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We all know that multiplication is the inverse of division, and therefore $x\div{x}\cdot{x}=x$ But what if $x=0$? $0\div0$ is undefined so $0\div0\cdot0$ should be too, but whatever happens when we divide that first $0$ by $0$ should be reversed when we multiply it by $0$ again, so what is the right answer? 0, undefined, or something else entirely?

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  • $\begingroup$ You didn't tell us what the set of numbers was. If you keep the set of numbers in mind, it becomes clearer that there is no such thing, since division on the natural numbers is VERY unlike division on the real numbers, especially in comparison to multiplication on the natural numbers and real numbers respectively. $\endgroup$ – Doug Spoonwood Jul 27 '15 at 21:36

17 Answers 17

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whatever happens when we divide that first 0 by 0 should be reversed when we multiply it by 0 again

I think this sentence reflects a fundamental misunderstanding of what it means when people say "$0 \div 0$ is undefined". It is not helpful to think of "undefined" as the name for an exotic, mysterious number-like object of some sort that is produced by the action of dividing $0$ by $0$. It is more helpful to think of "undefined" as a signal that you are trying to ask a question that has no answer.

When you write "$0 \div 0$" you are essentially asking the question: What is the unique number $r$ with the property that $0 \cdot r = 0$? That question doesn't have an answer at all, so $0 \div 0$ is a meaningless expression.

Now it is certainly possible to decide to introduce a new symbol into your mathematical world -- let's use 😱 (the emoji for "face screaming in fear") -- and declare that whenever we encounter the expression $0 \div 0$ we can replace it with the new symbol 😱 . Then your question is, "What is 😱 times 0?" And the answer to that question is, "We can't answer, because we don't know what 'times' means in this context." You may as well ask "What is pickle times lumberjack?" The question doesn't have an answer because "times" doesn't have a pre-existing meaning when one or both of the things being "multiplied" aren't numbers.

If that result is unsatisfying, you can try to extend your number system by defining a new rule: "From now on, I declare that 😱 times any real number is 😱 ." Or perhaps "From now on, I declare that zero times anything -- whether or not the anything is a number -- will always be zero." Or you can say "From now on, I declare that 😱 times zero is $1$." You can pick such a rule if you want to, but having made such a choice, there will be consequences. In particular it might turn out that in your new extended number system, you have to abandon the associative or commutative properties; alternatively, if you insist on keeping those properties, it might turn out that all numbers can be proven to be equal to each other, so the entire system collapses.

Ultimately, all mathematical definitions are conventions, and all conventions are local. So you can decide to define the expression $0 \div 0 \cdot 0$ any way you want to, if you really want to. But while you can define the expression any way you want, you have to live with the consequences, and the sad truth is that there really isn't a way to make the expression $0 \div 0 \cdot 0$ meaningful in a way to make it compatible with the other operations and properties of arithmetic.

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    $\begingroup$ "face screaming in fear" is the perfect emoji for this context. $\endgroup$ – preferred_anon Jul 27 '15 at 21:53
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    $\begingroup$ @DanielLittlewood Thanks, I felt pretty good about that choice myself. :) $\endgroup$ – mweiss Jul 27 '15 at 23:22
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    $\begingroup$ I'm a little bummed out that nobody has +1'd this for "all conventions are local". I'm awfully fond of that little aphorism. $\endgroup$ – mweiss Jul 27 '15 at 23:22
  • $\begingroup$ I think the term not defined works very well. For $\{ x | a * x = b\}$ only allows to write $x = f(a,b)$ IF $|\{ x | a * x = b\}| = 1$, and $|\{ x | 0 * x = 0\}| > 1$, whence $x = f(a,b)$ is not defined. $\endgroup$ – johannesvalks Jul 29 '15 at 21:22
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    $\begingroup$ @columbus8myhw I know about wheel theory, and I think it exemplifies my last sentence quite well: you can define division by zero, but you have to sacrifice many of the usual laws of arithmetic to do so. But I think if someone (perhaps you?) would like to write an answer to the OP that explains how wheel theory responds to the question, it would be a welcome addition to this thread. $\endgroup$ – mweiss Jul 29 '15 at 23:07
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[W]hatever happens when we divide that first $0$ by $0$ should be reversed when we multiply it by $0$ again[.]

