Re: Idle Thoughts

"Rudy Velthuis [TeamB]" <velthuis@xxxxxxxxx> wrote
> somebody wrote:
> > "Rudy Velthuis [TeamB]" <velthuis@xxxxxxxxx> wrote
> > > Lauchlan M wrote:

> > > > > No, it isn't. 0.33... is an approach, the best you can do with
> > > > > decimals, no matter how many you use. It is not equality. The
> > > > > only thing equal to 1/3 is 1/3.

> > > > and the limit as integer n tends to infinity of the series 3/10 +
> > > > 3/100 + . . . + 3/(10 to the power n)

> > > Yes, the limit.

> > and 0.3(bar) *is defined* as such.

> No, the limit (for the number of digits going to infinity) of 0.3(bar)

See, now, that makes no sense. As it is, 0.3(bar) encapsulates the limit.

0.3(bar) = = Lim{n:1->inf; Sum[ 3/10^n ] } = 1

In other words, 0.3(bar) *already* would have an infinite number of 3's if
you were to write it out in full. How are you now going to take a limit of
something that's already there? Please write out the expression you are
talking about, which is yet another limit of the limit:

Lim{n:1->inf ; 0.3(bar) }

Where's the n dependence of 0.3(bar) ?

> is defined as such, not 0.3(bar) itself. You seem to have a problem
> distinguishing between a sequence and its limit. I agree that this is
> not so easy to grasp, but there IS a difference.

No. Just like

if (...(((x=true)=true)...)=true is redundant, so is taking a limit of a
constant:

5 = Lim{n:1->inf; 5 } = Lim{n:1->inf; Lim{n:1->inf; 5 } } = Lim{n:1->inf;
Lim{n:1->inf; Lim{n:1->inf; 5 } } } = ...


.

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