Dividing by infinity, exactly, finally!

[jaketodd]

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

 

[pbuk]

Why not use these number systems, in place of the real number system.

Because, unlike $\mathbb R$, they do not satisfy the field axioms.

 

[pbuk]

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.

 

[FactChecker]

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.

 

[fresh_42]

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.

 

[Nugatory]

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.

 

[FactChecker]

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 $x=0$.

 

[jaketodd]

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.

 

[jaketodd]

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?

 

[fresh_42]

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.

Because, unlike $\mathbb R$, they do not satisfy the field axioms.

 

[jaketodd]

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.

 

[jaketodd]

All,

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

Thanks

 

[fresh_42]

If infinity is defined, I would expect that infinity/infinity equals 1.

No.

 

[phinds]

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).

 

[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.

 

[Mark44]

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.

 

[FactChecker]

Infinity is a complicated thing that requires different treatment in different situations. To tear down all those treatments and replace them with a simple thing like $\frac {\infty}{\infty} = 1$ violates the principle of Chesterton's Fence: Don't tear something down until you understand why it is there.

 

[fresh_42]

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.

 

[hutchphd]

Your thoughts please.

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

 

[jbriggs444]

If infinity is defined, I would expect that infinity/infinity equals 1.

Then $$2 = \frac{2}{1} = \frac{2}{1} \times \frac{\infty}{\infty} = \frac{2 \times \infty}{1 \times \infty} = \frac{\infty}{\infty} = 1$$

 

[PeroK]

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.

 

[WWGD]

Figure it out and you'll get a

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.

 

[WWGD]

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.

 

[jaketodd]

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.

 

[jaketodd]

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.

 

[jaketodd]

Figure it out and you'll get a

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

No, you first =)

 

[jaketodd]

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

 

[TeethWhitener]

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).

 

[WWGD]

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?

 

[Mark44]

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?

The multiplications would stay with it.

I think you're on a wild goose chase.

 

[jaketodd]

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.

 

[jaketodd]

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.

 

[TeethWhitener]

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.

Ok so then either $2\times\infty$ is infinity or it is a finite real number. So @jbriggs444 ‘s point is still valid.

 

[FactChecker]

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.

 

[jaketodd]

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.