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I came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.

  1. $\dfrac {x}{0}$ is Impossible ( If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both $1.$ and $2.$ are divided by zero, but only $2. $ has infinite solutions so as $1.$ has none solution, how and why ?)

  2. $\dfrac {0}{0}$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?)

  3. $\dfrac {0}{x}$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.

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    $\begingroup$ On 3, $0/x$, you need to say explicitly $x \ne 0$, to distinguish it from 2. $\endgroup$ – Henry Mar 11 '11 at 20:43
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    $\begingroup$ The following text was added in a suggested edit: "The 787 videos listed in the auto generated You-Tube channel: Division by zero is a good indication of the importance and common misunderstanding of this question. In my opinion, it is the obligation of this mathematical community to provide a greater variety of explanations to interested students at all levels." If this information is useful, it should be added as a comment, not by editing posts by other users. $\endgroup$ – Martin Sleziak Nov 8 '12 at 11:41
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    $\begingroup$ Yes, this makes sense! Clearly a question with +19 that has an answer of +50 hasn't received enough attention! $\endgroup$ – Asaf Karagila Nov 8 '12 at 13:33
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    $\begingroup$ It is undefined because it has infinite solutions. $\endgroup$ – Parth Kohli Feb 2 '13 at 15:42
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    $\begingroup$ en.wikipedia.org/wiki/Wheel_theory $\endgroup$ – Martin Brandenburg May 8 '14 at 7:42

17 Answers 17

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The first question you need to ask is: What does "$a/b$" mean?

The answer is: "$a/b$ is the unique solution to the equation $bz = a$." (I'm using $z$ as the unknown, since you are using $x$ for other things).

Given that answer, let's discuss your points out of order:

(3) is perfectly fine: $0/x$, with $x\neq 0$, is the solution to $xz = 0$; the unique solution is $z=0$, so $0/x = z$. The reason it's unique is because $x\neq 0$, so the only way for the product to be $0$ is if $z$ is $0$.

In (1), by "impossible" we mean that the equation that defines it has no solutions: for something to be equal to $x/0$, with $x\neq 0$, we would need $0z = x$. But $0z=0$ for any $z$, so there are no solutions to the equation. Since there are no solutions to the equation, there is no such thing as "$x/0$". So $x/0$ does not represent any number.

In (2), the situation is a bit trickier; in terms of the defining equation, the problem here is that the equation $0z=0$ has any value of $z$ as a solution (that's what the "infinite solutions" means). Since the expression $a/b$ means "the unique solution to $bx= a$, then when $a=b=0$, you don't have a unique answer, so there is no "unique solution".

Generally speaking, we simply do not define "division by $0$". The issue is that, once you get to calculus, you are going to find situations where you have two variable quantities, $a$ and $b$, and you are considering $a/b$; and as $a$ and $b$ changes, you want to know what happens to $a/b$. In those situations, if $a$ is approaching $x$ and $b$ is approaching $y\neq 0$, then $a/b$ will approach $x/y$, no problem. If $a$ approaches $x\neq 0$, and $b$ approaches $0$, then $a/b$ does not approach anything (the "limits does not exist"). But if both $a$ and $b$ approach $0$, then you don't know what happens to $a/b$; it can exist, not exist, or approach pretty much any number. We say this kind of limit is "indeterminate". So there is a reason for separating out cases (1) and (2): very soon you will see an important qualitative difference between the first kind of "does not exist" and the second kind.

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    $\begingroup$ Here's an example of how #2 is undefined. What makes it infinite is the fact that this can be repeated for any number. $\endgroup$ – James Mertz May 2 '13 at 16:14
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    $\begingroup$ I like this answer for the first part, but I don't like the part about calculus.. 1) $\infty$ is not anything and 2) stated like that, it's as you never know what happens to $a / b$.. you just need more informations about the exact nature of the variables $\endgroup$ – Ant Apr 4 '14 at 17:31
  • $\begingroup$ In your last paragraph you need to add that when you have these a,b "approaching something" that they are continuous at this point. $\endgroup$ – Matthew Levy Dec 23 '14 at 17:07
  • $\begingroup$ I have borrowed your starting paragraph for my answer :-) $\endgroup$ – arivero Aug 16 '15 at 20:11
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The key is to realize what a fraction $\frac ab$ really represents: $\frac ab$ is the number with the property that $\frac ab \cdot b = a$.

So, in the first case if $x \ne 0$, there is no number $\frac x0$ with the property that $\frac x0 \cdot 0 = x$ since anything times zero is zero. So $\frac x0$ is undefined. In the second, any number $y$ has the property that $y \cdot 0 = 0$, so $\frac 00$ could represent any number $y$ according to the above characterization of a fraction, so $\frac 00$ is said to be indeterminate. Finally, if $x \ne 0$ then $0$ has the property that $0 \cdot x = 0$, so $\frac 0x$ is the number $0$.

