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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
Firstly, I have to say that the talk of numbers such as 0.(0)1  taken to mean zero point zero zero... an infinite number of times... followed by a one  is entirely nonsensical. It doesn't mean anything to put a number 1 after an infinite series of digits. For a start, it tries to imply that the series of digits terminates (supposedly at the "1"), but it clearly does not as it is infinitely long.
Secondly, when you write down 0.999..., you may be asked to define what you really mean by that. Note that:
So, what we really mean when we write down 0.999... is: Which, if we were to be fully rigorous, means that: So the above is what we really mean by 0.999...  so the question is, is the above expression equal to one? Well... yes, it is.
Bad speling makes me [sic]
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
In either proof a guess is used. So false proof.
For instance, 0.999...*10=9.999... the guess is about the rule to handle multiplication with some form of a "number" with infinite digits, which is neither a natural logic of product involving a number with finite digits, nor testified by any human experience.
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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
George,Y wrote:In either proof a guess is used. So false proof.
For instance, 0.999...*10=9.999... the guess is about the rule to handle multiplication with some form of a "number" with infinite digits, which is neither a natural logic of product involving a number with finite digits, nor testified by any human experience.
Where is there a "guess" in my post, just before yours?
Bad speling makes me [sic]
 All_Is_Number
 Power Member
Re: 0.9999....(recurring) = 1?
George,Y wrote:In either proof a guess is used. So false proof.
For instance, 0.999...*10=9.999... the guess is about the rule to handle multiplication with some form of a "number" with infinite digits, which is neither a natural logic of product involving a number with finite digits, nor testified by any human experience.
A*B*A^C = B*A(C+1)
You can shear a sheep many times but skin him only once.
 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
For instance, 0.999...*10=9.999... the guess is about the rule to handle multiplication with some form of a "number" with infinite digits, which is neither a natural logic of product involving a number with finite digits, nor testified by any human experience.
So are you saying we don't know how to multiply and recurring rational number or irrational number? Cause both those have infinite digits.
Sound a bit irratioanl if you ask me.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
I guess you've got some misunderstanding, Ricky.
1/9 is rational. 1/9=0.111... with infinite digits is a guess. Sure you can reach a point where no matter how many 1 can appear as long as enough fraction steps have been taken, but that doesn't mean you can just say 1/9 is equal to 0.111...with infinite digits by any logic.
In regard to irrational numbers, I feel more unreasonablehas anyone every drinked up a cylinder cup of water? The answer should be no, because no material can be made circle or cylinder. Pi is used as an approximation, an approximation for some long digits, which is reality.
A number with infinite digits has not yet been testified, hence it's my choice to use a competitor system.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
1/9=0.111... with infinite digits is a guess. Sure you can reach a point where no matter how many 1 can appear as long as enough fraction steps have been taken, but that doesn't mean you can just say 1/9 is equal to 0.111...with infinite digits by any logic.
1/9 has infinite decimal digits of 1 because of the way we define division, 1, and 9. If you use the alogirthm of division and try to divide 9 into 1, you will always get a 1 as the result because you will always be doing the same thing:
_0___R1 9 ) 1 0  1 _0.1___R1 9 ) 1.0 .9  0.1
_0.11___R1 9 ) 1.00 .9  0.10 0.09  0.01
And you will always get remainders of 1's. Because of this, you will always have 10, and you will always have a 1's digit in the decimal place.
If that isn't an infinite number of 1's, I don't know what is.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
In regard to irrational numbers, I feel more unreasonablehas anyone every drinked up a cylinder cup of water? The answer should be no, because no material can be made circle or cylinder. Pi is used as an approximation, an approximation for some long digits, which is reality.
Can there be a real cylinder in the real world? The problem isn't pi, it's that we neither have the ability to create or measure to infinite accuracy.
But who cares about the real world anyways? I'm only interested in the math world.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
A number with infinite digits has not yet been testified, hence it's my choice to use a competitor system.
I think you need to study different base systems. For example, try representing 0.1 in base 2. Then try representing 1/9 in base 9.
Infinite digits occur only because of the relation to the denominator and the base. It is an artifact of not being divisble.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
"If that isn't an infinite number of 1's, I don't know what is."
Yep, you match what I saidyou just proved as many 1's so long as many steps. What I disagree is the infinite digits proposal, that because you cannot express it in finite digits, you guess that there is a state when infinite digits do exist, and make some seemingly reasonable rules to it.
The reason why I am against 0.999...=1 is that it's trying to equate some entity human never being able to testify on the left with something out of common experience on the right which is touchable in daytoday life , on an assumptionhidden means. The main purpose, I guess, is to stablish absolute correctness for results by infinite ways.
