## Infinity

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### Infinity

I'm having trouble in math and that bastard 'infinity' is to blame.

The first of its many scandelous behaviors involve the value of ∞*0. To me, multiplication between A*B means "A sets of B", which is the same as "B sets of A". Thus, 0*A is no sets of A and yeilds the null set, or A*0 means A sets of zero, which, again yields the null set as its value. Now, infinity is a bit different in as much as its an uncountable number - which means shit gets wierd when you try to apply normal operators, like add one to it. But with the aforementioned definition of multiplication, we run into one of two cirucmstances which both appear to be pretty decisively zero.

In the first circumstance, we have 0*∞. This should literally read "Zero sets of infinity", or "No sets of infinity". In this case, infinity's uncountable and rule bending nature is redundant, because whatever monstrocity it it might have been they're now none of it - and for any number, real, uncountable, imaginary, or else, to have a value when it does not exist is a matter entirely more sinister. Clearly, 0*∞ has gotta be zero.

In the second circumstance, ∞*0, things get a bit messier. By my definition of multiplication, this means "An infinite number of null sets". We can't add up all sets of zero, since that would either take forever or be impossible. But, again, the actual quantity of null sets being added is redundant because we have the pleasure of knowing the only thing that's happening in this series: 0+0. Thus, it would be impossibe for the value of ∞*0 to be anything except zero.

But the god of this world is a cruel one, and nearly every math professor in my institution tells me this value is undefined, and I'm pretty confident the rest of modern mathmatics is backing them up on it. If I'm wrong (I don't want to say I am), its probably because my definition of multiplication is in error. If you believe this is true, then please supply the actual definition of multiplication before you correct me. Aside from that, who can help me?

The first of its many scandelous behaviors involve the value of ∞*0. To me, multiplication between A*B means "A sets of B", which is the same as "B sets of A". Thus, 0*A is no sets of A and yeilds the null set, or A*0 means A sets of zero, which, again yields the null set as its value. Now, infinity is a bit different in as much as its an uncountable number - which means shit gets wierd when you try to apply normal operators, like add one to it. But with the aforementioned definition of multiplication, we run into one of two cirucmstances which both appear to be pretty decisively zero.

In the first circumstance, we have 0*∞. This should literally read "Zero sets of infinity", or "No sets of infinity". In this case, infinity's uncountable and rule bending nature is redundant, because whatever monstrocity it it might have been they're now none of it - and for any number, real, uncountable, imaginary, or else, to have a value when it does not exist is a matter entirely more sinister. Clearly, 0*∞ has gotta be zero.

In the second circumstance, ∞*0, things get a bit messier. By my definition of multiplication, this means "An infinite number of null sets". We can't add up all sets of zero, since that would either take forever or be impossible. But, again, the actual quantity of null sets being added is redundant because we have the pleasure of knowing the only thing that's happening in this series: 0+0. Thus, it would be impossibe for the value of ∞*0 to be anything except zero.

But the god of this world is a cruel one, and nearly every math professor in my institution tells me this value is undefined, and I'm pretty confident the rest of modern mathmatics is backing them up on it. If I'm wrong (I don't want to say I am), its probably because my definition of multiplication is in error. If you believe this is true, then please supply the actual definition of multiplication before you correct me. Aside from that, who can help me?

Last edited by mwace on Thu Apr 05, 2007 1:35 am UTC, edited 2 times in total.

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Zero times infinity is what's called an indeterminate form. If something has that form, you can find the limit as it approaches the indeterminate value, and find where the discontinuity of the function is.

Typically with zero*infinity forms you need to rewrite them as zero/(1/infinity) aka zero/zero and use L'Hopital's rule. If you're not in calculus, then don't worry about it, just realize that it could be any value, and that's why it's called indeterminate.

For example, let f(x) = (1/x) * sin(x). As x approaches zero, you get an infinity*zero form. Graph the function and you can clearly see the limit as x approaches zero is actually one.

Typically with zero*infinity forms you need to rewrite them as zero/(1/infinity) aka zero/zero and use L'Hopital's rule. If you're not in calculus, then don't worry about it, just realize that it could be any value, and that's why it's called indeterminate.

For example, let f(x) = (1/x) * sin(x). As x approaches zero, you get an infinity*zero form. Graph the function and you can clearly see the limit as x approaches zero is actually one.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

EvanED wrote:It's like what's 0*apple? You could argue that it's zero, but it's sort of hard to fit that into the definition...

