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# .99999... = 1 (again)

Discussion in 'Off-Topic' started by dantose, Mar 23, 2004.

1. ### dantoseIncGamers Member

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.99999... = 1 (again)

ok, here goes:

both .9999... and 1 are real numbers

Between any two real numbers, there is always another real number. (density property)

no real number exists between .9999... and 1

hence they can't be different numbers

2. ### FreemasonBanned

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Remember the scam they pulled in the movie Office Space? The rounded off penny was shipped off to their account. Figure in several million transactions a day and it starts to add up.

.0001 is the difference between the two numbers in question. 15,000,000 transactions where that is skimmed off per day = \$1,500 a day. There is a HUGE difference between the two numbers.

3. ### TechnetiumIncGamers Member

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He's talking about an decimal point with an infinite series of 9s after, not .9999.

As the thread title implies, we had a whole big thread on this a while ago already and came to the same solution (via slightly different proof, I think).

4. ### dantoseIncGamers Member

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not .9999

but .9999... the repeating decimal.

There is a number between .9999 and 1 (.99995) but none between .9999... and 1

5. ### dantoseIncGamers Member

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actually the old one just died with no agreement being reached

6. ### KawaiiIncGamers Member

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I'll ask my Math teacher tomorrow, but i believe that 0,9999.... is not = 1, since it approaches 1 but never actually reaches it. Practically, it's 1, but in theory it's not.
Draw a graph on your calc for the equation Y=1-(1/x)(since we can assume 0,9999...=1-(1/inf). You'll see that the line approaches 1 but never actually reaches 1.

7. ### FreemasonBanned

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Ah, I never read the first thread.

No, they are still a different number. With an infintisimally small difference, there is still a difference. Infinity divided by .999 repeating will result in a different number than infinity divided by 1.00

8. ### StevinatorIncGamers Member

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I think you'd get infinity. anything over 1 is itse;f, and infinity over anything is still infinity...it's not like you'd have infinity point zero zero zero zero zero zero zero zero one. you can't do that with infinity. you can't add to infinity...it's already infinity.

9. ### Wuhan_ClanIncGamers Member

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In the way you are describe recurring numbers, the number between 0.999.....9 and 1 is the number 0.000.....1

I'll say what I said in the other thread about circles: I doubt if anyone really understands what infinity means. Personally, I can't comprehend something that is infinitly small. This is because my common sense (by common sense I mean the way my brain has been thinking all my life) is telling me to quantify infinity. The problem is, infinity is not a defined quantity in our common logic. So when we treat it in such a manner, we come across problems such as this one and also the problem about perfect circles.

10. ### TechnetiumIncGamers Member

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Actually, one of the people arguing that they weren't equal went and asked his college math professor, and then relayed the explanation back in this thread of why they are equal.

They are equal because 1 - .999... would be .000..., followed by a 1. However, since the 0s are infinite, the "final" digit of the series is irrelevent as it is never reached (only approached), thus the difference is 0.

What it comes down to is just a different way of representing the definition of a whole number.

11. ### dantoseIncGamers Member

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.0000...1 would be 0 and is significantly less than 1 or .99999... you would have to define a number that fell between .9999... and 1 to show them to be two distict numbers.

Oh, and lets not use infinity in any arguments since it does not fall within the real number set

12. ### rodigeeIncGamers Member

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Time for magic math

x = 0.999...
--------------------------------
10x = 9.999.....
--------------------------------
10x - x = 9.999.... - 0.999.....
--------------------------------
9x = 9
--------------------------------
x = 1
--------------------------------
0.99.. = 1

case closed *shuts book*

13. ### Wuhan_ClanIncGamers Member

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I know the real explanation might have been more thorough and its going to be my word vs a college professor but ....

In the same sense, 0.999...9 is not equal to 1 because the "final" digit is never reached so it is always slightly less then 1.

Dantose, if you disregard infinity in this sense, 0.999.... cannot be a real number either.

(hey, you said that between any 2 real numbers is another real number but if you are not saying that 0.999... and 1 are the same number, then it was pointless to discuss in the first place)

Personally, I feel this is a fallacy of the human race. That our comprehension of numbers must involve recurring digits.

14. ### SuggestiveNameIncGamers Member

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Actually thats not true, x/2 is bigger than x/10 as x approaches infinity.

15. ### {KOW}SpazedBanned

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Take 1/2 and divide it by 1/2, keep doing this you will never reach zero. . .you will just very very close to it. Same applies here you get very very very close to one, but you never actually reach it.

16. ### Pierrot le FouIncGamers Member

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Incorrect math, technically, since 10x (with x as defined) would technically have a trailing 0.

Though the resolution to the last thread was that, by definition, .999... is 1. No fancy proof, just a tautological definition.

Sorry to rain in on the parade so early.

17. ### dantoseIncGamers Member

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I meant infinity as a number not the concept of infinate repetition.

As for my argument, I just made the case that both 1 and .999... are real numbers and since they do not have another real number between them they must be the same real number.

18. ### Jigga-ScroogeIncGamers Member

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1 - 1 = 0
1 - .999... = .000...1
.999... does not equal one

19. ### dantoseIncGamers Member

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actually that just proves .00000...1=0

20. ### asdfIncGamers Member

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that 1 at the end does not exist. it is at the end of an infinite number of 0's.

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