Do you remember the exact defination of a limit near some points? Note the difference in the domains of x and y y can reach y0, while x is not able to reach x0. Mr Leibniz's intention was very clear, as he tried to avoid the logic flaw of the reached point. (Search for "Berkley vs Newton" for more details)
Modern Mathematicians are so confident that they provide such kinda false proof, where they embeded reached infinity assumption such as infinit digits implicitly, trying to verify maths' correctness and accurateness, and its ability to gain a stable result, although none of them is able to write out 3.14159... or 2.7818... wholy.
Thanks for your advice, I know base systems, and I agree with you that is artificial. Human ideas are indeed artificial, some of which belong to religion, some belonging to arts, and some belonging to science. I guess the unique charactor of the last one mentioned is its courage to risk enough to be tested wrong but not yet.
Last edited by George,Y (20061003 22:05:24)
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
Yep, you match what I saidyou just proved as many 1's so long as many steps.
The division algorithm requires you to keep going until you reach 0. Since you will never reach 0, you will always keep going. Just because I didn't take the next step doesn't mean that it isn't flat out obvious I will get a 1, it is. In fact, I don't keep going because I always know what I will be getting, more 1's.
The reason why I am against 0.999...=1 is that it's trying to equate some entity human never being able to testify on the left with something out of common experience on the right which is touchable in daytoday life , on an assumptionhidden means.
Since when is math ever based on what we see in the real world, George?
Thanks for your advice, I know base systems, and I agree with you that is artificial. Human ideas are indeed artificial, some of which belong to religion, some belonging to arts, and some belonging to science. I guess the unique charactor of the last one mentioned is its courage to risk enough to be tested wrong but not yet.
Artifact and artificial are two very different words, however similar they may appear. A base may have an infinite number of digits for any given number only because it is unable to properly display such a number. Converting the base often leads to a finite number of digits. Still the same number, different representations. For example, 1/9 is .1 in base 9. When we convert from base 9 to base 10, we get .11111... You continue to claim these are different numbers, but all they are, are different representations of the same number.
I have a proof above showing that 1/9 = .111... if we never stop trying to divide. Do you have one which shows they are different? If not, how is my proof flawed?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Ricky wrote:I have a proof above showing that 1/9 = .111... if we never stop trying to divide. Do you have one which shows they are different? If not, how is my proof flawed?
I am sorry to remind you that "your proof" was first proposed by me, in an answer to your induction request in the other thread discussing 0.999...
Today's Base systems are built coherently infinite digits belief and real system background such as 0.999... So you provided a inbuilt proof.
My proof was stated clearly at the begining, yet I can show it more obviously here: Cannot be devided at any digits=>Can be devided or expressed at infinite level false logic. You need to assume that there is indeed an infinite level first, and assume the result.
Since when is math ever based on what we see in the real world, George? Since ancient times, Ricky. Since we humans first identified 2 apples.
Let us see how maths interpret the real world and practice. When you divide a pizza by nine pieces, you can say you get 9 more or less same ninths. You do get the chance of getting them same when the total amount of particles is devided by 9. But let's say what happens when you state 0.1111... first you divide it by 10, get 1 tenth, since 1 ninth is larger than 1 tenth but less than 2 tenths. So you pick up another 1 tenth, and devide it smaller by 10. Here you find that 1 ninth is larger that 1 tenth plus 1 hundredth but less than 1 tenth plus 2 hundredths and you scrible down 0.11 and go on. These process are what 1/9 and 0.111...work in practice. The difference between the former and the latter is that the latter implicitly holds a forever divisiblity assumption, that a pizza can be devided smaller and smaller without limitation, which has been tested wrong ever since the invention of microscope.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
My proof was stated clearly at the begining, yet I can show it more obviously here: Cannot be devided at any digits=>Can be devided or expressed at infinite level false logic. You need to assume that there is indeed an infinite level first, and assume the result.
That made no sense whatsoever.
Let us see how maths interpret the real world and practice.
I never said math didn't apply to the real world. Only that it doesn't always come from there.
that a pizza can be devided smaller and smaller without limitation, which has been tested wrong ever since the invention of microscope.
Which is one of the many differences between the math world and the real world. Infinity, as far as we know, does not exist in the real world. It does however exist in the math world.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Since you cannot accept verbal logic, I can just provide some proof with reference.
"The division algorithm requires you to keep going until you reach 0. Since you will never reach 0, you will always keep going. Just because I didn't take the next step doesn't mean that it isn't flat out obvious I will get a 1, it is. In fact, I don't keep going because I always know what I will be getting, more 1's."
