The New Sudoku Players' Forum

[Bigtone53]

I guess that this shows that to get a clear answer, you have to ask a clear question.

A good example of this is the broken stick question. Assume a straight stick is broken at random into three pieces. What is the probability that the pieces can form a triangle? You will find that you get different answers depending on how you define the random breaking process eg

break the stick at two randomly selected points on the stick

break the stick at a randomly selected point, pick the longer piece and break it at a randomly selected point.

break the stick at a randomly selected point, select a piece at random and break it at a randomly selected point

etc

[udosuk]

Chessmaster originally asked whether 0.99999 (repeating) ever reached exactly 1.

I've just noticed Chessmaster actually quoted .99999(repeating) and not 0.99999(repeating) which I have been incorrectly quoting. I don't think my mis-quoting makes any difference to my previous answer but thought I should mention this.

udosuk, in arriving at my answer I chose your above second option as to what "0.99999(repeating)" means which is to continually add 9's after "0." and so never reach 1.

does .999 repeating equal 1 excatly?

Cec, assuming "excatly" is a misspelling of the word "exactly", I'd say the first option is what he intended to ask and your interpretation of "whether 0.99999 (repeating) ever reached exactly 1" was a wrong interpretation. "Equal" and "ever reached" are 2 completely different concepts...

As to your first option being to regard 0.99999(repeating) as 'a real number with the representation "0." followed by an infinite number of 9s', I don't mind confessing I simply don't understand what this means even after reading about real numbers.

It's very hard to understand for me too... But it seems the mathematicians have pretty much all agree that the answer is yes (they're equal), so whether we understand or not we better accept the result...

[Hud]

Something has been on my mind after reading this thread for awhile. Would this problem be analogous to dropping a ball onto a flat surface. The distance between the object and surface would seem to never hit zero since the distance apart could always be halved. Since the ball obviously hits the surface, I would suppose the .9999... would eventually hit 1.
About placing the "0" in front of the decimal, I believe in some countries decimal dimensions are expressed that way. Feel free to correct me if I'm mistaken. In the US, we never prefix with the "0".

[re'born]

Something has been on my mind after reading this thread for awhile. Would this problem be analogous to dropping a ball onto a flat surface. The distance between the object and surface would seem to never hit zero since the distance apart could always be halved. Since the ball obviously hits the surface, I would suppose the .9999... would eventually hit 1.

It's a good analogy Hud, being more or less the same as the version of Zeno's paradox I presented above, just vertical instead of horizontal. I think the most intriguing thing about this whole problem is how it can pit your intuition against itself. On the one hand, our intuition tells us that different decimal representations correspond to different numbers. On the other hand, our intuition immediately rejects Zeno's paradox since most of us have experienced moving from point A to point B. But, logically, rejecting Zeno's paradox is the same as accepting .999...=1 and denying that .999...=1 is the same as accepting that motion is impossible. For me, denying motion is far worse than allowing .999...=1 so, at least intuitively, I accept .999...=1. To really understand why, one should study analysis (in particular, Cauchy sequences or Dedekind Cuts), but...

About placing the "0" in front of the decimal, I believe in some countries decimal dimensions are expressed that way. Feel free to correct me if I'm mistaken. In the US, we never prefix with the "0".

I wouldn't say never as far as the US is concerned. I'm from California and you will see groceries priced $0.99. In mathematics classes, we used whichever version caused the least amount of confusion at the time.

[udosuk]

About placing the "0" in front of the decimal, I believe in some countries decimal dimensions are expressed that way. Feel free to correct me if I'm mistaken. In the US, we never prefix with the "0".

I have a gut feeling what Hud observes here is from the statistical data from sports such as the win-loss percentages or batting averages that frequently we use .500 or .333 instead of 0.5 or 0.333...

But in finance or accounting as rep'nA mentioned I suppose the "0" should be there...

[re'born]

About placing the "0" in front of the decimal, I believe in some countries decimal dimensions are expressed that way. Feel free to correct me if I'm mistaken. In the US, we never prefix with the "0".

I have a gut feeling what Hud observes here is from the statistical data from sports such as the win-loss percentages or batting averages that frequently we use .500 or .333 instead of 0.5 or 0.333...

But in finance or accounting as rep'nA mentioned I suppose the "0" should be there...