Is it not equally "true" that if you multiply something by $x$, you can reverse whatever happens by dividing by $x$?

For example, if this sort of thing works, we should have $(3 \times 0) \div 0 = 3$ and $(17 \times 0) \div 0 = 17$. (The parentheses are there just to emphasize that we take a number, multiply it by zero, and then divide it by zero.)

But $3 \times 0 = 0$ and $17 \times 0 = 0$, so now we are saying that $0 \div 0 = 3$ and also $0 \div 0 = 17$.

This runs afoul of other commonsense notions, such as the notion that two things each equal to a third thing are equal, the notion that the result of a binary operation on two numbers should be another number, and the notion that there is one number named $3$ and that $3$ is always that same number and not equal to $17$.

You can probably think of some set of rules to avoid this mess, but in the end they will come down to saying that you can't compute $(17 \times 0) \div 0$ and have it come out to the "commonsense" answer. How about the rule "do not divide by zero"? It's very simple, and it solves the problem.

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    $\begingroup$ +1 for the last two sentences. Just don't do it, it's not rocket science. $\endgroup$ – fkraiem Jul 23 '15 at 3:31
  • $\begingroup$ Point is that $(\mathbb{Z}, \cdot)$, $(\mathbb{Q}, \cdot)$, $(\mathbb{R}, \cdot)$ are NO GROUPS - they do not have a neutral element, whence the inverse of multiplication is not general defined. $\endgroup$ – johannesvalks Jul 29 '15 at 21:31
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Just because one function is the inverse of another for infinitely many cases doesn't automatically mean that it's the inverse for all cases.

Here's a simpler example for you to consider: the Collatz problem. The Collatz function is defined thus: $f(n) = 3n + 1$ if $n$ is an odd integer, $f(n) = \frac{n}{2}$ if $n$ is an even integer. We can also define a doubling function $d(f(n)) = 2f(n)$ and a "thirding" function $c(f(n)) = \frac{n - 1}{3}$. If $f(n) \equiv 4 \pmod 6$, which is the inverse function of $f(n)$? The doubling function? Or the thirding function? If we're not keeping track of the iterations, then we just have no way of knowing.

Suppose $n = 7$ and $f(n) = 22$. If we put $f(n)$ through the doubling function we're not going to reverse $f(n)$ because the Collatz function was fed $7$, not $44$.

And so it is with the "zero function" $z(x) = 0x$. Whatever number $x$ is, $z(x) = 0$. If we only know what $z(x)$ is, we have no way of knowing what $x$ is, it could be any number at all.

But this zero function assumes you're giving it a number. Turns out $\frac{0}{0}$ is actually not a number. As others have already indicated, you can really wear down your brain constructing an argument that seems to define division by $0$ only to find it demolished by a slightly different argument. One of the arguments presented so far demonstrates three possible values for $\frac{0}{0}$. When you learn about complex numbers, you will see the possibilities become infinite, but no more valid.

So $\frac{0}{0}$ is undefined and invalid, and trying to multiply it by $0$ or any other number is pointless.

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Think about what it means for $x \div x$ when $x$ is any nonzero real, imaginary or complex number. We have $x \div x = 1$. This suggests that $0 \div 0 = 1$. But what about $2x \div x = 2$? We can then argue that $0 \div 0 = 2$. Whatever nonzero number we want $0 \div 0$ to be, we can argue for it.

Let's say that we agree that $x \div 0 = 0$, as if somehow that repairs the discontinuity in a $c \div y$ graph, where $c$ is some nonzero number. What does that gain us? Not much. If for nonzero $x$ we have $x \div y = w$, then $x = wy$, but this is false if $y = 0$ because $0w = 0$. Multiplication still can't be relied upon as a calculation history look-up function.