So are you using a guess as your proof? A guess cannot make up a proof no matter how obvious it looks, as you can consult to your math teacher.
If you are not using a guess, I assume you were using a deduction proof, agree?
A deduction proof can reach any step that you can can assign a sequence number. Step 1000, Step 198745 for instance. However, it does not give crevidence to Step Infinity, as you could consult to any textbook introducing deduction.
Then how is the infinity situation been proved? I'm afraid to say it is not proved unless you assume something first.
My points are: A) 0.999...=1 is not proved unless a new assumption is added, such as one that infinite digits can make up a number while this number share more or less the same +*/ rule with integers and rationals or one that 1/9 is assumed as 0.111... with infinite ones. B) You can reject this kind of assumption when you find it is of little use in reality or it contradicts reality when you believe you study math for its usefulness rather than its complexity. Accepting the assumption is not a must.
I agree with you that math is not alway a reflection of reality. I propose math is made of assumptions and their logical conclusions. That is why I've spent so much on figuring out an assumption within the proof.
Last edited by George,Y (20061005 19:35:22)
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 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
More about base systems:
A base system, as I see it, is an attempt to express/equate a given amount/quantity by a particular combination of a base's integer exponents. For example:
IIIIIIIIIIIIIIII can be expressed as I*[(II)^(IIII)] name I as 1, and you can get 10000 in base 2. can be expressed as I*[(IIIIIIIIII)^I]+(IIIIII)*[(IIIIIIIII)^0] name 1, 2, ...9 correspondingly, and you can get 16 in base 10.
However this attempt doesn't always work well with rationals, say, can you express any given rational by a combination of exponents of a particular integer? That's why they develop a combination of "infinite" entries. The strongest evidence supporting this is the as close as many entries approach and a following guess " wow, so after infinite entries that is larger than any "many" entries it can be the same"
Finally, another reference.
Zeno was arguing that it does not make sense to add infinitely many numbers together. But there are other situations in which we implicitly use infinite sums. For instance, in decimal notation,k the symbol 0.3^{}=0.333... means 3/10+3/100+3/1000+... and so, in some sense, it must be true that 3/10+3/100+3/1000+...=1/3 page 8, Calculus:Concepts and Contexts by James Stewart.
You should know the author intend to prove infinite sums' correctness by proving it is equivalent to infinite decimals and people's approval of the latter. One right, another also. Or one wrong, another also. If one needs an assumption, another needs a similar assumption. I guess many would agree 1+1/2+1/4+...=2 is one assumption/defination more than
Last edited by George,Y (20061005 20:29:07)
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
0.999...=1 is not proved unless a new assumption is added, such as one that infinite digits can make up a number while this number share more or less the same +*/ rule with integers and rationals or one that 1/9 is assumed as 0.111... with infinite ones.
If infinite digits can't make up a number, then what is e or pi? Or the square root of two?
You have to allow infinite digits to be a number. Otherwise, you are left with rationals that don't repeat an integers. Math isn't much fun if you chop off entire branches of it.
Do you deny that pi + pi = 2pi? Or that e*e = e^2?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
So are you using a guess as your proof? A guess cannot make up a proof no matter how obvious it looks, as you can consult to your math teacher.
If you are not using a guess, I assume you were using a deduction proof, agree?
Well this may explain quite a bit. Do you not accept the principal of mathimatical induction as being valid?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
If infinite digits can't make up a number, then what is e or pi? Or the square root of two? Those are real numbers. Real numbers are defined as infinite sets of rational numbers. It involves much more complex settings than the issue we discuss. Personally, I don't like the defination for the reason of untouchability.
You have to allow infinite digits to be a number Well, you are discussing a number's defination. What I was trying to prove is that there are two types of numbers, one is friendly with human experience, one is derived imagination. The Reals is the latter. Moreover, a defination cannot be arbitary freely. For example, you may deny 0.999...91 but accept 0.999.... But the reason for the denial may be irrational.
pi and e are derivatives of corresponding sequences. They are both created on a closingsoreaching basis. I don't deny their definations but their feasibility and rationality. I do not accept the statement " pi is true, but we never get a chance to measure it". Nor do I think it's any fun to fool a person around with a misleading 0.999...=1 proof .
You asked my acception about induction. I don't deny induction logic, it works well within its boundry. For example, you step southwards, then step again, finding your latitude has lowered. Again and again you predict the latitude will keep on lowering as long as you walk in the same direction. This is not the fault of induction, however. This process is called a naive induction or an incomplete induction, which means you induce by limited experience of the past, and that you don't prove a definate 100% reason for one step to go to the next. The 100% reason here, particularly in maths, is logic or support from some other theorems or definations. Note, a defination is always an assumption.