That is an interesting point. I suppose though that a stud pitcher's baseball card would list his E.R.A. as 0.87 even though if said aloud nearly everyone would say "point eight seven". Presumably the difference is that those stats are on a scale of 0 to 1, so adding the 0 to the front is even more redundant than usual. For stats (like E.R.A) where the integer part can be greater than or equal to 1, maybe it is more common to use the 0. notation.

[Hud]

rep'nA wrote:

I wouldn't say never as far as the US is concerned. I'm from California and you will see groceries priced $0.99. In mathematics classes, we used whichever version caused the least amount of confusion at the time.

I was basing my observation on engineering drawing which I did a lot of in another life. Sometimes I wish I was still working (especially after losing all my $ in the stockmarket today lol).

[Cec]

udosuk, in arriving at my answer I chose your above second option as to what "0.99999(repeating)" means which is to continually add 9's after "0." and so never reach 1.

Then you are answering a different question. There are two distint questions:

1) Does .999... = 1?

and

2) Does any term in the sequence .9, .99, .999, .9999,... equal 1?

The answer to the first question is yes. The answer to the second question is, of course, no.

Reading the two different answers above suggests (to me) that .999... has a different meaning to .9, .99, .999, .9999,.... My understanding is that both these notations mean the same thing ie. .9 (repeating) or .99(repeating) or .999 (repeating) and so on. Thanks in anticipation.

Cec

[re'born]

udosuk, in arriving at my answer I chose your above second option as to what "0.99999(repeating)" means which is to continually add 9's after "0." and so never reach 1.

Then you are answering a different question. There are two distint questions:

1) Does .999... = 1?

and

2) Does any term in the sequence .9, .99, .999, .9999,... equal 1?

The answer to the first question is yes. The answer to the second question is, of course, no.

Reading the two different answers above suggests (to me) that .999... has a different meaning to .9, .99, .999, .9999,.... My understanding is that both these notations mean the same thing ie. .9 (repeating) or .99(repeating) or .999 (repeating) and so on. Thanks in anticipation.

Cec

It is true that .9(repeating), .99(repeating), .999(repeating),... all mean the same thing. When I write .9, .99, .999, .9999,... I am writing down a sequence of numbers a_1, a_2, a_3,... where

a_1 = .9

a_2 = .99

a_3 = .999
.
.
.

Or to be even more precise

a_1 = .900000...

a_2 = .990000...

a_3 = .999000...
.
.
.

When you talk about continually adding 9's and not getting to 1, this is the sequence to which you are referring.

[Bigtone53]

From Wikipedia

A recurring or repeating decimal is a number which when expressed as a decimal has a set of "final" digits which repeat an infinite number of times. For instance 1/3 = 0.3333333... (spoken as "0.3 recurring"). One convention to indicate a recurring decimal is to put a horizontal line above the repeated numerals. Another convention is to place dots above the outermost numerals of the recurring digits. Where these methods are impossible, the extension may be represented by an ellipsis (...) although this may introduce uncertainty as to exactly which digits should be repeated:

1/9 = 0.111111111111...
1/7 = 0.142857142857...
1/3 = 0.333333333333...
1/81 = 0.0123456790...
2/3 = 0.666666666666...
7/12 = 0.58333333333...
Another notation, used for example in Europe, encloses the recurring digits in brackets:

2/3 = 0.(6)
1/7 = 0.(142857)
7/12 = 0.58(3)

The convention of using a horizontal line above the repeated digits is new to me; I was taught the dot convention. If the entry is correct though, it would seem that conventionally, the use of the ellipsis (...) indicates an infinite repetition, rather than an infinite series.

[Cec]

Thanks rep'nA and Bigtone53 for your replies. If I'm being a nuisance in not understanding or not making myself clear which seems to be the case in my previous post then tell me I'm being a nuisance or something similar and I'll call it quits on this thread.

The two notations 1) and 2) quoted by rep'nA and which I was referring to (or meaning to refer to) in my previous post as being the same thing (to me anyway) were ...

1) .999... = 1 and

2) .9, .99, .999, .9999,...= 1

If both these notations have the same meaning which I believe they do (being .9 repeating) then I can't see how one can conclude they are two distinct questions where a "yes" answer applies to 1) but a "no" answer applies to 2)

"...When you talk about continually adding 9's and not getting to 1, this is the sequence to which you are referring."

I still don't understand your meaning that "this is the sequence to which you are referring". By continually adding 9's (eg .9999999999999 or even more 9's if you like) the value becomes closer to 1 but never reaches 1 (or equals 1).