Haha, actually, I used the number 0*banana when arguing in support of my explanation of 0*∞. This is because *BIG-THESIS-MOMENT* the value of banana, apple, or infinity is redundant because their's either none of it, or theres the sum of that many null sets - which is gonna friggan be zero no matter what possible quantity of null sets, real or unreal, there is.

@ Warriorness: Now you're just chasing a divide by zero out of the problem, which can be done to pretty much anything. You identify the whole L'Hopital's rule thing as the "typical" procedure for multiplication, which, I might add doesn't seem particularly typical, consdering I don't ever divide by the reciprical whenever I want to multiply by two numbers. Is multiplication actually defined as dividing by the reciprical?

Last edited by mwace on Thu Apr 05, 2007 1:47 am UTC, edited 1 time in total.

mwace wrote:EvanED wrote:It's like what's 0*apple? You could argue that it's zero, but it's sort of hard to fit that into the definition...

Haha, actually, I used the number 0*banana when arguing in support of my explanation of 0*∞. This is becaues *BIG-THESIS-MOMENT* the value of banana, apple, or infinity is redundant because their's either none of it, or theres the sum of that many null sets - which is gonna friggan be zero no matter what possible quantity of null sets, real or unreal, there is.

But the problem is that it needs you to be either inconsistent or really weird. If I say what is 5*2, you can easily reduce that to 10. If I say what is 5*banana, you can't do anything with it... the best you can say is "5 bananas", which isn't any better than saying that 5*2 is 5 twos.

So saying that 0*banana is 0 requires saying that a*b is defined if b is "banana", but only if a equals 0.

Gelsamel wrote:5*a

What can I evaluate that as?

5a

5bananas.

Edit:

Similarly 0*a is 0.

How's that similar? Similarly would be 0*a is 0a, and 5*2 is 52.

Last edited by EvanED on Thu Apr 05, 2007 1:49 am UTC, edited 1 time in total.

Gelsamel wrote:5*a

What can I evaluate that as?

5a

5bananas.

Edit:

Similarly 0*a is 0.

Nonono, in 5*a, the value of a is pertinant and must be real. But in the special case 0*a, I'm arguing that the value of a becomes completly redundant. I think this is statement is going to apply to a lot of people who try to use the real numbers when explaining why 0*∞ is undefined.

Last edited by mwace on Thu Apr 05, 2007 1:53 am UTC, edited 1 time in total.

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5i

Edit: Or sqrt(-25)

Edit: Or sqrt(-25)

Last edited by Gelsamel on Thu Apr 05, 2007 1:57 am UTC, edited 1 time in total.

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Pick a definition of multiplication. You say "The first of its many scandelous behaviors involve the value of ∞*0. To me, multiplication between A*B means "A sets of B", which is the same as "B sets of A"", but your equivalence is no longer true.

If we replace infinity in your sentence with banana, we that "0 sets of apples" is the same as "apple sets of 0". But what's apple sets of zero? That sentence doesn't make any sense.

If you pick one direction, multiplication is no longer commutative; e.g. if you choose "A sets of B" then 0*banana makes sense, but banana*0 doesnt.

This means that if you want to keep standard arithmentic multiplication commutative, you need to do something like say "if 'a sets of b' makes sense, then that's the answer, otherwise it's 'be sets of a'."

Starting to see why your definition of multiplication is weird?

sqrt(-25)

Oh. Yea, I guess thats simplified. Shoulda seen that coming.

EvanED wrote:If you pick one direction, multiplication is no longer commutative

It sure as hell is cumulative! I'd like you to explain a circumstance in which A sets of B isn't B sets of A. Its like the difference between a 5 by 2 block and a 2 by 5 block. The only time in which things get a bit different is in 0*A, because in one circumstance you have a null set and in the other you have A null sets - which I'd like to point out is the same damn thing.

Last edited by mwace on Thu Apr 05, 2007 2:26 am UTC, edited 3 times in total.

EvanED wrote:"apple sets of 0". That sentence doesn't even make sense, at least IMO.

Yes, part of the sentance "Apple sets of 0" doesn't make sense, *BUT* the big point I'm trying to make here is that the part of the sentance that doesn't make sense is redundant because its just a strange quanitity of null sets, the sum of which cannot possibly ever not equal zero.

I gotta get some sleep, looking forward to replys from pros.

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mwace wrote:EvanED wrote:If you pick one direction, multiplication is no longer commutative

It sure as hell is cumulative! I'd like you to explain a circumstance in which A sets of B isn't B sets of A. Its like the difference between a 5 by 2 block and a 2 by 5 block. The only time in which things get a bit different is in 0*A, because in one circumstance you have a null set and in the other you have A null sets - which I'd like to point out the same damn thing.