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    $\begingroup$ One should not use "indeterminate" here. That applies only to functions, not numbers. $\endgroup$ – Bill Dubuque Mar 11 '11 at 20:56
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    $\begingroup$ @BillDubuque I am currently learning elementary arithmetic. In mostly all contexts it is said that $\dfrac 00$ is indeterminate. $\endgroup$ – user103816 May 25 '14 at 9:48
  • $\begingroup$ I say that $\frac{0}{0}$ does evaluate to a value and that it is not what you would expect yet the term $\frac{0}{0}$ has no value. The value that the term evaluates to would be $1$ since this would fit under the multiplicative identity property as in $\frac{anything}{anything} = 1$. The division of something by itself even if that something is nothing will return itself. This doesn't mean that $\frac{0}{0} = 0$ because this wouldn't return itself, anything that returns itself has an associated value of $1$. $\endgroup$ – Francis Cugler Mar 20 '17 at 23:51
  • $\begingroup$ We understand that anything multiplied by $1$ is itself which is true, and that anything multiplied by $0$ is $0$ is also true and when we try to explain what $1 \cdot 0$ or $0 \cdot 1$ is... the evaluation of this will result the same every time as the answer would be $0$ which fits both statements. So let's apply this to fractions $0 \cdot 1 = \frac{0}{1} \cdot \frac{1}{1} = \frac{0 \cdot 1}{1 \cdot 1}$ ... $\endgroup$ – Francis Cugler Mar 20 '17 at 23:57
  • $\begingroup$ (...continued) $\frac{0 \cdot 1}{1 \cdot 1} = \frac{0}{1} = 0$ so far so good. So what happens when we reverse this: Let's try it out. But first we need to understand that $\frac{1}{0} \neq 0$ we need to consider this the reciprocal of $0$ or $\frac{0}{1}$ and we know that the only value that has an equivalent reciprocal is $1$ or $\frac{1}{1}$. Then it would suggest that $\frac{1}{0} \cdot \frac{1}{1} = \frac{1}{0}$ and that $\frac{1}{0} \cdot \frac{0}{1} = \frac{0}{0}$ and since $\frac{0}{0} = 1$ from my above statement it would state that anything divided by $0$ would result in $1$. $\endgroup$ – Francis Cugler Mar 21 '17 at 0:10
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Definition of Division

For every real number a and every nonzero real number b, the quotient a$\div$b,

or $\dfrac{a}{b}$, is defined by:

$$a\div b=a \cdot \frac{1}{b}.$$

Dividing by zero would mean multiplying by the reciprocal of 0.

But 0 has no reciprocal (because 0 times any number is 0, not 1.)

Therefore, division by 0 has no meaning in the set of real numbers.

Multiplicative property of 0

Prove:

If $a$ is any real number, then $a\cdot 0 = 0$ and $0\cdot a = 0$.

Proof:

Statement _________________Reason

  1. $0 = 0 + 0$ ______________1. Identity property of addition

  2. $a\cdot0 = a(0 + 0)$ __________2. Multiplication property of equality

  3. $a\cdot0 = a\cdot0 + a\cdot0$ ________3. Distributive property of mult. with respect to add.

  4. But $a\cdot0 = a\cdot0 + 0$ _______4. Identity property of addition

  5. $\therefore$ $a\cdot0 + a\cdot0 = a\cdot0 + 0$ ____5. Transitive property of equality

  6. $a\cdot0 = 0$ _______________6. Subtraction property of equality

  7. $0\cdot a = 0$ _______________7. Commutative property of multiplication

Therefore, 0 times any number is 0, not 1.

(Source: Algebra: Structure and Method Book 1)

The two cases presented in an older edition of Book 2 of the above source are:

  1. Dividing a nonzero number by zero, violates the multiplicative property of zero and therefore the properties of the real numbers upon which it is proven, as shown above.

  2. Dividing zero by zero, which does not violate the multiplicative property of zero, but multiplication by zero is an operation that results in zero for every real number.

If $\dfrac{a}{0}$ = c, then $a = 0\cdot c$. But $0\cdot c = 0$. Hence, if $a$ is not equal to $0$, no value of $c$ can make the statement $a = 0\cdot c$ true, while if $a = 0$, every value of $c$ will make the statement true.

Thus, $\dfrac{a}{0}$ either has no value or is indefinite in value.

This separation into two cases, one of which results in no value satisfying the multiplicative property of zero and the other resulting in an indefinite value satisfying it, gives the impression that $\dfrac 0{0}$ is allowed.

The following argument starts with the equation: $$a \cdot \frac{1}{a} = 1$$ and notes that $a\neq 0$ because $0$ times any number is $0$.

Thus, the product of $0$ and no real number equals $1$.

This further reinforces the idea that $0$ has no number that when multiplied by it equals $1$.

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  • $\begingroup$ By the definition at the top, $\frac{1}{b} = 1 \cdot \frac{1}{b} = 1 \cdot 1 \cdot \frac{1}{b} = \dots$ which is not really a definition. I think $b^{-1}$ would be clearer than $\frac{1}{b}$, because $b^{-1}$ (pronounced "$b$ inverse") emphasizes that it's defined by the field axioms, not defined in terms of division. $\endgroup$ – Jordan Nov 13 at 5:59
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Just to give another point of view (purely algebraic one):

Suppose you have a ring (a mathematical structure with addition and multiplication satisfying a bare minimum of laws you need to call them that), and in it there is a number $x$ such that $x \cdot 0 = 1$. Then $1 = x \cdot 0 = x \cdot (1 - 1) = x - x = 0$, which means that for any $y$ in this ring $y = y \cdot 1 = y \cdot 0 = y (1 - 1) = y - y = 0$, so this ring has just one element. Such ring is called trivial, and it is clearly not an extension of the ring of integer numbers.

If you can divide by zero, then $1/0$ is defined and it has to satisfy the property above. Thus, division by zero is possible only in the trivial ring.

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    $\begingroup$ What if I don't have a ring? What if I have a wheel? $\endgroup$ – JMCF125 Jan 18 '14 at 14:57
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    $\begingroup$ @JMCF125 then you roll $\endgroup$ – Alexei Averchenko Sep 7 '15 at 9:12
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You will pretty much never see "0/0" or "x/0" in a situation where somebody expects you to actually interpret it as having a value. Other than simple mistakes, the times it comes up are usually of the following sort:

To solve the equation $ax=b$ for $x$, consider the expression "$b/a$".