Cec

[Bigtone53]

Cec,

Let me have a try.

1. The number 0.9999999 ... (ie recurring, with an infinite number of 9's) is a particular number somewhere on the real numbers line. The question is what that specific number is. Conventional wisdom is that the real number in question is 1

2. By writing down the numbers 0.9, 0.99, 0.999 etc, you are generating a series of individual numbers on the line which will approach (ie get as near to as you wish ) the real number that is 0.999... , but will never quite get there. No member of this series will equal 0.999 ... and hence no member of the series will equal 1.

I am sure that rep'nA will have no problem (as do I) in trying to explain further but the concept of infinity is a difficult one to grasp. As an example:-

Imagine a number line containing all possible real numbers starting out with 0 and stretching out to infinity. It can be shown that there are the same (infinite) number of real numbers in the section of the line from 0 to 1 as there are in the entire line from 0 to infinity. Weird!

[re'born]

The two notations 1) and 2) quoted by rep'nA and which I was referring to (or meaning to refer to) in my previous post as being the same thing (to me anyway) were ...

1) .999... = 1 and

2) .9, .99, .999, .9999,...= 1

If both these notations have the same meaning which I believe they do (being .9 repeating) then I can't see how one can conclude they are two distinct questions where a "yes" answer applies to 1) but a "no" answer applies to 2)

"...When you talk about continually adding 9's and not getting to 1, this is the sequence to which you are referring."

I still don't understand your meaning that "this is the sequence to which you are referring". By continually adding 9's (eg .9999999999999 or even more 9's if you like) the value becomes closer to 1 but never reaches 1 (or equals 1).

Cec

Bigtone's geometric explanation is dead on.

Consider the following argument as well:

Every term in the sequence .9, .99, .999, .9999,... has a finite number of 9's and so each term is less than 1. But as .999... has an infinite number of 9's, each term in the sequence is also less that .999... This doesn't show that .999... = 1, but it does show that your sequence of numbers isn't the same as .999... either.

Thanks rep'nA and Bigtone53 for your replies. If I'm being a nuisance in not understanding or not making myself clear which seems to be the case in my previous post then tell me I'm being a nuisance or something similar and I'll call it quits on this thread.

What is they say? "We give an inch and you'll take .999... miles".

[Cec]

"....What is it they say? "We give an inch and you'll take .999... miles".

Quite clever

"..but the concept of infinity is a difficult one to grasp..."

I agree and this is probably why I'm finding it difficult to accept that .99999(repeating) will eventually equal 1.

"2. .... By writing down the numbers 0.9, 0.99, 0.999 etc, you are generating a series of individual numbers on the line which will approach (ie get as near to as you wish ) the real number that is 0.999... , but will never quite get there. No member of this series will equal 0.999 ... and hence no member of the series will equal 1...."

Even accepting that the real number for .999... is 1, the above statement implies (to me anyway) that no matter how far you write down the numbers 0.9, 0.99, 0.999 etc. to get as near to this "imaginary" real number 1 you will still never quite get there.

"... Every term in the sequence .9, .99, .999, .9999,... has a finite number of 9's and so each term is less than 1. But as .999... has an infinite number of 9's, each term in the sequence is also less that .999... This doesn't show that .999... = 1, but it does show that your sequence of numbers isn't the same as .999... either. ..."

True but I would think it's somewhere between .999... and 1 and therefore still never reaches 1.

It looks like we still have different views on this and I again thank you for your patience. Browsing quickly through all the posts again I found it hard to dispute the logic of the following two particular posts which presented good arguments why .99999(repeating) =1.

One third plus one third plus one third = 1, exactly.

.333 repeating plus .333 repeating plus .333 repeating = .999 repeating,
which is one.

Let x=0.999999...

10x=9.999999...

9x=10x-x=9.999999...-0.999999...=9

Therefore x=1.

(Q.E.D.)

I've been persistently arguing that .99999 (repeating) doesn't equal 1 but now I'm not sure

Cec

[999_Springs]

0.99999... is clearly equal to 0.9+0.09+0.009+0.0009+0.00009...

This is a GP with first term 0.9 and common ratio 0.1 so the sum to infinity (i.e. the sum of all the terms) is

(first term)/(1-(common ratio)) when -1<common ratio<1

which is 0.9/(1-0.1)=1

Therefore, 0.99999... =1