Dividing by Zero

Okay, so everybody knows that anything multiplied by zero is zero. 6x0=0. what if you reverse that? 0/0=6. of course, all numbers do this. 3x0=0. so, 0/0=3. therefore, if zero divided by zero is every number, then should 0/0= {R} (all real numbers) or infinity?

0/0 should equal a set of all numbers, real or imaginary. Since that's meaningless and since that only works for dividing zero by zero, we just say it can't be done.

The division operator is defined on R × (R\\{0}). Division by zero is undefined.

It's a straight vertical line so of course the range will be all reals. This isn't really new.

Dividing anything by zero should equal infinity, but it doesn't,(partly) because you can't apply math to infinity.

No, not really. The limit of 1/0 is ∞ while the limit of -1/0 is -∞, and such. And you can too apply mathematics to infinity, it's just different.

Maybe I should rephrase that, you can apply math to infinity, you're just not going to get any meaningful results. Also, -∞ = ∞, because infinity is both positive and negative. ( So I'm still right. And how do you get the ∞ to come out? )

edited 27th Nov '10 5:04:41 AM by satannstuff

Like this: ∞

Also:

∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞
Check it out, it's a knotwork border for my post!
∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞

Edit: Dammit.

edited 27th Nov '10 5:04:36 AM by BlackWolfe

That's not very helpful, I just copied it from tzetze's post.

That's what I'd done, but putting it in a pothole converts it over. Which is why the "dammit" edit.

∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞
becomes
∞ over and over

edited 27th Nov '10 5:10:55 AM by BlackWolfe

Maybe I should rephrase that, you can apply math to infinity, you're just not going to get any meaningful results.

Oh sure you can. 2^ℵ₀ = ℵ₁, for example.

edited 27th Nov '10 10:14:01 AM by Tzetze

You really want to go there?

ℵ0 = ∞

ℵ1 = ∞

ℵ0 = ℵ1 = ∞

The whole idea that you can define sets of infinity is just plain wrong. Both sets of numbers are infinite, if ℵ0 is infinite, and ℵ1 is infinite, both must be the same size. Just because the numbers in ℵ1 occur more frequently than those in ℵ0 doesn't mean there are more of them. You can't even describe infinity in terms of size or numbers because it is all sizes and numbers at once.

( Nothing personal, by the way, it's just that this transfinite stuff really bugs me. )

edited 28th Nov '10 4:18:42 AM by satannstuff

Anything divided by zero is inherently a hyperbolic equation and thus not a mathematical function at the asymptote.

However the asymptote of such equations has real solutions even when divided by zero. Simply put, you have three solutions: ∞, -∞, and 0. All at the same time. Which is why things puke up an error on calculators and computers and teachers tell you don't bother calculating. It's the same answers no matter how the equation is designed.

You are wrong. There is no bijection between the set Z and the set 2^Z, therefore they have different cardinality.

∞ = ∞

No matter how you try to divide it up, it's still infinite, size doesn't apply to infinity, to think that it does is just naive.

Seriously, guys... I'm only taking algebra 1. I'm a freshman in high school.

Don't worry, you're not going to have to learn this in high school.

No matter how you try to divide it up, it's still infinite, size doesn't apply to infinity, to think that it does is just naive.

To think that it doesn't is denying about a century's worth of math. We can get useful results by recognizing different infinities.

A century's worth of math has failed to prove CH, if you want to try go right ahead, I'm not going anywhere.

The continuum hypothesis? If all infinities are equal as you insist, CH isn't even possible to formulate.

That's my point. ( Nice avatar by the way, that's like your third, this week? )

edited 2nd Dec '10 1:20:16 AM by satannstuff