Re: Problems I have with 1.999...=2

abe.buckingham@xxxxxxxxx wrote:

Thank you for the time you took, and for the consideration in using small words that a child or layman might understand. Still, I must have zoned out or missed something near the end, because I was aware that you lost me about there.
....Ah. I think I got it, after rereading it. Here:
<snip> 
So finally we can see how we've come full circle, we've defined what we
mean by these infinite decimal expansions and from the context of
geometry slowly built up the properties we wanted them to have using
the rational numbers and our own reasoning. Since the above defines
precisely what we mean by two of these infinite sequences being the
same, we now turn to the question of 1.999... =2. In this case we have
two sequences, the first being (1, 1.9, 1.99, 1.999, 1.9999, ....) and
the other being (2, 2, 2, 2, 2, 2, 2, 2, ...). Both of these are
convergent sequences, so we know they're allowable, the only question
left is do they both represent the same point on the real number line?
So we turn to our equivilence relation that says if we subract these
two term by term, given some arbitrarily small number e, can we find a
spot where this difference is less then e.

Lets consider what happens when we subtract these two sequences term by
term. We get a new list, that goes like this (1, 0.1, 0.01, 0.001,
0.0001, ...) Now no matter what e I chose to start with, I could always
find a point on that new list that is less then e since e is greater
then 0 and fixed.

Granted, as stated. My point is that no term in your new list, ever, will be equal to zero.

This means we would bundle (2, 2, 2, 2,...) and (1,
1.9, 1.99,...) into the same group and would indeed call them equal in
the context we're discussing. I hope this answers you're question.


This calls for more consideration. How does my objection square with your persuasion, or does it, and, if not, what does the difference look like?
Thanks again.
.

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