I've seen this snake diagonal proof and I can kind of understand the logic and how equivalent fractions eventually cancel out. However, why would something like this not disprove that the set of rational numbers is countable? Since (N/(N + 1)) will never create repeat/equivalent fractions and no fractions of that form will ever equal 1/3, how is the set of rationals countable? Shouldn't the cardinality be at least one greater than that of the naturals (infinitely more fractions that don't follow (N/(N+1)) that I could include)?
2026-04-01 09:45:28.1775036728
Why does this not disprove that the set of all rational numbers is countable?
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Since the set of natural numbers is infinite, adding a single extra fraction doesn't change its cardinality. The set of rational numbers is indeed countable, and if you look at one of the standard enumerations, you get that the subsequence using the fractions that you considered go $1/2, 1/3, 2/3, 3/4, 4/5, 5/6, ...$ See the Stern-Brocot tree and the Calkin-Wilf tree for two simple bijections.