Question

How does `0.99999....=1`?

Answer 1

See explanation.

But,this question was asked by me,But I realized the proof and wanted it to be known for others...

Explanation:

Let `0.99999....=x`

`rarr9.9999....=10x`

Subtract `x` both sides

`rarr9.9999....-0.9999....=9x`

`rarr9=9x`

`rArrx=1`

Answer 2

The number `0.9999 ...= sum_(n=1)^ oo9/10^n`,

that is the sum of the series starting at `n=1`

Explanation:

The number `0.9999 ...= sum_(n=1)^ oo9/10^n=9 * sum_(n=1)^ oo1/10^n= 9 * (1/10)^1/(1-1/10)`, since the sum of the geometric series `sum_(n=1)^ oo1/10^n=(1/10)^1/(1-1/10)`

Then the series `sum_(n=1)^ oo9/10^n= 9 * (1/10)^1/(1-1/10)=9*(1/10)*(10/9)=1`,

and so `0.99999 ...=1`