# Dividing by infinity, exactly, finally!

• B
• jaketodd
In summary, the extended real number line and projectively extended real line are alternative number systems that allow for division by infinity to result in an exact value of zero. However, these systems do not satisfy the field axioms and can cause more problems than they solve. Additionally, the idea of dividing by infinity is not always well-defined and can lead to different values depending on the growth rates of the numerator and denominator. Therefore, these systems are not commonly used in mathematics and should be approached with caution.
jaketodd
Gold Member
Why not use these number systems, in place of the real number system, when these allow us to divide by infinity exactly?

According to these, division by infinity equals exactly zero! No need for calculus limits, which only can say it approaches zero when tending towards infinity.

https://en.wikipedia.org/wiki/Extended_real_number_line
https://en.wikipedia.org/wiki/Projectively_extended_real_line

My motivation is resolving the argument between continuous and discrete/quantum notions.

Continuous says there is no smallest unit. So that would be 1 unit divided by infinity. And if that equals zero exactly, then it really doesn't exist, does it? So that would point to a smallest unit/quantum.

Your thoughts please.

Thanks,

Jake

jaketodd said:
Why not use these number systems, in place of the real number system.
Because, unlike ## \mathbb R ##, they do not satisfy the field axioms.

Demystifier, WWGD, PeroK and 2 others
jaketodd said:
My motivation is resolving the argument between continuous and discrete/quantum notions.
There is no argument, but it sounds like you would be interested in sets that include infinitessimals such as the hyperreals and surreals rather than the extended reals. But note that none of these can replace the reals (because they are not fields), they just have certain specific applications.

PeroK
jaketodd said:
Why not use these number systems, in place of the real number system, when these allow us to divide by infinity exactly?
Because, although it makes the answer of division by infinity very simple, there are too many situations where that does not make sense. Making division by infinity simple is not the ultimate goal of mathematics.

Demystifier, pbuk, PeroK and 1 other person
jaketodd said:
Why not use these number systems, in place of the real number system, when these allow us to divide by infinity exactly?
Infinitesimals cause more problems than they solve. E.g. the derivative is
$$\left. \dfrac{d\,y}{d\,x}\right|_{x=x_0}=\lim_{h \to 0}\dfrac{f(x_0+h)-f(x_0)}{h}$$
So both infinitesimals are explained by the same limit. How would you explain only one in this context?

Hewitt, Stromberg have used ##\mathbb{R}^{\#}=\mathbb{R}\cup \{\pm \infty \}## besides ##\mathbb{R}## in their book about analysis (GTM 25) which is heavily directed by measure theory. So it is possible and done where it makes sense. It is only not done during the early steps of graduation because the disadvantages - will say the strongly increased amount of possibilities of making mistakes - outnumber the advantages.

FactChecker and pbuk
jaketodd said:
Why not use these number systems, in place of the real number system, when these allow us to divide by infinity exactly?
One such system exists and is pretty much ubiquitous: the IEEE 754 floating point arithmetic used in modern computers.

Of course a 32-bit or 64-bit container can hold only a finite number of discrete values. I think of IEEE 754 as a mapping from some of these values (the NaNs and Infinities are excluded) to a subset of the reals along with definitions of operations on these bit patterns that preserve our intuition about how arithmetic “ought to” work.

PeroK
Suppose you were dealing with something that was like this: ##\lim_{x \rightarrow 0}\frac {1/x^3}{1/x}## in a more complicated and disguised form. Clearly, this is ##1/x^2## except that it is undefined at ##x = 0##. You would not normally want to say that the function is defined and equal to 0 at ##x=0##.

FactChecker said:
Suppose you were dealing with something that was like this: ##\lim_{x \rightarrow 0}\frac {1/x^3}{1/x}## in a more complicated and disguised form. Clearly, this is ##1/x^2## except that it is undefined at ##x = 0##. You would not normally want to say that the function is defined and equal to 0 at ##x=0##.
That seems to be dividing by zero. I'm talking about dividing by infinity, which the alternative number systems I mentioned are able to arrive at exactly zero.

