Re: 0.999... = 1? (I know, a beaten dead horse)



In <1125096661.612883.125870@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, on
08/26/2005
at 03:51 PM, shepherdmoon@xxxxxxxxx said:

>My question is about how this equality fits in with the concept of a
>real number.

That's not the key issue; the key issue is what you mean when you
write "0.999...".

>I always thought that the set of real numbers was such
>that they had a one-to-one correspondence to points on a line and
>that they had an order to them, i.e., any given real number is
>either greater than or less than any other real number.

Correct.

>But if 0.999... = 1 and both are real numbers, then that would mean
>that there are at least two real numbers that are equal to each
>other.

No. That would mean that there are two strings of characters that name
the same number. Does it distrub you that 2=two?

>I understand that some people say that 0.999... and 1 are just two
>ways of representing the same number. I get that, but I thought the
>point of the real numbers is that it is a set where each member is
>unique.

Well, yes, just as each student in a classroom is distinct, bout
Robert, Bob and Bobbie might all be the same student.

>But in the case of 0.999... = 1, I can't understand how both can be
>in the same set of real numbers and be equal to each other too.

The same way that Bob and Robert can be the same person.

>I think it has to do with the fact that 0.999... isn't really a
>number you can calculate with since the string of 9s is infinite

You're confusing a number with one of its names.

>so it is really more of a limit representation.

It's an abbreviation for lim n->oo Sigma i=1:n 9^{-i}; what is that
limit?

>But that makes me think, well, is there a
>discontinuity in what we think of as the real number line?

No.

>If 0.999... is equal to 1, then what value is given to the point just
>to the left of 1 on the real number line?

There is no such point. In fact, Euclid give a construction for a
point midway between any given pair of points.

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