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## Main Question or Discussion Point

Represent .999... by (1-10

Then

(1-10

(1+10

(1)

I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...

If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.

The above equation seems to converge with the number of digits of accuracy approximately equal to n.

^{-n}), limit as n>infinity.Then

(1-10

^{-n})^{10n}, limit as n>infinity = 1/e (binomial expansion)(1+10

^{-n})^{10n}, limit as n>infinity = e(1)

^{10n}, limit as n>infinity=1I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...

If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.

The above equation seems to converge with the number of digits of accuracy approximately equal to n.