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For every number $x$, $x\times 0=0$, hence $\dfrac{0}{0}$ can be any number! So $\dfrac{0}{0}$ "is knows as indeterminate" [1]. But what if we define it to be $0$? I already have an answer, but don't know how convincing it is:

$1=\dfrac{1}{1}=\dfrac{1}{1}+0=\dfrac{1}{1}+\dfrac{0}{0}=\dfrac{1\times 0}{1\times 0}+\dfrac{0\times 1}{0\times 1}=\dfrac{0}{0}+\dfrac{0}{0}=0+0=0$, a contradiction.

Is there any better explanation why not to define $\dfrac{0}{0}$ to be $0$ (or any other number)?

Thanks.

[1] http://en.wikipedia.org/wiki/Division_by_zero#In_algebra

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    $\begingroup$ Why do you need a better explanation? what's wrong with this one? $\endgroup$ – Dennis Gulko Nov 8 '13 at 15:21
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    $\begingroup$ If $\mathbb{Q}$ is viewed as a meadow, then the reciprocal of $0$ is zero. So $x/0 = x \cdot 0^{-1} = x \cdot 0 = 0$. $\endgroup$ – goblin GONE Nov 8 '13 at 15:22
  • $\begingroup$ Related thread: math.stackexchange.com/q/527613/264 $\endgroup$ – Zev Chonoles Nov 8 '13 at 15:31
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    $\begingroup$ That $0$ is a multiple of any number by $0$ is already a flawless, perfectly satisfactory answer to why we do not define $0/0$ to be anything, so this question (which is eternally recurring it seems) is superfluous. Your answer is not convincing because it assumes $\frac{a}{b}=\frac{ac}{bc}$ holds for $c=0$, but haven't justified this (even for $c\ne0$ it has to be justified in the beginning). $\endgroup$ – anon Nov 8 '13 at 15:31
  • $\begingroup$ Actually, $\frac 11+\frac 00$ should be \frac{1\cdot 0+0\cdot 1}{0\cdot 1}$. Anyway, if you define $\frac 00$, you must not use most of th eusual laws of arithmetic. $\endgroup$ – Hagen von Eitzen Nov 8 '13 at 15:33
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You can define $0/0$ if you want, but that would be missing the point. One considers quotients as an operation. So what you want to be able is to have the multiplicative inverse of $0$, which can be easily seen to not make sense.

The idea of considering the fraction $p/q$ is to be able to think about it as $p\times1/q$. But, what arithmetic can you do with $0/0$?

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    $\begingroup$ I disagree. Division is arguably a more basic and natural operation than multiplying with a reciprocal. In abstract algebra we define division as $p\cdot q^{-1}$ because that just happens to produce just the division operation we want to get and is technically simpler than having a binary division operation as primitive. But that technical coincidence is not in itself an argument that this division operation is the one we should want to get. Indeed if we're talking about integers or polynomials we're perfectly happy with considering a division (with remainder) that isn't $p\cdot q^{-1}$. $\endgroup$ – hmakholm left over Monica Nov 8 '13 at 22:00
  • $\begingroup$ Fine, then. So what properties should division have? Or you want to define division without any mention of arithmetic? $\endgroup$ – Martin Argerami Nov 8 '13 at 22:08
  • $\begingroup$ x @Martin: Oh, I'm happy enough with the properties of division we have, including the undefinedness of $0/0$. I'm just saying that $p\cdot q^{-1}$ doesn't work as an explanation for why we want those properties in particular. $\endgroup$ – hmakholm left over Monica Nov 8 '13 at 23:41
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If you define $\frac00$ to be $0$, then you either have to abolish some other basic rules of arithmetic or accept the following argument: Since $3\times 0=0$, divide both sides by $0$, thereby cancelling the $0$ factor on the left and leaving $3=\frac00=0$. Neither alternative looks desirable to me.

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    $\begingroup$ If we defines $\dfrac{0}{0}$ to be $0$ then by cancelling $0$'s from the two sides of $3\times =0$ we get to $0=0$, and not to $3=0$. $\endgroup$ – Behzad Nov 8 '13 at 15:35
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    $\begingroup$ @Behzad What makes you think we get $0=0$ rather than $3=0$? $\endgroup$ – anon Nov 8 '13 at 15:37
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    $\begingroup$ @anon Suppose that $\dfrac{0}{0}$ is defined to be $0$. Cancelling a number from the both sides of an equality in fact is dividing both parts to it. $(3\times 0)/0=3\times \dfrac{0}{0}=3\times 0=0$. $\endgroup$ – Behzad Nov 8 '13 at 15:46
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    $\begingroup$ @Behzad What you suggest is exactly what I meant by "abolish some other basic rules of arithmetic", in this case the rule that dividing $a\times b$ by $b$ produces $a$. $\endgroup$ – Andreas Blass Nov 8 '13 at 21:04
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    $\begingroup$ In your answer you assume $0/0=1$, not $0/0=0$, downvoted. $\endgroup$ – Anixx Jan 1 at 13:47
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Note: The below answer does not answer properly why we cannot define $\frac 0 0=0$ but only shows that it can be approached from different directions to get different limits. It is left on here as this is the answer I had been taught in school and is easy to understand but should be understood to be inadequate.

from the right: $$\lim_{x\rightarrow0}\frac xy = 0$$ from the left: $$\lim_{x\rightarrow0}\frac xy = 0$$ if y is not 0. However it is also true that.

from the right: $$\lim_{y\rightarrow0}\frac xy = \infty$$ from the left: $$\lim_{y\rightarrow0}\frac xy = -\infty$$ if x is not 0

$$\lim_{y\rightarrow0}\frac yy = 1$$

These widely different answers means we can't define 0/0 easily.

