0.9... = 1

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skeptical scientist

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« Reply #120 on: November 07, 2006, 09:11:11 PM »
Quote from: "Erasmus"
Quote from: "skeptical_scientist"
Hence, the best point on the river to get water is ...


Well done... here is another one which has just been sent to me: http://xkcd.com/blue_eyes.html (btw, the comic on that website is quite good as well).


I've seen similar before (a little less complicated, but much more violent!) Short answer: On the 100th night after the day when the guru makes his statement, all of the blue eyed people leave the island. Nobody else leaves, until the guru makes another pronouncement and triggers a second delayed-reaction mass exodus. As to why, you'll have to figure it out yourself. Bonus question! All of the islanders already knew that someone had blue eyes, so exactly what information was conveyed by the guru's statement?

P.S. this one's really good: http://www.xkcd.com/c123.html
-David
E pur si muove!

0.9... = 1
« Reply #121 on: November 07, 2006, 09:16:55 PM »
I never seemed to figure this one out:  http://www.highiqsociety.org/puzzles/puzzle1_1.php

Maybe I'm stupid, I just never got it.
ooyakasha!

0.9... = 1
« Reply #122 on: November 07, 2006, 09:23:11 PM »
Well, Erasmus, if you're interested here is part a) of the problem I was working on:

 captain is sailing through the arctic. The first mate runs up and says to him, "captain, there is an iceberg dead ahead. What should we do?" The captain looks at the iceberg, then glances at his map and says, "there's no iceberg here! Keep going!"

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skeptical scientist

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« Reply #123 on: November 07, 2006, 09:32:34 PM »
Quote from: "Knight"
I never seemed to figure this one out:  http://www.highiqsociety.org/puzzles/puzzle1_1.php

Maybe I'm stupid, I just never got it.

The triangles aren't really what you think they are, so it's really an optical illusion. The graph paper makes you think they pass through points which they don't actually pass through. If we call the skinny tip of the triangle (0,0) then you are led to believe that the upper border of the top triangle passes through (8,3) and (13,5), and the upper border of the bottom triangle passes through (5,2) and (13,5), when really the points should be (8,3.07) and (5,1.92) In other words, the lighter and lightest blue triangles on top are slightly bigger than the ones on bottom, which accounts for the missing area.
-David
E pur si muove!

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Erasmus

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« Reply #124 on: November 07, 2006, 09:43:15 PM »
Quote from: "skeptical_scientist"
All of the islanders already knew that someone had blue eyes, so exactly what information was conveyed by the guru's statement?


I think the information is, "If there is exactly one blue-eyed person, that person now knows that his eyes are blue, and should leave."  The rest of the information is conveyed by the passage of time.

(p.s. I can't take credit for this.... my girlfriend, who also brought me dinner, came up with it.)

Quote
P.S. this one's really good: http://www.xkcd.com/c123.html


Yeah, one of my favourites, definitely.
Why did the chicken cross the Möbius strip?

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jholman

0.9... = 1
« Reply #125 on: November 08, 2006, 01:04:25 AM »
Regarding the triangles puzzle, rather than the "triangles not being what you think they are", I would say that they do not form one large triangle, but rather a quadrilateral that looks a lot like a triangle.  And it should come as no surprise that two quadrilaterals with different shape happen to have different area.

A key point is that the two light blue triangles are not similar to one another: one has ratio 2:5, and the other 3:8.  This should help persuade you that neither of the large figures is a triangle, but rather a quadrilateral.



Regarding the highiqsociety.org site, it appears to me to be a scam, whereby one does a test, feels good about oneself, and is motivated to spend $79 USD for a "lifetime membership" of sight- (site- ?) unseen value.  They claim their IQ tests are blah blah so sophisticated blah, but they are the same every time (contrary to claims), and also I managed to get a score of 97 (almost as smart as an average human!) using a random number generator.  Given that one can retake the test until one is happier with one's results, it's not a very effective filter.

Of course, none of this makes their puzzles less puzzley.  Just a caution for those who think it's a much nicer website than mensa.org (which it is).



Regarding the blueeyes puzzle, if you want more thinking about it, consider solution.html at the same domain name.

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BOGWarrior89

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« Reply #126 on: November 09, 2006, 11:09:38 AM »
0.9... != 1.

9/9 = 1.

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Erasmus

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« Reply #127 on: November 09, 2006, 11:31:03 AM »
Quote from: "BOGWarrior89"
0.9... != 1.


