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Yes, but your argument goes circular with this. And someone could argue that .999.../2 = .499... <=> .999... = 2*.499... = .499...+.499... which "surely m

by jonsen 4y ago

Yes, but your argument goes circular with this. And someone could argue that .999.../2 = .499... <=> .999... = 2*.499... = .499...+.499... which "surely must" have 8 as the rightmost digit. Moving it further away from 1.

peterhalburt33 4y ago
Agreed this isn’t airtight, but if you accept the premise that real numbers are specified by their sequence decimal digits: .499…+.499… has 9 as its first digit after the decimal, 9 as its second and so on - the only other number that has this decimal expansion is .999… infinitely repeating, so these must be the same numbers.