Final 0.999... question I promise

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mackdaddy

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This is my last point on this topic I promise, but it is a good one

ok I want everyone to think logically here and not mathematically.

In a world where distance can be divided by infinity you are going to walk a mile

but you decide you are only going to walk 9/10 of the distance left to go

so at first you walk 9/10 of a mile then you walk 9/10 of 1/10 of a mile and so on,

but would you ever get to the end? no because even if you did forever and ever you would never go 10/10 of the way left, am I right?

so therefore, the total distance you will travel is 0.999... since 9/10 plus 9/10*1/10=9/100 plus 9/10*1/10*1/10=9/1000 and so on forever equals 0.999..., but you never reached one mile did you?????? So then how can 0.999...=1????????

Hows that for a logical proof that 0.999...does not equal 1? ;)
 
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Boys & Girls

Line up all the girls on one side of the classroom. Likewise, all the boys against the opposite wall. It is forbidden for a boy to touch any girl for any reason during school hours.

The teacher says, "Every time I blow this whistle everyone move 1/2 the distance between you and the center. They never do touch, but after a few whistles -- the boys and girls are close enough for all practical purposes :)
 
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Haha! I like it, but you must admit this is a valid argument for why 0.999... doesn't = 1.

really I think it just depends on how you view the number. I f one views as a progression like I did in my proof, then it does not equal 1, but if one sees it as a set number that is already set in place and is never growing, then it is equal to 1!:D
 
mackdaddy,

What follows is exactly what you have said (in different words)

Define an infinite ordered set of numbers Sn as,
S0 = 9/10 and Sn+1 = 9/10 + Sn/10

Two Important Facts:
1) 1 is not a member of S
2) You can find a value in S as close to 1 as you like, by choosing larger values for n

In Calculus you will call S a converging sequence.
You will call 1 the limit of Sn as n approaches +infinity.

So, if you mean S when you write 0.999... then it is not 1.
However, if you mean the limit of Sn as n approaches +infinity. then 0.999... is 1.

If when you write 0.999... you mean nothing, then it is undefined.
 
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Bob Brown MSEE said:
What follows is exactly what you have said (in different words)

Define an infinite ordered set of numbers Sn as,
S0 = 9/10 and Sn+1 = 9/10 + Sn/10

Two Important Facts:
1) 1 is not a member of S
2) You can find a value in S as close to 1 as you like, by choosing larger values for n

In Calculus you will call S a converging sequence.
You will call 1 the limit of Sn as n approaches +infinity.

So, if you mean S when you write 0.999... then it is not 1.
However, if you mean the limit of Sn as n approaches +infinity. then 0.999... is 1.

If when you write 0.999... you mean nothing, then it is undefined.
Click to expand...

I still don't quite understand but I'm sure I will eventually, but are you agreeing with me that it can or can't be equal to 1 depending on how you think of the number?

But also if you view 0.999... as a fixed number that does not change, then my original algebraic proof in the last thread cannot be true, this is why, 0.999... has infinite digits after the decimal, and 0.999... being a fixed number times 10, has infinity-1 digits after the decimal, ex: 10*0.999...=9.999...0, where zero is the infinitieth digit after the decimal. So 9.999...0-0.999... does not equal 9, thus flawing that original proof in the first thread.
And if you don't agree that 10*0.999... does not have infinity-1 digits after the decimal then you are contradicting the fact that you believe 0.999... is a fixed number.:)
 
Bob Brown MSEE said:
The thinking that you have done about this will be very useful when you take a Calculus class!
Click to expand...

I hope! but I enjoy thinking problems like these out. I believe that just by thinking about them one becomes smarter in some way.
 
The subtitles of the thread discussion is summed up in my last statement, undefined.
But I lean toward defining it to 1 for entirely different reasons than those stated by others, If you look back in my posts, I think the definition will of repeated 99's will be useful when defining the arithmetic operations for repeated decimal expressions.

Jeff said, "As daon's has now explained several times, the real number system is dense, and two real numbers are the same by definition if no real number intervenes between them." -- That is TRUE and a good proof that the limit of Sn as n approaches +infinity. then 0.999... is 1 iff it means a limit.

