This little guy craves the light of knowledge and wants to know why 0.999... = 1. He wants rigour, but he does accept proofs starting with any sort of premise.

Enlighten him.

  • AssortedBiscuits [they/them]
    ·
    4 months ago

    (I haven't done real math since forever)

    Part 1: Proving If x,y are distinct positive real numbers and x < y, there is a z in R such that x < z < y.

    Let y = x + e, where e is in R. e is also positive because:

    x < y
    x < x + e
    0 < e

    x < z < y becomes x < z < x + e. We can then choose z = x + 0.5e so that

    x < z < y
    x < x + 0.5e < x + e
    0 < 0.5e < e
    0.5e - 0.5e < 0.5e + 0 < 0.5e + 0.5e
    -0.5e < 0 < 0.5e

    Since we've shown earlier that e > 0, -0.5e < 0 < 0.5e is true no matter what e.

    Part 2: Decimal Notation

    Decimal notation is written as the sequence amam-1...a1a0.b1b2...bn, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence amam-1...a1a0.b1b2...bn itself is of the number am10m + am-110m-1 + ... + a0100 + b110-1 + b210-2 + ... + bn10-n.

    Part 3: The actual proof through proof by contradiction

    Let x = 0.999... and y = 1. This means there exists a z that is between 0.999... and 1. So let's construct z = a0.b1b2...bn. a0 has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a0=0. z = 0.b1b2...bn

    0.999... = 9*10-1 + 9*10-2 + 9*10-3 + ... while z = b110-1 + b210-2 + ... + bn10-n. If 0.999... < z, then

    0.999... - z < 0
    9*10-1 + 9*10-2 + 9*10-3 + ... - (b110-1 + b210-2 + ... + bn10-n) < 0
    (9*10-1 - b110-1) + (9*10-2 - b210-2) + (9*10-3 - b310-3) + ... < 0
    (9 - b1)*10-1 + (9 - b2)*10-2 + (9 - b3)*10-3 + ... < 0

    But as we've established in Part 2, b1, b2, b3, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - bn) > 0 for any n. Therefore, (9 - b1)*10-1 + (9 - b2)*10-2 + (9 - b3)*10-3 + ... cannot be less than 0.

    The theorem requires for x and y to be distinct positive real numbers and x < y, and since 0.999... and 1 can be trivially shown to be positive real numbers, this means that 0.999... and 1 are not distinct. In other words, 0.999... = 1.