On 0.999...

While trying to explain to someone that 0.999... aka 0.(9) is equal to 1 - not almost equal, but equal - one of the "arguments" he used was that recurring decimals are not really numbers, you can't do mathematical operations with them. While the idea is mind-boggling - 0.(1) is 1/9 which is a rational number - I thought of a proof that doesn't use recurring decimals: using base 9.

0.(1)10 = (0.1)9

(0.0)9 + (0.1)9 = (0.1)9
(0.1)9 + (0.1)9 = (0.2)9
(0.2)9 + (0.1)9 = (0.3)9
(0.3)9 + (0.1)9 = (0.4)9
(0.4)9 + (0.1)9 = (0.5)9
(0.5)9 + (0.1)9 = (0.6)9
(0.6)9 + (0.1)9 = (0.7)9
(0.7)9 + (0.1)9 = (0.8)9

Finally,

(0.8)9 + (0.1)9 = ?

Well... in any numeration base, when you finished the digits you restart at 0 and add 1 to the next order, so:

(0.8)9 + (0.1)9 = (1.0)9

Recasting the above in base 10, we have

0.(8) + 0.(1) = 1.0

However, in base 10 we still have a digit left, so it is equally valid to write

0.(8) + 0.(1) = 0.(9)

just as it is valid to write 0.(7) + 0.(1) = 0.(8).

Given that the left side is identical, it follows that 1 = 0.(9). QED.

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