How does #0.99999....=1#?

2 Answers
Mar 13, 2016

Answer:

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#

Jul 17, 2016

Answer:

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#