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3 minutes ago, uncool said:

Once again: The fact that the set of hyperreal numbers contains infinite numbers doesn't mean it can't contain finite numbers.

Not all hyperreal numbers are infinite. Infinitude is not a "requirement" for all hyperreal numbers. There are finite hyperreal numbers. Some hyperreal numbers are finite.

In mathematics, the system of hyperreal numbers is a way of treating infinite and infinitesimal quantities.

Briefly, the standard part function "rounds off" a finite hyperreal to the nearest real.

The reason why it is a standard part function is because they are not standard real numbers, like 31.  It rounds it off to the nearest real number.  The equation they use to do this is the same as the equation of a derivative.

It associates to every such hyperreal {\displaystyle x}, the unique real {\displaystyle x_{0}} infinitely close to it, i.e. {\displaystyle x-x_{0}} is infinitesimal.

Then the hyperreals are infinity close to the real numbers.

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5 minutes ago, Conjurer said:

"Nonstandard analysis deals primarily with the pair [real numbers, hyperreal numbers], where the hyperreals are an ordered field extension of the reals, and contain infinitesimals, in addition to the reals."

Edited by uncool

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39 minutes ago, Conjurer said:

The sheer notion of this completely boggles my mind.  Pie can also be expressed as a fraction of the circumference of a circle divided by it's diameter, but that doesn't make it a rational number.  You couldn't know the exact value of both the circumference and the diameter with 100% accuracy, so you could only approach a slightly better estimate for it by making more accurate measurements of a perfect circle.

Then there exist computer programs that can calculate it to a further degree of places.  That doesn't change the fact of it being irrational.  Everyone I have ever heard teach about this aspect  of math uses that type of lingo.

that contain anything greater in the form

1+1+...+1

31 doesn't contain anything greater in the form of adding an infinite number of 1's.  Therefore, it fails to meet the requirement.  It is trying to deal with infinities that are greater than an infinity of adding an infinite number on single digits or counting to infinity by integers.

!

Moderator Note

As you clearly have a limited understanding of the topic, I suggest you restrict yourself to reading others' posts and, possibly, asking questions.

Do not keep making confident assertions about topics that you do not fully understand.

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59 minutes ago, uncool said:

"Nonstandard analysis deals primarily with the pair [real numbers, hyperreal numbers], where the hyperreals are an ordered field extension of the reals, and contain infinitesimals, in addition to the reals."

I don't see how real numbers could possibly behave in the same way as hyperreals, because the equation for the standard part function would divide that number by zero.  Then the limit isn't taken for it to find the derivative.  The limit has to be found to divide by an infinitesimal for all real numbers.

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22 minutes ago, Conjurer said:

I don't see how real numbers could possibly behave in the same way as hyperreals, because the equation for the standard part function would divide that number by zero.

I assume you mean the equation for the derivative. The denominator is specifically assumed to be infinitesimal, and 0 is not infinitesimal.

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1 minute ago, uncool said:

I assume you mean the equation for the derivative. The denominator is specifically assumed to be infinitesimal, and 0 is not infinitesimal.

The standard part of any infinitesimal is 0. Thus if N is an infinite hypernatural, then 1/N is infinitesimal, and st(1/N) = 0.

And?

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Although Hewitt may have coined the term hyperreal, he did no introduce infinitesimals. They go way further back, but here is an interesting treatment from
Prefessor Franklin (MIT) in 1940.

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It may be worth noting, however, that that idea of infinitesimals is not the same as the hyperreal idea of infinitesimals.

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1 hour ago, uncool said:

It may be worth noting, however, that that idea of infinitesimals is not the same as the hyperreal idea of infinitesimals.

It is indeed worth noting this as it demonstrates that the subject is not cut and dried.

And there have been yet other versions of infinitesimals.

Just as some have othere versions of infinitesimals, Thurston has other versions of extended numbers which he calls 'super-cauchy' numbers and 'neighbourhood of infinity numbers'.

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22 hours ago, gib65 said:

What happens if you multiply the infinitely small hyperreal number e by the infinitely large hyperreal number R? Do you get a real number?

BTW, I don't think there is a way to work with numbers that explode into higher infinities.  I hear it is one of the leading problems in theoretical physics to find unification and describe what goes on in a black hole or the Big Bang.  It is a mathematical process that has been sought after, so you would have to invent it.

It just so happens that finding the derivative is the only mathematical situation that is known where you can actually divide by zero and get a correct reasonable finite answer.  Normally, they say you have to find the limit so it isn't actually zero when you cancel them out, but it actually does nothing to the mathematical operation differently than if you just canceled the zero.  Then the derivative can be found where it actually is zero.  In this one type of situation, you can get away with dividing by zero, and it makes no difference to the answer.