Much has been made of $\sqrt x$ being not technically a function. If $x = y^2$, then there are two possibilities for $y$. But that's only two possibilities, and one can be obtained from the other with multiplication by $-1$. If we define $x \div 0$ to be some specific number, then there are infinitely many possibilities for solutions to $x \div 0 = w$.

Another problem with defining $x \div 0$ to equal some specific number is that then we have to re-examine what $0x$ is. But since $0x = 0$ has worked for us so well for centuries, we have to ask ourselves if defining $x \div 0$ is really so worthwhile that we may possibly have to redefine $0x$.

If we agree that $0x = 0$, then we just have to accept that multiplication might not be a reversal of division when zeroes are involved. So even if we were to agree upon $x \div 0 = 0$ (which we don't) we'd have $0 \div 0 \times 0 = 0$ but we'd also have $1 \div 0 \times 0 = 0$, $2 \div 0 \times 0 = 0$, etc. ad infinitum (not to mention all the rational and irrational numbers).

The only way that multiplication can be a consistently reliable reversal of division is for it to be defined that way.

To use computer programming parlance, $0 \div 0$ is "not a number" and throws some kind of exception. If instead of dealing with the exception we try to multiply that "not a number" by $0$ then we get another instance of "not a number" and another exception.

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Due to a small discussion about the question posted, I was suggested to post the idea.


A more abstract way to handle the problem.

Magma

Let us consider a magma $(\textbf{M},\cdot)$. Whence the binary operation $\cdot$ is closed on $M$, which can be written as

$$ \textbf{M} \times \textbf{M} \ni (m,m') \mapsto m \cdot m' \in \textbf{M}. \tag 1 $$

Left division

Let $m, m' \in \textbf{M}$, the left divisible subset is defined as $$ D_l(m,m') = \{ d \in \textbf{M} | m \cdot d = m' \}. \tag 2 $$

IF $|D_l(m,m')| = 1$, we have a mapping $$ \textbf{M} \times \textbf{M} \ni (m,m') \mapsto d \in \textbf{M}, \tag 3 $$ and we can define the left division as $$ d = m \backslash m'. \tag 4 $$

Right division

Let $m, m' \in \textbf{M}$, the right divisible subset is defined as $$ D_r(m,m') = \{ d \in \textbf{M} | d \cdot m = m' \}. \tag 5 $$

IF $|D_r(m,m')| = 1$, we have a mapping $$ \textbf{M} \times \textbf{M} \ni (m,m') \mapsto d \in \textbf{M}, \tag 6 $$ and we can define the right division as $$ d = m' / m. \tag 7 $$


Important conclusion:

We can only define division IF $|D(m,m')|=1$.

A magma is called divisible IF $|D(m,m')|=1$.

Projective

A magma is left projective IF $$ \exists q \in \textbf{M} \forall m \in \textbf{M} : q * m = q. \tag 8 $$ A magma is right projective IF $$ \exists p \in \textbf{M} \forall m \in \textbf{M} : m * p = p. \tag 9 $$ A left projecitve element $q$ is also a right projective element $p$: $$ q = q * p = p. \tag {10} $$ Two projecitve element $p$ and $p'$ are identical: $$ p' = p' * p = p. \tag {11} $$


A magma is called projective, IF there is a projective element.

Conclusion

If a magma $\textbf{M}$ --- such that $|\textbf{M}| > 1$ --- is projective by the element $p$ then we obtain

$$ D(p,p) = \{ d \in \textbf{M} | p \cdot d = p \} = \textbf{M}. \tag {12} $$

Therefore $|D(p,p)| > 1$.

Whence the magma can not be divisible, as divisibility requires $|D(p,p')| = 1$.

A projective magma cannot be a divisible magma.