But as I showed in the repeating integer examples, where I define \(\displaystyle \overline{285714}3.0\) to be 1/7", the same limit argument does NOT apply.
It is not necessary that repeating digits mean some sort of limit.
 
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Bob Brown MSEE said:
The subtitles of the thread discussion is summed up in my last statement, undefined.
But I lean toward defining it to 1 for entirely different reasons than those stated by others, If you look back in my posts, I think the definition will of repeated 99's will be useful when defining the arithmetic operations for repeated decimal expressions.

Jeff said, "As daon's has now explained several times, the real number system is dense, and two real numbers are the same by definition if no real number intervenes between them." -- That is TRUE and a good proof that the limit of Sn as n approaches +infinity. then 0.999... is 1.

But as I showed in the repeating integer examples, where I define \(\displaystyle \overline{285714}3.0\) to be 1/7", the same limit argument does NOT apply.
Click to expand...

Ok sorry I'm sure you've got a great point but I do not understand so we will leave it at that.
 
Repeating integer Expressions (MackDaddy ignor if you like)

Consider \(\displaystyle \overline{6}7\)

S0 = 7
S1 = 67
S2 = 667
S3 = 6667
S4 = 66667

Clearly the limit as n->infinity is infinite.
However it is VERY interesting to define this to mean the rational value of 1/3.
If I encode the rationals using this system...
1) I also get 1 to 1 mapping between these expressions and rational numbers.
2) The Arithmetic is much easier.
3) No unary minus sign is necessary.
It is not used because the advantage of comparing magnitudes and ease of rounding is easier using repeating decimal expressions.
It is only of Theoretical use, as I am using it here to help understand the meaning of repeating decimals.
 
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1/3 ?

Consider \(\displaystyle \overline{6}7\)
Before you dismiss this too quickly, lets try one of the non-limit-type arguments.

... 6667 = n
... 6670 = 10 n
------------------
... 0003 = 9 n --> n = 3/9 = 1/3

Hmm ...........
What do you get if you multiply ...6667 by 3 ?
 
mackdaddy said:
This is my last point on this topic I promise, but it is a good one

ok I want everyone to think logically here and not mathematically.

In a world where distance can be divided by infinity you are going to walk a mile

but you decide you are only going to walk 9/10 of the distance left to go

so at first you walk 9/10 of a mile then you walk 9/10 of 1/10 of a mile and so on,

but would you ever get to the end? no because even if you did forever and ever you would never go 10/10 of the way left, am I right?

so therefore, the total distance you will travel is 0.999... since 9/10 plus 9/10*1/10=9/100 plus 9/10*1/10*1/10=9/1000 and so on forever equals 0.999..., but you never reached one mile did you?????? So then how can 0.999...=1????????

Hows that for a logical proof that 0.999...does not equal 1? ;)
Click to expand...
Mackdaddy. No one here is going to object to someone with intellectual curiosity. As Subhotosh Khan has pointed out, what you have written is a variant of Zeno's Paradox, which had mathematicians stumped for well over 2000 years. A solution to the paradox was proposed by Georg Cantor in the late 19th century. (Look up Cantor on wikipedia.) Most mathematicians have accepted that solution, but it requires exploration of the properties of transfinite (that is infinite) numbers. As Bob Brown has pointed out, your questions will prepare you for calculus, which started out as a practical study of the infinitely small, the infinitesimal. The mathematicians were not able to make that rigorous for 200 years, but again finally did so in the 19th century. (Look up Cauchy and Weierstrass.) A number of people have found the 19th century's solutions to these problems to be unattractive on esthetic or philosophical grounds; I do myself, but I accept them nevertheless. After all, the "logic" of Zeno's Paradox clearly leads to the wrong result: the slower turtle is passed by the faster Achilles despite the turtle's head start. If you accept "standard analysis" of the real numbers, then 1.000... = 0.999... follows logically as daon has explained: there is nothing between them within the system of standard real analysis so they are equal. Alternatives to the 19th century's standard analysis have since been proposed, and at least one (called non-standard analysis) is considered by mathematicians to be rigorous. That proposal requires a new kind of number, the hyper-reals, but I have not studied them. It may be that using non-standard analysis there is a distinction between 0.999... and 1.000... As I said, I have not studied non-standard analysis. So, as far as I can see, you can: (1) accept standard analysis, from which it follows logically that 1.000... = 0.999... and that Achilles can catch the turtle, (2) agree with Zeno that motion is impossible so Achilles cannot catch the turtle, or (3) create a logically rigorous alternative to standard analysis in which Zeno's Paradox fails but 1.000... and 0.999... are different numbers. I find option 2 absurd, and option 3 is something that I do not have the capacity to attempt.
 
