9
$\begingroup$

Several years ago I was bored and so for amusement I wrote out a proof that $\dfrac00$ does not equal $1$. I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$. As this is clearly false and if all the steps in my proof were logically valid, the conclusion then is that my only assumption (that $\dfrac00=1$) must be false. Unfortunately, I can no longer recall the steps I used to arrive at the contradiction. If anyone could help me out I would appreciate it.

$\endgroup$
5
  • 10
    $\begingroup$ Hint: $~a\cdot0=0,~$ for all a. $\endgroup$
    – Lucian
    Commented Nov 17, 2014 at 9:08
  • 14
    $\begingroup$ Unfortunately, it isn't possible to form a proof in the way you describe. $\frac{0}{0}$ is simply undefined, and so it is impossible to perform any manipulations on the expression you have given. $\endgroup$ Commented Nov 17, 2014 at 9:09
  • 9
    $\begingroup$ @BenjaminAlderson, your objection is not valid. The reason $0/0$ is undefined is that it is impossible to define it to be equal to any real number while obeying the familiar algebraic properties of the reals. It is perfectly reasonable to contemplate particular vales for $0/0$ and obtain a contradiction. This is how we know it is impossible to define it in any reasonable way. To say, it's simply undefined so this is invalid is not the way mathematics is done. OP is interested in why it can't be defined, not in blindly accepting authority. $\endgroup$ Commented Nov 17, 2014 at 9:32
  • 8
    $\begingroup$ @Ittay: While you're right, Benjamin is not wrong. $0/0$ being undefined is simply a matter of definition, and the question the OP asked doesn't really make sense. What you argue is that the OP really should be asking a different question: "what motivated mathematicians to define division in a way so as to leave $0/0$ undefined?". $\endgroup$
    – user14972
    Commented Nov 17, 2014 at 10:00
  • 4
    $\begingroup$ I think the title of the questions clarifies OP's intentions sufficiently well, though the question certainly could have been worded with more care. However, saying it's undefined cause it's undefined is a poor argument if it's an argument at all. $\endgroup$ Commented Nov 17, 2014 at 10:06

11 Answers 11

28
$\begingroup$

If $0/0$ were equal to $1$, then $1=\frac{0}{0}=\frac{0+0}{0}=\frac{0}{0}+\frac{0}{0}=1+1=2$.

$\endgroup$
9
  • 3
    $\begingroup$ @vuur Sure, $1=0+1=\tfrac00+\tfrac11=\tfrac{0+0}{0}=0.$ $\endgroup$
    – Hakim
    Commented Nov 17, 2014 at 16:31
  • 2
    $\begingroup$ @Hakim How do you go from $\frac00+\frac11$ to $\frac{0+0}0$? $\endgroup$ Commented Nov 17, 2014 at 21:41
  • 1
    $\begingroup$ @JaycobColeman $\tfrac{a}{b}+\tfrac{c}{d}=\tfrac{ad+bc}{bd}$. $\endgroup$
    – Hakim
    Commented Nov 17, 2014 at 21:49
  • 2
    $\begingroup$ I really think you need to work with a definition of division on the whatever set of numbers you choose (natural, integer, rationals or reals). Then it should be immediately apparent that you can neither prove nor disprove that $0/0=1$. $\endgroup$ Commented Nov 18, 2014 at 5:18
  • 2
    $\begingroup$ typically, in a ring, one defined $a/b$ as $a\cdot b^{-1}$, when $b^{-1}$ exists. Then $0/0=0\cdot 0^{-1}$, which would be $0$ if $0^{-1}$ exists. Of course $0^{-1}$ does not exist since by definition this element would solve $0\cdot x = 1$. But $0\cdot x = 0$ and most definitions of ring demand that $0\ne 1$. So, ring theoretically, this is the end of the story. $\endgroup$ Commented Nov 18, 2014 at 5:40
9
$\begingroup$

In lay terms, evaluating 0/0 is asking "what number, when multiplied by zero, gives zero". Since the answer to this is "any number", it cannot be defined as a specific value.

$\endgroup$
6
  • $\begingroup$ but perhaps there are compelling reasons then to choose one particular solution as the value of $0/0$. Your argument does not exclude $0/0=1$. $\endgroup$ Commented Nov 17, 2014 at 20:04
  • $\begingroup$ @IttayWeiss if you reduce an expression, say f(x) to 0/0 then that does not tell you that f(x) = 1 nor that it is not 1. 0/0 is undefined and could have any value (or even a different value in each place it appears) in your formula, so it gives you no information about the value of f(x). $\endgroup$ Commented Nov 17, 2014 at 23:06
  • $\begingroup$ your answer only shows that treating $0/0$ as the solution to $x\cdot 0 =0$ does not determine any unique value for $0/0$. That is correct, but does not answer OP's question. OP asked what is inconsistent with defining $0/0=1$. You did not answer that question. $\endgroup$ Commented Nov 17, 2014 at 23:13
  • 1
    $\begingroup$ @IttayWeiss You could define 0/0 to be 1, but you then lose the very useful property that division is the inverse of multiplication. I believe that also loses you the equivalence $\frac{a+b}{c}\equiv\frac{a}{c}+\frac{b}{c}$ that you rely on in your answer. $\endgroup$ Commented Nov 17, 2014 at 23:29
  • $\begingroup$ I'm not trying to convince you that $0/0=1$ is a good idea. I know why it does not work. I'm just pointing out that your answer does not address OP's question. $\endgroup$ Commented Nov 18, 2014 at 1:04
6
$\begingroup$