I think we've established that you believe this.  I think we've also established that several of us feel that you are provably wrong.  The only thing that isn't clear is why, at no point, have you provided anything resembling a proof that 0.9.... != 1.

I'll give you some starters!

1)  You can try to prove that 0.9... < 1.

2)  You can try to prove that 0.9... > 1.  I think (1) is more promising.

3)  Keeping (1) in mind, you can try to find a number x such that 0.9... < x < 1.

That sounds like a good place to work from.  Obviously if you can find some number between 0.9... and 1, you would have proven that they are equal.  Can you find such a number?
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BOGWarrior89

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« Reply #128 on: November 09, 2006, 01:44:21 PM »
Quote from: "Erasmus"
Quote from: "BOGWarrior89"
0.9... != 1.


I think we've established that you believe this.  I think we've also established that several of us feel that you are provably wrong.  The only thing that isn't clear is why, at no point, have you provided anything resembling a proof that 0.9.... != 1.

I'll give you some starters!

1)  You can try to prove that 0.9... < 1.

2)  You can try to prove that 0.9... > 1.  I think (1) is more promising.

3)  Keeping (1) in mind, you can try to find a number x such that 0.9... < x < 1.

That sounds like a good place to work from.  Obviously if you can find some number between 0.9... and 1, you would have proven that they are equal.  Can you find such a number?


By this logic, you don't exist, because I can't find you.

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cadmium_blimp

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0.9... = 1
« Reply #129 on: November 09, 2006, 01:45:26 PM »
Not being able to find something doesn't count if you're not looking.

Quote from: Commander Taggart
Never give up, never surrender!

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BOGWarrior89

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« Reply #130 on: November 09, 2006, 01:46:04 PM »
Quote from: "Erasmus"
Quote from: "BOGWarrior89"
0.9... != 1.


I think we've established that you believe this.  I think we've also established that several of us feel that you are provably wrong.  The only thing that isn't clear is why, at no point, have you provided anything resembling a proof that 0.9.... != 1.


Ok.

1/9 ~= 0.11111...
9(1/9) ~= 0.9999...
1 ~= 0.9999...

(btw, "~=" means "about equal to".)

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Erasmus

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« Reply #131 on: November 09, 2006, 02:37:13 PM »
Quote from: "BOGWarrior89"
1/9 ~= 0.11111...
9(1/9) ~= 0.9999...
1 ~= 0.9999...


Are you concluding anything from this?  Specifically, are you trying to conclude that 0.9... != 1?  That conclusion is invalid, since for all x, x ~= x, but x = x.  In other words, just because two numbers are "about equal" it doesn't mean that they are not equal.

You didn't like my suggestions for a proof?
Why did the chicken cross the Möbius strip?

0.9... = 1
« Reply #132 on: November 09, 2006, 03:29:34 PM »
Quote from: "Erasmus"
Quote from: "BOGWarrior89"
1/9 ~= 0.11111...
9(1/9) ~= 0.9999...
1 ~= 0.9999...


Are you concluding anything from this?  Specifically, are you trying to conclude that 0.9... != 1?  That conclusion is invalid, since for all x, x ~= x, but x = x.  In other words, just because two numbers are "about equal" it doesn't mean that they are not equal.

You didn't like my suggestions for a proof?

How about this:

.3....  != 1/3 (the more places you go, the closer the approximation, but it is never exactly equal  plotted on a graph the curve of .3..3 will approach but never touch 1/3)
Also  3 * .3...3  = .9...9 not 1

Given that x = .9...9:

for any finite number of repeats represented by the elipse:
 9.9...9  (10x)
- .9...9  (x)
= 8.9...1
therefore
9x = 8.9...1

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BOGWarrior89

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0.9... = 1
« Reply #133 on: November 09, 2006, 03:31:19 PM »
Quote from: "Curious"
Quote from: "Erasmus"
Quote from: "BOGWarrior89"
1/9 ~= 0.11111...
9(1/9) ~= 0.9999...
1 ~= 0.9999...


Are you concluding anything from this?  Specifically, are you trying to conclude that 0.9... != 1?  That conclusion is invalid, since for all x, x ~= x, but x = x.  In other words, just because two numbers are "about equal" it doesn't mean that they are not equal.

You didn't like my suggestions for a proof?