  1. If it is of the form "$0/0$", your equation has infinitely many solutions
  2. If it is of the form "$x/0$" for nonzero x, your equation has no solutions
  3. Otherwise, the division makes sense and the result is the unique solution

You might hear the word "form" mentioned to refer to the fact that we're talking about a formula, and not the number that might result from evaluating the formula. (especially in the context of limits)

It's important to note the difference, because the number system you use was created so that "$0/0$" and "$x/0$" are not allowed -- they are meaningless when viewed as arithmetic expressions. (the same is true for "$x/y$" if you do not already know that $y \neq 0$) There are lots of very good reasons why arithmetic was created in this way, and the other answers mention some of them.

I feel I should mention that there are other forms of arithmetic that behave differently. The projective numbers are often useful, and 1/0 makes sense in them (but $0/0$ does not). Wheel theory is somewhat more esoteric, but it provides an example where even $0/0$ makes sense. (and also clearly demonstrates the difficulties in accommodating its existence)

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Let me add this quick bit in addition to what has already been said (assuming you know a little set theory). Take the following semi-random example.

Suppose you have a function $f$ with $f(5) = 14$ and $f(6) = 11$. Going backwards, what do you get? Assuming $f$ is only defined for $5$ and $6$, $f^{-1}(14) = 5$ and $f^{-1}(11) = 6$, right? Now consider a different function, which like $f$ is only defined at two points... let's call it $g$ with $g(5) = 8$ and $g(6) = 8$. Now, what is $g^{-1}(8)$? Well, it wants to be both $5$ and $6$ if it's really the inverse, but defining it either way causes a contradiction so it must be undefined.

So what's the point? In algebra many operations (like multiplying by a nonzero number) are like $f$ and and many (multiplying by zero, for example) are like $g$. Here, draw the picture with arrows... with $f$ you know where the arrows came from, with $g$ two arrows point to the same place so you don't know where you 'came from'.

Now back to multiplication. Imagine the arrows if you multiply by $\frac{1}{2}$ versus if you multiply by $0$. You can see that essentially the same thing is going on, but with infinitely many numbers instead of just a few.

BTW: There is one partial 'fix' if you know a little set theory: (new notation here, NOTE THE CURLY 'SET' BRACKETS): $f^{-1}(\{8\}) = \{5,6\}$. Another example: if $h(x) = x^2$ then $h^{-1}(\{4,25\}) = \{2,5\}$. And back to were we started from, inverting division by zero: if you have $s(x) = 0$ then $s^{-1}(\{0\}) = (-\infty,\infty)$.

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If your domain consists of a two-element set, or a set with more than two elements (where any infinite set gets classified as having more than two elements) then Arturo's answer applies.

However, if your domain consists of {0}, then things work out a bit differently, as Bill Dubuque pointed out in a comment elsewhere.

For a system with domain of {0}, you'll still define a/b as the unique solution to (bz)=a. Now, (3) still works out as perfectly fine. x/0 means the unique solution to (0z)=x. Well, x can only equal 0 in this domain, so this means the unique solution to (0z)=0. z can also only equal 0 here, so the equation does have a unique solution of 0. 0/0 means the unique solution to (0z)=0. Well, z can only equal 0 here, and if z=0, the equation does hold, so (0/0)=0 here also.

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  • $\begingroup$ what does domain mean here? $\endgroup$ – Srivatsan Sep 2 '11 at 17:29
  • $\begingroup$ The domain consists of the set of elements under consideration. $\endgroup$ – Doug Spoonwood Sep 2 '11 at 17:37
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Because dividing by zero destroys everything.

Reuben Hersh in "What is Mathematics, Really?" gives the following explanation (quotes). It articulates the point that we invent rules, as we go along, in order to account for extended powers of calculation. However, not unexpectedly, this often comes at a price. And each time we have to decide whether the cost is worth bearing or not.

Note that the question "why do we not divide by zero?" is loaded with the psychology of the exception. Meaning, it seems as if have a perfectly nice rule for all integers to begin with, and all of a sudden we exclude zero from it. This prompts the question "why do we exclude zero?" as a reasonable one. Especially since that's what it comes down to - excluding just one number. But understanding the process of creation (of definition) is important in order to understand why a plausible (or even an excellent) question may be missing a point. It is not the case that we "have a perfectly nice rule for all integers to begin with". Addition and multiplication start only with the natural numbers (positive integers starting with $1$) and only later do we move on to the reverse processes of subtraction and division. Given multiplication, therefore, it is not that we exclude zero from this new process we call "division", but rather that we find no good reason to extend the process in order to include zero - other than a psychological sense of ease that something holds for all numbers which is fine but not if it destroys all arithmetic. "God made the Integers, all else is man's work" goes the saying. It shouldn't come as a surprise that there should be a few of these exceptions, to this work of rules. (It turns out we can extend division to the negative integers without a problem, and with great gains, thus enhancing the psychology of a special case for zero. It's only a special case, and one shouldn't expect otherwise.)

Back to math. Suppose it were handy to introduce division by zero. What would be the cost of this rule? Hersh explains.

1/0 Doesn't Work (0 into 1 Doesn't Go)

Division by $0$ is not allowed. Why not? If it's allowed to introduce a symbol $i$ and say it's the square root of $-1$, which doesn't have a square root, why not introduce some symbol, say $Q$, for $1/0$?

We introduce new numbers, whether negative, fractional, irrational, or complex, to preserve and extend our calculating power. We relax one rule, but preserve the others. After we bring in $i$, for example, we still add, subtract, multiply, and divide as before. I now show that there's no way to define $(1/0)\cdot0$ that preserves the rules of arithmetic.

One basic rule is, $0\cdot$ (any number) $=0$.

(Formula I) So $0\cdot(1/0) = 0$.

Another basic rule is $(x)\cdot(1/x) = 1$, provided $x$ isn't zero. (But if we want $1/0$ to be a number, this proviso becomes obsolete.)

(Formula II) So $0\cdot(1/0) = 1$

Putting Formulas I and II together, $1=0$.

