Do You Think .9~=1?
09-17-2005, 10:21 PM
#12
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[quote name='psyclo' date='Sep 17 2005, 06:38 PM']Infinity is not a number it is a concept. You can't technicaly do algebra with infinity.
[/quote]
were not doing any with infinity.
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were not doing any with infinity.
09-18-2005, 05:13 AM
#14
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You don't need modular arithmetic to solve this one. If you want to take that path, we know .333~ is a third, and .666~ is 2/3 - but on that basis we can't assume that .999~ is 3/3, i.e. if we try to perform operations on a number with recurring decimals because you're always going to be out by 1/infinity.
If you multiply something by 10, you're effectively moving the whole number across the decimal point one digit. Unless you convert .999~ to a fraction/different base etc. you're going to be out by 1/infinity when you try to do this, because you can't multiply a number and get a result that's more accurate than what you started out with. So the best you can do is the following:
0."infininte amount of 9's" x 10 = 9."(infinte - 1) amount of 9's"
so what you have is:
9.999...99
0.999...999 -
8.999...991
and 8.999...991/9 = 0.999~
Now your arguement states that "infinity" and "infinity - 1" are the same figure, which is essentially right - since they are both undefined, you have the problem psyclo talks about. Think of it in terms of distance, where we have 0.999~ millimeters, but instead of the 9's recurring to a negligible distance, have it the other way round where you have 1 millimetre as the negligible distance and the recurring figure is the units. So instead of the millimeter tending towards nothing, you have the units tending towards light years and multiples of that, you're always going to be out by 1 millimetre.
Mark
If you multiply something by 10, you're effectively moving the whole number across the decimal point one digit. Unless you convert .999~ to a fraction/different base etc. you're going to be out by 1/infinity when you try to do this, because you can't multiply a number and get a result that's more accurate than what you started out with. So the best you can do is the following:
0."infininte amount of 9's" x 10 = 9."(infinte - 1) amount of 9's"
so what you have is:
9.999...99
0.999...999 -
8.999...991
and 8.999...991/9 = 0.999~
Now your arguement states that "infinity" and "infinity - 1" are the same figure, which is essentially right - since they are both undefined, you have the problem psyclo talks about. Think of it in terms of distance, where we have 0.999~ millimeters, but instead of the 9's recurring to a negligible distance, have it the other way round where you have 1 millimetre as the negligible distance and the recurring figure is the units. So instead of the millimeter tending towards nothing, you have the units tending towards light years and multiples of that, you're always going to be out by 1 millimetre.
Mark
09-18-2005, 01:17 PM
#15
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infinity - 1 is still in finity. theres always an infinite amount of 9s to the right of the decimnal, regardless of how many of them you move to the left of it. and this number only exists because we use base10 in our number system. look up to where i posted the argument in base3 and it works out just fine.
09-18-2005, 03:23 PM
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it doesnt matter what u do .9~ does not = exactly 1 it is the closest you can get and there is no real number that can express infinite so it has to = 1 but rally it doesnt actually = 1 its just the closest you can get to 1.
09-18-2005, 10:29 PM
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Originally Posted by spaceman Spiff' post='760084' date='Sep 18 2005, 01:23 PM
it doesnt matter what u do .9~ does not = exactly 1 it is the closest you can get and there is no real number that can express infinite so it has to = 1 but rally it doesnt actually = 1 its just the closest you can get to 1.
It is a nonterminating repeating decimal. This means it is a rational number, which means it can be expressed in the form a/b where a and b are integers and b != 0.
The only possible way to satisfy those requirements for .999... is if a=b, and a/b = 1.
Proof 1:
The set of rational numbers is a member of the set of real numbers.
Any infinitely repeating decimal is a rational number.
0.999... is an infinitely repeating decimal.
0.999... is a real number.
The set of integers is a member of the set of real numbers.
1 is an integer.
1 is a real number.
0.999... and 1 are both real numbers.
A property of the real numbers is that between any two distinct real numbers A and B, B>A, there are an infinite number of real numbers greater than A and less than B.
If 0.999... and 1 are distinct real numbers (not equal to each other), there are an infinite number of real numbers greater than 0.999... and less than 1.
There are no numbers greater than 0.999... and less than 1.
Therefore, 0.999... and 1 are the same number.
Proof 2:
1 = 1
1/3 = 1/3
1/3 = 0.333...
3(1/3) = 3(0.333...)
3/3 = 0.999...
1 = 0.999...
Proof 3:
x = 0.999...
10x = 10*0.999...
10x = 9.999...
10x - x = 9.999... - x
9x = 9.999... - 0.999...
9x = 9
9x/9 = 9/9
x = 1
0.999... = 1
Proof 4:
0.999... is an infinite geomtric series of the form:
Code:
∞ ∑ a * r^n n=0
The value of an infinite convergent series is its sum.
An infinite geometric series of the above form converges if r < 1.
0.999... in series form is:
Code:
∞ ∑ .9 * (1/10)^n n=0
a = 0.9 and r=1/10.
Because r < 1, the series converges.
The sum of a convergent geometric series is a/(1-r).
The sum of this series is 0.9/(1-1/10) = .9/(1-.1) = .9/.9 = 1.
A repeating decimal can be converted to its fractional form like this (it's a shortcut derived from proof #3 for repeating decimals where everything to the right of the decimal is part of the repeating sequence) :
First, find the period of repetition and call it n:
.142857142857142857... has six digits that repeat, so n=6
.123451234512345... n=5.
.234234234... n=3.
.090909... n=2
.333... n=1
.999... n=1.
Take the repeating sequence and call it a:
.142857142857142857... a=142857
.123451234512345... a=12345
.234234234... a=234
.090909... a=09
.333... a=3
.999... a=9
Take n nines and call it b (mathematically, b=10^n-1):
.142857142857142857... n=6, 10^n=1000000, b=999999
.123451234512345... n=5, 10^n=100000, b=99999
.234234234... n=3, 10^n=1000, b=999
.090909... n=2, 10^n=100, b=99
.333... n=1, 10^n=10, b=9
.999... n=1, 10^n=10, b=9
Form a fraction with a as the numerator and b as the denominator:
.142857142857142857... = 142857/999999
.123451234512345... = 12345/99999
.234234234... =234/999
.090909... = 09/11
.333... = 3/9
.999... =9/9
And reduce:
142857/999999 = (3*3*3*11*13*37) / (3*3*3*7*11*13*37) = 1/7
12345/99999 = (3*5*823) / (3*3*41*271) = 4115/33333
234/999 = (2*3*3*13) / (3*3*3*37) = 26/111
09/11 = (3*3) / (11) = 9/11
3/9 = 3 / (3*3) = 1/3
9/9 = (3*3) / (3*3) = 1
But some people seem to think .9 repeating is a special case - the ONLY repeating decimal in the entire set of rational numbers where this doesn't work.
and like ive said probably three times already, if you change the number from base10 to base3 and do the problem, then it works out just fine. proofs are things that are 100% true, and here are four of them.