1 Divided by zero

Originally posted by shiv
facepalm

Originally posted by King Kandy
That's because both 1/0 and 0*∞ are undefined values, as explained. So no, nothing in that equation works.

Originally posted by lord xyz
If
x/0 = ∞
then
x/∞ = 0
and then 0*∞=x

But x is any number.

So what the ****?

Best explanation I ever seen of why divide by zero doesn't work.

Originally posted by lord xyz
My point.

Originally posted by King Kandy
But that "proof" proves nothing whatsoever. All it "proves" is that x is equal to 0*infinity, while being equal to any number. However since 0*infinity is undefined, x still equals any number in that scenario.

I'm merely explaining why you can't divide by 0.

its undifined because we are not smart enough to conceive it, but theres very well an answer. we might discover it in the future perhaps

Why do we need to find 1/0?

Do we ever use it?

You can't bilaterally define 1/0, provided that 0 is a real number. If you want to define 1/0 you need to do so unilaterally. If you approach 0 with positive numbers, it would give plus infinity, otherwise, it is minus infinity.

On the other hand, if 0 is a matrix, there could be a non-zero matrix whose multiplication with another non-zero matrix would equal 0.

Originally posted by Colossus-Big C
its undifined because we are not smart enough to conceive it, but theres very well an answer. we might discover it in the future perhaps

On the subject.

0^2/0^1=0^1 I'm sure no one can disagree with this.

but 0^2 and 0^1 both equal 0, since x^1=x and 0^2=0.

So, 0/0=0

and if that's the case, then 0^1/0^1=0^0=0

Even though x^0=1 for any other number. (Zero isn't really a number anyway)

So then, 0^0/0^1=0^-1
0/0=0
0^-1=1/0
1/0 therefore must equal 0.

And that, can go for any number, since 2/0 would equal 0 as well, because double 0 is 0.

This contradicts all 1/x graphs that show an increase the closer x goes to 0.

But my point is, this is another reason why it's undefined.

Originally posted by lord xyz
So then, 0^0/0^1=0^-1

no, it equals "undefined."

0^-1=undefined.

Also, you should simplify before the operation so you wouldn't even end up with 0^-1. Remember PEMDAS.

I honestly thought my closing statement of the post would prevent another person telling me it's undefined.

PEMDAS?

We did BODMAS in like key stage 3, but what is PEMDAS?

power, exponential, multi, divide, add, subtract?

what about brackets?

And why does it apply?

Originally posted by lord xyz
I honestly thought my closing statement of the post would prevent another person telling me it's undefined.

You can replace it with whatever you want. I figured I'd correct you right there, as any continuation beyond that point would not be logical.

Originally posted by lord xyz
PEMDAS?

We did BODMAS in like key stage 3, but what is PEMDAS?

power, exponential, multi, divide, add, subtract?

what about brackets?

And why does it apply?

Parans Expons Multi Divid Add Sub

It applies because you need to simply your expons before you divide. Basic algebra stuff.

Originally posted by lord xyz
On the subject.

0^2/0^1=0^1 I'm sure no one can disagree with this.

but 0^2 and 0^1 both equal 0, since x^1=x and 0^2=0.

So, 0/0=0

and if that's the case, then 0^1/0^1=0^0=0

Even though x^0=1 for any other number. (Zero isn't really a number anyway)

So then, 0^0/0^1=0^-1
0/0=0
0^-1=1/0
1/0 therefore must equal 0.

And that, can go for any number, since 2/0 would equal 0 as well, because double 0 is 0.

This contradicts all 1/x graphs that show an increase the closer x goes to 0.

But my point is, this is another reason why it's undefined.

a/x, where a is a number that is not 0, is thought to equal in infinty in calc. Just like a/infinity, where a is not infinity, is thought to be 0. This seems to be supported by the graph of a/x.

or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number.

Originally posted by ares834
a/x, where a is a number that is not 0, is thought to equal in infinty in calc. Just like a/infinity, where a is not infinity, is thought to be 0. This seems to be supported by the graph of a/x.

or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number.

Originally posted by ares834
a/x, where a is a number that is not 0, is thought to equal in infinty in calc. Just like a/infinity, where a is not infinity, is thought to be 0. This seems to be supported by the graph of a/x.

or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number.

Originally posted by dadudemon
You can replace it with whatever you want. I figured I'd correct you right there, as any continuation beyond that point would not be logical.

Well, I can understand what he's trying to do, it's a bit like a proof by contradiction. Though there's of course some issues with it.

Originally posted by Colossus-Big C
so your agreeing with me that 1/0=infinity
0=nothingness
theres an infinit amount of nothingness therefore 0 is also infinity
1/0=infinity
infinity*1= 0(which is also infinity)

No. 1/0 is not defined, it is not infinity.

Originally posted by King Kandy
No, not quite correct. Because, approaching the left, x=negative infinity. So the limit does not exist because left and right bound limits don't agree.

I to think that undifined is correct, as x approaches 0 in a/x the solution begins to go toward either infinity or -infinity.

1/x integrated is in theory, 1/0 (x^-1 becomes (x^0)/0)

but in reality it's ln(x).

I don't know what to make of this.