Re: Question


Virgil wrote:
In article <1146845383.856952.102270@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Virgil wrote:
In article <1146742858.929914.154140@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Virgil wrote:

You say that 0.9999........ equals one.

Let me pose a simple argument against that. In the process of dividing
one by one
now say it is zero the remainder will be one, now put zero to the right
of that remainder
and of coarse a dot after the zero, the number after the dot will be 9
and the remainder will be 1, the same thing will be repeated, and it is
always with a remainder of 1.

WRONG!

The first "remainder" is 0.1 since it is 0.9 + 0.1 = 1.0.
The second is 0.01, since 0.99 + 0.01 = 1.00
The next is 0.001, since 0.999 + 0.001 = 1.000
So the "remainders" a la Zuhair, converge to zero, and the finite
sequences of nines converge to 1.

Now do you think if you continue that infinitelly the remainder will be
zero.

It converges to zero.

I don't think so

The more fool you.

I agree that it converges to zero, but the question is would it reach
zero.
I don't think it will REACH zero.

The mathematical "value" of a sequence, insofar as any sequence has a
"value", is its *limit*, not necessarily any of the values that it ever
"reaches". Whether any term in that sequence actually equals the limit
is immaterial.

According to Zuhair¹s analysis, decimals exclude the majority of reals.

Not only decimals, in effect all number systems of the kind binary,
tertiay, ..., decimal,..
all these exclude the majority of reals.

Zuhair

.

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