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I was talking about division by zero with my discrete math instructor, and it was explained to me that dividing can be broken down into simpler terms, i.e: Consider 6 divided by 3. To reach the answer consider how many times 6 can repeatedly have 3 subtracted from it, such that it comes to zero. 6 - 3 once is 3, and twice is zero. The answer thus is two.

So, I said that zero divided by zero then is already zero, and no subtraction ever need take place. Therefore the claim is that $\frac{0}{0} = 0$

How can we show that this does not work well? My instructor claimed that there is a smart way of showing how this will not work, but could not remember what it was. Does anyone know what it might be?


marked as duplicate by Alexander Gruber Jan 12 '15 at 4:04

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    $\begingroup$ Using your method, one can show that $0/0$ is equal to any real number, because $0 - k\,0 = 0$ for any $k \in \mathbb{R}$. $\endgroup$ – Adam Saltz Jan 15 '13 at 4:07
  • $\begingroup$ You can prove as much that that is a true statement as Wichita Falls is USA's capital city: it just doesn't exist. Division by zero is undefined in mathematics, so a more interesting and profound discussion could probably be: why is division by zero undefined? $\endgroup$ – DonAntonio Jan 15 '13 at 4:08
  • $\begingroup$ @DonAntonio Yes, sorry I do not necessarily mean prove. I asked that since everything seems to revolve around proofs for me recently. I realize it is undefined. Perhaps show that it does not work is what I am interested in. $\endgroup$ – Leonardo Jan 15 '13 at 4:10
  • $\begingroup$ @DonAntonio I remember once getting scolded by Arturo because I said "division by zero is undefined". There are some structures called "wheels" where it is apparently. It is however true that in the usual places we know, it isn't defined. $\endgroup$ – Pedro Tamaroff Jan 15 '13 at 4:11
  • $\begingroup$ @PeterTamaroff Where? $\endgroup$ – ՃՃՃ Jan 15 '13 at 4:16

For $a, b \in \mathbb{Z}$, $b\neq 0$, the number $\frac{a}{b}$ is the unique solution to the equation $bx = a$. If we allow $a$ and $b$ to be both zero, we are trying to solve the equation $0x = 0$, but this is true for any $x$. In particular, there is not a unique solution which we could call $\frac{0}{0}$.

  • $\begingroup$ That is the best way it could be explained to me, thank you. $\endgroup$ – Leonardo Jan 15 '13 at 4:25
  • $\begingroup$ I think this is the case when you work with integers. It is more or less how they are defined rather than "why". $\endgroup$ – user123454321 Jan 15 '13 at 4:26
  • $\begingroup$ @GilYoungCheong Does this not also work for real numbers too? $\endgroup$ – Leonardo Jan 15 '13 at 4:27
  • $\begingroup$ Real number system is not the only number system that you can think about. $\endgroup$ – user123454321 Jan 15 '13 at 4:29
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    $\begingroup$ @user1965813: You made a suggested edit that changed "$a,b\in\mathbb Z\setminus\{0\}$" to "$a\in\mathbb Z, b\in\mathbb Z\setminus\{0\}$". While that's not incorrect, I think edits should not change the meaning of the original post, and the suggestion should have been left as a comment instead. $\endgroup$ – Rahul Jan 15 '13 at 7:07

Part of the problem, lies in the way you were exposed to addition, subtraction, multiplication and division.

When you were first taught to add, you were told that you can add integers: $$1+1 = 2$$

Then, you were told that you could subtract integers:


You were told to think of subtraction as 'the opposite of addition'. You were then introduced to this fancy concept of negative numbers, and told that

$$2 + (-1) = 1$$

But I bet they never told you why this works. The reason why this works, is because $-1$ is the additive inverse of $1$. Simply put, $1 + (-1) = 0$. (Of course, we then have to define the additive identity, but since they never did that either ....)

Next, you were told that you could multiply integers:

$$2 \times 2 = 4$$

You were also told to think of multiplication as repeated addition. I'm sure you've drawn pictures of 4 groups of 5 objects and counted, to show that $4 \times 5 = 20$. Next, you were told that you could divide integers:

$ 4 \div 2 = 2$

You were told to think of division as 'the opposite of multiplication'. That since $ 2 \times 2 = 4$, hence $4 \div 2 = 2$.

You were most probably warned (with more than a slap on the wrist I hope) to never 'divide by 0'. I bet they never told you why not. The reason why, is because $0$ has no multiplicative inverse. There is no number $x$, that allows us to say $ 0 \times x = 1$. If there was, then we can say that $ a \div 0 = a \times x$, and evaluate that.

What else did they not tell you? Probably a whole lot. For example, what is $ \sqrt{2} \times \pi$ through repeated addition? How am I supposed to draw those sets out and count? Also, fractions are defined by division. Why does that mean that we can add and subtract fractions? What is a real number, and why must it exist? I know what $1.4 = \frac {7}{5}$ is, but what is $\sqrt{2}$?

Furthermore, we can actually divide by 0 in special situations. For example, when we attach $\infty$ to the real line, we could say that $\frac {1}{0} = \infty$. However, that does come with its own set of problems, and you must learn addition and subtraction all over again. For example, what is $\infty - \infty$? It could be anything.

  • $\begingroup$ It is more of symbolic when you say $1/0 = \infty$. Adding infinity loses the rich algebraic structure of real line. $\endgroup$ – user123454321 Jan 15 '13 at 4:28
  • $\begingroup$ @GilYoungCheong Agreed. It screws up a ton of stuff, like subtraction. $\endgroup$ – Calvin Lin Jan 15 '13 at 4:31
  • $\begingroup$ Yes. I think really the meaning of "division" does not makes sense any more when we do such "completion." By the word division, I think one should think of algebraic view of number system, which is just my opinion. $\endgroup$ – user123454321 Jan 15 '13 at 4:34
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    $\begingroup$ I disagree. If you have a field structure, you are forced to say that $1 = 0 \cdot 1/0 = 0$. This is not very rich structure at all. In fact, in the case of field (as $\mathbb{R}$ is), it even causes a contradiction. $\endgroup$ – user123454321 Jan 15 '13 at 4:41
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    $\begingroup$ If you want to define $1/0$, it is fine but this cannot algebraically interact with other numbers without "screwing" the number system up. If you are not convinced, write $1/0 = 1 \cdot 0^{-1}$. This is the meaning of "division". $\endgroup$ – user123454321 Jan 15 '13 at 4:45

You can divide by $0$ but it is not interesting to do math like that. What we ususally want is to say $x \cdot 0 = 0$, for any number $x$. Suppose we can divide by $0$, so for an arbitrary number $a$ let $y = a/0$ for some number. Then we have $0 = 0 \cdot y = a$, so every number $a = 0$.

Therefore, it is perfectly fine if you allow division by $0$, but then in order to make algebraic property "rich" enough to do mathematics your number system should only contain $0$ but nothing else.

In fact, for integers, people do say that $0$ is divisible by $0$. The definition you have in mind in this situation should be the following. When $a, b \in R$ we say $a$ is divisible by $b$ if there is $x \in R$ such that $a = bx$, but again, this should be distinguished with the fraction notation.


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