I've done that with 0.27 recurring. It's quite a logical thought.

Neat and concise ... thanks ganesh.

I thought of it in a different way. I used the summation of a Geometric Progression and this is the result I got:-
0.9999999999999999999999999999.....................
= 0.9(1 + 1/10 + 1/100 + 1/1000 + 1/10,000 + 1/100,000 + 1/1,000,000.......)
The ∑ of the series within the brackets would be 1/(1-1/10) = 1/(9/10)
= 10/9
10/9 x (0.9) = 9/9 = 1
I know this is an approximation for an infinite series.
But 0.999999999999999999999999999999999.................... is a devilish number. You never encounter this number in any mathematical problem. I have never come across recurring 9s after the decimal in any division. It can be proved that there can be no recurring 9s in any division.
Let me prove this by a counter-solution.
Let there be a recurring decimal 0.2999999999999999999.......
When we try to convert this into a fraction, this is what happens.
x = 0.2999999999999999999............
10x = 2.999999999999999999999......
100x = 29.9999999999999999999999......
90x = 27
x = 27/90 = 3/10, which is 0.3 !
Therefore, I think 0.9999999999999..........exists only theoretically. Only in the imagination.

woah woah woah woah woah!!! slow down people! i'm thirteen! how the heck am i meant to have a clue bout what ur on about? can anyone explain SIMPLY? by the way my maths teacher says that if i like something on the scale of one to ten i can have 10.1!?!?!?! is that right? how? help!!!!!!!!!!

Well, evil, your question simply got a lot of us interested! Once started, a topic takes on a life of its own ...

The answer, in a nutshell, is that 0.9999....(recurring) does not equal 1, because it misses out by an "infinitesimal" amount.
______

On a scale of one to ten, the lowest value is 1, the highest is 10, I would have thought. Why not ask your maths teacher to come on to the forum one day and they can explain how you get 10.1!
______

(de·fer  v.  1. To submit to the opinion, wishes, or decision of another)

1/3 = 0.333...
3 x (1/3) = 1

I dare not go any further ...

If kyle is right, I have a  page to redo. Have a read of this. (I think this is the original page that "evil" found.)

I had thought it was right up until now.

Looks OK to me. You're still assuming that 0.999... ≠ 1, but it's OK this time because you're doing a proof by contradiction.
Also, you might want to turn smilies off.

wcy wrote:

a = 1
b=0.0000000000000000...1
a-b=0.99999999....
a²-b²=1²-0.000000....1=0.99999999...
(a+b)(a-b)=a²-b²
a+b = (a²-b²)/(a-b)=0.9999.../0.9999...=1
a=1
therefore b=0 !!!!
therefore a-b=0.999999...=1-0=1  !!!!!

If we are talking of infinite number of zeros after the decimal, before the 1, in b,
the proof is correct!
Because, a²-b² would be equal to a-b.
This is because we have defined 0.0000000000.......1 as 1 ater an infinite number of zeros after the decimal followed by 1,
as kylekatarn denotes 0.(0)1.


However, if the number of zeros is finite, (Don't worry, it is not)
a²-b² would contain more 9s than a-b.

I tend to agree with you, kyle, the proof I have on the website has that niggly little thing that assumes that:

If x=0.999... , and 10x = 9.999...

Then the fractional part of 9.999... is still the SAME 0.999... as x=0.999...

So perhaps that is a flawed proof.

I want to replace it with something new, but it needs to appeal to a wide audience, so possibly we could condense the interesting arguments here into something really good.

This proof will probably prove once and for all that 0.999..=1

0.9999...=0.9+0.09+0.009+0.0009+...
This is clearly a Geometric Progression, so we can use the formula Sum to infinity=a/(1-r), where a is the first term and r is the common ratio (0.1)

Hence, the sum to infinity=0.9/(1-0.1)=0.9/0.9=1 !!!

Proof of GP formula:
(1-r)(a+ar+ar^2+ar^3+...)=a
Therefore, a+ar+ar^2+ar^3+...=a/(1-r)

Basic proof for those of you who are confused:

x = 0.999...
10x = 9.999...
9x = 10x - x = 9
9x ÷ 9 = x
9 ÷ 9 = 1

∴ x = 1
∴ 0.999... = 1