How about this:

.3....  != 1/3 (the more places you go, the closer the approximation, but it is never exactly equal  plotted on a graph the curve of .3..3 will approach but never touch 1/3)
Also  3 * .3...3  = .9...9 not 1

Given that x = .9...9:

for any finite number of repeats represented by the elipse:
 9.9...9  (10x)
- .9...9  (x)
= 8.9...1
therefore
9x = 8.9...1


That was my next argument.  :\  Thief.

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Erasmus

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0.9... = 1
« Reply #134 on: November 09, 2006, 03:42:05 PM »
Quote from: "Curious"
.3....  != 1/3 (the more places you go, the closer the approximation, but it is never exactly equal  plotted on a graph the curve of .3..3 will approach but never touch 1/3)


It is true that any finite truncation of the number 0.333... -- i.e. any result you get from chopping off the end -- is not exactly equal to 1/3.  However, the written string of symbols "0.333..." means a certain thing, just as the written string of symbols "2006" means a certain thing.  It means a sum of products between digits and powers of ten.  The sum is infinitely long -- there are infinitely many terms in it -- but because there's a repeating pattern, we don't have to write down the whole thing; instead we just write down the part that repeats.

Thus to point out that any approximation of 1/3 that you get by writing down "0." followed by some finite number of 3s is not actually equal to 1/3, while correct, is a straw man -- when we say "0.333... = 1" we are not saying that some finite truncation of "0.333..." is equal to "1"; we are saying that in fact the infinite sum represented symbolically by "0.333..." is in fact equal to the number represented by "1".

Quote
Also  3 * .3...3  = .9...9 not 1


That notation doesn't mean anything.  If you'd like to define it, please feel free.

Quote
for any finite number of repeats represented by the elipse:


Again, the ellipse does not represent a finite number of repetitions; this is a misuse of the notation.

Do you agree that sum( 9/10^n, n>0 ) = 1?
Why did the chicken cross the Möbius strip?

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BOGWarrior89

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0.9... = 1
« Reply #135 on: November 09, 2006, 03:44:19 PM »
Quote from: "Erasmus"
Quote from: "Curious"
.3....  != 1/3 (the more places you go, the closer the approximation, but it is never exactly equal  plotted on a graph the curve of .3..3 will approach but never touch 1/3)


It is true that any finite truncation of the number 0.333... -- i.e. any result you get from chopping off the end -- is not exactly equal to 1/3.  However, the written string of symbols "0.333..." means a certain thing, just as the written string of symbols "2006" means a certain thing.  It means a sum of products between digits and powers of ten.  The sum is infinitely long -- there are infinitely many terms in it -- but because there's a repeating pattern, we don't have to write down the whole thing; instead we just write down the part that repeats.

Thus to point out that any approximation of 1/3 that you get by writing down "0." followed by some finite number of 3s is not actually equal to 1/3, while correct, is a straw man -- when we say "0.333... = 1" we are not saying that some finite truncation of "0.333..." is equal to "1"; we are saying that in fact the infinite sum represented symbolically by "0.333..." is in fact equal to the number represented by "1".

Quote
Also  3 * .3...3  = .9...9 not 1


That notation doesn't mean anything.  If you'd like to define it, please feel free.

Quote
for any finite number of repeats represented by the elipse:


Again, the ellipse does not represent a finite number of repetitions; this is a misuse of the notation.

Do you agree that sum( 9/10^n, n>0 ) = 1?


No, you Newton/Leibniz-lover.

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skeptical scientist

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« Reply #136 on: November 10, 2006, 07:58:59 AM »
Several people here keep writing decimal sequences as if they know what they mean, and clearly there is some uncertainty, as there is confusion as to whether .99999... and 1 represent the same number. So let's go to definitions:

The set of real numbers is (by definition) and ordered field containing the rational numbers and satisfying the least upper bound property: any bounded nonempty set has a least upper bound. (Rudin: Principles of Mathematical Analysis). Real numbers are not decimal sequences; they are numbers representing actual quantities. We often represent them with decimal sequences, but it should be understood that this is a representation of an object, rather than the object itself. There is also no reason to think that an object should have a unique representation: for example, the quantity one and a half can be written as 1 1/2 or 3/2 or 6/4, but all of these represent the same number.

