Report, edit, etc...Posted by PCFredZ on 2006-03-03 at 17:41:42
QUOTE(DT_Battlekruser @ Mar 1 2006, 02:29 AM)
Why use such lame reasoning? It can be algrebraically proven.
GIVEN: x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
THEREFORE: 0.999... = 1
Q.E.D.
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GIVEN: x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
THEREFORE: 0.999... = 1
Q.E.D.
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10x - x would not equal 9. That only takes reasoning, not calculus, lmao BeeR.
Report, edit, etc...Posted by BeeR_KeG on 2006-03-03 at 17:53:45
People started talking about limits so I gave them a substituted indefinite integral 
Want something more advanced to prove that 0.999... does not equal 1?
Want something more advanced to prove that 0.999... does not equal 1?
Report, edit, etc...Posted by Lithium on 2006-03-03 at 17:56:35
but if you used that into your math test. the teacher'd give you F-. gg'ed anyways.
this is proven by million math researchers in the world.
graduated in college. etc. more exp in math than you in many ways. etc. yeah.
this is proven by million math researchers in the world.
graduated in college. etc. more exp in math than you in many ways. etc. yeah.Report, edit, etc...Posted by CheeZe on 2006-03-03 at 18:03:25
QUOTE(Lithium @ Mar 3 2006, 05:31 PM)
this topic is silly. guys this is proven in science. X = 1. GET IT?
because (0.9 dot above 9) goes on forever, there can be no 000000000001 you can put in. Silly things. If that 9 goes on forever, there is no difference between a 1 and a (0.9 dot above 9.) It's mathematically proven. plus DKTB and few others proved it true anyways.
think of it logically. and easily. 0.9999999999999999... goes on. and never stops.
the whole point is, the gap is too small to be considered different. it's like a perfect another way to say 1. but longer. it's like human, you wouldn't see a difference between 2 same boxes but in atomic scale, they are different. so. in logic its not different. in science it's not different at all. really. and. since they go on forever... theres no 0 in the end, and all it will do is bring up a number one digit that will not do anything down there because its "FOREVER"
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because (0.9 dot above 9) goes on forever, there can be no 000000000001 you can put in. Silly things. If that 9 goes on forever, there is no difference between a 1 and a (0.9 dot above 9.) It's mathematically proven. plus DKTB and few others proved it true anyways.
think of it logically. and easily. 0.9999999999999999... goes on. and never stops.
the whole point is, the gap is too small to be considered different. it's like a perfect another way to say 1. but longer. it's like human, you wouldn't see a difference between 2 same boxes but in atomic scale, they are different. so. in logic its not different. in science it's not different at all. really. and. since they go on forever... theres no 0 in the end, and all it will do is bring up a number one digit that will not do anything down there because its "FOREVER"
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You're describing a limit. I already told you, a limit is not the same as equal to.
I've already discussed this with my Math teacher 2 years ago. We came to the conclusion that I was right in that .9999... != 1 but there is little difference since we're going to be taking the limit. But! .9999... is NOT 1.
Report, edit, etc...Posted by BeeR_KeG on 2006-03-03 at 18:09:19
In not one mathematics book have I seen that 0.999... = 1
My Calculus teacher studied Electrical Engineering and Pure Mathematics, he says that a variable can only be equal to another variable if they are the same variable.
x=y only if y=x meaning that x=x
My old Algebra II teacher studies Pure Mathematics and I proved her wrong on several problems that she gave, gave me A+.
Why would those Mathematicians be right and I wrong? I proved that 0.999... does not equal 1 and they proved that 0.999... = 1. But the transitive property states that if a=b
then b=a, but in this case b does not equal to a, making null the assumption that a=b.
If there was a way to prove that 0.999... = 1 you would not be using a limit, you would want an exact value, not an approximation to that value that might not exist.
My Calculus teacher studied Electrical Engineering and Pure Mathematics, he says that a variable can only be equal to another variable if they are the same variable.
x=y only if y=x meaning that x=x
My old Algebra II teacher studies Pure Mathematics and I proved her wrong on several problems that she gave, gave me A+.
