Infinity subtracted from infinity is undefined.

[Hippasos]

Let n be any number.

infinity + n = infinity

n = infinity - infinity

n = undefined

!...?

 

[mgb_phys]

If you're a mathematician - yes it's undefined
If you're a physicist then it can be zero or infinity depending on which makes your theory work!

 

[CRGreathouse]

Yes, lots of operations in the extended reals are undefined.

 

[DaveC426913]

infinity is not a "real" number and cannot be used in arithmetic calculations.


http://en.wikipedia.org/wiki/Real_number" ;
vs.
http://en.wikipedia.org/wiki/Extended_real_number_line"

 

[al-mahed]

Let n be any number.

infinity + n = infinity

n = infinity - infinity

n = undefined

!...?

the second statement doesn't make any sense, am I right?

If you're a mathematician - yes it's undefined
If you're a physicist then it can be zero or infinity depending on which makes your theory work!

Could you explain?

 

[MaWM]

Actually.. its a lot more complicated. You need to know how quickly each of the infinities is diverging in order to compare them.

 

[HallsofIvy]

Let n be any number.

infinity + n = infinity

n = infinity - infinity

n = undefined

!...?

the second statement doesn't make any sense, am I right?

NEITHER statement makes any sense until you specify what what number system you are working in. "Infinity" itself is not defined in either the real number field nor the complex number field and so neither statement makes any sense in them. There are several different ways to extend the number systems to include "infinity" and operations involving "infinity". Those statements may or may not make sense depending on the system.

 

[nicksauce]

Let n be any number.

infinity + n = infinity

n = infinity - infinity

n = undefined

!...?

Let n = be any number
cat + n = blue
n = cat - blue
n = undefined ?

Infinity is not a real number, so you cannot perform operations like infinity + n, just like you cannot perform operations like cat + n, or n = cat - blue.

 

[Gib Z]

Could you explain?

It was a joke, poking fun at the way physicists do mathematics.

 

[yasiru89]

You do realize mgb phys that your little tricks of trade have extrapolated from the mathematical theory of limits instead of adding to it? In functional form an expression of the form \infty - \infty may be reduced to the form \frac{\infty}{\infty} and evaluated by a method such as l'Hopital's rule(as long as putting into the required form does not make an answer obvious) to obtain a finite, infinite or null answer.

The space being worked on may be altered to fit the theory and methods of renormalisation introduced to deal with these forms. Even the most sloppy work can possibly be made rigorous but sloppy is never elegant...

 

[al-mahed]

It was a joke, poking fun at the way physicists do mathematics.

ohhh, all right!

 

[Gib Z]

You do realize mgb phys that your little tricks of trade have extrapolated from the mathematical theory of limits instead of adding to it? In functional form an expression of the form \infty - \infty may be reduced to the form \frac{\infty}{\infty} and evaluated by a method such as l'Hopital's rule(as long as putting into the required form does not make an answer obvious) to obtain a finite, infinite or null answer.

The space being worked on may be altered to fit the theory and methods of renormalisation introduced to deal with these forms...

I have no idea what any of that meant...

 

[yasiru89]

Say some expression is of the form \infty - \infty and defined by a limit on the variable(say x approaches a) by f(x) - g(x)
Since this may be changed into,
g(x).[f(x)/g(x) - 1]


We can deal with this case using l'Hopital's rule. This works when the limit exists of course but some 'physical' theories require more, the most famous being Quantum Field theory I believe. Then we work with alternate definitions and methods like going through extra dimensions to drop off infinities along the way. In such ways we are left with sensational results like \prod_{p} p = 4\pi^{2}, (product over all primes denoted)
The way physicists do mathematics needn't be wrong, radical or 'revolutionary', just insufferably poor.

I was speaking generally earlier but if this isn't straightforward enough for you I'd be at a loss to explain further.(without writing a book that is...on that note there are 2 great calculus books from the early 1900s by J. Edwards on the Internet Archive you might want to check)

 

[Gib Z]

Say some expression is of the form \infty - \infty and defined by a limit on the variable(say x approaches a) by f(x) - g(x)
Since this may be changed into,
g(x).[f(x)/g(x) - 1]

This thread wasn't about a function approaching infinity, that is a normal thing easily dealt with. The original poster talked about arithmetic operations with infinity as a number. Two distinct things.

