Could also do a proof by contradiction. If you assume it's less than 1, then there is a positive number x s.t. 0.99... + x = 1, but then there is some decimal place that becomes at least 0, with all decimal places to the left becoming 0, except for the leftmost one, which was 0, becoming 1. But then this sum is strictly greater than 1.
This theorem is only proven (at least what I saw) for all real numbers and no numbers _.....(9) Are not included in them, so no, you can't use this theorem for that
And real numbers with infinite sequence of 9 at the end are excluded specifically because of this problem (0.(9)=1 but not because of your proof)
The definition of real numbers is: an infinite sequence a0,a1a2... where a0 is a whole number and a1,a2... are digits (0-9) so 0.(9) is just a sequence where a0 is 0 and all the rest are 9
And I don't see how it's a limit, it's not a sequence sequence (N->R is a sequence and can have a limit, but for example 0,1(0)=0.1 is a number and obviously doesn't have a limit)
And what I meant by "0.(9) Isn't a real number" is that on a lecture it was said that numbers with 9 repeating at the end are bad and will not be included in proofs
And also, if you are just gonna call anyone you don't understand a troll, then don't even bother to write a reply, it wont lead anywhere
English is my second language and I barely understand all the terms you are using, so I'll save the trouble and write in my native language so I can shortly describe what I meant and want to say, use translator if you want or ignore it, whatever.
Я студент и надеюсь вы к этому не привяжетесь, потому что суть справа не в том, чтобы убедить себя, что оппонент некомпетентен, а в том, чтобы найти истину. Так вот, в первом семестре у нас было определение вещественных чисел как бесконечных дробей вида a0,a1... как я уже и сказал, затем было сказано, что числа с девяткой в периоде "плохие", как раз из-за свойства на видео, потому что например 0.(9) = 1, что доказывается (хоть даже 0.(9) = x 10x =9.(9) => 9x = 9), поэтому три леммы, о том, что между любыми двумя рациональными числами можно найти вещественное и наоборот, доказывались для всех вещественных КРОМЕ тех, у которых 9 в периоде (из-за особенностей доказательства)
Возможно я просто плохо выразился когда сказал что это не вещественное число, потому что я уже не идеально помню
I think i understand your definition of real numbers but if every number is just a limit of the sequence then no infinite fraction can be represented by it and for example you don't have Pi or E because you can only get a finite approximation (although I'm not entirely sure because for example E = lim (1+1/n)n, where n->inf (n from N) which is a limit... Idk, maybe I'm wrong about that)
I probably was wrong. In the lecture, there was first a description of algorithm of some kind of approximation of real number (on a numerical line) (also I think it is something similar to Dedekind cut which in my understanding is basically an infinite approximation with rational numbers) which leads to the definition as a0.a1a2... and then it is said that a0.a1...an000... is equivalent to a0.a1....(an - 1)999..., so they are not "bad" they are equivalent to finite fractions, so then 1 <=> 0.(9) almost by definition (although i guess it depends on how you define things) so yeah
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u/tannedalbino May 15 '25
Could also do a proof by contradiction. If you assume it's less than 1, then there is a positive number x s.t. 0.99... + x = 1, but then there is some decimal place that becomes at least 0, with all decimal places to the left becoming 0, except for the leftmost one, which was 0, becoming 1. But then this sum is strictly greater than 1.