Sujet : Re: Does the number of nines increase?
De : chris.m.thomasson.1 (at) *nospam* gmail.com (Chris M. Thomasson)
Groupes : sci.mathDate : 10. Jul 2024, 22:56:42
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6n02q$231iu$3@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
User-Agent : Mozilla Thunderbird
On 7/9/2024 2:49 PM, Moebius wrote:
Am 09.07.2024 um 23:34 schrieb Chris M. Thomasson:
On 7/9/2024 2:07 PM, Moebius wrote:
Am 09.07.2024 um 22:10 schrieb Chris M. Thomasson:
On 7/9/2024 3:11 AM, FromTheRafters wrote:
>
Does that mean there are as many rationals as there are reals?
>
I already told you that: The set of rational numbers is countable infinite while the set of real numbers is _uncountable_.
>
Again: One of my math professors once tried to express this state of affairs the following way: "There are (in a certain sense) much more real numbers than rational numbers."
>
Strange that any real can be represented by a rational up to infinite precision...
What is "up to infinite precision" and "represent"?
Hint: There is NO rational number such that r = sqrt(2).
Hence for each and every rational number r there is a "non zero" difference between sqrt(2) and r.
I guess you might have a _sequence_ of rational numbers in mind, say,
(1, 1.4, 1.41, 1.414, ...).
So we might say that this SEQUENCE represents the real number sqrt(2) - in a certain sense. :-P
Actually, its limit is sqrt(2).
Well, basically, I was thinking that for any element of:
(1, 1.4, 1.41, 1.414, ...)
there is a rational that can represent it. So, it kind of makes my brain want to bleed from time to time, shit happens! Uggg. Taken to infinity, there are rationals that can represent any element of the sqrt 2:
(1, 1.4, 1.41, 1.414, ...)
However, there is no single rational that equals sqrt 2. Humm... Fair enough?
…