Why would those Mathematicians be right and I wrong? I proved that 0.999... does not equal 1 and they proved that 0.999... = 1. But the transitive property states that if a=b
then b=a, but in this case b does not equal to a, making null the assumption that a=b.
If there was a way to prove that 0.999... = 1 you would not be using a limit, you would want an exact value, not an approximation to that value that might not exist.
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-03 at 19:00:19
QUOTE(CheeZe @ Mar 3 2006, 05:03 AM)
I never expressed .999... as a limit. I've always used .999...
I don't want it approaching any number in terms of limits. Thus, I'm not taking a limit and it does not equal 1.
Besides, are you going to tell me 1/8 is also a limit approaching some rational number? Or pi? Both of these functions can be expressed in terms of limits.
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I don't want it approaching any number in terms of limits. Thus, I'm not taking a limit and it does not equal 1.
Besides, are you going to tell me 1/8 is also a limit approaching some rational number? Or pi? Both of these functions can be expressed in terms of limits.
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Infinity is not a real number, so expressing it as 0.999... means take the limit as the number of nines goes to infinity. Such a number with truly an infinite number of nines and not as the limit of nines approaches infinity cannot exist.
QUOTE
1 - 0.999... = x
0.000...1 = x
1 and 0.999... are not equal.
0.000...1 = x
1 and 0.999... are not equal.
0.000...1 = 0. The .000...1 is infinitely small and therefore goes to zero as you go to an infinite number of zeroes preceding it.
1.999...6000...1 = 2. As the number of nines goes to infinity, the remaining digits do not matter. In this way, lim[sub]x→∞[/sub] x[sup]2[/sup]/x = ∞. It does not matter that you divide it by x, it still goes to infinity.
1.999...6 can be expressed as 1 + lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) + 6/(10[sup]x+1[/sup]). This degenerates into 2. I don't have the time to examine BeeR's calculus.
Report, edit, etc...Posted by BeeR_KeG on 2006-03-03 at 19:55:00
Edited my previous post with Calculus to better support my arguement.
Also know that the average between 1 and 0.999... is those two numbers added and divided by 2. It isn't 0.999... and it isn't 1.
Also know that the average between 1 and 0.999... is those two numbers added and divided by 2. It isn't 0.999... and it isn't 1.
Report, edit, etc...Posted by PCFredZ on 2006-03-03 at 21:29:25
BeeR knows his calculus. Are there any other things that need to be proven? If not, this topic is getting pointness.
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-04 at 00:59:41
Examining BeeR's calculus, the only conclusion I come to is this: People who say 0.999... ≠ 1 agree that lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) = 1, and to that I say this: A number with an infinite number of nines cannot exist. When we notate 0.999.... we indicate to take the limit as the number of nines approaches infinity. If f(x) = x[sup]2[/sup], f(∞) ≠ ∞, but for any realistic purposes when we think of f(∞), we mean lim[sub]x→∞[/sub] f(x), which is infinite.
If you wish to say that an inifinite number of nines is not taking the limit, I say this: Repeating decimals are only generated to approximate a fraction, and therefore exist to express taking the limit of the number of repititions going to infinity. You cannot have 0.333... pieces of pie if the .333... does not signify taking the limit; you can only have 1/3 pieces of pie.
If you wish to say that an inifinite number of nines is not taking the limit, I say this: Repeating decimals are only generated to approximate a fraction, and therefore exist to express taking the limit of the number of repititions going to infinity. You cannot have 0.333... pieces of pie if the .333... does not signify taking the limit; you can only have 1/3 pieces of pie.
Report, edit, etc...Posted by Mini Moose 2707 on 2006-03-04 at 01:31:39
So, if .99999 = 1, does 99,999 = 100,000 too, or better yet, does 100,000,000,000(.999999) = 100,000,000,000(1)? I'll take that extra $100,000 if you don't want it.
Report, edit, etc...Posted by Shadow-Killa_04 on 2006-03-04 at 01:46:21
Can we just accept that theres no answer? Actulaly nm. I like pointless topics 
Many of you have proven (I'm assuming by those calculations, way to lazy to check) that 9/9 = .9999999999.
