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1. I just realised something.

What is 4.4444... to 2 decimal places?
2. 4.44?
3. hmm. do you see where I'm coming from though. Usually when you see a 4 you look to the next number to see if it makes that 4 a 5, in which case you round up rather than down. but if it goes on forever, you'll never stop looking.
4. (Original post by mik1a)
hmm. do you see where I'm coming from though. Usually when you see a 4 you look to the next number to see if it makes that 4 a 5, in which case you round up rather than down. but if it goes on forever, you'll never stop looking.

It's along similar lines to 0.9999999... = 1 I guess.
5. yeah I did that in the olympiad handbook... I thought there's no way you can prove 0.999 recurring equals one but you do the algebra and it works :s
6. But what is the actual theory behind it though?
7. (Original post by mik1a)
yeah I did that in the olympiad handbook... I thought there's no way you can prove 0.999 recurring equals one but you do the algebra and it works :s
8. (Original post by bono)
Do the a = 0.999999.....

therefore 10a = 9.999999....

10a -a = 9a = 9

therefore a = 1

ie 0.999999.... = 1
9. The theory behind what? I just came across this when I was thinking about recurring decimals. How do you round a recurring decimal whose digits' rounding (up or down) depends on the next digit behind that.
10. (Original post by 2776)
Do the a = 0.999999.....

therefore 10a = 9.999999....

10a -a = 9a = 9

therefore a = 1

ie 0.999999.... = 1
Ahhh right, yes.
11. (Original post by mik1a)
The theory behind what? I just came across this when I was thinking about recurring decimals. How do you round a recurring decimal whose digits' rounding (up or down) depends on the next digit behind that.
I bet theres a deep math theory behind all of this. Or maybe I'm mistaken.
12. Another approach is to show that between any two real numbers there is another real number. So if .999999 =/= 1 then there is a number such that 0.9999... < x < 1. This contradicts itself, since no number can have a digit more than 9 without going over one, if you get my drift.
13. (Original post by theone)
Another approach is to show that between any two real numbers there is another real number. So if .999999 =/= 1 then there is a number such that 0.9999... < x < 1. This contradicts itself, since no number can have a digit more than 9 without going over one, if you get my drift.
i think the other one is more convincing
14. therefore 0.9999... must equal one.. heh.

I'm usually no good at proofs. I usually make up my own notation and have things like a.bnc where b -n- c are the repeated digits, therefore a, b, c, and n (the set of digits between b and c) are all integers between 0 and 10. But I didn't know how to show an integer. Then I made up some formula where

x = a + 0.1b + 0.1^(n+1)n + 0.1^(n+2)c

blah blah

i could go on forever with these strange things.
15. I just go on the basis that if the next number is five or more then round up.

1.046
1.045
1.044

Well 1.046 is closer to 1.05 than 1.04, and 1.044 is closer to 1.04 than 1.05... I just see it very simply...
16. (Original post by 2776)
Do the a = 0.999999.....

therefore 10a = 9.999999....

10a -a = 9a = 9

therefore a = 1

ie 0.999999.... = 1
But if a=0.9999... then 9a cannot equal 9. Not exactly anyway.
17. (Original post by ZJuwelH)
But if a=0.9999... then 9a cannot equal 9. Not exactly anyway.
But you times it by 10 FIRST so it is 9.9999999999999......

And so:

10a - a = 9.99999999999999....... - 0.99999999999999......... = 1

9a = 9

a = 1
18. Write it as an expression with a and n, the number of digits which you take of the recurring decimal 0.999...

Therefore when n = 2, a = 0.99; when n = 5, a = 0.99999

As n increases, a tends to 1. THerefore in a recurring decimal, when n nis infinite, a is one.
19. I guess you can extend it to 0.49999999

If 0.49999999..... is a recurring decimal then multiply it by two:

0.9999999999

and therefore 0.4999999 = 0.5
20. (Original post by 2776)
But you times it by 10 FIRST so it is 9.9999999999999......

And so:

10a - a = 9.99999999999999....... - 0.99999999999999......... = 9

9a = 9

a = 1
<corrected error>

I refuse to accept it, there's a flaw somewhere, that ain't supposed to work...

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