The reason division by zero meme is often potrayed

[secret]

as a black hole

Some reference:
www.physicsforums.com/showthread.php?t=327845&page=1
knowyourmeme.com/memes/divide-by-zero

My opinion:
a/a means a number x such that x*a=1

Therefore 1/0 means a number x such that x*0=1

If we tried to put 1/0 into the number system and keeping everything else unchanged
I.e. 0*n=0 for every n=/=1/0

Then you will always guarentee to end up with
1=0

Thus the entire number system collasped into the trival ring {0}, which is a number system with only one element: 0
This number system is very very boring as no matter what operation you done on it, it does not change

Attempt to interpret this physically, this means everything in the universe collapsed into just one thing

Therefore if you want to define division by zero with everything else intact, you have to find a way so that 1=0 or similar never show up in your steps

[Qwerty]

Hi Secret.

So, I see this is about the central paradox surrounding the indeterminate number 0/0?

[secret]

If 0/0 is treated as an ordinary number, you will have it equal to every number at once (which is not desirable)

0/0 in the context of limits, is an indeterminate form, since lim x->0 (x^2/x) is not equal to lim x->0 (x/x) but shared the same form 0/0 when to plug x=0 into the limits. Thus the behavior of the limit is unknown (i.e. the value is unknown at that moment and the limit might not even exist) unless one apply something called L' Hospital rule to find it out

n/0, on the other hand, in real and complex numbers, is undefined as there exist no complex number x such that x*0=n (a consequence of 0*n=0)
(real numbers are within the complex numbers/real numbers are a subset of complex numbers)

On other mathematical systems, n/0 can be treated as infinity and -n/0 can be treated as -infinity, however the operation involving them is not as straight forward as the real and complex numbers

en.wikipedia.org/wiki/Division_by_zero

Below is a detailed resource explaining why n/0 is undefined and 0/0 is indeterminate. It also explain how does 0*n=0 arise in the reals
sciforums.com/showthread.php?t=110019

[Qwerty]

Hm, but what if you take the limit of it instead?

[secret]

You cannot take the limit of the expression f(x)/g(x) (where f(a)=g(a)=0) directly as x->a, else you will end up with 0/0, which as you have stated, is indeterminate

To take the limit of those expressions, you need to apply something called L' hospital rule, which means you have to differentiate f(x) and g(x) separately and repeately until you no longer get 0/0 when you plug x=a into lim f'(x)/g'(x)

If however after differentiation and you still get 0/0 or even (non zero expression)/0 it means the limit does not exist

[Elmach]

You don't necessarily need to apply L'Hopital's Rule when taking the limit of f(x)/g(x) to x->a with f(a)=g(a)=0. For instance, if f(x)=x^2 and g(x)=x, and we want to find the limit of the ratio as x goes to 0, you could alternatively show that the ratio is equal to x at all points except at a removable discontinuity, showing that f(x)/g(x) as x->0 is x as x-> 0, which is x.

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WHAT! Surely you jest. That is a lie. Wait, no, you are just contradicting yourself.

Say, take x^2/(1-cos(x)) as x -> 0. Both are zero at x=0, so take L'Hopital.

2x/sin(x) as x->0. Both are zero at x=0, but that doesn't mean the limit is undefined! It means you take L'Hopital again.

2/cos(x) at x->0, giving 2.

As a matter of fact, there is no way you can take L'Hopital and then have both the numerator and the denominator be the zero function. That either means you started with 0/0, or that you applied it with both being a constant function, which means you can't apply L'Hopital in the first place.

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Another thing, I rant a bit when people post incorrect math stuff.

[Qwerty]

The math -> X
My head -> ☺

[secret]

@generic D&D Player

It's good to point out mistakes rather than leaving it behind
That's how we learn