However, back to what I said in my first post, can anyone explain why when you have 9 people and 9 pies each person would get a whole pie, not .99999999999999999999 of a pie? Or am I just stupid? (For some reason I think the stupid one is going to come up)
Many of you have proven (I'm assuming by those calculations, way to lazy to check) that 9/9 = .9999999999.
However, back to what I said in my first post, can anyone explain why when you have 9 people and 9 pies each person would get a whole pie, not .99999999999999999999 of a pie? Or am I just stupid? (For some reason I think the stupid one is going to come up)
Report, edit, etc...Posted by HolySin on 2006-03-04 at 02:28:39
Guys, it really is based on significant digits. If it's:
1
Then yes, 0.999... does in fact equal 1 because you have to round it so the equation is proper. However, 1.000... does not equal 0.999... since both have infinite significant digits.
1
Then yes, 0.999... does in fact equal 1 because you have to round it so the equation is proper. However, 1.000... does not equal 0.999... since both have infinite significant digits.
Report, edit, etc...Posted by Gigins on 2006-03-04 at 03:42:25
1st try
I say 1 = 1.000... > if 1.000... ≠ 0.9999... > then 1 ≠ 0.9999...
2nd try
0.9999 is not a real number, it has infinitive nines, but it's still a number. That means it still has the old math rules that other numbers have, doesn't it?
Anyway...
9 lacks 1 till 10
99 lacks 1 till 100
999 lacks 1 till 1000
0.9 lacks 0.1 till full 1
0.999 lacks 1 till full 1
0.999{billion nines}999 lacks 0.000{billion zeros}001 till full 1
Conclusion, however far we go it's still 0.00..01 till full 1. Even if the nines go infinitive the 0.0...01 is still hanging in the infinity. And this 0.0...01 make 1 ≠ 0.999...
3rd try
Every number has multiplier 1(infinitive times) behind it. example: x[sup]2[/sup]-x=0 > x(x-1)=0
So if you multiply any number by 1 infinitive times it won't change at all. To multiply number with 0.999... we must make it real, let's say we cut it down after trillion nines(it will be our x).
Now multiply number 10, trillion times with x > we got 9 instead of 10 > 9 ≠ 10 > 1 ≠ 0.999...
Even if we don't cut it down at all, multiply 10 {infinitive times} with 0.999...{infinitive nines} and you got 9...
If you agree, cool
If you don't you have to burry all 3 tries
I say 1 = 1.000... > if 1.000... ≠ 0.9999... > then 1 ≠ 0.9999...
2nd try
0.9999 is not a real number, it has infinitive nines, but it's still a number. That means it still has the old math rules that other numbers have, doesn't it?
Anyway...
9 lacks 1 till 10
99 lacks 1 till 100
999 lacks 1 till 1000
0.9 lacks 0.1 till full 1
0.999 lacks 1 till full 1
0.999{billion nines}999 lacks 0.000{billion zeros}001 till full 1
Conclusion, however far we go it's still 0.00..01 till full 1. Even if the nines go infinitive the 0.0...01 is still hanging in the infinity. And this 0.0...01 make 1 ≠ 0.999...
3rd try
Every number has multiplier 1(infinitive times) behind it. example: x[sup]2[/sup]-x=0 > x(x-1)=0
So if you multiply any number by 1 infinitive times it won't change at all. To multiply number with 0.999... we must make it real, let's say we cut it down after trillion nines(it will be our x).
Now multiply number 10, trillion times with x > we got 9 instead of 10 > 9 ≠ 10 > 1 ≠ 0.999...
Even if we don't cut it down at all, multiply 10 {infinitive times} with 0.999...{infinitive nines} and you got 9...
If you agree, cool
If you don't you have to burry all 3 tries
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-04 at 05:19:27
QUOTE(Mini Moose 2707 @ Mar 3 2006, 10:31 PM)
So, if .99999 = 1, does 99,999 = 100,000 too, or better yet, does 100,000,000,000(.999999) = 100,000,000,000(1)? I'll take that extra $100,000 if you don't want it.
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.99999 ≠ 1. 0.999.... where .999.... denotes a theoretically infinite number of nines is equal to 1. Wherever I use a .... after a number, it indicates infinite repetition.