We can deal with this case using l'Hopital's rule. This works when the limit exists of course but some 'physical' theories require more, the most famous being Quantum Field theory I believe. Then we work with alternate definitions and methods like going through extra dimensions to drop off infinities along the way. In such ways we are left with sensational results like \prod_{p} p = 4\pi^{2} (product over all primes denoted)

Not sensational results, just a load of dung. If the product of every element in the set of the prime numbers is equal to 4\pi^2, then I am a cat. That product is not equal to any real number. If physicists had made some alternative way to formulate primes, then please, do not use the traditional Product operators without even giving notice that a different operator was meant.

 

[yasiru89]

A half dead Schroedinger's cat if you ask me, and the result is not due to physicists, I merely used it to illustrate that WITHIN (all the 'notice' you need had you bothered to read)certain theories(here but a little more than the meromorphic extension of the zeta function) supposedly 'outrageous' results hold true. The case of the result I presented is an analytically acceptable one, regularised using the Euler product.

As for my treatment of the issue Garfield, I simply added to what the subsequent posters were hinting towards, eg- rapidity of divergence.
I have little to add to HallsofIvy so far as arithmetic operations are concerned. Without specifics things go without saying.

 

[Gib Z]

Not even physicists (at least at the level of those who create decent theories) have such terrible mathematics that they believe that product. Of course they know they are using some different operator. Perhaps this operators gives the right product for where the product of the usual operator exists, but its still different.

The only other poster who was hinting towards rapidity of divergence was MaWM, who is also misinterpreting the question. If you notice everyone elses posts, you'll realize we weren't even talking about functions, order of divergence or anything very analytical at all.

PS. Something can't be half dead.

 

[Hippasos]

Let n = be any number
cat + n = blue
n = cat - blue
n = undefined ?

Infinity is not a real number, so you cannot perform operations like infinity + n, just like you cannot perform operations like cat + n, or n = cat - blue.

Well, in my limited experience and knowledge, I would say then:

Arithmetics - as I know it - is a cat with a leg of an elephant(<-infinity) - detached or not I don't know...

Still confused...

 

[zhentil]

Easiest example I can think of: Let f(x)=x, g(x)=2x, h(x)=3x. As x goes to infinity, all three go to infinity as well, right? But if you try to do arithmetic with each "infinity," you get different answers (i.e. a contradiction). If we tried to calculate 3+2, and the answer was different than 3+1+1, we would have to be more careful about our addition rules. So it is with infinity.

 

[ramsey2879]

Well, in my limited experience and knowledge, I would say then:

Arithmetics - as I know it - is a cat with a leg of an elephant(<-infinity) - detached or not I don't know...

Still confused...

No need to be confused. Just accept the fact that infinity is just a concept not a number sinced there is no such thing as a biggest or smallest number. Only if you accept that can you appreciate the beauty of mathematics. Also you can liken an infinite series that has a limit as the sum to an endless do loop in which one gets closer and closer to the answer without actually reaching it, except that you can see the what the answer should be by looking at the computer screen.

 

[yasiru89]

Not even physicists (at least at the level of those who create decent theories) have such terrible mathematics that they believe that product. Of course they know they are using some different operator. Perhaps this operators gives the right product for where the product of the usual operator exists, but its still different.

As was acknowledged had someone bothered to read the post. Of course in my most humble view, which I do not force upon you, it is in definition that mathematics reigns, unlike physical theories- thereby lending little except upon choosing an all too psychological conviction to one's beliefs. A conceptual clarity is very much elusive for my tastes.

That aside,

The only other poster who was hinting towards rapidity of divergence was MaWM, who is also misinterpreting the question. If you notice everyone elses posts, you'll realize we weren't even talking about functions, order of divergence or anything very analytical at all.

There's analytics involved everywhere(part of why I illustrated with a number-theoretic example) but if we are to take a simple point of view on the matter it should be noted that one can come up with as many systems as imaginable to include infinity as an operable quantity, however the real number field and most that follow(complex, biplex, etc.)do not succumb to such. On the other hand we have a measure of infinities like the case where functions stand in allowing a certain sense to be made of simple arithmetics on them.

PS. Something can't be half dead.

Well there's the uncertainty; psychology and interpretation.

