Possible to Multiply or Divide Infinities?

[Raza]

Hi all,
Some people are arguing that it is possible to divide, multiply, add, subtract infinities, but I believe that impossible.

Am I right or wrong?

Thank you.

 

[rbj]

fundamentally, we add, subtract, multiply, and divide numbers. \infty is a mathematical concept, but is not a number.

but, sometimes it can be meaningful to discuss these operations acting on numbers where one or more numbers are increasing without bound ("going to infinity"). for example, dividing a finite constant by such a number that is increasing without bound would result in something getting closer and closer to zero (to within any given \epsilon > 0).

\lim_{n \rightarrow \infty} \frac{1}{n} = 0

sometimes is stated crudely as

\frac{1}{\infty} = 0 .

while this sematic is icky, it does have meaning. but the following don't:

(+\infty) + (-\infty) = ??

or

\frac{\infty}{\infty} = ??

 

[morphism]

It depends on what you mean by "infinities". Try looking up "cardinal arithmetic", and see if that helps you.

 

[Raza]

Yes, I know that you can use limits to know where they are approaching. Explicitly put, there is no answer for:
\infty + \infty= Undefined
\infty - \infty= Undefined
\infty \times \infty= Undefined
\infty \div \infty= Undefined

 

[sutupidmath]

Yes, I know that you can use limits to know where they are approaching. Explicitly put, there is no answer for:
\infty + \infty= Undefined
\infty - \infty= Undefined
\infty \times \infty= Undefined
\infty \div \infty= Undefined

Like it was said arithmetic operations like, +,-, /, *, etc are applied with numbers, but \infty is not a number so you cannot impose these operations when one deals with infinity. however\infty+\infty=\infty this basically is written this way to state the fact that the sum of any two values that increases without bound is still a value that increases without bound. also \infty*\infty=\infty states that the product of two such quantities that increase without bound is still a quantity that increases without bound. THis is what it implies. While \frac{\infty}{\infty} is undefined for the fact that we have one quantity that increases without bound over another quantity that increases without bound, we do not know which one of them grows faster, so it is undefined. Also, the quantity on the denominator might represent say the total number of points in a line, while the quantity in the numerator the total number of points in a plane, so we have two such quantities which quotient we do not actually know. that's why it is undefined. etc...

 

[andytoh]

If a and b are two infinite cardinals, then ab = max{a,b} and a+b = max{a,b}, assuming the Axiom of Choice.

 

[Dragonfall]

I'm pretty sure you can define division on cardinals, though it would be pretty trivial in most cases. a / b = c if c is the smallest cardinal such that a can be written as a disjoint union Ub of c-sized sets. Or something to that effect.

 

[tiny-tim]

Hi Raza!

For multiplying by infinity, have a look at http://en.wikipedia.org/wiki/Dirac_delta_function

 

[HallsofIvy]

In other words, infinity is not a real number and operations on the real numbers to not apply to infinity. There are other ways to define number systems that include "infinity", such as the cardinals, mentioned above, in which you can do arithmetic on "infinities" but they do not, in general, have the same properties as arithmetic operations on the real numbers.

tiny-tim, I see no mention of "multiplying by infinity" on the page. And while physicists mgiht think of the delta function as involving "infinity", no mathematician would!

 

[tiny-tim]

tiny-tim, I see no mention of "multiplying by infinity" on the page. And while physicists mgiht think of the delta function as involving "infinity", no mathematician would!

Hi HI!
http://en.wikipedia.org/wiki/Dirac_delta_function:

Informally, it is a function representing an infinitely sharp peak bounding unit area: a function δ(x) that has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1.

Strictly speaking.of course, the delta function is not a function, but something called a distribution.

Its use is essential in evaluating the very complicated multiple integrals in (perturbative) Quantum Field Theory.

Whether one regards \delta(0) as being infinite or as being meaningless is a matter of taste - and my taste is to prefer some meaning to no meaning!

I admit this is a trick - but I think Raza knows that you can't really multiply by infinity without some trick being involved!

 

[Hurkyl]

Hi all,
Some people are arguing that it is possible to divide, multiply, add, subtract infinities, but I believe that impossible.

In all seriousness, it depends on precisely what meaning you choose for each of the words "divide", "multiply", "add", "subtract", and "infinity".

