emptyset| = |\emptyset| = 0. $$

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1  
I'd love to see how that is relevant to multiplication of numbers. – gnasher729 Feb 23 at 11:01
    
@gnasher729, natural numbers are cardinals of finite sets. – Martín-Blas Pérez Pinilla Feb 23 at 11:15
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Why the assumption that we are working with natural numbers? – Morgan Rodgers Feb 23 at 13:22
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@Martín-BlasPérezPinilla 0 is the additive identity in any ring. – Morgan Rodgers Feb 23 at 13:25
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@MorganRodgers, true, but I don't think that the OP was thinking of rings. – Martín-Blas Pérez Pinilla Feb 23 at 13:31

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.

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$f(x) = x*x = x^2$

$f'(x) = 2*x$

Since f(x) is differentiable at x = 0 it means that the original function is continuous at x = 0.

Limit as x -> 0 the function also approaches 0.

Thus, $x*x = 0$

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2  
Note the linear algebra and abstract algebra tags... – Benjamin Dickman Feb 23 at 7:51

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.
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How do you know that $1\times 0 \times 0$ and $1\times 0$ are the same numbers? – Vincent Feb 23 at 15:20
    
By Hypotheses, I am assuming that the inequality is supposed to be true: 1*0*0>1*0 and 1*0*0<1*0. Then, multiply by 1 in the both sides, don't change the equation. In that way, "The numbers" refereed in one equation are the same numbers in the other equation. But those equations cannot be true, at the same time. So, they have to be equal. 1*0*0=0*0==1*0=0=> 0*0=0 is true. – oagostinho Feb 23 at 17:02
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But then your assumption itself is a contradiction. You cannot assume $a<b$ and $b<a$ at the same time. You can assume that one of them is true, and then derive a contradiction, but you can't combine the two, since that is by definition a contradiction and doesn't give you new information. Suppose I want to prove $sin(0)=77$. If $sin(0)>77$ and $sin(0)<77$, then $2 \times sin(0) > 2 \times 77$ and $2 \times sin(0) < 2 \times 77$. But these cannot be true at the same time. Hence $sin(0)=77$. – Vincent Feb 24 at 8:59
    
@Vincent Yes, that's right! – oagostinho Mar 22 at 15:44
    
@Vincent Yes, that's right! what about this: (a) 0x0>0 => multiply 0, both sides => (b) 0x0x0>0x0. Additionally, for (a) and (b) We have that 0x0x0>0x0>0 => 0x0x0 >0 => 0^3>0. And, this equation could not be possible because if we apply LOG in the both sides that would be undefined. What do you think guys? – oagostinho Mar 22 at 16:50

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