# Solve this math?

• B
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

Therefore 0.999... = 1 is a contradiction

Mark44
Mentor
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.
Fala483 said:
By linear relation all g(n) end in 8
No
Fala483 said:
Assume 0.999... = 1

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

1 x 1 does not end in 8

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