Total posts:
[21]

1
Hnnng

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?

Above all, always remember to stay positive.

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.But soft! What rock through yonder window breaks? It is a brick! And Juliet is out cold.

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.

^{5}satannstuff26th Nov 2010 06:09:16 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

DUMB

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.

^{7}satannstuff27th Nov 2010 04:57:51 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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!

∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞

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

∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞

**Edit:**Dammit.edited 27th Nov '10 5:04:36 AM by BlackWolfe

But soft! What rock through yonder window breaks? It is a brick! And Juliet is out cold.

^{9}satannstuff27th Nov 2010 05:06:14 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

becomes

∞ over and over

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

But soft! What rock through yonder window breaks? It is a brick! And Juliet is out cold.

DUMB

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}= ℵ

_{1}, for example.

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

^{12}satannstuff28th Nov 2010 04:14:23 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

Eye'm the cutest!

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.

"Allah may guide their bullets, but Jesus helps those who aim down the sights."

You are wrong. There is no bijection between the set

**Z**and the set 2^{Z}, therefore they have different cardinality.^{15}satannstuff29th Nov 2010 01:40:37 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

∞ = ∞
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.

Hnnng

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

Above all, always remember to stay positive.

^{17}satannstuff30th Nov 2010 01:02:19 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

DUMB

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.

^{19}satannstuff1st Dec 2010 01:02:55 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

DUMB

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

^{21}satannstuff2nd Dec 2010 01:18:35 AM from squishy parts unknown , Relationship Status: I like big bots and I can not lie

the devil you know

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

The system doesn't know you right now, so no post button for you.

You need to Get Known to get one of those.

You need to Get Known to get one of those.

Total posts: 21

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