Poll: 0.999... = 1
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Well, I stated why i dont believe that in a post that was buried in the thread earlier, as you using two separate X values, not the same X value for the same problem. When you said implying, that was a separate math problem and you found a different value.havass said:
Either way, you're still getting the same number, which yes, rounds up to 9, but this is a thread about not rounding up and more about perfect equaling.Rabid Toilet said:
I already fixed my inadequate assertion in this [http://www.escapistmagazine.com/forums/read/18.252127-Poll-0-999-1?page=4#9363044] reply.Rubashov said:WolframAlpha disagrees:benzooka said:
http://www.wolframalpha.com/input/?i=summation%289%2F%2810^n%29%29+from+n+%3D+1+to+infinity
Its unproven assertions are probably at least as reliable as yours.
This thread is highly amusing. It's like mathematical schadenfreude!
Yes, just like 5/5, 75-74, and 0.2 * 5 are three separate numbers.emeraldrafael said:
'x = 0.999...' is not an assumption, it's a definition. A definition does not require proof, as it is always right, by, uhm, definition. The assumption was '1 = 0.(9)' and it was reached legally from a true statement. QED.Shadowkire said:
I apologize, it was wrong of me to say you failed at math, I didn't intend to offend you.havass said:
The notation is kind of hard to get across in text, since there's a lot of subscripts and superscripts involved.tthor said:
He's modeling the number .99... as a geometric series (basically .9 + .09 + .009 ...). Given what we know about these series, the sum of all of the numbers in the series (which would equal .99...) is equal to the first number (9/10 = .9) over 1 minus the number it's being multiplied by to the power of its place in the series (the first place is zero) ((9/10) * (1/10) ^ 0 , (9/10) * (1/10) ^ 1 , (9/10) * (1/10) ^ 2 , etc.).
Thus, the sum of all the numbers in the series equals the fraction (9/10) divided by (1 - (1/10)), or (9/10) divided by (9/10), which is 1.
Not sure if that explains it any better or if it just made it more confusing, but the math is sound, and it does indeed prove that .99... = 1.
No, they're all numbers with the same equivical value. each = 1. But .999 =/= 1.Coldie said:
The easiest way to put is this. Say you had... um... oka, I'm going ot have to change the numbers a bit, but it will stil be the same.
Say you had $9.99. and in this perfect world that you will live in for sake of argument, everything costs exactly has advertised (no tax). So an item is $10. You're 9.99 does not equal the 10, so you do not get what you wanted to be (a DVD lets say just to make it realistic). 0.999..... =/= 1 in much the same way.
You apparently have no idea how logic/mathematical proofs work so here I go:Coldie said:
claiming x=0.999... is an assumption, 1=0.999... is what he intended to prove.
the reason his proof is wrong is because he failed to prove the assumption.
Understanding it requires college level math.
I won't bother replying to the second point, since I'm not even sure what the original point was.emeraldrafael said:
As to your first point, where you assert that there were two different equations in his proof? That is false.
You said that 10x - x = 9 and 9x = 9 were two different equations. What is 10x - x? If you have 10 "x"es and you take one away, you get 9 "x"es.
Damn, that should have been a nine instead of an eight when it rounds up.Rabid Toilet said:
he said it Implies. he's changing the equation and the value of X. X as variable can not hold the same value. and again, if you want to prove it, look into the concept of graphical limits.
I honestly have no idea what you did there, but your algebra is most certainly not correct.acer840 said:
If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)
For practical purposes, yes - in practice, there will never be a significant difference between both.Hail Fire 998 said:
The reason for this is that in practice, we can never do something infinitely accurate. Let's say, we want to bring the temperature of a can with water to 1 degrees celsius. The water will never be "exactly" 1 degrees anyways, so the difference between 0.999999999... and 1 really makes no practical difference (because our accuracy at regulating the water temp is lower, than the difference).
In theory, there is however a difference between both.
Many people think of infinity as a number. But it isn't really a number. Rather, it is a loop... for example, for 0.999... what it does is this:
Step 1: add another 9 at the end
Step 2: go to step 1
So, it is more like a command "do this over and over forever".
They keyword is "forever".... this "task" will never finish, because it has no end (it's a loop).
When infinity is used for practical purposes, for example, in a computer program, they just let the above loop run for a while, until there are enough 9's to be satisfied with the accuracy..... for example, they may say "stop when it reaches 9 digits after the comma".
0.999999999
Thats not infinity, because the loop didn't run forever
He shouldn't have used the word "implies", but it doesn't change the fact that it's the same equation.emeraldrafael said:
I cannot even begin to comprehend how you have reached '0.999... != 1' from '9.99 != 10'. 0.(9) does equal 1. 9.99 does not equal 10. The only common theme is that here there be nines.emeraldrafael said:
The only argument you could have here is that numbers 0.(x) are a canonical representation of periodic numbers, where x is the repeating pattern, so 0.(9) and 1 are different canon forms and therefore are different numbers. However, since 0.(9) and 1 both represent the same value, (9) notation is expressly forbidden. Thus x cannot be 9 (or consist only of nines, i.e. 99, 999, for obvious reasons) and 0.(9) is a non-canon form of 1.
