1. ## Infinite periodic decimal.

Guys chill,
I know this is probably the dummest question that have been ever uploaded in this forum, but I have to know that answer!

So if 1/3=0.3333333(and so on)
then how can 1/3 *3 = 0.33333333333 *3 = 1
why isn't it 0.9999999(and so on) since 3*3=9, not 10.
I know that if we do it in a fraction we will get 1 as 1*3/3 = 3/3 = 1
but how? is the answer really one or is it 0.9999999999999999999999999 (infinity) rounded to 1?

thanks

here it is:

An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer.
The bartender says "I see where this is going", and pours two beers.

so if 1+0.5+0.25+0.125 goes on to infinity, will it reach 2, the numbers get smaller as we go, hmmmm

3. Originally Posted by igcsestudent
Guys chill,
I know this is probably the dummest question that have been ever uploaded in this forum, but I have to know that answer!

So if 1/3=0.3333333(and so on)
then how can 1/3 *3 = 0.33333333333 *3 = 1
why isn't it 0.9999999(and so on) since 3*3=9, not 10.
I know that if we do it in a fraction we will get 1 as 1*3/3 = 3/3 = 1
but how? is the answer really one or is it 0.9999999999999999999999999 (infinity) rounded to 1?

thanks
\displaystyle \begin{align*} x &= 0.999\,999\,999\dots \\ 10x &= 9.999\,999\,999\dots \\ 10x - x &= 9.999\,999\,999\dots - 0.999\,999\,999\dots \\ 9x &= 9 \\ x&= 1\end{align*}

Therefore $\displaystyle 0.999\,999\,999\dots = 1$

4. The notation 0.99999... means 0.9+ 0.09+ 0.009+ ...= 0.9(1+ .1+ .01+ ...), an infinite geometric sum. It is well known that the geometric sum, $\sum_{n=0}^\infty ar^n= a\sum_{n=0}^\infty r^n$, as long as |r|< 1, is $\frac{a}{1- r}$. In this case, a= 0.9 and r= .1< 1 so the sum is $\frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1$.

There is no rounding. That "infinite decimal" is exactly equal to 1.