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In the "Is This Really True?" thing it said 10X=9.999, but X=0.999 and 0.999X10 does not equal 9.999. Also, later it said X=1. If that is true then 10X would equal 10. Could anyone who knows the answer please reply? Thanks!
Ah, the old 0.999... = 1 debate again. This has turned out to be a very controversial topic.
The proof goes like this:
Let X = 0.999...
Then, 10X = 9.999...
Subtracting the first equation from this gives 9X = 9
Therefore, X=1
0.999... = 1.
You're right that 0.999 * 10 ≠ 9.999, but X has an infinite amount of nines after its decimal point, so after multiplying by 10 will still have an infinite amount of nines.
You're also right that 10X = 10, but that's OK because 9.999... = 10 as well.
Why did the vector cross the road?
It wanted to be normal.
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Does this happen in binary system, senary system or any other.
also
Numbers are the essence of the Universe
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Does this happen in binary system, senary system or any other.
Yes. In binary:
0.111...
Is the equivalent of:
1/2 + 1/4 + 1/8 + ...
Which of course converges to 1. In ternary:
0.222...
Is the same as:
2/3 + 2/9 + 2/27 + ...
Which also converges to 1. In fact:
Prove it.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Last edited by Stanley_Marsh (2007-04-30 02:31:12)
Numbers are the essence of the Universe
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Fixed.
Why did the vector cross the road?
It wanted to be normal.
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Uh uh uh, can't reorder terms until you prove convergence.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Why?
If so , I should use partial sum n , then prove n-->infinity converge?
Last edited by Stanley_Marsh (2007-04-30 12:55:22)
Numbers are the essence of the Universe
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Nevermind, you must only do that when it's a question between absolute and conditional convergence. If a series is divergent, it will always be divergent upon a permutation of its terms.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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This cant be true because when u multiply by 10 the no. of nnes in the dicimal part will redce by 1. so the diffrence will be 8.9999999999999..............................1 ???
That only works if there are a finite amount of nines after the decimal point.
When there are infinitely many of them, reducing their amount by one doesn't have meaning.
Why did the vector cross the road?
It wanted to be normal.
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Here's a useful purpose for the fact that 0.999.....=1.
I was at the casino looking at the Big Six wheel, there are 54 slots with 2 of them taken up by twenties which pay off at 20 to 1. What is the casino's advantage at this bet?
Well, since you don't lose your original bet when you win, you simply add 1 to the payoff, multiply the payoff times the probability and subtract from 1 to get the casinos advantage.
So, 21*(1/27) = 7/9 = 0.777777777...
We want to subtract this from 1 so 0.9999999... - 0.777777..... = 0.22222.....= 2/9
So the casino's advantage is a whopping 22.222...%. Horrible bet!
In this case, using the fact that 0.99999... = 1 makes things easier.
Last edited by Fruityloop (2009-07-05 10:51:18)
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I agree with Xyz, once the infinite decimal (0.9999) gets multiplied, it's not infinite any more, so
0.9999 X 10= 9.999 ( 1 "9" less)≠ 9.9999
9.999-0.9999=8.9991
8.9991÷9= 0.9999
P.S. This works for any amount of nines
EYE AM FRIENDLY, THAT'S O U NEED 2 NO,
psst,
Don't trust strangers, EYE AM FRIENDLY, THAT'S O U NEED 2 NO ..........psst, Don't trust strangers......
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9.999-0.9999=8.9991
This would only work if there are string of zeros at the end. For example..
9.9990000000.....- 0.999900000000......
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