The accepted definition of division on the natural numbers is something like:

For all natural numbers $x, y, z$ where $y\ne 0$, we have $x/y = z$ iff $x=y\times z$. (Also works for the integers, rational numbers and reals.)

Using this definition, you can neither prove nor disprove that $0/0=1$. You wouldn't be able to draw any inferences from your assumption that $0/0=1$. If $y$ (the divisor) is $0$, this definition tells you nothing.


Suppose we did not have the restriction $y\ne 0$ and that, instead, we simply defined $x/y = z$ iff $x=y\times z$ for any natural numbers $x, y$ and $z$.

Then, consider two cases: $x=0$ and $x\ne 0$.

If $x=0$, then the definition would be inconsistent with our definition of the natural numbers.

$0/0$ could be $0$ because $0\times 0 =0$

$0/0$ could be $1$ because $0\times 1 = 0$

$0/0$ could be $256$ because $0\times 256 = 0$

All natural numbers would have to be equal (a contradiction). This alone would be enough to reject our restriction-free alternative definition. It is inconsistent.

If $x\ne 0$, then no natural number would work for $x/0$. For any natural number $z$, we could not have $x=0\times z$. Zero times any number is always zero.

Either way, the alternative definition simply doesn't work.

$\endgroup$
3
$\begingroup$

Let's view the problem within ring theory, i.e., an algebraic structure with addition and multiplication following familiar axioms. Usually when we write $0$ in this context, it means an additive neutral element, i.e. $x+0=0+x = x,\ \forall x$. In any ring it follows from distributive laws that for any $x$ we have: $$x\cdot 0 = x\cdot(0+0) = x\cdot 0 + x\cdot 0\implies x\cdot 0 = 0$$ Now, when we write $\frac xy = z$, in this context we mean that there is unique $z$ such that $x = zy$. If there are more than one $z$ meeting this condition, $\frac xy$ wouldn't be well defined.

Let us assume that we have a ring $R$ with additive neutral element $0$, such that division by $0$ is well defined for some $x$, that is, there is unique $y\in R$ such that $y\cdot 0 = x$. First we note that $x = y\cdot 0 = 0$, so only $\frac 00$ might be defined. Now, assuming $\frac 00 = y$, as we said before, it means that $y$ is the unique element in $R$ such that $y\cdot 0 = 0$. But since for any $y\in R$ we have that condition, we conclude that $R$ has exactly one element, namely $R=\{0\}$. We call this ring zero ring.

TLDR: Division by $0$ is possible, but only in a trivial (zero) ring, and nowhere else.

$\endgroup$
2
  • $\begingroup$ Why has this been downvoted? $\endgroup$
    – Karl
    Commented Feb 20, 2016 at 16:47
  • $\begingroup$ I agree it is important to start with assuming a structure such as a ring and deducing the fact from the given axioms. I'm honesty not sure why your answer has been downvoted. $\endgroup$
    – Karl
    Commented Feb 20, 2016 at 16:55
1
$\begingroup$

Yes. There is a algebraic reason. In a field there is no reasonable way we can divide by zero, because one cannot have both the identities $(a/b)\times b =a$ and $c\times0=0$ hold simultaneously if $b$ is allowed to be zero.

Note that the cancellation law depends of non-zero divisors:

Proposition (Integers have no zero divisors). Let $a$ and $b$ be integers such that $ab=0$. Then either $a=0$ or $b=0$ (or both).

Corollary (Cancellation law for integers). If $a, b, c$ are integers such that $ac=bc$ and $c$ is non-zero, then $a=b$.


EDIT. By other hand, is possible to construct a algebraic structure with $0/0=1$ (similar a ring, but adding another axioms, maybe as $0/0=1$). But in that case, we must consider that we are no working with the rationals $\Bbb Q$ or the reals $\Bbb R$, since they are fields, so their theorems couldn't be true.

