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:{{re|Krzysztof1137}} Unfortunately, your table contains only numbers with finite decimal representation. Arbitrarily long, but finite. So it does not even contain 1/3 or 3/7. As such it has nothing to do with a proof of uncountability of a set of all possible decimal strings. --[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 19:40, 20 April 2025 (UTC)
::The matrix was created only to contradict the diagonal method. It arranges numbers with all possible combinations of digits after the decimal point. You write: "Your table contains only numbers with a finite decimal representation." - Yes, but if you analyze the process of creating numbers in the diagonal method step by step, it turns out that each subsequent digit added creates a sequence of rational numbers, the same sequence of rational numbers as in any row of the matrix. You write: "Any length, but finite." - No. There is an infinite number of columns in the matrix, so in each row there is an infinite sequence of rational numbers, which means the matrix is not finite. You write: "So it does not even contain 1/3 or 3/7." - I wrote earlier that if you need irrational numbers, we can assume that the matrix contains the initial digits, and the remaining ones can be added, e.g.; in row 3, column A is 0.333... , the diagonal method does not check if all the numbers are there, it only creates a new number that is not on the list. A question for you; if in the following columns A, B, C... the number 0.3 , 0.33 , 0.333 ... etc. appears, does it mean that it approaches 0.333...? Can we say that it is 1/3? If so, then 1/3 exists, if not, the diagonal method will not create an irrational number. You write; "As such, it has nothing to do with the proof of the uncountability of the set of all possible decimal sequences." - I did not write anything about countability, I only mean the diagonal method itself. [[User:Krzysztof1137|Krzysztof1137]] ([[User talk:Krzysztof1137|talk]]) 21:18, 20 April 2025 (UTC)
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