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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.
You will probably not read this, but even so I would like to call your
attention to the fact that 0,11111111... = 1/9. On the other hand, you
wrote that half of an infinite number is that same number. It follows
that (1/9)/2 = 1/9. Don't you see anything strange here?
Best regards,
Jose Carlos Santos
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