Reasons why division by zero is not infinity or it is infinity.

So, here is a fundamental question, division by zero is undefined, there are some ways to prove this. Lets take the following limit

$$\lim_{x \to 0^+} \frac{1}{x} = +\infty $$

but

$$\lim_{x \to 0^-} \frac{1}{x} = -\infty $$

So the limit does not exist, thus division by zero is undefined.

But let's go way more basic. What is division? Division can be though of as a number of succesive substractions until you reach zero or before you hit a negative number.

Say you want to divide 8/3, you start by

$$8-3 = 5$$

Do it once more

$$5-3 = 2$$

And then you stop, because if you substract 2-3 = -1 and we said we were going to stop at zero or before it got negative.

So how many substractions did we had to perform before we reached zero or a negative number? 2 of them, what number was left after the last substraction? also 2, so 8/3 is 2 with a remainder of 2. Simple as that.

Let's do the same but now lets divide 8/0

So we start

$$8-0=8$$

One more time

$$8-0 = 8$$

and again...

$$8 - 0 = 8$$

Do you see were I am going with this? So we can say, ok, substracting zero from eight will never reach zero so division by zero is not defined, or we can go crazy and count the number of times we can keep doing this and the result is infinity, so could we say that 8/0 is infinity with a remainder of 8?

Here are four different counter-examples/arguments to disprove what you said:

$1:$

Division by $0$ is literally impossible and you cannot use this operation. Why? Suppose $\frac{n}{0}=k$. This would result in $n=0$, which is a contradiction (I don't even want to talk about why $\frac{0}{0}$ doesn't exist)

$2:$

AssuminG you can indeed work with $\frac{k}{0}$, by the same algorithm, you could say that for any $n$, $8$ divided by $0$ is $n$ with a remainder of $8$, thus leading in a contradiction.

Why is this a contradiction? Well (in our case), division is a bijection, a function, from $\mathbb{N}$ to $\mathbb{Q}$ (actually from $\mathbb{N^*}$ but we assume dividing by $0$ is possible) and if $\frac{k}{0}$ could take multiple values, this would defeat the purpose of a function.

Remember, division was first and then people found out we cannot divide by $0$

$3:$

The algorithm you defined is not at all division. It is finding the integer part of $\frac{m}{n}$ where $m,n>0$. Just to put in evidence another flaw of your so called division, look at this:

What is $\frac{8}{-1}$? (it is $-8$)

Well $8-(-1)=9$. We need to do it again... $9-(-1)=10$ uh we need to do it again... $10-(-1)=11$ oh wait we never reach a negative number!

So does this imply that $-8=\frac{8}{-1}=+\infty$?

Just as another observation, people often belive that division is some sort of subtraction and multiplication is some sort of adittion. This is wrong, exactly because of $0$ (which led to people struggling with this in the ancient time), because of negative numbers, because of irrational numbers, because of complex numbers etc.

$4:$

Assuming you still belive in division by $0$, you don't care that division is a bijection and you still belive in your algorithm, then take a look at this:

$$\frac{1}{0}=\infty=\frac{2}{0}\Rightarrow\frac{1}{0}=\frac{2}{0}\Rightarrow\frac{1}{0}\cdot 0=\frac{2}{0}\cdot 0\Rightarrow 1=2$$

I am sorry if at this points your eyes are bleeding, reader