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by jonsen 1514 days 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.
1 comments

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.