.
(Because I'm bored)
Consider a zero followed a decimal point and an infinitely long string of nines: 0.999999 . . . .
What is the value of that unending series? The answer, strangely enough, is that it equals 1.
Here's why.
1. define X as 0.999999 . . . .
2. multiply X by 10: ( 10X )
3. We know that 10 times 0.999999 = 9.999999 . . . So this means 10X = 9.999999. . .
4. Now subtract X from 10X : 9.999999 . . .  0.999999. . . . = 9. (the parts to the right of the decimal point disappear because they are the same)
5. So 9X = 9
6. and if 9X = 9, then X must = 1
7. Substituting for X from 1. above, 0.99999 . . . . . = 1
.
0.999999 . . . . = 1
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OnlineTcg
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Re: 0.999999 . . . . = 1
Post #2If X is .9999999999..... as you've claimed in step 1.Miles wrote: ↑Sat Apr 24, 2021 11:14 pm .
(Because I'm bored)
Consider a zero followed a decimal point and an infinitely long string of nines: 0.999999 . . . .
What is the value of that unending series? The answer, strangely enough, is that it equals 1.
Here's why.
1. define X as 0.999999 . . . .
2. multiply X by 10: ( 10X )
3. We know that 10 times 0.999999 = 9.999999 . . . So this means 10X = 9.999999. . .
4. Now subtract X from 10X : 9.999999 . . .  0.999999. . . . = 9. (the parts to the right of the decimal point disappear because they are the same)
5. So 9X = 9
6. and if 9X = 9, then X must = 1
7. Substituting for X from 1. above, 0.99999 . . . . . = 1
.
Then 9X = 8.99999999999....
There is no reason to consider step 4 as a valid part of this calculation.
I thought you were going to discuss the sometimes inaccuracy of using floating point numbers in computer code. This is simply a slight of hand trick that fails to attract attention away from step 4. You've got to point to something that distracts attention for this to work.
Look over there...
...therefore 9X = 9.
Unless I've missed something which is highly likely.
Tcg
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Re: 0.999999 . . . . = 1
Post #3But that's the whole point of this exercise: to consider steps 4, 5, 6, and 7
Now, if you don't want to, fine, but for those of us who do it presents a mathematical conundrum. At least to my brain that never got past calculus I in college many years ago.
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Re: 0.999999 . . . . = 1
Post #4Like I said, I might have missed something and apparently I did.
Tcg
To be clear: Atheism is not a disbelief in gods or a denial of gods; it is a lack of belief in gods.
 American Atheists
Not believing isn't the same as believing not.
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I must assume that knowing is better than not knowing, venturing than not venturing; and that magic and illusion, however rich, however alluring, ultimately weaken the human spirit.
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 American Atheists
Not believing isn't the same as believing not.
 wiploc
I must assume that knowing is better than not knowing, venturing than not venturing; and that magic and illusion, however rich, however alluring, ultimately weaken the human spirit.
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Re: 0.999999 . . . . = 1
Post #5In a sense, he is.
A base system can only represent a fraction as a terminating decimal if the denominator and the base share all of their prime factors. Base 10 can only represent a fraction accurately if its denominator's only prime factors are 2 and 5. It can represent 1/2, 1/5, 1/10, 1/4, and 1/25, but can't represent 1/3, 1/6, or 1/15. Base 2 can represent fractions whose denominators only have a prime factor of 2, so it can accurately represent 1/2 or 1/4, but can't represent 1/5 or 1/10.
Fractions that can't be represented exactly in a particular base, end up as repeating decimals if one does the long division. The deal with the .9 repeating is that no long division problem will ever end up with that as an answer. That's where the "sleight of hand" is.
Any math problem that could end up as .9 repeating is actually 1:
 1/3 = .3 repeating, 2/3 = .6 repeating, 1/3 + 2/3 = ?
 1/9 = .1 repeating, 2/9 = .2 repeating, 3/9 = .3 repeating, ..., 8/9 = .8 repeating, 9/9 = ?
 1  .7 repeating = .2 repeating, 1  .8 repeating = .1 repeating, 1  .9 repeating = ?
My preferred pronouns are he, him, and his.

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