The only way you can find a value from an infinite number is to take the integral, which is the reverse process of taking the derivative.  That finds the area under a function and the x axis.  Even though variables are infinite in the function, you can have a finite number of area to work with still.

The only known method would be to analyze the hyperreal on a coordinate plain as a part of a function and use calculus on it.

Edited by Conjurer

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18 hours ago, Strange said:
!

Moderator Note

The rules require a level of politeness that this post does not reach. Just the second sentence would have been enough.

My apologies.

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2 hours ago, Conjurer said:

It just so happens that finding the derivative is the only mathematical situation that is known where you can actually divide by zero and get a correct reasonable finite answer.

I think that if you look carefully enough, you will find that the derivative is calculated using hyperreals by dividing by an infinitesimal $$\Delta x$$ which is nonzero. Only after dividing do you extract the standard part.

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11 hours ago, taeto said:

I think that if you look carefully enough, you will find that the derivative is calculated using hyperreals by dividing by an infinitesimal Δx which is nonzero. Only after dividing do you extract the standard part.

Then it wouldn't be the derivative.  The derivative is the line that intersects a curve at exactly one single point.  Then if the change in x was not zero, it would be the equation for a line that intersects at two points on the curve.

It is mathematically correct to divide by an infinitesimal in this situation, because it has been mathematically proven that you can divide by an infinitely small number or zero in only this one specific situation.

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17 minutes ago, Conjurer said:

It is mathematically correct to divide by an infinitesimal in this situation, because it has been mathematically proven that you can divide by an infinitely small number or zero in only this one specific situation.

I suspect that according to the rules of the forum, since you state here that something has been proven, then you have to provide some kind of reference or other solid evidence. Is this alleged fact something which has been proven by you personally?

Edited by taeto

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6 minutes ago, taeto said:

I suspect that according to the rules of the forum, since you state here that something has been proven, then you have to provide some kind of reference or other solid evidence. Is this alleged fact something which has been proven by you personally?

!

Moderator Note

I will make that official.

@Conjurer Either provide a reference to support this or stop posting in this thread where (yet again) it is pretty obvious you are operating with delusions of competence.

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Going back to Wiki

Note that it specifies

"let dx be a non-zero infinitesimal"

dx is not zero.

However the rest of the procedure is just a smokescreen for the fact that in the last line the writer is going to ignore  (drop) the dx.
Talking about the 'standard' part is no more justifiable than choosing only to take the real part of a complex number.

The alternative I was taught was to use limits and perform algebraic manipulations (including using the rules for manipulating limits) until the last line was reached

$\mathop {\lim }\limits_{\delta x \to 0} \left( {2x + \delta x} \right)$

$= \mathop {\lim }\limits_{\delta x \to 0} \left( {2x} \right) + \mathop {\lim }\limits_{\delta x \to 0} \left( {\delta x} \right)$

Then and only then was the limit taken

$= 2x + 0$

We developed these for quite a few derived functions.

It can immediately be seen that the algebra is the same until the end but then nothing is discarded.

It also shows why the principle of working out a few appropriate fundamental examples and then relying on developing an algebra of derivatives is the usual method taught.

Edited by studiot

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2 minutes ago, studiot said:

Talking about the 'standard' part is no more justifiable than choosing only to take the real part of a complex number.

About that point, I think that the "standard part" of a hyperreal is only defined to be the nearest real number if the hyperreal is finite. In the context of derivatives that just means that if the hyperreal differential quotient is infinite, then you conclude that the derivative does not exist.

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6 minutes ago, taeto said:

About that point, I think that the "standard part" of a hyperreal is only defined to be the nearest real number if the hyperreal is finite. In the context of derivatives that just means that if the hyperreal differential quotient is infinite, then you conclude that the derivative does not exist.

I think it is truncated not rounded.

I am having trouble with a conflict between the editor here and the Wiki extract I am trying to fix in my post.

The paragraph following in Wiki clearly calls it a 'double number' and claims this to be rigourous.

Quote

Wiki

The use of the standard part in the definition of the derivative is a rigorous alternative to the traditional practice of neglecting the square[citation needed] of an infinitesimal quantity. Dual numbers are a number system based on this idea. After the third line of the differentiation above, the typical method from Newton through the 19th century would have been simply to discard the dx2 term. In the hyperreal system, dx2 ≠ 0, since dx is nonzero, and the transfer principle can be applied to the statement that the square of any nonzero number is nonzero. However, the quantity dx2 is infinitesimally small compared to dx; that is, the hyperreal system contains a hierarchy of infinitesimal quantities.