 

[HallsofIvy]

Informally, it is a function representing an infinitely sharp peak bounding unit area: a function δ(x) that has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1.

I still see nothing about "multiplying by infinity".

 

[tiny-tim]

I still see nothing about "multiplying by infinity".

Ah, but lower down, the delta-function gets multiplied by other functions and integrated.

And the whole is equal to the sum of its parts, so if you multiply the whole, you must multiply each bit …

As Hurkyl sagely says:

In all seriousness, it depends on precisely what meaning you choose for each of the words "divide", "multiply", "add", "subtract", and "infinity".

 

[ice109]

I'm pretty sure you can define division on cardinals, though it would be pretty trivial in most cases. a / b = c if c is the smallest cardinal such that a can be written as a disjoint union Ub of c-sized sets. Or something to that effect.

yea that definitely wouldn't work for infinite sets

 

[Dragonfall]

Why not?

 

[ice109]

Why not?

meh i guess it would but it would be the trivial a/b = a for all infinite sets

 

[Dragonfall]

No, it may be for all regular infinite cardinals. For singular cardinals, it is non-trivial.

 

[rbj]

Strictly speaking.of course, the delta function is not a function, but something called a distribution.

Its use is essential in evaluating the very complicated multiple integrals in (perturbative) Quantum Field Theory.

Whether one regards \delta(0) as being infinite or as being meaningless is a matter of taste - and my taste is to prefer some meaning to no meaning!

better watch out, tiny. them's maths guys take this pretty seriously and will beat you up. they beat me up (or at least beat on me a bit) for defending the Neanderthal engineering POV of \delta(t). i can't remember who was involved, maybe wonk was.

 

[tiny-tim]

better watch out, tiny. them's maths guys take this pretty seriously and will beat you up.

I'm not afraid.

Physics guys 'n' gals can use force. Maths guys can only rearrange the coordinates.

 

[CRGreathouse]

I'm pretty sure you can define division on cardinals, though it would be pretty trivial in most cases. a / b = c if c is the smallest cardinal such that a can be written as a disjoint union Ub of c-sized sets. Or something to that effect.

Of course if you don't have AC that's not well-defined. What about just using the indirect image? Then you can have
\aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\{\aleph_0\}
\aleph_0 / \mathfrak{c} = \emptyset

 

[Hurkyl]

my taste is to prefer some meaning to no meaning!

Oh good, then maybe you can do something with

Flower accomplished bash sine related category inflict.​

(courtesy of http://www.zokutou.co.uk/randomword/ )

 

[Dragonfall]

Of course if you don't have AC that's not well-defined. What about just using the indirect image? Then you can have
\aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\{\aleph_0\}
\aleph_0 / \mathfrak{c} = \emptyset

Mmm... I'm not sure what you mean by "indirect image".

 

[tiny-tim]

Oh good, then maybe you can do something with

Flower accomplished bash sine related category inflict.​

No … should be cosine! ​

Try again!

 

[CRGreathouse]

Mmm... I'm not sure what you mean by "indirect image".

Preimage? Inverse image? What do you call the set f^{-1}(X)=Y with y\in Y \Leftrightarrow f(y)\in X?

 

[sutupidmath]

Preimage? Inverse image? What do you call the set f^{-1}(X)=Y with y\in Y \Leftrightarrow f(y)\in X?

Simply the image!

 

[Dragonfall]

Yup, that looks like the image. But of what function?

 

[CRGreathouse]

Yup, that looks like the image. But of what function?

The original question was about division by cardinals. I suggested using the image as division. That is, take a / b as the set {n: b * n = a} for cardinals a and b. So with that definition, \aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\aleph_0, for example, with \mathbb{Z}^+=\{1,2,3,\ldots\}. An alternate definition in ZFC would take the least of these, so \aleph_0 / \aleph_0 = 1 in that case. Neither could 'handle' \aleph_0 / \mathfrak{c}; the first would give the empty set and the latter would be undefined.

 

[Hurkyl]

No … should be cosine! ​

Try again!

As you might have guessed, the point is that it's silly to think everything you can write has a meaning.


The raison d'être of a distribution is to be convolved with a test function to produce a number. The thing that makes interesting distributions possible is that you can define them by how they convolve. Even the boring distributions that are representable by ordinary functions are still only defined in this way -- you can change the representing function on any set of measure zero without changing which distribution it represents.