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    $\begingroup$ But the question wasn't about limits at all. Division is part of arithmetic, which comes before limits conceptually. $\endgroup$ – hmakholm left over Monica Nov 8 '13 at 15:27
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    $\begingroup$ Limits explain only why it is indeterminate, not why it is undefined. $0^0$ is defined to be $1$ even though $0^0$ is an indeterminate form by similar reasoning to your answer. $\endgroup$ – Thomas Andrews Nov 8 '13 at 15:28
  • $\begingroup$ Please take note how some minor edits improved the typesetting of your post. $\endgroup$ – Lord_Farin Nov 8 '13 at 15:29
  • $\begingroup$ I thought that if limits approaching 0/0 widely vary, it helps show the absurdity of defining it as such for all cases. $\endgroup$ – kaine Nov 8 '13 at 15:31
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    $\begingroup$ Although it is indeterminate form, if $\frac{0}{0}$ was defined as a number, it would be $\frac{0}{0}=b$ for some real number $b$ and hence we had $0=0\cdot b$. But this holds true for any $b$. $\endgroup$ – daulomb Nov 8 '13 at 16:01
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Problematic expressions like $\frac{0}{0}$ (and many others) were considered in detail by Cauchy in his Cours d'Analyse. The term "indeterminate form" itself was coined by his would-be student Moigno.

To Cauchy what the expression $\frac{0}{0}$ meant was the process of determining the nature of the ratio of two infinitesimals, say $\alpha$ and $\beta$. Here the problem is that the ratio $\frac{\alpha}{\beta}$ can itself be infinitesimal, or infinite, or neither. For example, if $\beta=\alpha^2$ then the ratio is infinite. If $\beta=\sqrt{\alpha}$ then the ratio is infinitesimal. If $\beta=\alpha$ then the ratio is appreciable (not infinitesimal).

From this point of view it is clear that one cannot assign any definite value to $\frac{0}{0}$ without possessing additional information about the numerator and denominator.

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  • $\begingroup$ @egreg, oops, thanks. I think I answered a question about Fermat at about the same time and therefore the name was at the tip of my tongue. $\endgroup$ – Mikhail Katz Dec 18 '14 at 16:04
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With this welter of lengthy and diverse answers, you must be thoroughly confused by now; so I apologize for adding yet another answer to the pile:$$\dfrac00 \text{is meaningless}.$$That is all you need to know. If anyone challenges this, ask him then what it does mean, and why. You will easily be able to show that, using his own terms, a different meaning can be deduced.

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    $\begingroup$ The question wasn't whether $\frac00$ has meaning, but why we choose not to assign a meaning to it. Just repeating that it doesn't have one is not an answer to the question. $\endgroup$ – hmakholm left over Monica Nov 8 '13 at 21:53
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    $\begingroup$ @HenningMakholm: The OP did not refer to meaninglessness but rather to indeterminacy. Instead of considering $\frac00$ to be indeterminate---like $x$ or $\pm 1$, which do have a role to play in meaningful mathematical statements---we should not use it at all. To take a more obvious analogy, we don't need to demonstrate why $3-\sqrt/~ +=\int$ is meaningless; the burden of proof lies with the person who claims it to be meaningful. Writing $0/0$ to refer to a type of limit creates confusion, because unprepared people latch on to the form. We would be better off without it. $\endgroup$ – John Bentin Nov 9 '13 at 0:14
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Informally, $0/0$ is indeterminate, so it would be wrong to give it a value. It would be like o items out of 0 things. Is it none of the divisor or all of it, or maybe halfway? Since there's nothing to compare, all of these could be seen as valid viewpoints, so you cannot assign it a specific value.

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  • $\begingroup$ Other indeterminate forms have a value, for instance, $0^0=1$ $\endgroup$ – Anixx Jan 1 at 14:15
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$\frac{0}{0}$ is indeterminate because it is not a number.
If it was a number say $a$ then we could prove $$a\cdot 0=\frac{0}{0}\cdot 0=\frac{0\cdot 0}{0}=\frac{0}{0}=a$$
So, $$a\cdot 0=a\Rightarrow a=0$$
Also, $$1- a=\frac{1}{1}-\frac{0}{0}=\frac{1\cdot 0-0\cdot 1}{0\cdot 1}=\frac{0}{0}=a$$
So, $$1-a=a\Rightarrow a=\frac{1}{2}$$ Combining the above , $0=\frac{1}{2}$ which is a contradiction.
All the above show us that if we accept that $\frac{0}{0}$ is a real number we cannot use all basic operations.

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  • $\begingroup$ You wrote: $$\frac{1}{1}-\frac{0}{0}=\frac{1\cdot 0-0\cdot 1}{0\cdot 1}$$ - no, you cannot do this, you are multiplying thenomenator and numerator by zero, this does not prove anything (similar way you can prove that 2=1, so 2 is not a number) $\endgroup$ – Anixx Dec 18 '14 at 17:03
  • $\begingroup$ @Anixx yes this can be proved also by accepting that 0/0 is a number.So, you also agree that this is impossible.I used 1 as an example. $\endgroup$ – Konstantinos Gaitanas Dec 18 '14 at 17:09
  • $\begingroup$ what you are doing is impossible, it is irrelevant to whether 0/0 is number or not. $\endgroup$ – Anixx Dec 18 '14 at 17:11
  • $\begingroup$ @Anixx 1/2 is a number and 1 is a number also.try to prove using them that 0=1/2. $\endgroup$ – Konstantinos Gaitanas Dec 18 '14 at 17:54
  • $\begingroup$ you "proved" by multiplying all by $0$. $0$ is a number. $\endgroup$ – Anixx Dec 18 '14 at 17:57

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