QUOTE
However, back to what I said in my first post, can anyone explain why when you have 9 people and 9 pies each person would get a whole pie, not .99999999999999999999 of a pie?
Assuming you mean, 0.999.... pies, they would. 0.999.... = 1.
QUOTE
However, 1.000... does not equal 0.999... since both have infinite significant digits.
In the limit of the number of nines going to infinity, the number goes to 1. Nobody has so far denied this. 1 = lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])), and 0.999... = ∑[sub]m=1,m=∞[/sub](9/(10[sup]m[/sup])). For all purposes because of the derivation of the notation of repeating decimals, 0.999... = ∑[sub]m=1,m=∞[/sub](9/(10[sup]m[/sup])) = lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) = 1.
QUOTE
Conclusion, however far we go
No, if we go infinitely far, the gap becomes infinitely small and goes to zero. If you disagree that lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) = 0.999..., then you must also disagree that 0.333.... = 1/3.
Report, edit, etc...Posted by Gigins on 2006-03-04 at 05:23:16
And what about other 2 tries? You cheated
and yes 0.333.. is not 1/3 for the same reason. Since when numbers turn to 0 when they are to small? o.o
and yes 0.333.. is not 1/3 for the same reason. Since when numbers turn to 0 when they are to small? o.o
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-04 at 05:28:39
Since lim[sub]x→∞[/sub] 1/x[sup]2[/sup] = 0.
Your "first try" is not mathematically valid, 1.000... = 0.999...
To cut it down is cheating. If you cut it down, it's not 0.999... anymore.
The entire purpose of repeating decimals is to write fractions as a decimal.
Your "first try" is not mathematically valid, 1.000... = 0.999...
QUOTE
To multiply number with 0.999... we must make it real, let's say we cut it down after trillion nines(it will be our x).
To cut it down is cheating. If you cut it down, it's not 0.999... anymore.
The entire purpose of repeating decimals is to write fractions as a decimal.
Report, edit, etc...Posted by Gigins on 2006-03-04 at 05:32:50
Ok don't cut it down. Multiply 10, infinitive time with 0.999 > eventually it turns out 9.
Multiply 10, infinitive time with 1 > it stays 10 no matter what.
Multiply 10, infinitive time with 1 > it stays 10 no matter what.
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-04 at 05:34:20
0.999...(10) = 9.999..... = 9 + 0.9999..... = 10
The core of the argument is that I say, by definition of a repeating decimal, 0.999... = lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])).
Undesputidely, lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) = 1.
Agreed?
The core of the argument is that I say, by definition of a repeating decimal, 0.999... = lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])).
Undesputidely, lim[sub]x→∞[/sub] ∑[sub]m=1,m=x[/sub](9/(10[sup]m[/sup])) = 1.
Agreed?
Report, edit, etc...Posted by Gigins on 2006-03-04 at 05:44:44
Except this...
9(0.999...) = 8.999...991 not round 9
QUOTE
0.999...(10) = 9.999..... = 9 + 0.9999..... = 10
9(0.999...) = 8.999...991 not round 9
Report, edit, etc...Posted by DT_Battlekruser on 2006-03-04 at 05:46:30
8.99999......anythinghere = 9. It doesn't matter what happens after an infinite number of nines. You never get there.
Report, edit, etc...Posted by Gigins on 2006-03-04 at 05:51:58
I don't get it. So all your logic is that everything that ends with infinitive nines eventually rounds up?
If yes, then theres nothing to argue about thats just an opinion not math based.
If yes, then theres nothing to argue about thats just an opinion not math based.