And that's why math fucks up when you put the unmovable wall and the irresistible force together. It's indeterminate. You can't assign it a proper value because it'd be any number of values. Take these two examples:

Code: Select all

` 1 `

----- * (x-1) * 1337

(x-1)

Code: Select all

` 1 `

----- * (x-1) * 42

(x-1)

Now plug x = 1 into them. You get zero for that second term, and since 0*1337 is 0, and 0*42 is 0, they both simplify to the same thing. And they are both of the form infinity times zero. So they equal each other, right? Wrong. Look at it differently. You can cancel out the term (x-1), and therefore the values of each should approach 1337 and 42 respectively as x approaches one.

That's why forms like zero*inf, zero/zero, inf - inf, etcetera are called "indeterminate". They can be any number of terms depending on the situation. See Wikipedia for examples of other indeterminate forms.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

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mwace wrote:@ Warriorness: Now you're just chasing a divide by zero out of the problem, which can be done to pretty much anything. You identify the whole L'Hopital's rule thing as the "typical" procedure for multiplication, which, I might add doesn't seem particularly typical, consdering I don't ever divide by the reciprical whenever I want to multiply by two numbers. Is multiplication actually defined as dividing by the reciprical?

It's not, but as division can be defined as multiplying by the reciprocal, and the identity that the reciprocal of the reciprocal of x is equal to x (except in the special case where x = 0, in which case limits are necessary), you can prove that multiplication is the same as dividing by the reciprocal.

My point here is, when you see a zero*inf form, the generally accepted way to solve it (if you can't do some canceling or something) is to manipulate it into a zero/(1/inf) form which is the same as zero/zero, on which you can use L'Hopital's Rule.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

mwace wrote:EvanED wrote:"apple sets of 0". That sentence doesn't even make sense, at least IMO.

Yes, part of the sentance "Apple sets of 0" doesn't make sense, *BUT* the big point I'm trying to make here is that the part of the sentance that doesn't make sense is redundant because its just a strange quanitity of null sets, the sum of which cannot possibly ever not equal zero.

I disagree that this makes "apple sets of 0" make sense, but I don't think I can really argue that one.

I had another weirdness that it would introduce, but I forget what it was.

Remember that whatever you use for your definition of multiplication has to work for "banana" just as well as it does for "infinity".

warriorness wrote:My point here is, when you see a zero*inf form, the generally accepted way to solve it (if you can't do some canceling or something) is to manipulate it into a zero/(1/inf) form which is the same as zero/zero, on which you can use L'Hopital's Rule.

L'Hopital's rule is only applicable inside limits though.

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EvanED wrote:warriorness wrote:My point here is, when you see a zero*inf form, the generally accepted way to solve it (if you can't do some canceling or something) is to manipulate it into a zero/(1/inf) form which is the same as zero/zero, on which you can use L'Hopital's Rule.

L'Hopital's rule is only applicable inside limits though.

I know, and the reason I'm mentioning it is because using limits is the only way to represent the value of a function at a point where it's indeterminate. You can't evaluate it.

Division is the only basic arithmetic operation that's not closed over the set of real numbers, because of the number zero. You can't divide by zero. Multiplication IS closed over the set of real numbers, but infinity is not a number. It is a concept, a limit. It's essentially the same as one divided by zero. A concept, a limit. So (going from the reciprocal stuff from my last post), just as you can't divide by zero, you can't multiply by infinity.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

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### Re: Infinity

mwace wrote:Now, infinity is a bit different in as much as its an uncountable number...

Unless you're considering Georg Cantor's transfinite arithmetic, with a line drawn between countable and uncountable infinities. Then again, Cantor died in an insane asylum because he couldn't wrap his head around the equations he was getting.

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warriorness wrote:...but infinity is not a number. It is a concept, a limit. It's essentially the same as one divided by zero. A concept, a limit.

Which is exactly why the original question, why isn't 0*inf just 0, is just really simple to answer; multiplication isn't defined on infinity any more than it is on apples and bananas.

Mathematics is a language that is very clear and different than the way english is spoken. I believe it is your definition of multiplicationthat is giving you trouble. What class are you in and I will try to give you a definition based on that.

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EvanED wrote:warriorness wrote:...but infinity is not a number. It is a concept, a limit. It's essentially the same as one divided by zero. A concept, a limit.

Which is exactly why the original question, why isn't 0*inf just 0, is just really simple to answer; multiplication isn't defined on infinity any more than it is on apples and bananas.

