4 ms·

Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and t

by esrauch 14y ago

Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and that any two distinct numbers have a number between them that is not equal to either (infinitely many in fact). You can construct alternate number lines that do allow nonzero infinitesimals and then .9999... actually is not equal to 1 under that number line; this number line is a strict superset of Reals just as Reals are a strict superset of Rationals. Most of the practical properties of real numbers would still apply to this new number line; it is actually only a historical convention note that .999... is actually equal to 1.

alok-g 14y ago
Do you know of something that I can read further on this? I wonder about the impact this may have on mathematics. I did not know of this property of real numbers of nonzero infinitesimals. (Possibly there is where my confusion is starting from.)
- Evbn 14y ago
  Look up nonstandard analysis. It is a formalism that can define calculus, but doesn't have any practical difference on physical computations on real world entities.
  - Someone 14y ago
    For a more playful and, some would say, practical view, read "Winning Ways for Your Mathematical Plays" (http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathemati...)
gizmo686 14y ago
There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system. Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using basic algebra, that for any 2 distinct numbers, their is a third number between them. If that were not the case, then at least one of my steps must have been invalid.
- ColinWright 14y ago
  > There do exist number systems where .999.../=1, > however, they are not a strict superset of the reals. The hyper-reals of non-standard analysis are, in fact, a strict superset of the reals. > If it were, then any operation involving only real numbers > would behave identically to the real number system. Why is that a problem? Seems to me that that's desirable.
  - gizmo686 14y ago
    > The hyper-reals of non-standard analysis are, in fact, a strict superset of the reals. And .9999==1 in the hyper-reals >>If it were, then any operation involving only real numbers >>would behave identically to the real number system. >Why is that a problem? Seems to me that that's desirable. Consider the expression ".999... - 1" in the real number system. As has been established earlier, this simplifies to 0. Consider the same expression ".999... - 1" in the hyper-real number system. As all the numbers are real, this expression simplifies just as it would in the real number system. Meaning that ".999... - 1 =0" in the hyper-real system. Adding 1 to both sides ".999... = 1" in the hyper-real system. One could also argue that because the statement ".999... = 1" is true in the real number system, and the hyper-reals are a superset of the reals. Then ".999... = 1" is true in the hyper-reals.
    - ColinWright 14y ago
      Ah, I see your point. I was trying to address the more general "intuition" that people have about this, that 0.9999... falls short of 1.0, and that there's some sort of infinitesimal between them. You can use that to talk about the hyper-reals, but, as you say, it still doesn't "solve" the "problem" that people perceive. So you're right, and I was answering a different (although related) question.
- esrauch 14y ago
  Thanks for clarifying the strict superset part, I'm not actually a mathematician. To clarify my convention comment; it is only convention that mathematicians have decided that any given set of properties are useful or interesting to be used pervasively and alternate number systems are not. To any person who is asking why .9 repeating is 1 those reasons are entirely outside the scope of their knowledge and so entirely unrelated to the question; is it not true that the answer is "based on the axioms of which this number line is created due to complex reasons that you can't possibly know at this stage, this is effectively decided to be true as an axiom". If one of these alternate number lines was in common use and reals were only uncommonly used then I'm pretty sure there would be tons of people saying "why is .99.. NOT equal to 1 in ($reals-replacement)" and all the people who have been taught it would sigh condescendingly because they believe it is just an intuitively obvious feature of numbers and not simply an axiom of the number line they are choosing to use. I'm interested in whether you disagree with that sentiment. Re: your proof, trivially could be refused based on the other number line not being closed over division, then there would simply be no such number Z=(X+Y)/2 for X=.999.. and and Y=1. Then the proof where be like those silly ones where someone uses a /0 and arrives at a contradiction. Actually it seems that reals themselves already are not closed over division since x/0 doesn't result in a real.

by esrauch 14y ago

alok-g 14y ago

Do you know of something that I can read further on this? I wonder about the impact this may have on mathematics. I did not know of this property of real numbers of nonzero infinitesimals. (Possibly there is where my confusion is starting from.)

Evbn 14y ago
Look up nonstandard analysis. It is a formalism that can define calculus, but doesn't have any practical difference on physical computations on real world entities.
- Someone 14y ago
  For a more playful and, some would say, practical view, read "Winning Ways for Your Mathematical Plays" (http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathemati...)

gizmo686 14y ago

There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system. Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using basic algebra, that for any 2 distinct numbers, their is a third number between them. If that were not the case, then at least one of my steps must have been invalid.

ColinWright 14y ago
> There do exist number systems where .999.../=1, > however, they are not a strict superset of the reals. The hyper-reals of non-standard analysis are, in fact, a strict superset of the reals. > If it were, then any operation involving only real numbers > would behave identically to the real number system. Why is that a problem? Seems to me that that's desirable.
- gizmo686 14y ago
  > The hyper-reals of non-standard analysis are, in fact, a strict superset of the reals. And .9999==1 in the hyper-reals >>If it were, then any operation involving only real numbers >>would behave identically to the real number system. >Why is that a problem? Seems to me that that's desirable. Consider the expression ".999... - 1" in the real number system. As has been established earlier, this simplifies to 0. Consider the same expression ".999... - 1" in the hyper-real number system. As all the numbers are real, this expression simplifies just as it would in the real number system. Meaning that ".999... - 1 =0" in the hyper-real system. Adding 1 to both sides ".999... = 1" in the hyper-real system. One could also argue that because the statement ".999... = 1" is true in the real number system, and the hyper-reals are a superset of the reals. Then ".999... = 1" is true in the hyper-reals.
  - ColinWright 14y ago
    Ah, I see your point. I was trying to address the more general "intuition" that people have about this, that 0.9999... falls short of 1.0, and that there's some sort of infinitesimal between them. You can use that to talk about the hyper-reals, but, as you say, it still doesn't "solve" the "problem" that people perceive. So you're right, and I was answering a different (although related) question.
esrauch 14y ago
Thanks for clarifying the strict superset part, I'm not actually a mathematician. To clarify my convention comment; it is only convention that mathematicians have decided that any given set of properties are useful or interesting to be used pervasively and alternate number systems are not. To any person who is asking why .9 repeating is 1 those reasons are entirely outside the scope of their knowledge and so entirely unrelated to the question; is it not true that the answer is "based on the axioms of which this number line is created due to complex reasons that you can't possibly know at this stage, this is effectively decided to be true as an axiom". If one of these alternate number lines was in common use and reals were only uncommonly used then I'm pretty sure there would be tons of people saying "why is .99.. NOT equal to 1 in ($reals-replacement)" and all the people who have been taught it would sigh condescendingly because they believe it is just an intuitively obvious feature of numbers and not simply an axiom of the number line they are choosing to use. I'm interested in whether you disagree with that sentiment. Re: your proof, trivially could be refused based on the other number line not being closed over division, then there would simply be no such number Z=(X+Y)/2 for X=.999.. and and Y=1. Then the proof where be like those silly ones where someone uses a /0 and arrives at a contradiction. Actually it seems that reals themselves already are not closed over division since x/0 doesn't result in a real.