Addition gives $2=0$, $3=0$, and so on, $n=0$ for every integer $n$. Since all numbers equal zero, all numbers equal each other. There's only one number — $0$.

The supposition that $1/0$ exists and satisfies the laws of arithmetic leads to collapse of the number system. Nothing is left, except - nothing.

In the case of the imaginary $i$, it turns out that not only is there nothing destroyed (other than our sense of familiarity with numbers), but that things fall into place (Fundamental Theorem of Algebra).

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  • $\begingroup$ If we consider that $0 = \frac{0}{1}$ and that $\frac{0}{1} \neq \frac{1}{0}$ We would see that $\frac{0}{1} \cdot \frac{1}{0} = \frac{0 \cdot 1}{1 \cdot 0} = \frac{0}{0}$. What should $\frac{0}{0} = $... If you ask me I say it should equal $1$ since anything divided by itself would yield itself. This would allow $\frac{1}{0}$ to be the reciprocal of $\frac{0}{1}$. $\frac{1}{0}$ or any fraction of the form $\frac{n}{0}$ is a special form. It is not equal to neither $1$ nor $0$. If $\frac{0}{1}$ or $\frac{n}{1}$ is horizontal slope then $\frac{1}{0}$ or $\frac{n}{0}$ represents vertical slope. $\endgroup$ – Francis Cugler Mar 21 '17 at 1:51
  • $\begingroup$ This can bee seen in the linear equation $y = mx + b$ where $m$ is defined as $\frac{rise}{run} = \frac{y_2 - y_2}{x_2 - x_1} = \frac{\delta y}{\delta x} = \frac{\sin\theta}{\cos\theta} = \tan\theta$ where $\theta$ is the angle above or below the horizon. When we have $0$ slope as in $\frac{0}{d}$ it is the $\sin\theta$ that evaluates to $0$ at the angles $0°,\pm 180°, \pm 360°$ since $\delta y$ or $rise = 0$. No change in height or height is constant and when we apply this to $y = mx + b$ we see that $y = 0 \cdot x + b \implies y = b$ where $y$ is constant and $\delta y = 0$.... $\endgroup$ – Francis Cugler Mar 21 '17 at 1:57
  • $\begingroup$ We also know that every line has an infinite amount of perpendiculars to it. We can find the slope of the line that is perpendicular by the formula $-1 = \frac{rise}{run} \cdot -\frac{run}{rise}$. So if any line has a slope of $0$ then it's perpendicular must be $-1 = \frac{\sin\theta = 0}{\cos\theta = 1} \cdot -\frac{\cos\theta = 1}{\sin\theta = 0} = \frac{0}{1} \cdot -\frac{1}{0}$. Thus vertical slope must be defined!... $\endgroup$ – Francis Cugler Mar 21 '17 at 2:01
  • $\begingroup$ This makes complete sense because when we look at the range and domain of both the $\sin\theta$ or $\delta y$ component of the fraction or the numerator and the $\cos\theta$ or $\delta x$ component of the fraction or the denominator independently their domain and ranges are the same where their domain is $\mathbb R$ and their ranges are $[-1,1]$. So this means they both can accept all numbers and they output a valid range from $[-1,1]$... $\endgroup$ – Francis Cugler Mar 21 '17 at 2:02
  • $\begingroup$ So when we have vertical slope or division by $0$. It can not be $0$, nor can it be $1$. Now since both the $\sin$ and $\cos$ functions are rotational, periodical, sinusoidal continuous functions it would be err to think that either the $\tan$ or $\cot$ function should have any discontinuity at all! So if the slope is $0$, then $\tan\theta = 0$. And the reciprocal of $\tan\theta$ is $\cot\theta$ thus when referring to slope if the slope is defined by $\tan\theta$ then its perpendicular must be $-\cot\theta$.... $\endgroup$ – Francis Cugler Mar 21 '17 at 2:07
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This is an attempt at some intuition for someone who doesn't know a lot of math.

The equation $\frac{a}{b}=x$ is the same as the equation $a=b\cdot$ x. You can look at $a=b\cdot x$ as asking the question "if I want to build $a$ out of $b$'s, how many $b$'s do I need?" (the $x$ is how many pieces of $b$ you need to build $a$)

Example

$\frac{4}{0.1}=x$ is the same as $4=0.1\cdot x$.

How many $0.1$'s do you need to build 4 out of $0.1$'s?

Well you need $\frac{4}{0.1}$ or $40$ pieces. Indeed, $40\cdot 0.1 = 4$.

So what about dividing by 1?

Example

$\frac{7}{1}=x$ is the same as $7=1\cdot x$.

How many $1$'s do you need to build 7 out of $1$'s?

Obviously you'll need seven pieces of one, beause $1+1+1+1+1+1+1=7$

So what about dividing by zero?

Example

$\frac{3}{0}=x$ is the same as $3=0\cdot x$.

How many $0$'s do you need to build 3 out of $0$'s?

Well, how can you build a number out of zeros? No matter what you do with your zeros, you're never going to be able to build any number out of zeros. If you could build a number out of zeros, you should be able to write a number, say 9, as a sum of zeros, which would mean that $9=0+0+0+0+0+0+\dots$, which can never be true no matter how many zeros you use.

So it doesn't really make sense to divide something by zero.

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We can divide by $0$.

Let $X$ be the set of numbers we are debating on.

The multiplicative inverse of $x\in X$, namely $x^{-1}$ or $\frac{1}{x}$, is defined as

$x\cdot x^{-1} = id$, where $id$ does not change any element in $X$ when we multiply, for instance, $1 \in \mathbb{Q}$.

For any $y \in X$, $0 \cdot y=0$, so we require $id = 0$, so let $id = 0$.

