Re: Repeating decimals are irrational

wij <wyniijj5@gmail.com> writes:

On Tue, 2024-03-26 at 17:53 -0700, Keith Thompson wrote:

wij <wyniijj5@gmail.com> writes:

On Tue, 2024-03-26 at 17:28 -0700, Keith Thompson wrote:

[...]

So you're saying that 0.333... is not exactly equal to 1/3.
 
It seems odd that you agree that 0.999... is exactly equal to 1, but
0.333... is not exactly equal to 1/3.
 

 
I say the limit of 0.999... is 1, not 0.999... is 1. (this is also what you asked)
Read the definition carefully from trustworthy website.

 
So you're distinguishing between "the limit of 0.999..." and "0.999...".
 
I see no difference between them.  To me, the "..." notation *means* the
limit.  Can you explain what difference you see?
 
When I write "0.999...", I mean the limit as the number of 9s increases
without bound.  That limit, I think we both agree, is equal to 1.  And
perhaps we also both agree that the limit of 0.333... as the number of
3s increases without bound is equal to 1/3.
 
Are you saying that:
 
- 0.999... is something other than the limit as the number of 9s
  increases without bound?

>
Don't know what you are asking for

I don't know how to explain it more clearly.

We've both been using notations like "0.999...".  I've been using it to
mean exactly the limit as the number of 9s increases without bound.
That particular limit is exactly equal to 1.

You apparently mean something other than that limit when you write
"0.999...".  I'm asking you what you mean by "0.999...", and in
particular how that differs from the described limit.

- 0.999... is a real number?

>
Yes

Agreed.

- 0.999... is less than 1?
 

Yes

Incorrect, assuming we're talking about real numbers as that term is
universally used in mathematics.

If so, what is the real value of 1 - 0.999...?

>
Infinitesimal

"Infinitesimal" is not a real number.  No real number is infinitesimal.

If 0.999... and 1 are both real numbers, and they're not equal, then
their difference is a non-zero real number.

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
Working, but not speaking, for Medtronic
void Void(void) { Void(); } /* The recursive call of the void */