3
$\begingroup$

I've heard about this and I know that division can be used in one way like this:

For example, if I want to do $30$ divided by $3$, how many times can I subtract $3$ from $30$ to get to $0$? Well, I can do it this way: $30-3=27-3=24-3=21-3=18-3=15-3=12-3=9-3=6-3=3-3=0$As you just saw, I subtract $3$ from $30$ 10 times to get to $0$. So, $30$ divided by $3$ is $10$. Well, what if I divide by a negative number or what if I divide a negative number by a positive number? Well, I do know how to solve $15$ divided by $-5$ this way. Just do it like in the mentioned way, but $15$ will get bigger if I subtract $-5$. So, what should I do. That's easy; do it backwards by subtracting the opposite of $-5$, or $5$: $15-5=10-5=5-5=0$. As you can see, I just did it backwards, and because I did it backwards, I need to subtract the number of times I subtract $5$, since it's the opposite of $-5$. So, I subtracted $-5$ -3 times. So, $15$ divided by $-5$ is $-3$.

As you have seen, I have used subtraction while doing division. Let's cut to the chase now. What if we divided any number by $0$? Well, this would happen: If I want to solve $10$ divided by $0$, I would just subtract $0$ from $10$ until I make the $10$ a $0$. So, $10-0=10-0=10-0=10-0=10-0=10-0=10-0=10...$ and it just keeps going on and on forever. So, this must explain why any number divided by zero has no answer at all. Also, for right here, right now, I can say it's infinity since it just goes on and on. So, is this why any number divided by $0$ never has an answer? Good answers at the bottom down there, if possible!

|cite|improve this question
$\endgroup$
3
$\begingroup$

It is an interesting attempt, but you rather cannot induce, that any number $A$ divided by 0 is infinity. We know, that $-0=0$ and from your considerations $A$ divided by 0 must be equal to $-\infty$ in the same time. Moreover, your attempt gives $0/0=0$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ So, it looks like any negative number divided by zero gives you no number, but negative infinity. $\endgroup$ – Mathster Nov 8 '14 at 4:12
  • 1
    $\begingroup$ @Mathster Not so either. Division by $0$ is undefined, always. There are situations where you might interpret $0$ as $0^+$, a infinitesimally small positive number. Then $1/0^+$ could be interpreted as $+\infty$. But you could also interpret $0$ as $0^-$, an infinitesimally small negative number, and then $1/0^-$ would be interpreted as $-\infty$. So $1/0$ cannot be interpreted simply as $\infty$, even though the $1$ is positive. Likewise, $-1/0$ cannot be interpreted as $-\infty$. $\endgroup$ – alex.jordan Nov 8 '14 at 4:53
  • $\begingroup$ This is what I mean: I think dividing by $0$ is undefined. $\endgroup$ – Mathster Nov 8 '14 at 15:27
  • $\begingroup$ Why 0/0 should be 0, look here: math.stackexchange.com/a/1073101/2513 $\endgroup$ – Anixx Dec 18 '14 at 9:10
2
$\begingroup$

It is not infinity, take your example. Splitting 30 marbles between 3 people equally, will be 10 each. How about splitting 30 marbles between 0 people equally? It isn't infinity, it isnt even defined

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Well, I do know that infinity isn't a real number. $\endgroup$ – Mathster Nov 8 '14 at 4:12
2
$\begingroup$

The reason division by zero is not defined is simply that it's not possible to define it and still preserve the usual rules of algebra.

For example we know that if $x$ is a nonzero real number, and if $ax = bx$, then $a = b$. This is called the cancellation law and it follows from the fact that each nonzero real x has a multiplicative inverse $x^{-1}$ with the property that $xx^{-1} = 1$. So if $ax = bx$ then $axx^{-1} = bxx^{-1} \Rightarrow a = b$.

The cancellation law is very handy when we're solving equations and doing other algebraic manipulations, so we want to keep it around.

On the other hand, for example $2(0) = 3(0)$ but $2 \neq 3$. What have we learned? We can not cancel zero. And why is this? It's because zero does not have a multiplicative inverse. That is, there is no real number $x$ with $0x = 1$. That's why we can't cancel it.

Ok no problem, we just make a rule that you can't cancel zero. But if you think about it, what is division in the first place? Division is just another name for the cancellation rule. If you tell me that $x$ is nonzero and $ax = bx$ then I am allowed to conclude that $a = b$. But this is just regular old division.

In other words the cancellation law is division. We can cancel any number that has a multiplicative inverse. Therefore we can not cancel 0, because 0 does not have a multiplicative inverse. So we can't cancel zero ... which is just another way of saying that we can't divide by zero.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

I don't think a precise answer can be offered if the wording & language we use is not precise. Thus any attempt towards precise understanding is futile until all relevant terms are precisely defined.

FIRST possible approach: define '$\div$' as a binary operation over some subset of complex number ordered pairs were by definition $$a\div b = a \cdot (\text{the multiplicative inverse of } b)$$

it can be gracefully proven, using the standard field axioms, that for any complex number $b$, $$b\cdot 0 = 0$$ thus $$0\cdot x=1$$ has not solution in the complex field, therefore, 0 has no multiplicative inverse in the complex field, therefore "$\div$" can not be defined as such over pairs where the second number is zero, such subset must be excluded.

SECOND approach, use limits and look at numbers as equivalence classes of convergent sequences, in that case we can very well divide 0/0 meaning we can divide by sequences where each goes to zero as in $$\lim \frac{\sin(1/n)}{(1/n)} = 1$$ clearly in that case it is NOT true that we can not divide by zero

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.