Then, for any $x \in X,$ $x\cdot id = 0$.

We get $X = \{0\}$.

We can divide by $0$ on $\{0\}$, since $0\cdot 0 = 0$. But this is not an interesting case.

So, we usually assume $X\not =\{0\}$, so that we cannot divide by $0$.

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I usually argue as the other answerers have but recently I have started using a slightly different argument.

I argue that multiplying by $1/x$ undoes the action of multiplying by $x$. So for example $11*2=22$ and $22*(1/2)$ brings you back to $11$.

Now we have $0*11=0$ but you can't get back to $11$ by multiplying as zero times anything is zero.

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The first question you need to ask is: What does "a/b" mean?

The answer is: "a/b is a representative of a class of pairs of numbers $(a,b) \in Z \times Z $ such that two pairs $(a,b)$ and $(c,d)$ are equivalent if $ad=cb$."

This equivalence fails -is not transitive- if we include the pair $0/0$, because any pair should be equivalent to it, even two arbitrary pairs which are not equivalent between them. So we need to exclude the "indeterminate" pair, correcting the above definition to say instead $(a,b) \in Z \times Z - \{ (0,0) \}$.

All the elements of the form $0/a$ are equivalent. And similarly all the elements of the form $a/0$ are equivalent. Call them the classes $0/1$ and $1/0$

We can define multiplication operating in each $Z$ set, i.e. $(a,b).(c,d) \equiv (ac,bd) $. With this definition, the identity is the class represented by $(1,1)$.

We find that our multiplication is not defined for a product of elements in the classes $0/1$ and $1/0$. The product is precisely the excluded pair $(0,0)$. So if we want multiplication to be always well defined in our set of all the fractions, we need to exclude one of the two classes. Traditionally we exclude the class $1/0$, the class that the original post calls the "Impossible class".

Arriving here, we see that really the motivations to exclude the "undefined element" and the "impossible class" are very different: the former, being able to match any other fraction, completely destroys the transitivity of the equivalence relation, while the later simply is a nuisance to close the multiplication.

Both classes had the potential to do the role of a multiplicative zero: $$1/0 . (a,b) = (a,0) \propto 1/0 $$ $$0/1 . (a,b) = (0,b) \propto 0/1 $$ But now that we have declared the first one "impossible" it is clear that we now must kept the second one to be used as the traditional zero.

The last steps should be to define a sum operation such that $(a,b) + 0/1 = (a,b)$ and a mapping of the integer numbers into the fractions, $Z \longrightarrow Z \times Z - \{ (0,0) \}$, preserving the operations of sum and multiplication already defined in the integers, and mapping the zero and identity of the integers to the corresponding classes in the fractions. But as far as the OP questions goes, we do not need even to finish the construction :-)

EDIT it could be interesting to try to argue that, in the same way that with the usual definition of sum we have $(a,b) + 0/1 = (a,b)$, we should have that $(a,b)+ 1/0 = 1/0$. So we would be in accord with the usual meaning of "infinity" that we assign to this class. In fact it is so if we define $$ (a,b) + (c,d) \equiv (ad+bc,bd)$$

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Among the rules for arithmetic with fractions are the formulas for adding and subtracting,

$${a\over b}\pm{c\over d}={ad\pm bc\over bd}$$

Now what if we allow division by $0$? Keeping in mind the properties $x+0=x$, $x\cdot0=0$, and $x-x=0$ for all numbers $x$, we find first that

$$0={1\over 0}-{1\over 0}={1\cdot0-0\cdot1\over0\cdot0}={0\over0}$$

and then that, for all $x$,

$$x={x\over1}={x\over1}+0={x\over1}+{0\over0}={x\cdot0+1\cdot0\over1\cdot0}={0\over0}=0$$

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  • $\begingroup$ You failed to express 0 as a fraction initially when incorporating it into the expression of adding it to $\frac{x}{1}$ $\endgroup$ – Francis Cugler Mar 3 '17 at 13:07
  • $\begingroup$ Also $\frac{0}{0}$ can evaluate to different answers depending on the context in which it is used. It can yield 0, 1, or $\infty$ $\endgroup$ – Francis Cugler Mar 3 '17 at 13:09
  • $\begingroup$ One last thing 0 is not a number! It is only a place holder or it represents the null or empty set! Consider an arbitrary Point in any dimensional space: The space can be 1D, 2D, 3D, ... ND, etc. The Point itself is 0D space for it is arbitrary. Now because the Point itself has 0 dimensionality does not mean that the point doesn't exist. It only has locale. It does not have any sense of direction or awareness of any other object, it doesn't recognize angles or rotations. Once you introduce a 2nd point and associate the two, then you have 2 points that gives you a line, 1D space or distance. $\endgroup$ – Francis Cugler Mar 3 '17 at 13:15
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If you try to define $x=\frac a0$, you arrive at the contradiction that $x\cdot0 =a$ for nonzero $a$.

Also, take a look at the graph of $y=\frac1x$... it goes to $\infty$ as $x\to0^+$, and to $-\infty$ as $x\to0^-$. Thus there is no way to fill in the value at $x=0$ that isn't arbitrary...

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José Manuel Rodríguez Caballero (https://www.researchgate.net/profile/Jose_Rodriguez_Caballero2) Added an answer In the proof assistant Isabelle/HOL we have x/0 = 0 for each number x. This is advantageous in order to simplify the proofs. You can download this proof assistant here: https://isabelle.in.tum.de/ (https://www.researchgate.net/deref/https%3A%2F%2Fisabelle.in.tum.de%2F)

Close the mysterious and long history of division by zero and open the new world since Aristotelēs-Euclid: 1/0=0/0=z/0= \tan (\pi/2)=0. Sangaku Journal of Mathematics (SJM) c ⃝SJMISSN 2534-9562 Volume 2 (2018), pp. 57-73 Received 20 November 2018. Published on-line 29 November 2018 web: http://www.sangaku-journal.eu/ (http://www.sangaku-journal.eu/) c ⃝The Author(s) This article is published with open access1. Wasan Geometry and Division by Zero Calculus ∗Hiroshi Okumura and ∗∗Saburou Saitoh

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"$x \div y$" is actually just shorthand for "$x \cdot \lambda$, where $y \cdot \lambda = 1$".