Consider the magma's $(\mathbb{Z},\cdot)$, $(\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ and $(\mathbb{C},\cdot)$.

They are ALL projective as $x \cdot 0 = 0$.

Whence the magma's $(\mathbb{Z},\cdot)$, $(\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ and $(\mathbb{C},\cdot)$ are NOT divisible, i.e. division cannot be defined on $(\mathbb{Z},\cdot)$, $(\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ and $(\mathbb{C},\cdot)$.

Thus $x \div 0$ is not defined and therefore meaningless.

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What you are saying sounds reasonable enough, and so it is worthy of serious consideration.

You're saying that $a \div b \times b = a$, which is in fact true for all nonzero values of $a$ and $b$.

Then, if $a = b = 0$, we have $0 \div 0 \times 0 = 0$. That looks correct. So far so good.

What if $a = 1$ and $b = 0$? Then $1 \div 0 \times 0 = 1$. Whoa, that doesn't look right. This would mean there is a solution to $x \times 0 = 1$. But we can prove that $x \times 0 = 0$ no matter what number $x$ is.

Considering your idea, we have arrived at a contradiction. (This is not the only contradiction possible, though, there are others).

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IF this has any algebraic value (not defining by limits), it should be $0$. Simply because:

$0.0^{-1}.0=0.(0^{-1}.0)=0.1=0$

However, if $0^{-1}$ exists, then:

$0=0.0^{-1}=1$

Therefore, the only place where $0.0^{-1}.0$ makes (canonical) algebraic sense is in the trivial ring.

If you want to evaluate $0.0^{-1}$ elsewhere, you first have to define what $0.0^{-1}$ means, and there is no satisfactory way to do this in $\mathbb{R}$, whence we leave it undefined.

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    $\begingroup$ The probability that someone who uses the $\div$ sign in their question knows about the algebraic properties of a ring is approximately zero. Although this is a good answer for someone who does. $\endgroup$ – Simon S Jul 22 '15 at 17:41
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    $\begingroup$ The contradiction lies in the term $0^{-1}$. By definition $0 \cdot 0^{-1} = 1$. However $1$ is not defined as a neutral element as the definition is $1 \cdot x = x$, which does not hold for $0$ as $1 \cdot 0 = 0$. So $0^{-1}$ has no meaning at all. $\endgroup$ – johannesvalks Jul 30 '15 at 1:44
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    $\begingroup$ @johannesvalks: Rather, $0^{-1}$ has meaning (though not much) if and only if we're talking about the trivial ring, which has exactly one member, serving as both additive and multiplicative identity (i.e.: $0=1$). $\endgroup$ – Cameron Buie Aug 2 '15 at 19:02
  • $\begingroup$ True - a trivial magma in general has one operation and the element is projective, neutral and the trivial magma is divisible. The operation is also commutative and also associative. ;) $\endgroup$ – johannesvalks Aug 2 '15 at 19:15
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Here's the key: Don't even think of division as an operation. $a \div b$ is really just a shorthand for $a \cdot b^{-1}$, so it only makes sense to use this shorthand if $b^{-1}$ exists (where $b^{-1}$ is defined to be some element making $b \cdot b^{-1} = 1$). So $\div 0$ just doesn't make sense in any equation, ever.

As others have mentioned, you can have formulas where you divide by an indeterminate $x$, then take a limit as $x \to 0$, but that is different as you are never explicitly dividing by $0$.

Consider the following: $\frac{0}{0} \cdot 0 = \frac{3 \cdot 0}{0^2}\cdot 0 = \frac{0}{3 \cdot 0^2} \cdot 0$ (since $0 = 3 \cdot 0 = 0^2 = 3 \cdot 0^2$).

Allowing these corrupted operations you are trying to use, we would have $0 = 3 = 1/3$.

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Why is $0 \div 0$ undefined?

