Why is $\frac n 0 =$ undefined but $ \frac 0 n = 0$ is not?

Question

Any number $n$ divided by $0$ is undefined. This is because if we approach this from both sides (assuming $n$ is positive) $$\lim_{x\to0^+}\frac{n}{x}=+\infty$$ but $$\lim_{x\to0^-}\frac{n}{x}=-\infty$$ Thus $\frac{n}{0}$ is indeterminate.

But what about $\frac{0}{n}$? It is often said that if one starts with $0$ and divides it into $n$ parts, then each part will also be $0$, thus preventing the expression from being undefined. However, with some algebra: $$\frac{0}{n}=0$$ $$0=0n$$ $$\frac{0}{0}=n$$ This causes us to end up with the indeterminate form $$\frac{0}{0}$$ again.

So why is $\frac{n}{0}$ undefined, while $\frac{0}{n}$ isn't?

Answer 1

Here is a more detailed breakdown of why $0$ has no multiplicative inverse in the real numbers (and thus why we can't divide by zero, see my comment). To see why $ab=ac$ implies $b=c$ when $a$ has a multiplicative inverse, note that dividing both sides by $a$ is the same thing as multiplying both sides by $\frac{1}{a}$ and using the fact that multiplication is associative. In the course of this answer I will also address why we don't just "artificially" add in a multiplicative inverse of $0$, as I occasionally see people asking why we can just "define" $i$ to be the square root of $-1$ but can't just "define" some number $\varepsilon$ to be $\frac{1}{0}$.

A multiplicative inverse of $a\in\mathbb{R}$ is an element $b\in\mathbb{R}$ such that $a b=1$. For $a\neq 0$, this is given by $\frac{1}{a}$. Since $0\bullet a=0$ for all $a \in\mathbb{R}$, it follows that there is no $b\in\mathbb{R}$ such that $0b=1$, thus $0$ has no multiplicative inverse in the real numbers.

Okay, suppose now we want to "artifically" add in a multiplicative inverse for $0$, that is, we define $\varepsilon$ to be a quantity such that $0\varepsilon=1$ while keeping all the familiar rules of the real numbers (i.e. the distributive property). With this $\varepsilon$, we have: $$ 2=0\varepsilon+0\varepsilon=(0+0)\varepsilon=0\varepsilon=1$$ Which is a contradiction! Thus we cannot just "define" some $\varepsilon$ to be the multiplicative inverse of $0$ while keeping all the same properties of the real numbers.

Okay, let's keep playing this game then. To "force" there to be a multiplicative inverse of $0$, we have two options: either get rid of the distributive property or demand that $1=2$. The first option I don't know too much about, so I won't go into an explanation here, but I believe Wheel Theory may be related to this. For the other option, I will go into some basic Ring Theory.

A ring is something where you can add, subtract, and multiply. More formally, a ring is a set $R$ with two associative operations $\times$ and $+$ such that the following conditions hold. There exists $0\in R$ such that $0+a=a$ for any $a\in R$, there exists $e\in R$ such that $e a=a$ for any $a\in R$, and for each $a \in R$ there exists $b\in R$ with $a+b=0$. We also require that $+$ is commutative and that $a(b+c)=a b+a c$ and $(a+b)c=a c+b c$ for all $a,b,c \in R$. From these it follows that for any $a\in R$, $0\times a=(0+0)\times a=0\times a+0\times a$, thus by adding the additive inverse of $0\times a$ to both sides (call this element $-0\times a$), we have from associativity: $$0=0\times a+(-0\times a)=(0\times a+0\times a)+(-0\times a)=0\times a+(0\times a+(-0\times a))=0\times a+0=0\times a$$ Thus anything times zero is zero in a ring! We say that $a\in R$ has a multiplicative inverse if there exists some $b\in R$ such that $a \times b=b \times a=e$. Suppose now that $R$ has a multiplicative inverse of $0$, that is, there exist some $c\in R$ such that $0 \times c=c \times 0=e$. By our previous claim, $0\times c=0$, thus $0=e$, so for any $a\in R$, $a=a\times e=a\times 0=0$, thus the only element of $R$ is $0$! We see then that if we force there to be a multiplicative inverse of zero, then the resulting ring is what's called trivial, that is, the ring only has the zero element. But this ring is not particularly interesting, hence why mathematicians generally don't artificially make it so you can divide by zero. On the other hand, there are ways of defining rings that contain $\sqrt{-1}$ that don't result in the trivial ring, hence why the complex numbers are still interseting to study.

Answer 2

This is a more entry-level Calc-1 answer.

