My Math Forum Hyperreal vs Real numbers. What's the difference?

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 January 10th, 2016, 06:08 AM #1 Newbie   Joined: Jan 2016 From: House Posts: 7 Thanks: 0 Hyperreal vs Real numbers. What's the difference? Since this came up while reading a calc book I figured I post this question here. Apparently the text I read indicates that the field of hyperreal numbers is bigger than that of real and the difference is the former includes infinitesimals. But here's the problem with my thinking: I've always thought the real line is a continuum and infinitely divisible. So at zero, we can always approach arbitrarily close to it and at each attempt, we can always zoom in and approach with an even smaller step. Isn't that being infinitesimally close?
 January 10th, 2016, 06:13 AM #2 Newbie   Joined: Jan 2016 From: Numberland Posts: 6 Thanks: 0 The hyperreals are an extension of the real numbers to include infinitesimal and infinite quantities. In the real numbers you can't have a number infinitesimally close to 1 but not , but in the hyperreals you can. Also, it does makes sense that the hyperreals have a larger cardinal than the reals since the hyperreals includes (but is not limited to) every number infinitesimally close to every real number. What you have described is not being infinitesimally close. In order to be infinitesimally close to 0 one would have to zoom in an infinite amount of times. Last edited by Mobius; January 10th, 2016 at 06:19 AM.
 January 10th, 2016, 08:27 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra Yes. In the reals you can find a number that is arbitrarily close to a given number, but it is still finitely close. In the hyperreals, there are numbers that are closer to the given number than all other real numbers, and these are infinitely close.
 January 11th, 2016, 03:57 AM #4 Newbie   Joined: Jan 2016 From: House Posts: 7 Thanks: 0 Thanks for the response. Ok, I just want to confirm the properties of the real line: 1. it's a continuum and has no gaps 2. it is infinitely divisible 3. it spans from negative infinity to positive infinity 4. infinities are allowed but not infinitesimals Since an arbitrarily small real number is still finite and any point on the real line corresponds to a number, then does it mean an infinitesimal has no numerical representation? Otherwise it will correspond to a point on the real line very close to zero.
 January 11th, 2016, 04:13 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra All correct apart from 4. There are no infinite real numbers. Only arbitrarily large ones. 3 is correct as long as the infinities are not included.
 January 11th, 2016, 08:10 AM #6 Newbie   Joined: Jan 2016 From: House Posts: 7 Thanks: 0 Thanks! So as long as I treat infinitesimals as some abstract animals with no corresponding numerical values, I should be fine, right?
 January 11th, 2016, 11:36 AM #7 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 Is 0.999... exactly equal to 1.000...? Intuitively, it seem just a little bit less. But in the reals it can be proved to be EXACTLY equal. However, in the hyperreals, 0.999... is an infinitesimal less than 1.000... (harmonizing with our intuition). Thanks from greg1313 Last edited by Timios; January 11th, 2016 at 11:39 AM.
January 14th, 2016, 12:23 PM   #8
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 Originally Posted by Timios Is 0.999... exactly equal to 1.000...? Intuitively, it seem just a little bit less. But in the reals it can be proved to be EXACTLY equal. However, in the hyperreals, 0.999... is an infinitesimal less than 1.000... (harmonizing with our intuition).
It's my understanding that due to the transfer principle, .999... = 1 is a theorem of the hyperreals. https://en.wikipedia.org/wiki/Transfer_principle

If someone who knows more about this than I do can put the transfer principle into context for me, I'd appreciate it. Like I say, it is my understanding that .999... 1 is a theorem even about the hyperreals; and that (in my opinion) the hyperreals are therefore a tremendous distraction in the online discussions of whether and why .999... = 1.

Last edited by Maschke; January 14th, 2016 at 12:42 PM.

 January 14th, 2016, 01:43 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra In the hyperreals they are different, but the transfer principal would mean that they are equal in the reals. But the transfer principal is a huge fudge. It's just a formal statement of the idea that you ignore the infinitesimal part of your answer. Admittedly, there's a proof that goes with it, but I find the whole idea just as unsatisfactory as the non-formalised version. Of course, people don't bother with the result you state because people who think the two are representations of the same number don't need the hyperreals to prove it to themselves.
January 14th, 2016, 04:00 PM   #10
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 Originally Posted by v8archie IAdmittedly, there's a proof
You say that like it's a bad thing!! LOL.

We are agreed then that the statement ".999... = 1 is true in the hyperreals," is true. Even if its meaning (that the statement only refers to standard reals embedded in the hyperreals) makes you unhappy.

But that is true about many things in standard set theory too. Maybe I don't like the Axiom of Choice, or I don't like the powerset axiom. (Believe it or not this is a bit of a controversial axiom. It's very powerful, many of the so-called paradoxes really arise from accepting powersets).

Nevertheless, I still agree that various theorems are logical consequences of those axioms, and I don't spend any emotional energy being unhappy about it.

That's my point about the hyperreals. I'm making a pedagogical point. If someone is asking about .999..., and they've heard of the hyperreals, in my opinion the best thing to say is, ".999... = 1 is a theorem in the hyperreals too, so let's see if we can just gain some understanding of the standard real numbers -- which are complicated enough!"

The people confused about .999... are exactly the people who need a better grounding in the standard reals, and who are not capable of understanding the context of the hyperreals. For one thing they need more than ZF, since you need some form of choice to invoke a nonprincipal ultrafilter to show the existence of the hyperreals. This is Robinson's construction. And I note in passing that in the 40 years since Keisler's calculus text based on nonstandard analysis, nobody else has taken up the cause, and calculus is still taught via the traditional concept of limits. And I did see a study that shows that teaching calculus via infinitesimals is no better than limits. The students get confused by calculus no matter what you do.

So my point is that taking the hyperreals seriously isn't helpful to people seeking to understand why .999... = 1. Rather, the mention of hyperreals should be dismissed virtually out of hand. Not because it's not interesting from a more sophisticated point of view; but because beginners should be led to a proper understanding of the plain old vanilla reals first.

My two cents, speaking as a grizzled veteran of many online .999... discussions.

Last edited by Maschke; January 14th, 2016 at 04:05 PM.

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