 

[mikepr@mac.co]

n is still n

n + infinity does not just give you infinity back again, it gives you a different (ever so slightly larger) infinity.

if you are to subtract the two infinities, you'd still get back n. n will always be n. you defined it, it's not going to change.

not all infinities are the same. think about all the numbers between 1 and 2. there's an infinite set of fractions between 1 and 2. there's also an infinite set of whole numbers 1,2,3,4... you could consider these infinities the same since the fractions between 1 and 2 are created by the set of whole numbers. you can't write more fractions than you can whole numbers, and you can't write more whole numbers than you can write fractions. these kinds of infinities are equal.

If you consider a set of numbers that includes all the fractions between all the whole numbers and the whole numbers themselves though (all real numbers) though, that infinity IS larger than the others, because each whole number allows you to generate an infinite number of fractions (one for every other whole number).

so if you apply such thinking to your example, you have n, infinity, and n + infinity. for each number n, there will always be an n + infinity, and this term is not the same as plain old infinity.

so really you just need to be more careful with your algibra:

we start with:

n + infinity = (infinity + n)

and you can write:
n = (infinity + n) - infinity
infinity = (infinity + n) - n

so:
n = (infinity + n) - ((infinity + n) - n)
n = (infinity + n) - (infinity + n) + n
n = n

-mike

 

[CRGreathouse]

n + infinity does not just give you infinity back again, it gives you a different (ever so slightly larger) infinity.

Depends on what you mean by "infinity". In ordinal arithmetic, omega + 1 > omega, but 1 + omega = omega, where omega is the smallest infinite ordinal. In cardinal arithmetic, aleph_0 = aleph_0 + 1 = 1 + aleph_0, where aleph_0 is the smallest infinite cardinal. In extended real arithmetic, +infinity = +infinity + 1 = 1 + +infinity, where +infinity is the largest extended real number. In the projective line, infinity = infinity + 1 = 1 + infinity, where infinity is the distinguished element.

There's really no good way to do subtraction in infinite ordinal arithmetic. Subtraction of equal infinite cardinals is undefined (though there's no reason you couldn't subtract a smaller cardinal from a larger infinite cardinal; the difference would be the larger one). In the extended reals +infinity - +infinity is undefined. In the projective line infinity - infinity is undefined.

So you're clearly not talking about any of these standard systems, so you're creating your own form of numbers. That's perfectly fine; abstract algebra will tell you about the system you create. I'll give you one tip to start off: don't say infinity. You clearly have an infinite number of infinities, which are not the same. Call the basic one is w. You've said that w + 1 is not equal to w, so don't give them the same name. You should study the system and see which properties of the real numbers hold in your system.

 

[mikepr@mac.co]

well i thought the whole x + 1 and (x + 1) was self explanitory, but if you'd like i'll add in (x + 1) = w = x + 1. i thought my way was a bit easier to follow for the original poster since people tend to get confused when you go tossing new variables around. or maybe (x + 1) = x'. whatever.

i'm not making up my own number system. the example i used to demonstrate different kinds of infinities is over 100 years old. to be sure it doesn't have anything to do with this topic, other than if you can understand it, you can easily understand that in the orginal poster's example, n is still going to be n.

-mike

 

[CRGreathouse]

i'm not making up my own number system. the example i used to demonstrate different kinds of infinities is over 100 years old. to be sure it doesn't have anything to do with this topic, other than if you can understand it, you can easily understand that in the orginal poster's example, n is still going to be n.

Well, your system* is not any of the four standard systems I mentioned. Perhaps it can be related to a (proper?) subset of nonstandard analysis -- but you'd have to define it more carefully before I could tell you.

In your system, what is infinity * 0? Is your system commutative? Associative? Distributive? Is it a additive group? A ring? A field?

* Or the system you describe. By the way, if it has been around, do you know of a name it goes by or who developed it/worked out its properties?