It's not "just zero" because you actually CAN'T multiply by infinity. As infinity is not a number, but just a concept, you can't use it in basic arithmetic. More formally, the concept of infinity is not within the domains of the basic arithmetic operations.

SpitValve wrote:warriorness wrote:I know, and the reason I'm mentioning it is because using limits is the only way to represent the value of a function at a point where it's indeterminate. You can't evaluate it.

You can't replace 1/infinity with "limit of 1/x as x approaches infinity". They're different concepts.

But it's the only way to express it. You can't compute 1/infinity. It's like saying "what's the square root of banana?" The function is simply not applicable to the argument.

That's why we have limits. They make infinity make more sense; they let us express stuff that we previously wouldn't be able to express, such as the 1/x limit you just said.

EDIT: @sin2beta: We're talking about objects, not sets. (The Java programmer in me cries out in anguish.) The Empty Set, or Null Set, or {}, or the zero with a slash through it, is not a proper way of expressing an undefined or indeterminate form.

Last edited by warriorness on Thu Apr 05, 2007 4:08 am UTC, edited 1 time in total.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

I think it is good to point out that infinity is an extremely useful symbol. However it is not an element of the set of real number (or natural or rational,,,) therefore, axioms involving the real numbers do not apply. These axioms include a * the identity = a. Therefore, 1/infinity does not hold since you seem to be talking the axiom of inverses. 1/infinity does not make sense. People just have to agree on its value. Most of the time it is 0.

The limit as x approaches infinity of 1/x, is just analysis. It is just an approximation or it just describes the end behavior.

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sin2beta wrote:1/infinity does not make sense. People just have to agree on its value. Most of the time it is 0.

The limit as x approaches infinity of 1/x, is just analysis. It is just an approximation or it just describes the end behavior.

People don't agree on its value, as it doesn't have a value. You can't divide by infinity, as infinity is a concept, a limit. The point of it is for use in limits, hence taking the limit of 1/x as x->infinity. And taking the limit is our way of seeing how it behaves.

EvanED wrote:sin2beta wrote:The limit as x approaches infinity of 1/x, is just analysis. It is just an approximation or it just describes the end behavior.

There is no end behavior, because there is no end...

End behavior means "behavior TOWARDS" the end, or (assuming we're referring to the domain of a function f(x) ) more formally, the limit of f(x) as x->infinity.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

warriorness wrote:EvanED wrote:sin2beta wrote:The limit as x approaches infinity of 1/x, is just analysis. It is just an approximation or it just describes the end behavior.

There is no end behavior, because there is no end...

End behavior means "behavior TOWARDS" the end, or (assuming we're referring to the domain of a function f(x) ) more formally, the limit of f(x) as x->infinity.

Yeah, I'm a moron. I should put a filter on that will stop me from commenting after midnight. I keep saying this too, and never do...

warriorness wrote:sin2beta wrote:1/infinity does not make sense. People just have to agree on its value. Most of the time it is 0.

The limit as x approaches infinity of 1/x, is just analysis. It is just an approximation or it just describes the end behavior.

People don't agree on its value, as it doesn't have a value. You can't divide by infinity, as infinity is a concept, a limit. The point of it is for use in limits, hence taking the limit of 1/x as x->infinity. And taking the limit is our way of seeing how it behaves.

You will get little argument from me. I agree completely given the type of math you are doing. In analysis I completely agree. In fact, in general I would agree. However, many books will create an extended real number line (as infinity is clearly not included in the real number) where it is the reals adjoin infinity. Arithmetic properties associated with this will include x/infinity = 0. This was how it was defined in my abstract algebra book. Furthermore, 0* infinity was defined as indeterminate. (thought I'd add that to further answer the original post since I took time to look it up.)

Do these forums allow tex commands?

//////////

// Added later

//////////

One other thing that this brought up is if x/infinity = 0, it follows why 0*infinity is indeterminate

If x/infinity = 0 for all x then infinity * 0 = x for all x. However, when defined it was given that these are clearly not equivelent. Thus, 0*infinity is indeterminate given those properties (real adjoin infinity). (this may be worded very poorly, I questioned putting it up to avoid confusion)

Also, set theory defines infinity slightly different than just the limit. No upperbound would be more appropriate.