Thanks, but is there not just one answer to this problem, I mean why is it so hard to prove this wrong or right? Is this proof not satisfactory and why does it come to a different conclusion than the algebraic proof where you use x as 0.999...? I just don't understand how one number can equal a completely different number, I mean why does 0.999... even exist if it's the same as one? I mean 1 doesn't equal 1.0...1 does it? And if so then 1.0...1=1.0...2 and so 1=1.0...2, but it DOESN'T! but how is that not so, but 0.999... =1 is true?
 
mackdaddy said:
Thanks, but is there not just one answer to this problem, I mean why is it so hard to prove this wrong or right?
Click to expand...
It's not- not if you know the definitions well- in particular, do not know very well the difference between "numbers" and "numerals" or do not grasp the difference between a "sequence" and the limit of a sequence.

Is this proof not satisfactory and why does it come to a different conclusion than the algebraic proof where you use x as 0.999...? I just don't understand how one number can equal a completely different number, I mean why does 0.999... even exist if it's the same as one? I mean 1 doesn't equal 1.0...1 does it? And if so then 1.0...1=1.0...2 and so 1=1.0...2, but it DOESN'T! but how is that not so, but 0.999... =1 is true?
Click to expand...
One number can't "equal a completely different number" but one number can be represented by many different numerals- for example, 0.5 can be represented by 1/2, 2/4, etc.
 
mackdaddy said:
Thanks, but is there not just one answer to this problem, I mean why is it so hard to prove this wrong or right? Is this proof not satisfactory and why does it come to a different conclusion than the algebraic proof where you use x as 0.999...? I just don't understand how one number can equal a completely different number, I mean why does 0.999... even exist if it's the same as one? I mean 1 doesn't equal 1.0...1 does it? And if so then 1.0...1=1.0...2 and so 1=1.0...2, but it DOESN'T! but how is that not so, but 0.999... =1 is true?
Click to expand...
The difficulty has to with anything that incorporates the concept of infinity. Our intuition about what is logical with finite numbers turns out not to be logical with infinite (transfinite) numbers. Think about this. Your argument (and Zeno’s some 2500 years ago) is:
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^1 0 * 10^{-i}\right)-\left(\sum_{i=1}^1 0.9 *10^{-i}\right) = 1.0 - 0.9 = 0.1 = 10^{-1} ≠ 0.\)
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^2 0 * 10^{-i}\right)-\left(\sum_{i=2}^2 0.9 *10^{-i}\right) = 1.00 - 0.99 = 0.01 = 10^{-2} ≠ 0.\)
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^3 0 * 10^{-i}\right)-\left(\sum_{i=1}^3 0.9 *10^{-i}\right) = 1.000 - 0.999 = 0.001 = 10^{-3} ≠ 0.\)
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^4 0 * 10^{-i}\right)-\left(\sum_{i=1}^4 0.9 *10^{-i}\right) = 1.0000 - 0.9999 = 0.0001 = 10^{-4} ≠ 0.\) So by induction you arrive at
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^\infty 0 * 10^{-i}\right)-\left(\sum_{i=1}^{\infty} 0.9 *10^{-i}\right) = 1 + 0.000...\ - 0.999...\ = 1.000...\ - 0.999...\ = 10^{-\infty}.\)
.
You then presume that \(\displaystyle 10^{-\infty} ≠ 0 \implies 1.000... \ne 0.999... .\) But is that last step sound? We have no personal experience with the infinite.
.
\(\displaystyle \dfrac{1}{3}= 0.333... \implies 3 * \dfrac{1}{3} = 3 * 0.333... \implies 1 = 0.999...\)
.
Two lines of sound logic cannot arrive at a contradictory conclusion. And I for one am confident that 3 times one third is one. So I reject your line of argument as leading to a contradiction. And that means:
.
\(\displaystyle \displaystyle1 + \left(\sum_{i=1}^\infty 0 * 10^{-i}\right)-\left(\sum_{i=1}^{\infty} 0.9 *10^{-i}\right) = 1 + 0.000...\ - 0.999...\ = 1.000...\ –\ 0.999... = 10^{-\infty} = 0 \implies 1.000... =0.999....\) Welcome to the strange world of infinity.
.
One last thought. (Edit: Halls of Ivy expressed this thought much better than I did.) There is nothing strange in having two different symbolizations represent the same number:
.
\(\displaystyle 3 + 7 =10 = 12 - 2.\)