The definition of $\div$

$$ a \times b = c \Longrightarrow b = c \div a $$

We would get

$$ 0 \times b = 0 \stackrel{?\ }{\Longrightarrow} b = 0 \div 0 $$

But for $b' \ne b$ we would also get

$$ 0 \times b' = 0 \stackrel{?\ }{\Longrightarrow} b' = 0 \div 0 $$

If this would be true, we get a contradiction as

$$ b \ne b' \wedge b = b' $$

So $0 \div 0$ is simply not defined.

Why is $0 \div 0 \cdot 0$ undefined?

Now consider

$$ x \times x = x \times x \Longrightarrow x = \big(x \times x\big) \div x = x \div x \cdot x. $$

But now consider

$$ 0 \times x = 0 \times 0 \stackrel{?\ }{\Longrightarrow} x = 0 \div 0 \cdot 0. $$

And for $x' \ne x$ we consider

$$ 0 \times x' = 0 \times 0 \stackrel{?\ }{\Longrightarrow} x' = 0 \div 0 \cdot 0. $$

If this would be true, we get a contradiction as

$$ x \ne x' \wedge x = x' $$

So $0 \div 0 \cdot 0$ is simply not defined.

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Any expression involving "$\div 0$" is undefined because $0$ is not in the domain of the binary operation of division (of real numbers), no mater how it is combined with other mathematical symbols. You can't "pretend" it is and cancel it -- the expression is already meaningless, so you can't proceed further.

Your statement that "$x\div x \cdot x = x$" is not true as it stands -- it carries the implicit proviso that $x\neq 0$.

Amendment: I should have said that $(a,0)$ is not in the domain for any value of $a$, of course, because the domain consists of certain pairs of numbers.

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    $\begingroup$ This explanation -- essentially, "$0 \div 0$ is undefined because $(0,0)$ is not in the domain of the binary function $\div$" -- has things completely backwards. Domains are not handed down from heaven or enacted by some legislative body. The binary function $\div$ has the domain that it does because $0 \div 0$ cannot be defined, not the other way around. $\endgroup$ – mweiss Jul 22 '15 at 18:03
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    $\begingroup$ @mweiss: Nonsense. The domain here is assumed to be the largest it can be in the context of real numbers. It completely determines what numbers you are allowed to divide by. When one learns a function, one learns not only the rule of operation but also the legitimate inputs (domain) and the type of outputs (range) of the function. Without all three, one has not specified a function. This function can very easily be extended to allow division by zero: define $a \div 0 = 42$ for all $a$. $\endgroup$ – MPW Jul 22 '15 at 21:21
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    $\begingroup$ This function can very easily be extended to allow division by zero... Then why don't we? Any answer that does not include an explanation of why we don't assign a value to $a \div 0$ is not really answering the question -- it amounts to nothing more than "It's undefined because we don't define it." $\endgroup$ – mweiss Jul 22 '15 at 22:41
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    $\begingroup$ @mweiss: In the end, the answer is that the domain can't be extended to include $(a,0)$ if the extended function is to enjoy the usual algebraic properties of ordinary division. That's it, nothing more. $\endgroup$ – MPW Jul 23 '15 at 1:42
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    $\begingroup$ I agree completely. I think your answer would be strengthened by including this. There is a big difference between saying "The domain doesn't include $(a,0)$" and saying "The domain can't include $(a,0)$ because [reasons]." The former only restates the fact that $0 \div 0$ is undefined, the latter actually provides an explanation why. $\endgroup$ – mweiss Jul 23 '15 at 1:48
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With the set of real numbers you have several operations. For example, you have addition ($+$) and division ($\div$). You can think of each operation as functions: $$ \begin{align} \mathbb{R}\times \mathbb{R} &\to \mathbb{R} \\ (a,b) &\mapsto a + b\\ \newline \div: \mathbb{R}\times \mathbb{R}\color{red}{\setminus\{0\}} &\to \mathbb{R} \\ (a,b) &\mapsto a\div b \end{align} $$ The point with this is that you can add any two real numbers, but you can't divide any two numbers. The $\color{red}{\mathbb{R}}$ says exactly that. The denominator has to come from the real numbers $\mathbb{R}$, but can't be in the set consisting of $0$. The number is the denominator has to be non-zero. This is all a question about definition.