 

[mikepr@mac.co]

to continue with your thread jack, that'd be cantor. some quick links from google:

http://scidiv.bcc.ctc.edu/Math/diag.html
http://people.bath.ac.uk/jp253/Project.html
http://www.sciam.com/article.cfm?id=strange-but-true-infinity-comes-in-different-sizes
http://mathcentral.uregina.ca/QQ/database/QQ.09.03/plober1.html

"ever so slightly larger" was a poor but demonstrative word choice. the infinity might not be larger, but it is different. now would you like to comment on if n = n or what?

i'm not trying to argue that n + infinity can be defined, only that it doesn't change the definition of n, and that infinity and (infinity + n) are not one in the same. in n + x = w, it really doesn't matter what x is or if w is defined. if you do the opposite operation knowing the definition of n, you'll always get n back. w - x = n. this tells you nothing about x or w other than the relation that exists between them and n. this is very different from assuming that because you don't understand the nature of w or x, that they must then be equal, and that the two terms somehow confuse the nature of the term n that you defined.

-mike

 

[Hurkyl]

mike: CRGreathouse knows what he's talking about. You, apparently, do not. We'd be happy to try and explain one or more of the various kinds of arithmetic that include infinite elements (your references refer to the cardinal numbers); ideally you can start a new thread on the topic.

However, it is not acceptable for you to speak authoritatively on the subject.

 

[CRGreathouse]

CRGreathouse knows what he's talking about.

Thanks, Hurkyl. Sometimes I wonder if I'm explaining myself poorly... wouldn't be the first time.

If someone wants to split these posts off into their own thread that would be great; they're barely on-topic at best.

in n + x = w, it really doesn't matter what x is or if w is defined. if you do the opposite operation knowing the definition of n, you'll always get n back.

Okay, so your take on numbers has cancellation, so it's a quasigroup -- a loop, really, since I imagine you'll have x + 0 = 0 for all x. Are your numbers associative? Commutative?

 

[al-mahed]

I think as infinity is not a defined amount (you cannot count it), the ordinary algebra is not applicable

 

[CRGreathouse]

that'd be cantor. some quick links from google:

http://scidiv.bcc.ctc.edu/Math/diag.html
http://people.bath.ac.uk/jp253/Project.html
http://www.sciam.com/article.cfm?id=strange-but-true-infinity-comes-in-different-sizes
http://mathcentral.uregina.ca/QQ/database/QQ.09.03/plober1.html

Each of those four links is about cardinality, the second of my four examples. Cardinal arithmetic does not have an obvious subtraction operation. There are two natural ways that sprint to my mind:

1. Each cardinal is the smallest ordinal of that cardinality; call that primary_cardinal(ord) where ord is some ordinal. Then addition can be defined by A + B = primary_cardinal({(0, a) for all a in A} U {(1, b) for all b in B}), and subtraction by A + B = primary_cardinal(A \ B).
2. Each cardinal is the equivalence class of all ordinals of that cardinality. A + B = U{(0, a) for all a in A'} U {(1, b) for all b in B'} where the outer union is over all A' in A and B' in B. Then subtraction is the multivalued inverse of this function.

Let w denote the smallest infinite number in the system.*

In #1, w - w = {}, which I would identify with 0.

In #2, w - w is the class of all countable sets (finite and infinite).


* Note: I have not defined an order over these number systems; if this bothers you, because existence is not clear, just take it as omega for #1 and the class of infinite countable sets in #2.

 

[CRGreathouse]

I think as infinity is not a defined amount (you cannot count it), the ordinary algebra is not applicable

Well, okay, but abstract algebra can still deal with it. I mean, you also can't count to "hammer", but the rules (x universally quantified)

x + 0 = x = 0 + x
x * 0 = 0 = 0 * x
x + x = 0
x * x = x

over the set {0, hammer} is a field and has all the ordinary properties of a field. < over {rock, paper, scissors} with scissors < rock, rock < paper, paper < scissors is a binary relation (but not a partial order!).

Okay, you probably knew that, but I thought it was relevant at this point in the thread.

 

[Hurkyl]

I think as infinity is not a defined amount (you cannot count it)

What do you mean by "amount"? You would be correct to assert that there are no infinite natural numbers, nor are there infinite real numbers.

the ordinary algebra is not applicable

Of course -- different number systems require different arithmetic. e.g. we cannot use natural number arithmetic when working with the cardinal numbers, because natural number arithmetic doesn't tell us anything about infinite cardinals.

