Question

Why 0.9999... = 1

Answer 1

Suppose \(x = 0.99999...\)

multiply both sides by 10
\(10x = 9.99999...\)

subtract x from both sides

\(~10x ~~~~= ~~~~9.99999...\)
\(~- x ~~~~~~~~-0.99999...\)
_______________________________
\(~~~9x~~~~ =~~~~ 9\)

divide both sides by 9

\(x = 1\)

Answer 2

I just learned it the other day and it blew my mind

Answer 3

don't know why we don't learn more cool stuff like that in math classes :P

Answer 4

That's so cool
Never heard of that before

Answer 5

The weird part is, my teacher randomly brought that up a year or 2 back x'D

Answer 6

Abstractly speaking, that is not contingent to theoretical mathematics. In theoretical mathematics ventures to omit from deducing that x = 1, for such that is it equal to. Rather, can be any number we cannot grasp, and therefore is not possible to state x = 1; wherein, this become 9x = 9.999.. - x.

Instead, factoring would make it 9(x) = -(-9.999 + x). I reckon you could reconcile the arbitrary of the 'x' combination.

Answer 7

Oh mai gawd its magical

Answer 8

Amazing stuff.

\(\frac{1}{3}\ = 0. \dot 3\)

\(3 \left( \frac{1}{3}\ = 0. \dot 3 \right) \implies 1 = 0. \dot 9\)

is it a limit thing?

i mean if we say that \(0.9_1 = 0.9, 0.9_2 = 0.99, 0.9 _ 3 = 0.999\) then we can say that \( \lim\limits_{n \to \infty} 0.9_n = 1\)