Therefore: "what is $x \div 0$?" is just shorthand for "what is $x \cdot \lambda$, where $0 \cdot \lambda = 1$?"

In truth, $0 \cdot \lambda = 0$ for any $\lambda$. Therefore, there is no $\lambda$ such that $0 \cdot \lambda = 1$.

Because "$\lambda$, where $0 \cdot \lambda = 1$" does not exist, "$x \cdot \lambda$, where $0 \cdot \lambda = 1$" does not exist.

Same statement, just written in shorthand: "$x \div 0$" does not exist.

To answer the question: $0 \div 0$ does not have infinite solutions. It is the same as $x \div 0$ for any $x$ -- it simply does not exist.

I think the confusion about infinite solutions comes from the fact that there are infinite solutions to $x \cdot 0 = 0$. But saying $x \cdot 0 = 0$ is not the same thing as saying $x = 0 \div 0$.


Three sidenotes:

  1. At least when $y$ is a real number and $y \neq 0$, there is a unique real number $\lambda$ such that $y \cdot \lambda = 1$. This is called the "inverse law" of multiplication, and there are two basic approaches for determining that the inverse law is true. The first approach is to carefully define the real numbers and the arithmetic operations on them, and write a proof that they always behave this way. The second approach is to begin with the assumption that the inverse law is true (treat it as an axiom). By the second approach, you need a whole collection of axioms to characterize the real numbers, including the inverse law and some other axioms from which you can prove that $0 \cdot x = 0$ for any $x$. If you drop the "$y \neq 0$" condition from the inverse law, then it becomes inconsistent with the other axioms. An inconsistent set of axioms is useless.

  2. In a system where $0=1$, there does exist $\lambda$ such that $0 \cdot \lambda = 0$ and $0 \cdot \lambda = 1$. But if $0 = 1$, then you can prove that every number is equal to every other number. Rather: there is only one number in this system, and $0$, $1$, $\lambda$, $\pi$, $42$, $-1$, etc. are all different names for the same number. Let's give it the name $x$. In this system, $x + x = x$, $x \cdot x = x$, $x \div x = x$, etc. That's not a very useful system.

  3. There's a common misconception that there's a rule that says $x \div x = 1$ for all $x$. The rule is actually $x \cdot 1 = x$. "$x \cdot y = z$" is not the same statement as "$x = z \div y$". "$x = z \div y$" is the same statement as "$x = z \cdot \lambda$ where $y \cdot \lambda = 1$". Therefore, "$x \div x = 1$" is really equivalent to "$x \cdot \lambda = 1$ where $x \cdot \lambda = 1$".

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  • $\begingroup$ What do you think the expression 1+1 = 2 represents? The expression here has one object being added to itself which is simply doubling it, so we use another digit to represent the value of 2. In truth there are only 2 distinct digits, the digit that has a weighted value (1) and the digit that doesn't (0). This gives us the simplest of all number systems the binary number system in which Boolean Algebra and Boolean Logic can be applied. Yes we can increase the amount of digits in a number system to make more complex number systems such as octal, decimal and hexadecimal... $\endgroup$ – Francis Cugler Nov 12 at 0:06
  • $\begingroup$ But in this I'll use decimal since it is the one we most commonly associate with... Other than 0 for all + integers we can express them in the sums of 1s. For example 1+0 = 1 (additive identity), 1+1 = 2, 1+1+1 = 3, 1+1+1+1 = 4... etc. In the moment that you apply the operation addition which is a linear translation you instantaneously and implicitly introduce multiplication. Because 1*1 = 1, 1*2 = 2, 1*3 = 3, and 1*4 = 4 and so on (all of these are multiplicative identities). However, as you can see there is no expression of 1*(0) in the former as this yields 0... $\endgroup$ – Francis Cugler Nov 12 at 0:11
  • $\begingroup$ A number has a weighted or quantitative value where a place holder doesn't. Think of a light switch on your wall when it is in the "Off position" there is no light. As long as there is electricity being applied to the circuit, if the switch is in the "On position" then there is light. $\endgroup$ – Francis Cugler Nov 12 at 0:13
  • $\begingroup$ @FrancisCugler is there a specific part of my answer that you're trying to clarify? (see "When Shouldn't I Comment?") $\endgroup$ – Jordan Nov 13 at 2:06
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The responses of division by 0 that we have been taught is undefined I actually find to be unintelligible and completely unacceptable. Take this into consideration, when we look at a fraction $\frac{n}{d}$ where $n$ is the numerator and $d$ is the denominator and $d$ happens to be $0$ let's apply this concept to a combination of linear equations and some trigonometric functions. Before I begin with that I will clearly state that some functions or equations that are normally used in different contexts mean exactly the same thing in all contexts. And I will show that the actual operation of division and fractions are nothing more than the slope of a line or the tangent of some angle with regards to the horizontal and that the slope of a vertical line or $\tan(90°)$ is completely defined!