First of all, you are confusing an undefined operation with a limit that does not exist (normally abbreviated "DNE") and with an indeterminate form of a limit. These three concepts are totally different and should not be confused.

To begin with, forget about limits.

Division by zero is undefined because... it is not defined. Nobody defined what $\frac{n}{0}$ is. (Well, several people tried to but mathematicians at large didn't agree).

There are good reasons for that. Starting with a naive one, what does it even mean to divide 6 chocolates among 0 friends? You can think about dividing 6 chocolates among 6 friends (1 each), 3 friends (2 each) or 18 friends (1/3 chocolate each). You can even conceptualize dividing zero chocolates among 5 friends: Each of them receives zero chocolates (hence 0/5 is 0). But dividing something among zero people doesn't make sense. Hence, there is no accepted definition for dividing by zero. For more mathematically strict reasons, see the previous comment.

Now, going into limits... Remember, when you calculate the limit of a function of x as x tends to zero you are asking what happens to the function as you get closer and closer to zero, but not all the way to zero.

That means that when you evaluate $\lim \limits_{x \to 0} \frac{n}{x}$, $x$ may be as close to $0$ as you want but never $0$ itself. So you are not dividing by zero.

The reason why $\lim \limits_{x \to 0} \frac{n}{x}$ DNE (does not exist) (NOT "it's not defined") has nothing to do with dividing by zero, and everything to do with the fact that when you approach the limit from the left (x= -1, -0.1, -0.01, -0.000001) the function tends to minus infinity, while when you approach the function from the right (x= 1, 0.1, 0.01, 0.00001) the function tends to plus infinity. By definition (or rather as a corollary of the epsilon-delta definition of limit), the limit will only exist if the limit from the left approaches the same value as the limit from the right.

As a counter-example, take $\lim \limits_{x \to 0} \frac{n}{x^2}$.

You are still dividing by something that tends to zero, but this limit does indeed exist and it is plus infinity (assuming that n is positive), because now the denominator is always positive so the quotient is always positive as x approaches zero from either side.

The indeterminate form of a limit is just a limit that you can't know "at first sight" what the result will be, if it happens to exist at all. It just means that you have to keep working to see if it exists and what it is, or if it does not exist. A classic example is $\lim \limits_{x \to 0} \frac{\sin x}{x}$. Both $\sin x$ and $x$ tend to zero when x tends to zero. Furthermore, both $\sin x$ and $x$ are positive for positive values of x close to zero, and negative for negative values of x close to zero. That means that the limit can be plus infinity if the denominator shrinks at a rate orders of magnitude faster, 0 if the numerator shrinks at a rate orders of magnitude faster, a given number if the numerator and denominator shrink at a similar rate, or it may even not exist if the relative shrink rate of the numerator and denominator is different as we approach zero from the left or from the right. We just need more work, and more work indeed will show that this limit exists and it is equal to 1.

Now, 0/0, not the limit, but the actual division, is not indeterminate. It is undefined because, again, dividing by zero is not defined.

So the last part of your argument actually has several mistakes:

0/n=0... so far so good

0=0n... still good, you just multiplied both sides by n

0/0=n is not good. You just tried to divide both sides by zero, but you can't divide 0n by zero, or 0 by 0 because, again you can't divide by zero, it is not defined.

And finally, when you said 0/0 is indeterminate, no it is not. It is very well determined that dividing by zero is not defined, so it is not indeterminate but undefined.

Again, don't confuse that with having a LIMIT where both the numerator and the denominator TEND to zero, which means that they are NOT actually zero. They can be as close to zero as you want, but they cannot reach all the way to actual zero.

So, finally, as a summary, and risking being repetitive, remember:

$\frac{n}{0}$ is UDEFINED, because division by zero is not defined

$\lim \limits_{x \to 0} \frac{n}{x}$ DNE (does not exist). You are not actually dividing by zero so it is not undefined. The limit of that function when x approaches to zero does not exist because when you approach from the left the limit does exist but you get a different result than when you approach from the right, where the limit also exists. Other limits where the denominator tends to zero do exist.

$\lim \limits_{x \to 0} \frac{\sin x}{x}$ is neither undefined nor "does not exist (DNE)". And it's not 0/0 either. The numerator tends to zero and the denominator tends to zero too so we call it a 0/0 form, but that's just a name, we are not actually dividing 0/0. We say such limit has and UNDETERMINED FORM because you cannot apply the simple rules for when you divide by zero to tell if the limit exists and what it is if it does. You need more work (use the definition of limit, or use the L'Hopital's rule to see if the limit exists or not and, if it does, what it is (in this case it exists and it is 1).

Undefined, "does not exist (DNE)", and "undetermined form". Three very different concepts that must not be confused.