FactChecker said:
Because, although it makes the answer of division by infinity very simple, there are too many situations where that does not make sense. Making division by infinity simple is not the ultimate goal of mathematics.
But if there are number systems that can handle division by infinity, shouldn't we use them? What says which one is the best?

PeroK
jaketodd said:
But if there are number systems that can handle division by infinity, shouldn't we use them? What says which one is the best?
Because you cannot define ##\dfrac{\infty }{\infty }.## Those extended reals are not a number system in its classical meaning. It makes more trouble to manage it correctly than it has advantages to use it.
pbuk said:
Because, unlike ## \mathbb R ##, they do not satisfy the field axioms.

FactChecker
fresh_42 said:
Because you cannot define ##\dfrac{\infty }{\infty }.## Those extended reals are not a number system in its classical meaning. It makes more trouble to manage it correctly than it has advantages to use it.
If infinity is defined, I would expect that infinity/infinity equals 1.

All,

Is there a way to merge number systems, so that we get the best of all worlds?

Thanks

jaketodd said:
If infinity is defined, I would expect that infinity/infinity equals 1.
No.

pinball1970 and FactChecker
jaketodd said:
If infinity is defined, I would expect that infinity/infinity equals 1.
No. You are trying to treat infinity as though it were a number. It is not and pretending that it is just leads to problems (such as your expectation).

pinball1970 and FactChecker
If you have ##\frac {\infty}{\infty}##, considered just as numbers, you have to be careful about different orders of infinity (some are larger than others, indicated by ##\aleph_0, \aleph_1,## etc.
In applications where the numerator and denominator are functions growing unbounded, you have to worry about how fast the numerator and denominator grow. If the numerator grows much faster than the denominator, then the ratio gets large. If it is the other way around, the ratio goes to zero. You can make examples with growth rates where the ratio approaches any number you want. There are similar issues with zero, where you need to consider how fast the numerator and denominator go to zero.

fresh_42, topsquark and phinds
jaketodd said:
If infinity is defined, I would expect that infinity/infinity equals 1.
Then your expectation would be incorrect. Here are three simple examples of limits of the indeterminate form ##[\frac \infty \infty]##:
##\lim_{x \to \infty} \frac x x##
##\lim_{x \to \infty} \frac x {x^2}##
##\lim_{x \to \infty} \frac {x^2} x##
Only the first example has a limit of 1. The other two limits are zero and ##\infty## respectively.

PeroK
FactChecker said:
Infinity is a complicated thing that requires different treatment in different situations.
I just wrote an example in another thread about what can go wrong when dealing with infinities:
https://www.physicsforums.com/threads/infinite-series-of-infinite-series.1052865/#post-6901476

One possible way to deal with infinities is the Riemann sphere. Here we have a situation where infinity corresponds to a certain point on the sphere. However, the "corresponding" is a precisely defined function, so that we can go back and forth to "normal". It is no replacement, it is a different point of view.

More generally, the technique of the Riemann sphere is the basic concept of projective geometry.

This all shows that people did consider how to deal with infinity, and none of it led to an easy answer.

PeroK, phinds and FactChecker
jaketodd said:
Your thoughts please.

This was a chapter heading in my Freshman physics book (Sears and Zemansky 4th Edition). Still true

DaveE, FactChecker and fresh_42
jaketodd said:
If infinity is defined, I would expect that infinity/infinity equals 1.
Then $$2 = \frac{2}{1} = \frac{2}{1} \times \frac{\infty}{\infty} = \frac{2 \times \infty}{1 \times \infty} = \frac{\infty}{\infty} = 1$$

pinball1970, dextercioby, Nugatory and 3 others
jaketodd said:
All,

Is there a way to merge number systems, so that we get the best of all worlds?

Thanks
The nearest we have is ##\mathbb C##. Perhaps you could study the complex numbers in all their glory.

Figure it out and you'll get a
jaketodd said:
All,

Is there a way to merge number systems, so that we get the best of all worlds?

Thanks
Figure one out and go collect your Fields/Abel medal.