Equivalent Equations

  • slope of a line $m = \frac{\left( y_2 - y_1 \right)}{\left(x_2 - x_1\right)}$ is equivalent to the $\tan\theta$ where $\theta$ is the angle above the horizon when the angle is in standard form.
  • The Pythagorean Theorem and the Equation of the Circle are the same exact thing: $A^2 + B^2 = C^2$ and $X^2 + Y^2 = C^2$
  • The $\cos\theta$ between two vectors is equivalent to $\frac{\vec A \cdot \vec B}{|\vec A||\vec B|} $

Assertions

  • Both the $\sin$ and the $\cos$ functions have the same domain and range: Domains are the set of $\mathbb R$ and the ranges are $\left[-1,1\right]$ and they have a period of $2\pi$. They are continuous circular or rotational functions.
  • $\tan\theta = \frac{\sin\theta}{\cos\theta}$
  • When plotting on the Unit Circle where the radius has a value of 1: The $(x,y)$ pair is defined as $(\cos\theta,\sin\theta)$ where the radius is above the horizontal $x$ axis in standard position.

Assessments

  • Let's consider that we have 3 points $a,b,c$ where $a = (0,0)$ and $b = (1,0)$ and these points are fixed and point $c$ starts at point $b$'s location $(1,0)$. These 3 points will form vectors or line segments between each other. Initially there are only 2 valid vectors $\vec A$ and $\vec B$ where $\vec A$ is $b - a$ and $\vec B$ is $c - a$. $\vec C$ doesn't exist yet or is the $\vec 0$ since $\vec C$ is defined as $c - b$ and both points coincide.
  • We can technically rotate either direction but we will rotate $CCW$ and we will do a few observations in the process.
    • 1 - We will observe point $c$ as it rotates around the unit circle.
    • 2 - We will observe vector $\vec C$ as point $c$ rotates around the circle.
    • 3 - We will observe the area of the triangle that is generated from vectors $\vec A, \vec B,$ and $\vec C$
    • 4 - We will observe the slope of $\vec B$ as this is the line that rotates.
    • 5 - We will observe the angles between $\vec A$ and $\vec B$

Intuitive Declarations

  • When the point $c$ rotates to the position $(-1,0)$ the slope is $0$, the area of the triangle is $0$, but the length of $\vec C$ is at its longest which is $2$. Here you have nothing but horizontal translation with $0$ or no incline or change in height or elevation.
  • When the angle is $45°$ or $\frac{pi}{4}$ radians the slope $\frac{rise}{run}$ and the $\tan\theta$ or $\frac{\sin\theta}{\cos\theta}$ are 1. You have an equal amount of vertical translation as you do horizontal translation.
  • The $\cos\theta$ represents the $x$ value on the unit circle but also your horizontal displacement.
  • The $\sin\theta$ represents the $y$ value on the unit circle but also your vertical displacement.
  • When the Angle is $0°,180°,360°$ or an multiple of them the Slope and the Tangent are also $0$. This means we have $0$ rise and the $\sin$ or the $y$ has an output of $0$ for its range at that angle.
  • When the Angle is $45°$ both the $\cos$ and the $\sin$ have equivalent values since both legs of the triangle here are equal thus giving you both a slope and a tangent of $1$ and if you graph both the sine and cosine functions they will intersect at this point.

Considerations - Considering that both the sine and cosine are continuous circular functions neither of them at any point in their range nor their domain become undefined nor has any discontinuity in it.

Generalization - Imagine yourself walking down an alley between two skyscrapers and the alley starts off being level. You have $0$ slope or no incline but you are traveling either $N,E,W,S$ which doesn't matter because the ground you are walking on is a 2D plane. So you do have 2 degrees of dimension to travel upon. However since you are in a tight alley in this demonstration you are only heading in one arbitrary horizontal direction. Then the alley or the road has a hill that you have to walk up, now you have slope because you are rising in elevation. Then it levels off again and your slope is back to $0$ at the new elevation. This makes sense because two horizontal lines at different heights are parallel so both their slopes will be the same. Finally the alley comes to an end as their is another skyscraper in front of you and you can not go left nor right and there is no turning back, but the building in front of you has a ladder and you begin to climb it rung for rung. When you start to climb straight up your angle is $90°$ which is perpendicular and orthogonal to the ground. This means that you no longer have any horizontal displacement but you have incremental and continuous vertical translations. So in this case your elevation or height is forever increasing until you reach the roof and start to walk across a horizontal or sloped plane.

The argument - In the case where the $\sin$ component of the tangent or the $(y_2 - y_1)$ component is evaluated to $0$ you have no incline and the slope is $0$ because here $n = 0$ and $d = (x_2 - x_1)$ or $d = \cos\theta$ where this means you only have horizontal translation and this is valid because a numerator can be $0$. Let's reverse the case. This time the $\cos\theta$ component of the tangent or the $(x_2 - x_1)$ is $0$ which simply means just the opposite where we have no run yet we do have continuous rise, yet from the fallacies if what we were taught this is undefined because of division of $0$ because here the $d$, $(x_2 - x_1)$, or the $\cos\theta$ evaluates to $0$. I argue that these are valid outputs and acceptable domains of fractions or division. Division by $0$ is completely defined.

Conclusion - If $\frac{0}{d}$ means $0$ slope or no rise with infinite run then the opposite must be valid as well where $\frac{n}{0}$ means no run with infinite rise. The correct answer for division of $0$ would be $\infty$ since when the numerator is $0$ the slope is $0$ because there is no change in elevation. We can not evaluate $\frac{d}{0}$ as $0$ because here we have no run but we do only have rise and slope is defined as the change in the elevation and here the slope is ever increasing vertically up without any horizontal displacement and this does make complete sense. In trigonometry we do know that the tangent's graph has vertical asymptotes at periods of $\frac{pi}{2}$ or $90°$. We are taught that the tangent is undefined here. I think this is a wrong assessment because those vertical asymptotes are parallel vertical lines to the vertical $y$ axis and these are perpendicular and orthogonal to the horizontal or $x$ axis. When a line has a slope $\frac{a}{b}$ a line that is perpendicular to it is $-\frac{b}{a}$ Let's try this with a slope of $0$ then find its perpendicular.

$$\frac{0}{d} \implies 0$$ slope or a horizontal line therefor $$\frac{-d}{0} \implies \infty$$ vertical slope or a vertical line.

