Math Is Fun Forum

Hi guys, well this is a really long thread. I can see why, I remember I had a really hard time convincing my friends of this fact as its quite hard to imagine an infinitely long representation for a number which can be represented using one digit!!
I read the first and last page and I think I might just jump in and like Khushboo, play around with an infinite sum.

Now the proof's I've seen using the geometric series look perfectly valid and concise to me but here is another way which is longer but is a little more transparent so in my opinion it will hopefully provide some insight to those of you who are of a curious nature.

We will try to give a formal proof for the following sketch proof:
10*0.99... = 9.99...
so 10*0.99... - 0.99.. = 9.99.. - 0.99..
so 9*0.99.. = 9
so 0.99.. = 1

It is aimed at those of you who have studied sums and limits.
(Sorry in advanced about the normal text, I am new to the site and couldn't find how to add latex code lol)

let an=9/(10^1) + 9/(10^2) + 9/(10^3) + ...... + 9/(10^n)    (i)

Note that the above is just stating that an = 0.999.... where there is n 9's after the 0.
So an=0.9999... as n approaches infinity.
So lim(an) = 0.99999..... where lim(an) is the value of the sum an as n approaches infinity.

10an = (9*10)/(10^1) + (9*10)/(10^2) + (9*10)/(10^3) + ..... + (9*10)/(10^n)
So 10an = 9/(10^0) + 9/(10^1) + 9/(10^2) + ..... + 9/(10^(n-1))     (ii)

Now we subtract equation (i) from equation (ii)

10an - an = [9/(10^0) + 9/(10^1) + 9/(10^2) + ..... + 9/(10^(n-1))] - [9/(10^1) + 9/(10^2) + 9/(10^3) + ...... + 9/(10^n) ]

so 9an = 9/(10^0) + 9/(10^1) + 9/(10^2) + ..... + 9/(10^(n-1)) - 9/(10^1) - 9/(10^2) - 9/(10^3) - ...... - 9/(10^n)

so 9an = 9/(10^0) + [9/(10^1) - 9/(10^1)] + [9/(10^2) - 9/(10^2)] + ..... + [9/(10^(n-1) - 9/(10^(n-1)] - 9/(10^n)

so 9an = 9/(10^0) + [0] + [0] + ..... + [0] - 9/(10^n)

so 9an = 9/1 - 9/(10^n)

so an = 1 - 1/(10^n)

Now we just find the limit as n approaches infinity
lim(an) = lim(1) + lim(1/(10^n))
so lim(an) = 1 - 0
so lim(an) = 1
but remember lim(an) = 0.9999....
So 0.9999.... = 1

Cool, I hope thats all correct, I know its not the most elegant way of doing it but my aim was just to provide a little insight.

If you want to give it a try, You can use a similar technique to show 0.33... = 1/3.