You only need to check through ## \aleph_0## models of the Reals , per Lowenheim -Skolem. Outside of the standard Reals, you lose " standard " metrizability, since metric are Real-valued.

dextercioby
Mark44 said:
Then your expectation would be incorrect. Here are three simple examples of limits of the indeterminate form ##[\frac \infty \infty]##:
##\lim_{x \to \infty} \frac x x##
##\lim_{x \to \infty} \frac x {x^2}##
##\lim_{x \to \infty} \frac {x^2} x##
Only the first example has a limit of 1. The other two limits are zero and ##\infty## respectively.
You're using limits, which the two alternative number systems I mentioned, don't use. At least I don't think they do. I think one of the motivations for them is to be rid of limits.

jbriggs444 said:
Then $$2 = \frac{2}{1} = \frac{2}{1} \times \frac{\infty}{\infty} = \frac{2 \times \infty}{1 \times \infty} = \frac{\infty}{\infty} = 1$$
No, if infinity is defined, then 1 or 2 multiplied by infinity would not reduce to just infinity. The multiplications would stay with it.

WWGD said:
Figure it out and you'll get a

Figure one out and go collect your Fields/Abel medal.
No, you first =)

WWGD said:
You only need to check through ## \aleph_0## models of the Reals , per Lowenheim -Skolem. Outside of the standard Reals, you lose " standard " metrizability, since metric are Real-valued.
Please provide a link, thanks

jaketodd said:
No, if infinity is defined, then 1 or 2 multiplied by infinity would not reduce to just infinity. The multiplications would stay with it.
No, if you define infinity as your link in OP (##-\infty<a<+\infty,{}\forall a\in\mathbb{R}##), then ##2\times\infty\leq\infty## must be true by definition (if multiplication by infinity is even defined).

jaketodd said:
Please provide a link, thanks
Sure,
https://en.wikipedia.org/wiki/Compactness_theorem

Through the concept of Elementary Equivalence , aka, the Transfer Principle, 1st -order properties are preserved "upwardly", from lower to higher cardinality, between models of different cardinality.

Edit: Is that what you were asking?

Last edited:
jaketodd said:
No, if infinity is defined, then 1 or 2 multiplied by infinity would not reduce to just infinity.
Then what would, say, ##2 \cdot \infty## or ##5 \cdot \infty## reduce to, if as you maintain, they would be different?

jaketodd said:
The multiplications would stay with it.
I think you're on a wild goose chase.

berkeman
Mark44 said:
Then what would, say, ##2 \cdot \infty## or ##5 \cdot \infty## reduce to, if as you maintain, they would be different?

I think you're on a wild goose chase.
You'd have to ask the people who came up with the alternative number systems I've mentioned. I'm just concentrating on their ability to divide by infinity, mainly. I think what you're asking is kind of like asking what does 2x or 5x reduce to.

PeroK
TeethWhitener said:
No, if you define infinity as your link in OP (##-\infty<a<+\infty,{}\forall a\in\mathbb{R}##), then ##2\times\infty\leq\infty## must be true by definition (if multiplication by infinity is even defined).
This one: https://en.wikipedia.org/wiki/Projectively_extended_real_line
It only adds one notion of infinity, defined, instead of two. It doesn't have +/- infinities, just one encompassing one.

jaketodd said:
This one: https://en.wikipedia.org/wiki/Projectively_extended_real_line
It only adds one notion of infinity, defined, instead of two. It doesn't have +/- infinities, just one encompassing one.
Exactly. In one very important application, it is useful to think of ##\infty## as a single point. Clearly, there are other important applications where you would not want ##-\infty## and ## + \infty## to be the same.
Infinity is complicated.

TeethWhitener said:
Ok so then either ##2\times\infty## is infinity or it is a finite real number. So @jbriggs444 ‘s point is still valid.
I think we need to study the alternative number systems more. I just know that infinity is defined in both I mention, instead of being undefined and just an idea. You say "finite real number." Both of the alternative number systems I mention are different than the real number system, as mentioned in those two Wikipedia articles. Related, but different.

PeroK

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