Because of this expressions like $0\div 0$ makes no sense. Neither does $7\div 0$. If $0\div 0$ doesn't make sense, then neither does any more complex expression if $0\div 0$ appears as part of that expression.

You indeed have a rules that says $$ (x\div x) \cdot x = x, $$ but this rule is only true for $x\neq 0$.

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    $\begingroup$ This explanation -- essentially, "$0 \div 0$ is undefined because $(0,0)$ is not in the domain of the binary function $\div$" -- has things completely backwards. Domains are not handed down from heaven or enacted by some legislative body. The binary function $\div$ has the domain that it does because $0 \div 0$ cannot be defined, not the other way around. $\endgroup$ – mweiss Jul 22 '15 at 18:02
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    $\begingroup$ @mweiss: With the viewpoint of abstract algebra it is a matter of definition. $\endgroup$ – Thomas Jul 23 '15 at 13:51
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    $\begingroup$ Considering group theory, $(\mathbb{R}, \cdot)$ is NOT a group. Whence the inverse is NOT defined. $\endgroup$ – johannesvalks Jul 29 '15 at 21:25
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    $\begingroup$ @johannesvalks Yes, of course $(\mathbb{R},\cdot)$ is not a group. But as I said before, giving this as the reason why division by zero is not defined has things backward. How do you know it's not a group? You know it's not a group because division by zero is not defined. So answering the question "Why is it not defined?" by simply appealing to the fact that $(\mathbb{R},\cdot)$ is not a group is engaging in a circular argument. At some point you have to explain why it is not defined, rather than just reassert the fact that it isn't. $\endgroup$ – mweiss Jul 29 '15 at 22:58
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    $\begingroup$ No - it is not a group as it has no neutral element - whence division is not defined as division requires a neutral element. The sequence is: closed (magma) $\rightarrow$ associativity (semigroup) $\rightarrow$ identity element (monoid) $\rightarrow$ invertibility (group)... Once you include zero - there is no longer an identity element - whence not invertible, i.e. no division defined... $\endgroup$ – johannesvalks Jul 29 '15 at 23:18
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This is a fun question and really scratches the mind. First of all we need to understand the term "undefined". Most people would think of it as an inconsistency in math, which is understandable because that is what we usually associate it with and which is usually shown as an error on a calculator. But it's not an inconsistency. Undefined simply means the answer isn't defined or described by a single number, but can rather be any possible number in our range of infinite numbers, and we therefore say that it is not defined or undefined. Sounds weird, I know, but keep reading.

We know this by looking at the equation: $$x \times 0 = 0$$ No mater what number $x$ is, the answer will always be equal to $0$. Therefore $x$ can be any possible number, because the answer will always be $0$. Cool right? Take the zero over and we get $x = 0 \div 0$ which justifies our statement of $x$ being undefined. Substitute that into your original equation and we get $x \times 0 = 0$. Therefore $0 \div 0 \times0$ will be equal to zero.

However move the original equation to $0\times0\div0$, follow it right to left and we see the answer is undefined (being any possible number). This leads us to believe that the problem in the two equations lay in placement. Pretty strange for a single term equation, but that's how the gravy boat floats.

But for your original equation of $0 \div 0 \times 0$ the answer is $0$. I'm not saying I'm correct, in fact my statement may be complete hullabaloo, but the moral of the story is this: don't be afraid of undefined. It's fun and is pretty awesome part of our abstract idea of math :)

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  • $\begingroup$ $0\times (0\div 0)$ could reasonably zero by this logic, but $0\times 0\div 0$ is definitely undefined when using the left-to-right convention. $\endgroup$ – Eric Stucky Aug 3 '15 at 6:27
  • $\begingroup$ I feel like you have jumped to a conclusion at some point, @martin, but I can't quite put my finger on it. Anthony's answer suggests your statement probably leads to a contradiction, making it "complete hullaballoo." $\endgroup$ – Robert Soupe Aug 4 '15 at 3:47
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In general, when you're doing math, you have several options.

