Solve Math Puzzle: 0.999... ≠ 1

[Fala483]

Let f(n) be of the form F(1) = 0.999999 F(2) = 0.9999999 F(3) = 0.99999999 Etc

For arbitrary N >= 1

Let g(n) = f(n)^2

G(1) and g(2) end in 8

By linear relation all g(n) end in 8

Assume 0.999... = 1

0.999... X 0.999... must end in 8 per (4)

1 x 1 does not end in 8

By (5),(6),(7), contradiction

Therefore 0.999... = 1 is a contradiction

 

[Mark44]

Let f(n) be of the form F(1) = 0.999999 F(2) = 0.9999999 F(3) = 0.99999999 Etc

For arbitrary N >= 1

Let g(n) = f(n)^2

G(1) and g(2) end in 8

No. How are you getting 8 as the final digit?
Let's start in a more organized fashion.
$f_1 = 0.9$
$f_2 = 0.99$
$f_3 = 0.999$
Then $f_1^2 = 0.81$
$f_2^2 = 0.9801$
$f_3^2 = 0.99801$
and so on. The sequence $\{f_1, f_2, f_3, \dots \}$ is approaching 1, which is easy to prove,
and it can easily be shown that the sequence $\{f_1^2, 4_2^2, f_3^3, \dots \}$ also has 1 as its limit.

By linear relation all g(n) end in 8

No

Assume 0.999... = 1

0.999... X 0.999... must end in 8 per (4)

1 x 1 does not end in 8

By (5),(6),(7), contradiction

Therefore 0.999... = 1 is a contradiction

No.
Thread closed.