It seems you did not understand my proof. My proof is that you cannot prove it by induction( formerly misspelled as "deduction") alone. The last step must involve an assumption or a defination.
Last edited by George,Y (20061007 16:59:14)
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Re: 0.9999....(recurring) = 1?
I just think that: 0,9999(9) = 1  1/infinity
1. So, in physics or chemistry we will think like this > 0,99999(9) = 1 But mathematicians won't be satisfied, and I think this is the number that ends the (infinity;1) .
2. but if we consider about limits:
1/infinity = 0 => 0,9999(9) = 10=1
3. And, at last there is algoritm to find the proper fraction:
0,9999(9) = 0,9 + 0,09 + 0,009 .... x= 0,9 + x/10 .10 10x= 9+x 9x=9 x=1
I assume that x= 0,9999(9) is equal to x=1 at every equation, exept some cases like this 0/(1x)
x=1 => 0/0=? x=0,9999(9) => 0/x=0
Last edited by akademika (20061007 23:20:56)
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Dross's and akademika's similar prooves involve a more complex structure. I will explain them with lots of detail the other day.
Proof via 1/9=0.111... or 1/3=0.333... is comparatively easier to defy. In previous posts the former as an example is questioned from the begining. That is to say no proof can make 1/9 equal to 0.111... with infinite decimals.
It may be some cruel instead of fun in my proof, but a proof is either true or false, a zerosum game. It was also cruel for those who thought 0.999...=1 was wrong previously.
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 Ricky
 Moderator
Re: 0.9999....(recurring) = 1?
So what is 1/9 equal to in decimal notation, George?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
It does not equal to any decimal notation, if by strict means.
Unstrictly, it can be written as 0.111... . I used to see it on some site when it is explained as when you go on dividing, you will simply get more and more 1 s, where "more" , I interpret, is still within "many" and thus finite framework.
Unstrictly, you can imagine 1/9 does equal to 0.111... when the latter imaginarily has infinite, larger than any "many", decimals, but still on a finite "closer and closer" or "more and more" experience basis.
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 Dross
 Power Member
Re: 0.9999....(recurring) = 1?
George,Y wrote:It does not equal to any decimal notation, if by strict means.
Unstrictly, it can be written as 0.111... .
Every real number has a (not necessarily unique) decimal expansion. That includes pi, e, 1/9 and 1.
Read my above post (top of the page, as I see it) for a clear meaning of what we really mean when we write a recuring decimal. I can assure you, it is entirely well defined and unambiguous.
Bad speling makes me [sic]
 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
Post 26#, just another "closing so reaching" guess, though you've made a great identity which I agree as well.
I admit i am a little dumb at figures. I can calculate 0.9+0.09, and I can count 0.9+0.09+0.0009+0.00009+0.000009 and any sum with finite entries. But I am not sure about what it will be after infinite, larger than any "many", additions. So who is sure? and how?
Perhaps the most rational way to know it or to guess it is use a limit. The limit, though, can only tell it will get closer and closer, rather than its ultimate arrival. This is pretty natural because the N∈ is within the finite concept, meaning as close as you tell as long as enough steps, where "enough" is finite.
The limit is one guess less than the summation, and the guess is about infinity, another "closing so reaching" inference.
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 George,Y
 Super Member
Re: 0.9999....(recurring) = 1?
The most Tricky Proof:
10*0.999...=9.999... ) 0.999...=0.999... 9*0.999...=1
For the first step I get 10*0.999...=9.99... step by step, digits by digits so far. Thus the proof left a bothering remainder and is not so convincing.
An arguement against this is that 0.999... is a all developed state where are infinite 9's. Such state does exist in the lecturer's head. But it is sheer imaginery. Could I imagine a "well developed" state like 0.999...91? They may argue it doesn't make sense. Well, I don't know if I am not permitted to think of the infiniteth digit another way, but I demand the right. Further, if everyone comes up with different ideas about infinity, infinity is an art terminology.
Another argument against it is the "The impossibility of infinite steps" one again how do I know it's a state? So far I can do is to write 0.9, 0.99, 0.999,..., 0.9999999, which are all numbers and constant, stable, and unvarying. I don't know if infinity vary or not, and if it does not, what is it? It's just the amount you cannot count. Okay, how complex!
One proof that 0.9+0.09+0.009+... is a number is a limit convergence proof, which could be disproved by Post 49.
The last proof that 0.9+0.09+0.009+... is a number is the " A strict increasing series must have a least upper bound" proof, definately having something to do with the Real defination created by Georg Cantor. Actually the Reals are defined as approaching something, which I should write a little bit the other day.
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