Re: Question


José Carlos Santos wrote:
zuhair wrote:

The only representation of 1 is 1.0000......... and to me
0.9999........... < 1.
Then you are wrong. If that expression represented a number strictly
smaller than 1, call it x, then (1+x)/2 would be a number strictly
smaller than 1, and strictly larger than x.

What, pray tell, is the decimal representation of this number?
How come I didn't read that post?

Anyhow I should answer it.

I will change your terminology.

Let x = 1- 0.9999..............
That's a rather strange way to answer to Arturo, since he told you to
call _x_ to 0.9999... and you choose to call _x_ to 1 - 0.9999....

Now 1+ 0.99999... = 2 -x

Now (2 - x)/2 = 2/2 - x/2

now x/2= x
How did you get that? But if you think (as I do) that this is a true
statement, then the conclusion is, of course, that x = 0. Since you
defined _x_ as 1 - 0.9999..., it follows that 1 = 0.999....

Now, what about answering Arturo's question. If x = 0.9999..., what is
the decimal representation of (1 + x)/2?
I answered Arturo's question.

Let me repeat it again but I will use Arturo's terminology.

x= 0.9999.......

Let y = 1-x

it follows that x= 1-y

Now 1 + x = 1+ 1- y = 2 - y

Now y/2 = y
Two questions here:

1) How do you know that y/2 = y?

because y is an infinite number since, 0.9999...... is infinite, and if
you divide this
infinite number by any finite number you will get the same number.

I don't know what an "infinite number" is. Do you have a reference?

By the way, is 0,111111111111... an infinite number too?

According to my way of thinking , Yes. But these are only my dam
nonsensical misconceptions as standard mathematicians say. The whole
idea is not so serious.

1- 0.999........ = 1/(10^Aleph-0)

[1/(10^Aleph-0)]/n = 1/(10^Aleph-0)

This comes from the equation

[x/n] + [x/(n^2) ] + [x/ (n^3)] +.....+ [x/(n^k) ]= {x/(n-1)} - {
x/[(n^k)(n-1)]}

for n=2,3,4,5,........
K=1,2,3,4,.........

if we regard k as the number of terms in the sequence above.

Then when that sequence is infinite then = Aleph-0

so for x=9, n=10 and k=Aleph-0 we will have 0.9999...........= 1-
[1/(10^Aleph-0)]

Now whatever is the division operator in 1/(10^Aleph-0) is, the result
cannot equal zero.

and 1/(10^Aleph-0) > 0

So 0.9999........ <1

I have no idea what most of this means.

Best regards,

Jose Carlos Santos

Dear Santos. All of these ideas are only mine, they are not accepted by
standard mathematicians like Arturo , they consider them nonsense, Even
I don't have a solid believe in them, I am just figuring them out. So
don't take them seriouselly.

Yours
Zuhair

.

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