 
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JeffM said:

.
\(\displaystyle \dfrac{1}{3}= 0.333... \implies 3 * \dfrac{1}{3} = 3 * 0.333... \implies 1 = 0.999...\)
.
Two lines of sound logic cannot arrive at a contradictory conclusion. And I for one am confident that 3 times one third is one. So I reject your line of argument as leading to a contradiction.
Click to expand...


But how can you assume 1/3=0.333... without first assuming that 0.999... is equal to 1. So this argument is invalid because in order to prove that 0.999... is equal to one, you must prove 0.333... is equal to 1/3 but in order to do that you must prove that 0.999... is equal to one. In other words 1/3 of 0.999... is 0.333... but do we know for sure that 1/3 of 1 is equal to 0.333...? no we don't not unless we know for sure that 0.999... is equal to 1.
 
1/3 = 0.33333333333.... comes from definition of division (of fractions) and decimal numbers (i.e. old fashioned way).
 
JeffM said:
.
You then presume that \(\displaystyle 10^{-\infty} ≠ 0 \implies 1.000... \ne 0.999... .\)

Click to expand...


Ok but say this 10^-infinity does equal 0 so 1/10^infinity=0

B^-e=1/B^e

so 1/10^infinity=0
now using 10^infinity as x, and don't get mad at me but I'm not talking about 0.999... anymore I'm talking about 1/10^infinity=0

say you have a function f(x)=1/x
where f(x) equals the y coordinate
this function never ends up where f(x)=0, never ever, apparently, according to the laws of math.
But from what your saying 1/10^infinity=0 so if the coordinate pair of (1/10^infinity, 0) is on the graph of the function then the function does reach f(x)=0. That is if you are doubting the laws that this function follows. But if you aren't then 10^-infinity cannot equal 0.

Please see this as simply a completely different proof where I am trying to prove that 10^-infinity does not equal 0
 
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mackdaddy said:
I mean 1 doesn't equal 1.0...1 does it? And if so then 1.0...1=1.0...2 and so 1=1.0...2, but it DOESN'T!
Click to expand...

Ha,ha ... good one mackdaddy. If I understand, this kind of example is what I would use to attack my repeating integers notation. Dealing with infinity is tricky remember what happened to Georg Cator.

Setting that a side, If you are thinking of limits inspired by these examples ... then both limits equal 1.
I must make assumptions about your notation so let's use the sequence S again to communicate.

For 1.0...1 do you mean the limit of Sn as n -> +infinity
S0 = 1.1
S1 = 1.01
S2 = 1.001
S3 = 1.0001
S4 = 1.00001
The Limit here is 1

For 1.0...2 do you mean the limit of Tn as n -> +infinity
T0 = 1.2
T1 = 1.02
T2 = 1.002
T3 = 1.0002
T4 = 1.00002
The Limit here is 1

Sorry if I have miss understood.
However, if you mean by 1.0...1 notation, to be representing the LIMIT of the implied SEQUENCE then,
1 = 1.0...1 and
1 = 1.0...2
 
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