(And for completeness, I should point out that we have no a priori reason to asume that natural number arithmetic should coincide with cardinal number arithmetic for finite cardinals. This is a fact that needs to be proven (or assumed) before we can use it)

 

[Hurkyl]

Thanks, Hurkyl. Sometimes I wonder if I'm explaining myself poorly... wouldn't be the first time.

It can be hard to explain something new to someone. But it's much harder to explain something to someone who has already seen it, but learned it wrongly.

 

[CRGreathouse]

It can be hard to explain something new to someone. But it's much harder to explain something to someone who has already seen it, but learned it wrongly.

Yes, unlearning is hard.

Nice sig, by the way.

 

[al-mahed]

What do you mean by "amount"? You would be correct to assert that there are no infinite natural numbers, nor are there infinite real numbers.

I think my english wasn't good enough... I was trying to say that a set with infinite elements cannot be counted in a ordinary fashion

 

[sihag]

i came across lim x.logx, x --> 0
as equal to 0.
The logx component part would go to - infinity, and that multiplied with a quantity tending to 0 is zero. Does that mean 0 times any quantity (infinite or finite) is always 0?

Slightly off topic, but what's the proof for the sum of two irrational numbers to be irrational?

 

[Hurkyl]

i came across lim x.logx, x --> 0
as equal to 0.
The logx component part would go to - infinity, and that multiplied with a quantity tending to 0 is zero. Does that mean 0 times any quantity (infinite or finite) is always 0?

Before I answer your question, please answer mine: why would you think that might mean 0 times an infinite quantity is always 0?

 

[arildno]

I am getting mightily tired of these "infinity" -threads and what can, or cannot do with them, and that it is somehow strange that some operations remain undefined.

Let us look at the number system consisting of the numbers:

0, 1 and "many"

We let 0 be distinct from 1, and we have relations like 1+1=many, many+1=many and so on.

Numerous undefined relations will naturally occur within such a number system.

 

[zhentil]

i came across lim x.logx, x --> 0
as equal to 0.
The logx component part would go to - infinity, and that multiplied with a quantity tending to 0 is zero. Does that mean 0 times any quantity (infinite or finite) is always 0?

Slightly off topic, but what's the proof for the sum of two irrational numbers to be irrational?

Sometimes it helps to check if there are trivial counterexamples to what you've claimed. For instance, 1/x^2 tends to infinity as x->0 from the right. x tends to 0 as x->0 from the right. Does the product tend to zero?

For the second question, what is sqrt(2)+(-sqrt(2))? Is it irrational?

 

[sihag]

I meant besides the trivial case, for irrational sums.

I stand corrected on the 0 times arbitrarily large quantity case.
Thank you.

 

[zhentil]

How about (pi + e + 6) + (-pi - e)? If you want to exclude that case as well, it's going to be impossible to define what "non-trivial" means. I think what you want is that an irrational plus a rational is irrational, which is easy to prove.

 

[arildno]

Well, it might not be impossible for him to define what he means by "the trivial cases " in a mathematically meaningful way.

Until he does, though, we can't judge whether the sum of two irrationals in his "non-trivial" cases must equal an irrational.

 

[zhentil]

The only way to define "non-trivial" so that his statement is true is that the sum of the two numbers is not a rational, but then it's vacuously true.

 

[arildno]

The only way to define "non-trivial" so that his statement is true is that the sum of the two numbers is not a rational, but then it's vacuously true.

Not too sure about that.
It might be that we can separate irrationals into two fairly large classes (by some interesting criterion) and that surprisingly, this criterion would imply, throuygh lengthy proofs, that the sum of two such irrationals necessarily would be irrational.

This could well an important result, but we shouldn't hope for, or expect, its materialization.

 

[zhentil]

Could you be more specific? I have a feeling that's still circular.

 

[Hurkyl]

I thought we didn't know if (pi + e) was rational or irrational.

 

[sihag]

by non-trivial i meant 'not using the negative inverse of an irrational to cancel it out in the expression'.
For example sqrt 2 + sqrt 3 is a non trivial case.
How does one prove that
a + b = irrational, whenever a not equal to b, a,b both being irrational.

 

[Hurkyl]

by non-trivial i meant 'not using the negative inverse of an irrational to cancel it out in the expression'.

The irrational number (1 - sqrt 2) is not the additive inverse of sqrt 2. Nor is
\sqrt{3 - 2 \sqrt{2}}.