Let's apply the above with the trig functions again starting with $0$ slope.

$$\frac{\sin\theta}{\cos\theta} = \frac{0}{\cos\theta} \implies \tan\theta = 0$$

therefor it's perpendicular must be:

$$-\frac{\cos\theta}{\sin\theta} \implies -\cot\theta$$

since we have no run in vertical slope the $\cos\theta$ component must be $0$; then it suggests that:

$$-\cot\theta = \frac{0}{-\sin\theta}$$

which is also the same as:

$$\frac{0}{\sin\theta}$$

however, this does not evaluate to $0$ when regarding slope because we do have a change in height that is shown by the $\sin\theta$. The slope here is defined as being $\lim_\infty$ because the tangent has a vertical asymptote at $90°$ and cotangent is $0$. The cotangent has a vertical asymptote at $180°$ and the tangent is $0$ at $180°$.

It is these associations and relationships of division, fractions, slopes, trig functions and reciprocals with the use of the dot and even the cross products that define how two vectors are perpendicular and orthogonal to each other when they create a separation of $90°$ or $\frac{pi}{2}$ radians from each other. If you were to take just the $x$ and $y$ axis of a 2D Coordinate Cartesian Plane we know that the $x$ axis has $0$ slope because it is horizontal or level and that the $y$ axis has vertical slope. Vertical slope is NOT undefined! If we were to rotate these two axis together by $1°$ there is defined slope. Slope is always defined since one can always change their perspective to that system.

I think that people confuse what $0$ really is! $0$ is really not a number, it is a place holder, it also represents the empty or the null set, it has no value. So if you can divide any number by the empty set and it returns back to you the empty set. Is it not conceivable to divide anything into the empty set?

In this particular case and context division by $0$ here yields infinity because the problem pertains to slope or the change in height over distance.

In other contexts division by $0$ could mean, $0$ as the result is $0$ and is no different than when it is in the numerator.

It could also yield D.N.E. meaning that the function that it is being applied to just Does Not Exist at that location or context.

There is 1 special case and that is when both the numerator and denominator are $0$. This could yield $0$, $1$, and or $\infty$. This does satisfy that $0$ divided by any number equals $0$. It also satisfies the multiplicative identity that any number divided by itself is $1$. The infinity part also comes from the concept that if you have $0$ run and $0$ rise you are stationary and you are infinitely not moving in any direction which is no different than $0$. To try to make this understanding a little clearer then ask yourself this: Why does any number $n$ raised to the $0$ power always equal $1$? $n^0 = 1$.

For The Reader - It would be advisable to draw the unit circle and the points and vectors as described above in the assessments and to do several of them where the rotation around the unit circle are at different positions. Or you can visit this web page interactive graph of a graph that I made that shows all the relationships between these 3 points along with the tangent, area of the triangle, and even the volume if you increase the height factor, the coordinate pairs $(\cos\theta,\sin\theta)$ along the unit circle etc. And you will notice that when the angle is $0°$ the slope, area and volume are all $0$. They are also $0$ when the angle is $180°$ and $360°$. When the angle is 90° or $270°$ of course the "slope is labeled as undefined because that is what people have programmed it to be because of the fact that we were taught that division by $0$ is undefined!" However the Area and Volume of the triangle is at its maximum value. I have it set to where you can press the play button for the "t" value which stands for $\theta$ because this website at the time has not incorporated the use of the variable $\theta$ to be allowed into their functions or expressions to make graphs.

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  • $\begingroup$ "0 is really not a number"; how do you justify this statement? Zero is as much a number as 5 or 2918. As a further question, while an interesting interpretation, how does this address the formal definition of division? Shouldn't that be the baseline for how we proceed with addressing expressions such as $0/0$? $\endgroup$ – masiewpao Nov 11 at 14:53
  • $\begingroup$ @masiewpao It is a digit, it is a place holder but it has no value or magnitude. It is the empty or null set. Take for example the Binary number system where there are only 2 digits: 0 & 1 and its use within digital circuits. Either there is a voltage or there isn't and thus the memory cells preserve either a 1 or 0, on or off, true or false within the feed back loop of the circuitry. It's an abstract statement just as the idea of 0 itself is abstract. For example; write a mathematical proof that 1 == 1 or 2 == 2... $\endgroup$ – Francis Cugler Nov 11 at 20:15
  • $\begingroup$ But the same thing could be said about any 'digit'. It's a number in so much as 5 is a number. In a set theoretic construction of the natural numbers for example, why would you consider the set of the null set anymore of a number than the null set? $\endgroup$ – masiewpao Nov 11 at 21:32
  • $\begingroup$ Think of 0 and 1 as vectors in the binary number system. Think of 0 as being the 0 vector and 1 as being the unit vector. The 0 vector has no magnitude nor direction or infinite direction where as the unit vector the complete opposite has both magnitude (distance) and direction. This concept doesn't just apply to mathematics but also to physics especially when it concerns forces and displacement. $\endgroup$ – Francis Cugler Nov 11 at 23:03
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    $\begingroup$ This answer gets things backwards. It starts with the author's intuition about geometry, and then works backwards, forcing the definitions of $\mathbb{R}$, $\infty$, and $\div$ to fit that intuition. The correct place to start is with the algebraic properties of $\mathbb{R}$, because $\mathbb{R}$ is, in fact, an algebraic structure. If the algebraic result contradicts your geometric intuition, then either you're misapplying $\mathbb{R}$ or your intuition is incorrect. $\endgroup$ – Jordan Nov 13 at 2:55

protected by Michael Albanese Jul 7 '14 at 6:27

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