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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
I've never seen any logical argument that can prove it.
Just because a computer rounds up, it doesn't change the value of a number.
9/9 is 1, not 0.99999999...
I don't really know how to represent 0.9999... as a fraction, but that doesn't matter too much, since it can be represented as a decimal.
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digpoeJoin Date: 2008-11-02
Post Count: 9092 |
99999999/100000000
There you go. |
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
That would terminate eventually.
" but in some of these number systems, the symbol "0.999..." admits other interpretations that contain infinitely many 9s while falling infinitesimally short of 1."
It seems as if you cannot necessarily argue it. It just depends on the number system you're using. While in one system, people have come to the conclusion that 0.9999... is 1, in others, that make more logical sense to me already, allow for them to unequal.
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algebraic proof
let x = 0.999... (whereas ... is repeating)
(multiply both sides by 10)
10x = x * 10 = 0.999... * 10 = 9.999...
(subtract both sides by x, or 0.999...)
10x - x = 9x = 9.999... - 0.999... = 9
9x = 9
(divide both sides by 9)
x = 1 |
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It works the other way around, Sun.
1 - 0.999.... would equal zero. Why?
Well, there are infinite repeating zeros after that, but a one should be at the end. However because of the infinite repeat of zeros, the number is zero.
x - 0 = x
so from that we can conclude again that 1 = 0.999.... |
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sorry, correction
x - x = 0 |
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TheMyrcoJoin Date: 2011-08-13
Post Count: 15105 |
You can't just subtract multiplying.
10x - x = 9.9_ - x
10x - x = 9 |
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UithamJoin Date: 2007-06-24
Post Count: 667 |
0.999... * 10 = 9.999...990.
10x - x = 9x = 9.999...990 - 0.999... = 8.999...
9x = 8.999...
divided by 9, that would be
0.999... |
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TheMyrcoJoin Date: 2011-08-13
Post Count: 15105 |
Nvm confused something else. |
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
0.999... * 10 = 9.999...
(It helps if you put things on separate lines)
(0.9...*10) = (9.9...)
(0.9...*10) - 0.9...= (9.9...) -0.9...
9 = 9
Since x = 0.9...
If you do the same thing to both sides, all of the time, then you'll always get an equivalent statement.
x = 0.9...
10x = 9.9...
10x - 0.9... = 9.9... - 0.9...
At this point in time, you must substitute the value of x in for x, rather than subtract by x on one side and subtract by 0.9... on the other. If you're going to substitute, then you must do so everywhere.
9.9... - 0.9... = 9.9... - 0.9...
9 = 9
Otherwise,
10x - x = 9.9... - x
9x = 9.9... -x
9*0.9... = 9.9... - 0.9...
8.9... = 9
I didn't get an equivalent statement by subtracting both sides by x, which is strange.
9x = 9.9... -x
You're not able to divide x out of both sides either, since the right side of the equation is a binomial. Usually, proofs take advantage of things that are incorrect to make something appear correct. 1 does not equal 2, but people believe that the proof is right, even though at one point they are dividing by 0.
In the number system we learn in school, it has been decided that 0.9... is 1, so using that number system to prove it is redundant. Use a system that says they are unequal, and then prove that they are equal. When something is infinite, then it's not finite.
Also. why didn't you just multiple 1/3, or 0.3... by 3?
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
If there is a 1 at the end, then there is a 1 at the end.
Just today in math class, my teacher said we can't just erase something we don't like, don't understand, or don't want to work with in mathematics. It just doesn't work that way, you must account for it. Although, we were discussing derivatives, the same philosophy still applies.
0.0000...1
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TheMyrcoJoin Date: 2011-08-13
Post Count: 15105 |
The short way of writing that down would be 0.0. |
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
Anyway, we all know how this will end.
With a disagreement. Why bring it up? :/
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booingJoin Date: 2009-05-04
Post Count: 6594 |
Some people just don't accept that 4 bytes can only contain so much data... |
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maybe you didn't read it correctly, i'll say it again
let x = 0.999...
(multiply by 10)
10x = 9.999...
(subtract x, or 0.999... which would cancel out the infinite repeat of 9)
10x - x = 9.999... - 0.999...
(the result)
9x = 9
(divide by nine)
x = 1
x has become 1
the value itself did not change |
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there is indeed a 1 at the end
but there is no end to an infinite repeat of zeros
and rounded, it would equal zero |
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HaxHelperJoin Date: 2009-12-19
Post Count: 1208 |
all you've proven is that you can make 0.999... equal 1 through mathematical operations
get out |
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jobro13Join Date: 2009-08-05
Post Count: 2865 |
On binary level it's equal to 1, yes.
Same reason why in base-10 we cannot write 1/3 as a "whole" number, while in base-3 you can. |
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AdvorsusDuo's algebraic proof is right, and if it doesn't convince you, there are many much more elaborate proofs available on the internet.
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Twitter: https://twitter.com/MarkOtaris |
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sdfgwTop 50 PosterJoin Date: 2009-01-08
Post Count: 41681 |
2 years later and this thread still pops up |
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
Rounding makes it wrong.
There doesn't need to be an end. But there's a one at the end. Paradox, maybe.
Also, subtracting by x on one side and 0.9... on the other isn't the same exact thing. It's either you add -x to both sides or -0.9... to both sides.
Also, I am not sure if that is a proof. Typically a proof doesn't have a value for the variables. You take some equation that is accepted as true, and then you derive another equation from it.
Using similar logic to yours, I can state that,
Let x be some real number,
x = x --you can agree this is true
x/1 = x --you can agree this is true
x * 1/1 = x --you can agree this is true
(x-x)/(x-x) = 1 --Right? You can cancel both binomials and get 1.
Therefore,
x*((x-x)/(x-x)) = x * 1
Now, using this, you can go and state that all numbers everywhere are undefined, since (x-x)/(x-x) = 1, and if x is 1, then 0/0 is 1, except 0/0 is undefined, so 1 is undefined, and you can iterate that over just about any function, number, expression, etc...
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sdfgwTop 50 PosterJoin Date: 2009-01-08
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"Now, using this, you can go and state that all numbers everywhere are undefined, since (x-x)/(x-x) = 1, and if x is 1, then 0/0 is 1, except 0/0 is undefined, so 1 is undefined, and you can iterate that over just about any function, number, expression, etc... "
you've assumed algebraic laws hold over division by 0
the proof that 0.999...=1 does not depend this
you are silly
please leave |
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8SunTzu8Join Date: 2011-09-30
Post Count: 8199 |
When you're trying to use numbers to perform some sort of a trick, it's usually easy to see it. It only works because he subtracts x from one side and 0.9... from the other. x is being used as a finite value, while 0.9... is not. x should be used as if it continues infinitely, but that would any attempts to use it in math impossible. You cannot multiply forever.
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"Also, subtracting by x on one side and 0.9... on the other isn't the same exact thing. It's either you add -x to both sides or -0.9... to both sides."
Since you know that x = 0.999..., you can substitute 0.999... to x.
So if you prefer, we can just do it by substituting:
10x - x = 9.999... - x
10x - x = 9.999... - 0.999...
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Twitter: https://twitter.com/MarkOtaris |
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