  1. Use the existing definitions. Those were developed through experience in doing math. In this case, the simplest rule is that you never divide by zero, and that any division by zero gives you an undefined result. This rule is useful, because it saves you from getting wrong results. If you allow yourself to divide by zero you get things like this: $$ 3\cdot(1-1)=2\cdot(1-1) \longrightarrow 3=2$$ and it's unclear how you can do arithmetic when this sort of thing happens. So you rule out division by zero altogether.

  2. There are occasions where you really need to relax the rules. That's what happens when you do infinitesimal calculus. Suppose you have three functions - $a(x)$, $b(x)$, and $c(x)$. They all converge to zero at some point, and you want to understand the behavior of $z(x) = a(x) \div b(x) \cdot c(x)$ at that point. Then, you need to understand the rules of working with limits. Infinitesimal Calculus enables you to do so, sometimes, and it depends on the functions - you can actually get any numerical result - from zero to infinity, and there will be cases where the new function $z$ won't converge.

  3. Another possibility is to redefine the rules. After all, it makes sense to define $1/0 = \infty$. You can have any set of clear rules to make sense of multiplying infinities and zeros. But it won't be easy to make the rules work. Any consistent set of rules which gives us an interesting and useful version of arithmetic is a legitimate math.

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The division arithmetic operation is undefined for zero divisor, therefore I perceive the question as not well-defined.

One can try to pose similar question in terms of limits of sequences and/or functions, in which case one will find many tools suitable for investigating the problem, e.g. L'Hôpital's rule and numerous convergence tests. However, in the form the question posed right now it is impossible to answer rigorously.

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  • $\begingroup$ I was going to make exactly this point, but did not finish my answer before Vlad posted. L'Hôpital's rule is absolutely the most natural way to go from vague interest (and vaguely-asked questions) about division by zero to a better mathematical understanding. Don't knock questions for being ill-defined! Getting to the right question is very much part of studying maths. $\endgroup$ – HTFB Aug 3 '15 at 10:09
  • $\begingroup$ The question is well-defined, "Is $0÷0\cdot 0$ nonsense?" $\endgroup$ – Squirtle Aug 3 '15 at 17:52
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"We all know that multiplication is the inverse of division, and therefore $x \div x \cdot x = x$. But what if $x = 0$?"

No, I do NOT know that multiplication is the inverse of division. In truth, I categorically REJECT that inverse is the multiplication of division for what you have stated.

Multiplication, on the real numbers without $0$ or rational numbers without $0$, is the inverse of division. But, if $0$ gets included, then multiplication and division are not inverses.

Additionally, you might want to note that on the natural numbers multiplication is not the inverse of division. Division of two natural numbers $nat_1$ and $nat_2$ on the natural numbers, in fact, only exists for instances where there exists a common divisor of $nat_1$ and $nat_2$ other than $1$. On the other hand, multiplication exists for all natural numbers.

On the integers, division is also not the inverse of multiplication.

"$0 \div 0$ is undefined so $x \div x \cdot x = x$ should be too, but whatever happens when we divide that first $0$ by $0$ should be reversed when we multiply it by $0$ again, so what is the right answer? "

No, it should NOT be reversed when we multiply it by $0$ again, because division is not defined all real or rational numbers since $0$ is a real or rational number.

If we divide a number by $0$, we should return "not a number." If we multiply "not a number" by $0$, we should return "not a number".