So if your intuition that it makes sense to evaluate a distribution at a point... then your intuition is (probably) wrong.

 

[ice109]

The original question was about division by cardinals. I suggested using the image as division. That is, take a / b as the set {n: b * n = a} for cardinals a and b. So with that definition, \aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\aleph_0, for example, with \mathbb{Z}^+=\{1,2,3,\ldots\}. An alternate definition in ZFC would take the least of these, so \aleph_0 / \aleph_0 = 1 in that case. Neither could 'handle' \aleph_0 / \mathfrak{c}; the first would give the empty set and the latter would be undefined.

you mean the cardinality of the set " {n: b * n = a} " ? and what is *? the cartesian product?

 

[wildman]

In other words, infinity is not a real number and operations on the real numbers to not apply to infinity. There are other ways to define number systems that include "infinity", such as the cardinals, mentioned above, in which you can do arithmetic on "infinities" but they do not, in general, have the same properties as arithmetic operations on the real numbers.

tiny-tim, I see no mention of "multiplying by infinity" on the page. And while physicists mgiht think of the delta function as involving "infinity", no mathematician would!

Huh? Almost all real numbers are irrationals and they are infinite decimals. How do you define real numbers?

And why can't you divide two infinite numbers? If you divide .6666... by .22222... would you not get .33333...? An irrational would take infinitely long to divide by another irrational, but in principle there is no reason it can't be done. Is there?

 

[ice109]

Huh? Almost all real numbers are irrationals and they are infinite decimals. How do you define real numbers?

And why can't you divide two infinite numbers? If you divide .6666... by .22222... would you not get .33333...? An irrational would take infinitely long to divide by another irrational, but in principle there is no reason it can't be done. Is there?

repeating decimals aren't irrational and there's a difference between a number who's magnitude is infinite and a number who's representation is infinite.

 

[tiny-tim]

So if your intuition that it makes sense to evaluate a distribution at a point... then your intuition is (probably) wrong.

Hi Hurkyl!

My intuition is evaluated as a distribution.

It is worthless almost everywhere, but becomes of value when I get to the point.

(On its own, it is meaningless, but the more convolved it gets … )

 

[Hurkyl]

and what is *? the cartesian product?

More or less. Cardinal arithmetic is defined by

|A| + |B| = |A \amalg B|
|A| \cdot |B| = |A \times B|
|A|^{|B|} = \left|A^B\right|

(\amalg is disjoint union)

 

[HallsofIvy]

Huh? Almost all real numbers are irrationals and they are infinite decimals. How do you define real numbers?

And why can't you divide two infinite numbers? If you divide .6666... by .22222... would you not get .33333...? An irrational would take infinitely long to divide by another irrational, but in principle there is no reason it can't be done. Is there?

The numbers you give, 0.666..., 0.222..., and 0.333... have an infinite number of decimal places. The number of decimal places is an artifact of the base 10 numeration system and not a property of the numbers themselves. They are not "infinite" themselves. There are no "infinite" real numbers whether rational or irrational.

And I know several ways of defining "real numbers"- Dedekind cuts, equivalence classes of Cauchy sequences, equivalence classes of increasing, bounded sequences, etc. In none of those are there "infinite" real numbers.

 

[Dragonfall]

The original question was about division by cardinals. I suggested using the image as division. That is, take a / b as the set {n: b * n = a} for cardinals a and b. So with that definition, \aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\aleph_0, for example, with \mathbb{Z}^+=\{1,2,3,\ldots\}. An alternate definition in ZFC would take the least of these, so \aleph_0 / \aleph_0 = 1 in that case. Neither could 'handle' \aleph_0 / \mathfrak{c}; the first would give the empty set and the latter would be undefined.

Your former definition says, if I'm not mistaken, a/b = the number of ways that we can write a as b*n for some n. I don't think that this corresponds to what we mean by "division". The latter way makes more sense, in that \aleph_0 / \aleph_0 = 1. Seeing as cardinals are a sort of generalization of integers, we don't need to handle things like \aleph_0 / \mathfrak{c}, in the same sense that integer division does not need to handle 1/2.

 

[Dragonfall]

Also, in your former definition, 0/anything = class of all cardinals. While in the latter 0/anything = 0.