Report, edit, etc...Posted by Lithium on 2006-03-04 at 07:30:52
these are called RATIONAL NUMBERS. IT IS BASED ON MATH. THE INFINATE NINES ARE INFINATE. IS IT TOO HARD FOR YOU TO NOT KNOW WHAT AN INFINATE IS? JESUS BAJEEZUS YOU GUYS THINK TOO HARD.
are you all in elementary? just bluffing that you're in some high school, except beer. maybe. this is middle school stuff =/
0.99999999.... forever which is actually 0.9 with a dot above which points out that 9 goes on forever. theres no limit to how many nines before the decimal. the 9's go on forever, so forever until theres no difference between a 1 and 0.9 between it.
ok, say this.
are you all in elementary? just bluffing that you're in some high school, except beer. maybe. this is middle school stuff =/
0.99999999.... forever which is actually 0.9 with a dot above which points out that 9 goes on forever. theres no limit to how many nines before the decimal. the 9's go on forever, so forever until theres no difference between a 1 and 0.9 between it.
ok, say this.
CODE
10x = 9.9 dot above.
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
Report, edit, etc...Posted by Gigins on 2006-03-04 at 07:40:16
QUOTE(DEAD @ Mar 4 2006, 01:44 PM)
Except this...
9(0.999...) = 8.999...991 not round 9
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9(0.999...) = 8.999...991 not round 9
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Report, edit, etc...Posted by Lithium on 2006-03-04 at 08:46:45
"Infinate Nines". There is no 1. You cannot have ones when something divides to infinite nines. There is no one. Who says there is one? WHO SAYS? theres no 0 theres no any number beyond the infinate nines. no.
0.9 dot above is 1 even without a round up. why?
---
10x = 9.9 dot above.
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
and to explain it a logical sense. the nine goes on forever. WHATEVER YOU DO TO THE NUMBER, IT WILL BE 99999999 forever, even if you add something, the numbers that are not changed will go on forever and ever. since the nines are forever. in a logical sense of course. There is no gap between 1 and 0.9dot above.
for more reference "Rational Numbers"
ADDITION:
"Infinate Nines". There is no 1. You cannot have ones when something divides to infinite nines. There is no one. Who says there is one? WHO SAYS? theres no 0 theres no any number beyond the infinate nines. no.
0.9 dot above is 1 even without a round up. why?
---
10x = 9.9 dot above.
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
and to explain it a logical sense. the nine goes on forever. WHATEVER YOU DO TO THE NUMBER, IT WILL BE 99999999 forever, even if you add something, the numbers that are not changed will go on forever and ever. since the nines are forever. in a logical sense of course. There is no gap between 1 and 0.9dot above.
for more reference "Rational Numbers"
0.9 dot above is 1 even without a round up. why?
---
CODE
10x = 9.9 dot above.
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
and to explain it a logical sense. the nine goes on forever. WHATEVER YOU DO TO THE NUMBER, IT WILL BE 99999999 forever, even if you add something, the numbers that are not changed will go on forever and ever. since the nines are forever. in a logical sense of course. There is no gap between 1 and 0.9dot above.
for more reference "Rational Numbers"
ADDITION:
"Infinate Nines". There is no 1. You cannot have ones when something divides to infinite nines. There is no one. Who says there is one? WHO SAYS? theres no 0 theres no any number beyond the infinate nines. no.
0.9 dot above is 1 even without a round up. why?
---
CODE
10x = 9.9 dot above.
MINUS x = 0.9 dot above.
---------------------------
9x = 9 ( since it is subtracted by another infinate 9, theres no decimals.
9/9 = 1. there.
10x - x = 9x
and to explain it a logical sense. the nine goes on forever. WHATEVER YOU DO TO THE NUMBER, IT WILL BE 99999999 forever, even if you add something, the numbers that are not changed will go on forever and ever. since the nines are forever. in a logical sense of course. There is no gap between 1 and 0.9dot above.
for more reference "Rational Numbers"
Report, edit, etc...Posted by PCFredZ on 2006-03-04 at 09:25:30
Lithium, you didn't read the other posts; if you did, you didn't understand them.
If x = lim a->0 (10-a)
10x = lim a->0 (100-10a)
9x = lim a->0 (90-9a)
10x - 9x = lim a->0 (10-a)
You're skipping steps, therefore you're getting the wrong answer. This discussion is turning into Null.
If x = lim a->0 (10-a)
10x = lim a->0 (100-10a)
9x = lim a->0 (90-9a)
10x - 9x = lim a->0 (10-a)
You're skipping steps, therefore you're getting the wrong answer. This discussion is turning into Null.