Edited by studiot

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8 minutes ago, studiot said:

I think it is truncated not rounded.

What's the difference, in your opinion?

9 minutes ago, studiot said:

The paragraph following in Wiki clearly calls it a 'double number' and claims this to be rigourous.

I'm not sure what "it" refers to here, but both hyperreals and dual numbers are rigorous (though the "dual number" has some trouble with properly defining higher derivatives); a version of "dual numbers" is actually used in algebraic geometry.

25 minutes ago, studiot said:

It can immediately be seen that the algebra is the same until the end but then nothing is discarded.

That shouldn't be especially surprising; the hyperreals aren't really meant to be a fundamental change in outcome, only in outlook. As such, most formal processes should look essentially the same.

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36 minutes ago, uncool said:

What's the difference, in your opinion?

I'm not sure what "it" refers to here, but both hyperreals and dual numbers are rigorous (though the "dual number" has some trouble with properly defining higher derivatives); a version of "dual numbers" is actually used in algebraic geometry.

That shouldn't be especially surprising; the hyperreals aren't really meant to be a fundamental change in outcome, only in outlook. As such, most formal processes should look essentially the same.

The difference comes if the infinitesimal is multiplied by something suitable large in the final expression before it is discarded.

Rigourous ?
The dx that Wiki used is a 'dummy' that could just have easily been the 'symbol' I originally referrred to, for instance Ю.

My point about the limit is that the limit of (A+B) = lim(A) +lim(B) exactly and

$\mathop {\lim }\limits_{\delta x \to 0} \left( {2x} \right)$

is exactly 2x since it is independent of δx

and

$\mathop {\lim }\limits_{\delta x \to 0} \left( {\delta x} \right)$

is exactly zero

So the result is exact with nothing discarded.

I also realise that there is a correspondence (said to be) established between the reals and the hyperreals, in super advanced algebraic structure theory, which I hoped you would enlighten us about.

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What about the imaginary number, the square root of -1? <-- That's said to be imaginary, therefore not one of the reals. Does that make it a hyperreal? Do the hyperreals include imaginary numbers?

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6 minutes ago, gib65 said:

What about the imaginary number, the square root of -1? <-- That's said to be imaginary, therefore not one of the reals. Does that make it a hyperreal? Do the hyperreals include imaginary numbers?

No, they do not. Hyperreal is not simply "not real"; it is a specific extension of the reals. One of the major properties is that it is an ordered field extension; that is, it allows addition, subtraction, multiplication, and division (except division by 0), and that these operations act nicely with respect to "Which of these is bigger".

One result from that is that all squares are positive (as x^2 = (-x)^2, and one of x and -x is positive, and the product of two positive things must be positive). So if i were a hyperreal, then -1 would be positive - which it can't be.

That said, it is possible to construct the hypercomplex numbers, in the same way the complex numbers are constructed from the real numbers.

Edited by uncool

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4 hours ago, taeto said:

I suspect that according to the rules of the forum, since you state here that something has been proven, then you have to provide some kind of reference or other solid evidence. Is this alleged fact something which has been proven by you personally?

It was proved by Isaac Newton in the Principia Mathematica.

He developed calculus, because the mathematics they had before that was unable to complete his work to discover the equations of gravity.

4 hours ago, Strange said:
!

Moderator Note

I will make that official.

@Conjurer Either provide a reference to support this or stop posting in this thread where (yet again) it is pretty obvious you are operating with delusions of competence.

My confidence is completely shattered.

Edited by Conjurer

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2 hours ago, studiot said:

The difference comes if the infinitesimal is multiplied by something suitable large in the final expression before it is discarded.

I'm not sure I understand what you are trying to say. What's the difference between "rounding to the nearest real number" and truncation?

2 hours ago, studiot said:

So the result is exact with nothing discarded.

The result on the hyperreals is also exact. The "standard" function is specifically defined to give a result in the real numbers, not the hyperreal numbers.

2 hours ago, studiot said:

I also realise that there is a correspondence (said to be) established between the reals and the hyperreals, in super advanced algebraic structure theory, which I hoped you would enlighten us about.

There are two correspondences I can think of, but the one I think you are referring to is that the reals are a subset of the hyperreals (i.e. there is a natural embedding of the reals in the hyperreals). This isn't "super advanced algebraic structure theory"; it's just...a fact. I'm not sure what you think needs enlightening.

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