# 1 Divided by zero

Colossus-Big C
should 1 divided by zero=infinity? because nothingness can fit into somethingness an infinit number of times correct?

Discuss

King Kandy
1 divided by 0 is undefined.

Symmetric Chaos
The answer depends on the sort of math your doing. In your everyday experience it's simply undefined, if you ever go into higher mathematics there are a ways of defining what the answer is but these involve things like taking a number and psuedodividing it by its psuedoinverse.

Colossus-Big C
Originally posted by Symmetric Chaos
The answer depends on the sort of math your doing. In your everyday experience it's simply undefined, if you ever go into higher mathematics there are a ways of defining what the answer is but these involve things like taking a number and psuedodividing it by its psuedoinverse. yes, so by doing that you get infinity?

King Kandy
Originally posted by Symmetric Chaos
The answer depends on the sort of math your doing. In your everyday experience it's simply undefined, if you ever go into higher mathematics there are a ways of defining what the answer is but these involve things like taking a number and psuedodividing it by its psuedoinverse.
You can define it, but only if it's attached to a function of some sort... just a straight out 1/0 is always undefined.

King Kandy
Originally posted by Colossus-Big C
yes, so by doing that you get infinity?
No, you typically get some mundane number like 3/4 or 0 if you define it using calculus. It depends on what function it's part of.

Colossus-Big C
im assuming a(1) reality has infinit space,
so basically "1" can contain infinity

Symmetric Chaos
Originally posted by Colossus-Big C
yes, so by doing that you get infinity?

No, the answers vary. In linear algebra the answer happens to be zero.

King Kandy
http://img714.imageshack.us/img714/3883/codecogseqn2.gif

That's probably the closest you'll get to anything supporting your point. However, to be clear about this, that is NOT the same thing as 1/0=infinity.

shiv
Originally posted by Colossus-Big C
because nothingness can fit into somethingness

Discuss Colossus-Big C
Originally posted by shiv Explain

King Kandy
How is it you have taken such a supposedly advanced astronomy class, but haven't taken calculus?

Colossus-Big C
Originally posted by King Kandy
How is it you have taken such a supposedly advanced astronomy class, but haven't taken calculus? i took it and im telling you im Lucky i some how passed.....

King Kandy
Then why are you asking this question, half the stuff you do in derivative based calculus is figuring out how to compute divide by zero problems.

shiv
Originally posted by Colossus-Big C
Explain
facepalm

The Butcher
You can't divide by zero,there is no way.

Colossus-Big C
Originally posted by The Butcher
You can't divide by zero,there is no way. you missed the whole point Lord Lucien
The useless cases some people make are ridiculous. Essentially it's our verbiage enabling philosophical discussion.

lord xyz
If
x/0 = &#8734;
then
x/&#8734; = 0
and then 0*&#8734;=x

But x is any number.

So what the ****?

King Kandy
That's because both 1/0 and 0*&#8734; are undefined values, as explained. So no, nothing in that equation works.

Colossus-Big C
Originally posted by shiv
facepalm facepalm

lord xyz
Originally posted by King Kandy
That's because both 1/0 and 0*&#8734; are undefined values, as explained. So no, nothing in that equation works. My point.

Symmetric Chaos
Originally posted by lord xyz
If
x/0 = &#8734;
then
x/&#8734; = 0
and then 0*&#8734;=x

But x is any number.

So what the ****?

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

King Kandy
Originally posted by lord xyz
My point.
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.

lord xyz
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. It is undefined because of the evidence in that post.

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

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

lord xyz
Why do we need to find 1/0?

Do we ever use it?

FE Expert
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.

King Kandy
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
No, it's undefined because it has been mathematically proven to be so. That's the difference between math and, say, physics. It's possible to have absolute proof in math.

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.

dadudemon
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.

lord xyz
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?

dadudemon
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.

King Kandy
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.
Wrong in oh so many ways. Just because you got the correct answer, does not mean your proof was logically valid in any way. 0^2/0^1=/=0^1

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.
http://library.thinkquest.org/2647/media/odd1ox.gif
or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number.

Colossus-Big C
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.
http://library.thinkquest.org/2647/media/odd1ox.gif
or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number. 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)

King Kandy
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.
http://library.thinkquest.org/2647/media/odd1ox.gif
or as King Kandy showed us by limits.
0/0 is even stranger, as 0/0 could, theoretically be any number.
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.

Bardock42
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.

ares834
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.
Very true. It's been a while since I've used limits.

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

lord xyz
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.

King Kandy
Originally posted by lord xyz
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.
Where in the world did you get that notion? the antiderivative of X^-1 would just be X^0, or zero. Not X^0/0. However because all constants derive to zero, x^0 does as well, so ln(x) is the integral.

lord xyz
x^0=1 for all numbers but 0.

Bardock42
Kid's right.

Anyways, this explains why the derivative of ln(x) is 1/x

http://www.ltcconline.net/greenl/courses/116/ExpLog/logDerivative.htm

King Kandy
Originally posted by lord xyz
x^0=1 for all numbers but 0.
That was a typo, I meant to say X^0 or ONE, not X^0 or zero.

lord xyz
The way I was taught calculus was the rule for integrating:

add one to the power, then divide by the power.

differentiation is the inverse.

King Kandy
That somewhat works (however not in this situation, the dividing part is to cancel coefficients which doesn't apply here), but that doesn't work for 1/x, because even though you could conceivably see the integral as 0, the derivative of 0 is most certainly not 1/x.

Bardock42
Originally posted by lord xyz
The way I was taught calculus was the rule for integrating:

add one to the power, then divide by the power.

differentiation is the inverse.

Yeah, that's not a general rule though. Just a convenient way to find many integrals.