$\endgroup$
5
  • $\begingroup$ $0/0=1$ is consistent with the equalities you write. Moreover, it is incorrect to think of L'Hopital's rule as "dividing by a quantity that approaches zero. In fact, there is no such thing as "a quantity that approaces zero". Limits are numbers, not approximations or anything vague such as quantities approaching anything. $\endgroup$ Commented Nov 17, 2014 at 18:53
  • $\begingroup$ You're right, L'Hospital wasn't a good example. On the other hand, if we define $c:=a/b$ and $b:=0$, we have $(a/b)\times0=0$ or $(a/b)\times0=a$, but no both simultaneously. I don't say that $0/0=1$ is impossible, we can construct a algebraic structure with $0/0=1$. I'm saying that case, we're not working with a "field", and so we are not working with the rationals $\mathbb Q$ or the reals $\mathbb R$. $\endgroup$ Commented Nov 17, 2014 at 19:33
  • $\begingroup$ that is true only if $a\ne 0$. But $0/0=1$ is not inconsistent with the equalities you had written. $\endgroup$ Commented Nov 17, 2014 at 20:01
  • $\begingroup$ as for your later edit, $0/0=1$ is not the case in any ring. $\endgroup$ Commented Nov 17, 2014 at 20:02
  • $\begingroup$ OK. I meant "the ring axioms and another axioms, maybe $0/0=1$ as an axiom". $\endgroup$ Commented Nov 17, 2014 at 20:13
1
$\begingroup$

Assume $\frac{0}{0} = 1$ and define $x := \frac{0}{0^2}$. Then $$ 0 = 0\cdot x = 0 \cdot \frac{0}{0^2} = \frac{0}{0} = 1. $$

$\endgroup$
0
$\begingroup$

Here is one of the famous "fake" proof that $0=1$.

Let $a=b=1$ (But here you can choose another number is you want).

$a²=ab$

$a²-b²=b(a-b)$,

$(a-b)(a+b)=b(a-b)$

Then you divide each term by $(a-b)$ (here is the clincher, of course...) and you get

$a+b=b$, hence $a=0$ (But remember that $a=1$...).

Of course in order to "get" this result, you have to divide by $0$, which as NO meaning whatsoever... And you can get whatever result you want, which will be meaningless, since you did something meaningless.

$\endgroup$
3
  • $\begingroup$ OP was asking why $0/0$ can't be defined. The whole question is "how do you know you can't meaningfully define $0/0$?" $\endgroup$ Commented Nov 17, 2014 at 9:41
  • $\begingroup$ @IttayWeiss Well, one can deduce easily from the aforementionned demonstration that dividing by $0$ is not possible and not defined, hence $\frac{0}{0}$ is not defined as well. But I agree that this is not using a possible $\frac{0}{0}=1$ assumption... $\endgroup$
    – Martigan
    Commented Nov 17, 2014 at 9:49
  • 1
    $\begingroup$ exactly. The argument for why $a/0$ for $a\ne 0$ is not definable is different than why $0/0$ is not definable. $\endgroup$ Commented Nov 17, 2014 at 9:58
0
$\begingroup$

The other answers are quite helpful.I only want to add that $\frac00$ is just not defined any value.It is one of the several indeterminate forms of mathematics-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

$\endgroup$
0
$\begingroup$

Here is my theory: if anything times zero is zero, than zero divided by anything is zero, and zero divided by zero is anything. When I say "anything" I mean, any number that exists, even an imaginary number. 0*anything=0, so that should mean 0/anything=0, and 0/0=anything. So if I say 0/0=0, that would be true, and if I say 0/0=1, that would be true as well. 0/0=infinity would also be true, 0/0=(-infinity) would be true. Think of it like this: nine divided by three is three, because three goes into nine three times. But how many times must zero be added to get to zero? If zero is added once, it'll be zero. If zero gets added twice, it'll be zero. If it's added three times, four, five, six, seven, as much as infinity, a negative number, a decimal/fraction, a whole number, an integer, a rational number, an irrational number, a real number, an imaginary number, ALL NUMBERS, it'll still be zero. So basically zero divided by zero can be anything, in my theory.

$\endgroup$
0
$\begingroup$

Here is an argument for the kindergarten notion of division, i.e., to compute $a/b$ for natural numbers $a,b$, pretend you take $a$ cookies and divide them among $b$ kids, and ask yourself how many cookies will each child get. Theorem: $a/0$ does not have a unique value, and therefore it does not equal anything. Proof: Take $a$ cookies and divide them among $0$ children. It is vacuously true that each child gets $5$ cookies. It is also vacuously true that each child gets $m$ cookies, for any value of $m$. QED.

$\endgroup$
0
$\begingroup$

We know that

$$ 0 \times 1 = 0 \times 2 $$

If we assume $0/0 = 1$, then we can divide both side by zero, and get

$$\begin{align*} \frac{0}{0} \times 1 &= \frac{0}{0} \times 2 \\ 1 &= 2\end{align*} $$

We know that $1 \ne 2$, and so $\frac{0}{0} \ne 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .