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Andy my friend claims that $0=X\cdot 0$ where $X$ is an arbitrary value, $0\div 0=X$, and thus $0/0$ can be assigned to an arbitrary value. But he mistakenly considers $0÷0=1$ or the multiplication by $0$ is always reversible. As far as I know, $0/0$ is undefined and cannot be assigned to any arbitrary value $X$ by any valid mathematical operations.

But why doesn't math introduce an axiom $0\div 0=0$ as a means to claim the multiplication by $0$ is irreversible? The mathematical definition of fractions does exclude cases of $0$ dividers, so it seems that this axiom doesn't violate the established math rules.

What paradoxes, flaws, or inconsistencies will arise from this pseudo-axiom $0\div 0=0$ in whole established math system? Or does it change the true value of any equation?

ps: I am sorry if I have made the wrong statements.

Edits:

I think the pseudo-axiom should be clarified as, $$0∈φ$$, where every element of $φ$ is every solution to $0/0$, and $0$ is not necessarily the unique solution.

With the following perspectives:

  1. The division is meant to be the inverse of multiplication. Multiplying X by 0 is a multiple-to-1 function, so $0$ divider map $0$ to every real number, or $0/0$ have the range set containing elements of every real number.
  2. The law of associative is satisfied if $0/0$ is assigned to $0$ or $k=1$, so that $(k×0)/0=k×(0/0)$, where k is a constant.
  3. [deleted]
  4. It holds true if $0/0$ is assigned to $0$, so that $$k×(0/0)=m×(0/0)$$, where $k$ and $m$ are constants.
  5. In conclusion, $0/0$ have solutions of every real number in the perspective of the inverse of the multiplication by $0$, or mathematically called $undefined$. But if a non-1 constant is factored out from $0/0$ while holding the true value of equating with the initial value, $0/0$ is assigned to $0$ by this behaviour.

ps: I am sorry if I have the wrong statements or logical falsities.

Edits:

However, in the most general sense, $0/0$ is undefined mathematically, as the pseudo-axiom states that dividing $0$ by $0$ is a futile endeavour. From a strict mathematical perspective $0/0$ is undefined.

User
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  • Welcome to MSE! You can use MathJax for formulas, see the tutorial here: https://math.meta.stackexchange.com/questions/5020 I have already submitted an edit for your question to do that for you. – Samuel Adrian Antz Jun 06 '23 at 14:10
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    Binary operations have one value, so "can be anything" means multiplication would no longer be a binary operation. Defining $0/0=0$ would make it a binary operation, but it would have no purpose. The ultimate reason we can't divide by $0$ is that $0$ has no multiplicative inverse, and division is defined in terms of multiplicative inverses. – Thomas Andrews Jun 06 '23 at 14:14
  • Consider two functions: $f(x)$, $g(x)$. If we have $\lim_{x \to k} f(x) = a$ and $\lim_{x \to k} g(x) = b$ then $\lim_{x \to k} \frac{f(x)}{g(x)} = \frac{a}{b}$, with the exception of the case $a = 0, b = 0$ . That case is an indeterminate form, the value of the limit can't be determined from the value of the components. There we apply other rules, such as L'Hôpital's rule.

    https://en.wikipedia.org/wiki/Indeterminate_form https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

    Undefined values in operations (such as $0/0$ or $0^0$) correspond to indeterminate forms.

     – user3257842 Jun 06 '23 at 14:15
  • "and thus $0/0$ can be assigned to an arbitrary value" I would say, if you forced me to assign anything, that it should be assigned every value. Simultaneously. That's the correct conclusion of your friend's argument, in my opinion. – Arthur Jun 06 '23 at 14:16
  • @ user3257842 Hi, the asker here. We may not discuss limit values, because lim(x->0){x} is a value extremely approaching 0, not exactly 0. – User Jun 06 '23 at 14:18
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    Of the dozen times this exact question has been asked over the past week or two, this version is at least well phrased and respectful. Thank you for that. In the end, established mathematics that is used and taught is defined the way it is because it is useful. If you wish to suggest an alteration or modification, it would need to serve some purpose and be considered useful. I do not see any benefit to altering the definition of division by zero for the purposes of any real world applications, and would instead find attempting to change people's habits too harmful. – JMoravitz Jun 06 '23 at 14:22
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    In a similar vein, it would be really great if we were to work in a dozenal system rather than a decimal system, and it would be great if we preferred using $\tau$ as the circle constant rather than $2\pi$. Even if redefining division by zero may not cause logical inconsistencies, some things are so ingrained in culture that it will be nearly impossible to convince anyone to change. – JMoravitz Jun 06 '23 at 14:24
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    With just the tiniest change, you could make your question completely precise in a mathematical sense. Instead of referring vaguely to the whole established mathematical system, one could refer to the field axioms together with the standard definitions of subtraction and division, and all of the theorems derived from them. And then your question becomes whether the addition of one more axiom, namely $0 \div 0 = 0$, allows one to derive a contradiction. – Lee Mosher Jun 06 '23 at 14:31
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    The catch is that $\frac{0}{0}$ could be any real number because of $x\cdot 0=0$ for every real number. The consequence is that all real numbers are equal. Should be a good reason to consider $\frac{0}{0}$ undefined. – Peter Jun 06 '23 at 14:43
  • To me at least, if you are asking for $0 \div 0 = 0$ to be set as a definition, and the operation $\div$ is extended arbitrarily (i.e. not as an "inverse" of multiplication) to fit this, then I cannot think of a single true statement that is a consequence/logical implication of $0 \div 0 = 0$. The problem being that you can't do anything with the $\div$, it's been arbitrarily defined for this pair. The point is, to me at least, this definition is a logical dead end, and if statements are dead ends then they can be right but won't be useful. – Sarvesh Ravichandran Iyer Jun 07 '23 at 06:55
  • @User Sorry, but point 5. is beating a dead horse. On one hand, you say "$0/0$ have solutions of every real number" so for example $(x \cdot 0) / 0 = 0/0$ is undefined. On the other hand, you say "if a non-$1$ constant is factored out from $0/0$ ... $0/0$ is assigned as $0$" so for (the same) example $(x \cdot 0) / 0$ $= x \cdot (0/0)$ $= x \cdot 0 = 0$. You can't assign a meaningful (consistent with the rest of arithmetics) value to $0/0$ no matter how many rules and exceptions you try to add to your definition, which is precisely why math left $0/0$ undefined. – dxiv Jun 08 '23 at 06:34
  • @dxiv The statement that $0/0$ have solutions of every real number might not nullify that $0/0=0$ – User Jun 08 '23 at 06:41
  • @User But if you allow that $0 = \text{every real number}$ then your proposed "system" is useless. Which is what I meant in the previous comment. Attempting to evaluate the same expression $(x \cdot 0)/0$ two different ways - according to the rules you proposed - resulted in two different values, either undefined or zero. Whether you say that it's fine that an evaluation can have different values, or you say that zero can be any value, either claim violates fundamental principles of basic arithmetics. – dxiv Jun 08 '23 at 06:48
  • @dxiv The solution of x^2-3x-4=0 is x=4 or -1, but it doesn't mean 4=-1. – User Jun 08 '23 at 06:49
  • @User That's not the same $x$, obviously. Please don't. – dxiv Jun 08 '23 at 06:51
  • @dxiv The operation of $0/0$ is by the perspective of the definition that division is the inverse of the multiplication. In this perspective, the solutions of $0=0X$ is the same thing as those of $0/0$. $X$ can be any real number, but it doesn't mean any real number are equating. – User Jun 08 '23 at 06:58
  • @dxiv $X=0/0$. If factoring $X$ out a non-1 constant while being equal to the initial value, then you are assigning that $X$ can only be 0. – User Jun 08 '23 at 07:05
  • "$X$ can be any real number, but it doesn't mean any real number are equating" $;-;$ Sorry, but I have no idea what you even mean by that, and I have nothing more to add here at this point. – dxiv Jun 08 '23 at 07:06
  • @dxiv If the absolute value of Y is 7, Y can be 7 or -7, but it doesn't mean 7=-7. – User Jun 08 '23 at 07:15
  • @dxiv Multiplication by 0 is multiple-to-1 function. – User Jun 08 '23 at 07:22
  • @dxiv "If a non-1 constant is factored out from $0/0$ while holding the true value, $0/0$ is assigned as 0." For $x≠1$, $(x⋅0)/0≟x⋅(0/0)$. The associativie law holds true if assigning a unique value $0$ to $0/0$. If multiplying by 0 ($0=0x$) have the domain set that contains every real number, then the range of $0/0$ ($x=0/0$) also contains every real number. – User Jun 08 '23 at 09:21
  • Does this answer your question? How do we know our definitions don't lead to contradictions - see the third paragraph. This definition breaks the distributive law. – Ethan Bolker Jun 08 '23 at 11:17
  • @EthanBolker Exuse me, do you mean the falsity of being unable to expand (1-1)/0 as (1/0)-(1/0)? It is indeed one falsity. – User Jun 08 '23 at 11:36
  • @User Your post does not say what you want $1/0$ to be. The point of my linked answer is that any way you decide to define $x/0$ for any number $x$ (zero or not) will lead to a violation of some identity of ordinary arithmetic. Usually the distributive law. – Ethan Bolker Jun 08 '23 at 15:02
  • @EthanBolker You're right—defining x/0 for any number x (zero or not) will lead to a violation of some identity of ordinary arithmetic. While 0/0 is mathematically undefined, in perspective that division is inversed multiplication and 0 divider maps 0 to multiple real numbers, 0/0 can be defined as 0 and also meet the law of associative. – User Jun 08 '23 at 23:40

1 Answers1

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My favorite answer to this is: $0$ doesn't belong to the multiplicative group so that the question does not even arise.

A bit longer is the answer, that multiplication and addition are only connected via the distributive law. There is no other connection. This law allows us to conclude that $$0\cdot x=(1+(-1))\cdot x= 1\cdot x+ (-1)\cdot x=0.$$ Now, what is division? Division isn't really a mathematical concept. It is only the abbreviation of "multiply by the inverse element". Hence the question is, what would $0^{-1}$ be? Any inverse $x^{-1}$ of an element $x$ is defined as the solution to $x\cdot x^{-1}=1.$ Finally assume that there would be a reasonable solution to $0\cdot 0^{-1}=1.$ Then $$ 1=0\cdot 0^{-1}= 0 $$
by the above argument with the distributive law. But if we add the equation $1=0$ to our number system, then it will be pretty useless, or in mathematical terms: the real numbers wouldn't be a field anymore.

Marius S.L.
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    Despite the checkmark, the question was not what would $0^{-1}$ be, the question was what contradictions would ensue from adding one single formula to the laws of arithmetic, namely the formula $0 \div 0 = 0$. – Lee Mosher Jun 06 '23 at 14:54
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    Also, division certainly is a mathematical concept. We use abbreviations all the time in mathematics, they are called definitions. – Lee Mosher Jun 06 '23 at 14:56
  • The notation $\div 0$ needs an explanation because it wasn't given one. Hence, I gave it a meaning via $\div 0 =\cdot 0^{-1}.$ Based on that I illustrated a consequence. It is only a contradiction if we require $1\neq 0$ which was also not stated. I, therefore, explained where that formula would lead to, namely $1=0.$ – Marius S.L. Jun 06 '23 at 15:00
  • @LeeMosher "Also, division certainly is a mathematical concept. We use abbreviations all the time in mathematics, they are called definitions." Yes, that's why I used the definition of it. But it is an abbreviation, not a concept. The concept is inverses. – Marius S.L. Jun 06 '23 at 15:02
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    I disagree with this answer. The question as I interpreted it was saying "What if we define $a\div b$ to be $a\cdot b^{-1}$ in all cases where $b\neq 0$ but override the definition and instead define $a\div b = 0$ whenever $b=0$." The question of $0^{-1}$ is irrelevant and is well understood that $0^{-1}$ doesn't exist and that $0\cdot 0^{-1}$ is a meaningless expression and should not equal $1$. The question is "what benefits or negatives are there for doing this?" You have only covered why $0^{-1}$ shouldn't exist and why division by zero under the usual definition doesn't work. – JMoravitz Jun 06 '23 at 17:14
  • $a\div b=a \cdot (1\div b)$ and "whenever $b=0$" makes it $a\div 0=a \cdot (1\div 0)=a0^{-1}.$ I don't follow the nitpicking here. The question is all about $0^{-1},$ just not stated as clearly as it should have been. – Marius S.L. Jun 06 '23 at 17:33
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    @MariusS.L. S.L.

    Hi, the asker here. Thanks for your good insights.

    The established definition expects that any divider must be the inverse of a multiplication by X, or X(1/X)=1. But if 0(1/0)=1 is mathematically valid, then multiplying by (1/0) on both sides of the equation 20=30 gives (20)(1/0)=(30)(1/0), 2{0(1/0)}=3{0(1/0)}, and 2=3. Thus, the regular mathematical definition that X*(1/X)=1 can lead to paradoxes if the domain X includes 0. Therefore, this established rule of division may not prove the falsity of the pseudo-axiom.

     – User Jun 06 '23 at 19:15
  • Yes, exactly. You run into requirements like $2=3$ which can be made, but they don't represent our number systems anymore. The essential message is the distributive law $a \cdot (b+c)=a\cdot b +a\cdot c.$ Everything you can get by using this law is allowed. It connects multiplication and addition. Otherwise, they are not connected. E.g. you can multiply an area with the sum of two lengths and get a volume, but you cannot add length and area. The two operations are very different when you look at it. Only the formula above connects them. – Marius S.L. Jun 06 '23 at 19:44