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  • $\begingroup$ Considering magma's - division is the inverse of multiplication by definition... Consider a magma $M$. A magma $M$ is divisible, IF $\exists!x \in M : a*x = b$ AND $\exists!y \in M : y*a = b$, only due to the $\exists!$ we can write $ x = a \backslash b$ and $y = b / a$. $\endgroup$ – johannesvalks Jul 29 '15 at 21:39
  • $\begingroup$ quote: But, if 0 gets included, then multiplication and division are not inverses. The inverse requires a neutral element, once we include $0$ - there is no neutral element - whence inverse and division are not general defined. $\endgroup$ – johannesvalks Jul 29 '15 at 21:42
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The problem with $\frac{0}{0}$ is that $0$ is a multiple of itself. In fact, $0$ can be several multiples of itself. Some zeros are 3x bigger than other zeros, others are $\frac{\pi}{e}$ times bigger. The numerator and denominator of $\frac{0}{0}$ are both different flavors of $0$, but there is no indication in the expression what they taste like. Without that information, such as "the numerator is twice as big as the denominator", we can only assume that result is the universal omnivalue, a value that is equal to every number simultaneously, even infinite ones.

This concept isn't as ridiculous as it sounds. Consider possible replies to the question "Where are you right now?". Possible replies may include 704 Hausser Street, Queens, NYC, New York State, America, or simply just the Universe. People all over the world may give a different street address, but all would agree that they live in the universe. This begs the question, "Where, precisely, is the Universe?". The short answer: it's everywhere. It's a blur of all locations. It's an Omnivalue. Similarly, $t=\frac{0}{0}=[-\infty,\infty]$ is the solution to the equation $$0t=0$$

Before continuing. lets consider the omnivalue $\sin\infty$. Because $\mid\sin x\mid\le1$, this would evaluate to the omnivalue $[-1,1]$. If we multiply that value by $0$, we could just multiply each of it's components by $0$ and return the singularity $0$. $$0\sin\infty=0\cdot[-1,1]=[0,0]=0$$

Now that we can see that multiplying certain omnivalues by zero can merge the result back into a singuality (i.e. $0$), we would like to know if $$\frac{0}{0}\cdot 0 = 0$$ I don't think that it does because while the real components of the omnivalue $\frac{0}{0}=[-\infty,\infty]$, the infinite components reblur the result when they are multiplied by zero. So I guess my answer is that the result is indeterminate (another omnivalue).

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$$\frac{0}{0} \cdot 0 = \mathrm{undefined} \cdot 0 = \mathrm{undefined} $$

However, if you know limits, we can say that

$$\lim_{x \to 0} \frac{x}{x}\cdot x = \lim_{x \to 0} x = 0$$

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  • $\begingroup$ I might be missing something, but it seems like the first part of your answer says that its undefined and the second with limits says it is 0. $\endgroup$ – user169330 Jul 22 '15 at 15:58
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    $\begingroup$ That's only one possible limit of the form $0/0\cdot 0$. Another would be $\mathrm{lim}_{x\rightarrow 0} \frac{ax}{x^2} \cdot x =a$ for any $a\in\mathbb{R}$. Or $\mathrm{lim}_{x\rightarrow 0} \frac{x}{x^3} \cdot x $ which diverges. This is why $0/0 \cdot 0 $ is indeterminate while (some) limits can be evaluated. $\endgroup$ – GFR Jul 22 '15 at 16:03
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    $\begingroup$ @NicoA The difference is that we are working in the first one just with $\frac{0}{0}$, which is undefined. However in the limit, we do not care that it is undefined, because we only look to what happens around 0, and in that case it is 0. The answer to your question is undefined. The second equation with the limit is just a addition, to show you how mathematicians deal with this if they want it to equal 0. $\endgroup$ – wythagoras Jul 22 '15 at 16:03
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    $\begingroup$ @GFR Yes, but the OP explicitly talked about $x \div x \cdot x$. $\endgroup$ – wythagoras Jul 22 '15 at 16:04
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    $\begingroup$ @GFR, I think you should recheck that first limit. $\endgroup$ – Marcus M Jul 22 '15 at 16:05

protected by Zev Chonoles Jul 25 '15 at 21:44

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