Last edited by sin2beta on Thu Apr 05, 2007 4:42 am UTC, edited 1 time in total.

warriorness wrote:Code: Select all

`1`

----- * (x-1) * 1337

(x-1)Code: Select all

`1`

----- * (x-1) * 42

(x-1)

Now plug x = 1 into them. You get zero for that second term, and since 0*1337 is 0, and 0*42 is 0, they both simplify to the same thing. And they are both of the form infinity times zero. So they equal each other, right? Wrong. Look at it differently. You can cancel out the term (x-1), and therefore the values of each should approach 1337 and 42 respectively as x approaches one.

Wrong. If you let x=1, it is unknown because it would make 1/0, which is unknown. So you could get either unknown or (1337 or 42). In this function, it never actually reaches 0 (0 is the asymptote). [I think I'm right.]

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You mean LaTeX? No, you probably have to bust out your LaTeX program, save an image, upload it, etc. phpBB supports BBCode ([b]bold[/b] etc) and smilies, and that's about it.

EDIT:

I know it's wrong to try to evaluate them at x = 1. The point of the demonstration was to show that you can't do this sort of thing without taking limits, otherwise you end up breaking math (1337 = 42). Also, those functions have a "hole" (aka removable) discontinuity at x = 1. No asymptotes (and the function is most certainly defined at x = 0 !). Aside from the hole, both functions are constant functions.

EDIT:

lewis1350 wrote:warriorness wrote:Code: Select all

`1`

----- * (x-1) * 1337

(x-1)Code: Select all

`1`

----- * (x-1) * 42

(x-1)

Now plug x = 1 into them. You get zero for that second term, and since 0*1337 is 0, and 0*42 is 0, they both simplify to the same thing. And they are both of the form infinity times zero. So they equal each other, right? Wrong. Look at it differently. You can cancel out the term (x-1), and therefore the values of each should approach 1337 and 42 respectively as x approaches one.

Wrong. If you let x=1, it is unknown because it would make 1/0, which is unknown. So you could get either unknown or (1337 or 42). In this function, it never actually reaches 0 (0 is the asymptote). [I think I'm right.]

I know it's wrong to try to evaluate them at x = 1. The point of the demonstration was to show that you can't do this sort of thing without taking limits, otherwise you end up breaking math (1337 = 42). Also, those functions have a "hole" (aka removable) discontinuity at x = 1. No asymptotes (and the function is most certainly defined at x = 0 !). Aside from the hole, both functions are constant functions.

Iluvatar wrote:Love: Gimme the frickin' API.

yy2bggggs, on Fischer Random chess wrote:Hmmm.... I wonder how how a hypermodern approach would work

lewis1350 wrote:warriorness wrote:Code: Select all

`1`

----- * (x-1) * 1337

(x-1)Code: Select all

`1`

----- * (x-1) * 42

(x-1)

Now plug x = 1 into them. You get zero for that second term, and since 0*1337 is 0, and 0*42 is 0, they both simplify to the same thing. And they are both of the form infinity times zero. So they equal each other, right? Wrong. Look at it differently. You can cancel out the term (x-1), and therefore the values of each should approach 1337 and 42 respectively as x approaches one.

Wrong. If you let x=1, it is unknown because it would make 1/0, which is unknown. So you could get either unknown or (1337 or 42). In this function, it never actually reaches 0 (0 is the asymptote). [I think I'm right.]

The main thing is the cancellation can only take place if it is assumed that x-1 =/= 0

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Though I doubt they will answer your question, you might want to look at

http://en.wikipedia.org/wiki/Real_projective_line

and

http://en.wikipedia.org/wiki/Extended_real_number_line

Edit: and to answer your question think about

inf*(1-1) and inf-inf

You would want the first to be 0, and I would want the second to be undefined (it could be anything at all). I also want them both to be equal, because expanding the brackets shouldn't change the result.

http://en.wikipedia.org/wiki/Real_projective_line

and

http://en.wikipedia.org/wiki/Extended_real_number_line

Edit: and to answer your question think about

inf*(1-1) and inf-inf

You would want the first to be 0, and I would want the second to be undefined (it could be anything at all). I also want them both to be equal, because expanding the brackets shouldn't change the result.

jestingrabbit wrote:Though I doubt they will answer your question, you might want to look at

http://en.wikipedia.org/wiki/Real_projective_line

and

http://en.wikipedia.org/wiki/Extended_real_number_line

Edit: and to answer your question think about

inf*(1-1) and inf-inf

You would want the first to be 0, and I would want the second to be undefined (it could be anything at all). I also want them both to be equal, because expanding the brackets shouldn't change the result.

Nice links and a very nice proof in the edit. Seems your links did answer the question regarding inf*0 = indeterminate

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