So, what is the number represented by a decimal number 0.d1d2d3d4d5..., where d1, d2, etc. are digits between 0 and 9? It is the quantity d1/10+d2/100+d3/100+...+dn/10^n+.... So, what we should really be asking is whether the sum of the series 9/10+9/100+9/1000+9/10000+... is 1, (i.e. for any epsilon, if we sum enough terms we are within (1-epsilon, 1+epsolon)) or not. It turns out that the answer is yes: you can prove this by any of several means discussed above.

You could perhaps propose an alternative definition, in which case .99999... might not represent the same number as 1, but this is the definition we use, and if you were to use a different one, you would be talking about something other than the real numbers.
-David
E pur si muove!

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Nomad

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0.9... = 1
« Reply #137 on: November 10, 2006, 08:10:27 AM »
10 pages.  Wow.

Can I ask something?  What the hell kind of practical use is comparing 1 to .99...?  Why the hell has this gone for ten pages?  Who really gives a shit?

>.>
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BOGWarrior89

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0.9... = 1
« Reply #138 on: November 10, 2006, 09:23:36 AM »
Sorry thedigitalnomad, but I'm going to further the argument.

0.9999... does not equal 1, because the proof you gave is mathematically incorrect.  Here, allow me to show you why:

x = 0.999...999
10x = 9.999...990  (subtract "x")
9x = 8.999...991  (divide by "9")
x = 0.999...999

0.9... = 1
« Reply #139 on: November 10, 2006, 11:19:04 AM »
Quote from: "Erasmus"
Quote from: "Curious"
.3....  != 1/3 (the more places you go, the closer the approximation, but it is never exactly equal  plotted on a graph the curve of .3..3 will approach but never touch 1/3)


It is true that any finite truncation of the number 0.333... -- i.e. any result you get from chopping off the end -- is not exactly equal to 1/3.  However, the written string of symbols "0.333..." means a certain thing, just as the written string of symbols "2006" means a certain thing.  It means a sum of products between digits and powers of ten.  The sum is infinitely long -- there are infinitely many terms in it -- but because there's a repeating pattern, we don't have to write down the whole thing; instead we just write down the part that repeats.

Thus to point out that any approximation of 1/3 that you get by writing down "0." followed by some finite number of 3s is not actually equal to 1/3, while correct, is a straw man -- when we say "0.333... = 1" we are not saying that some finite truncation of "0.333..." is equal to "1"; we are saying that in fact the infinite sum represented symbolically by "0.333..." is in fact equal to the number represented by "1".

Quote
Also  3 * .3...3  = .9...9 not 1


That notation doesn't mean anything.  If you'd like to define it, please feel free.

Quote
for any finite number of repeats represented by the ellipse:


Again, the ellipse does not represent a finite number of repetitions; this is a misuse of the notation.

Do you agree that sum( 9/10^n, n>0 ) = 1?


Errors aside,
Quote
when we say "0.333... = 1"
I understand, and I have taken some time in reading up on the concept.  I will have to bow to the fact that according to the principles that mathematicians have set for the "Set of Real Numbers"(in this case that every whole number has an equivalent number that is one less followed by repeating decimal 9's), that the idea of 0.999... = 1 is in accordance with those rules.  

You win.

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Erasmus

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« Reply #140 on: November 11, 2006, 11:41:51 AM »
Quote from: "BOGWarrior89"
Sorry thedigitalnomad, but I'm going to further the argument.


I'm not sorry.

Quote
x = 0.999...999
10x = 9.999...990  (subtract "x")


This is not an infinite sequence of digits, so you're not talking about the same thing.  The notation you're using is in fact not ever used.  "..." doesn't mean what you think it means, as skeptical_scientist has explained in more detail.

Quote
9x = 8.999...991  (divide by "9")
x = 0.999...999


All you've "proven" is that x = x.  You have not proven that it is not equal to 1.

To thedigitalnomad: Well, it engages the minds of at least some of us so I guess it's reasonable to allow the conversation to continue.  Also, the notion of infinite decimal representations is useful in other proofs as well, so it's important that people not think what BOG thinks about them, if they care about those other proofs.

To Curious: errors?
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Nomad

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« Reply #141 on: November 11, 2006, 01:18:13 PM »
I didn't say anything about continuing.  I'm just genuinely curious as to what kind of practical application this subject has.
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skeptical scientist

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« Reply #142 on: November 11, 2006, 01:27:47 PM »
Quote from: "thedigitalnomad"
I didn't say anything about continuing.  I'm just genuinely curious as to what kind of practical application this subject has.

Math itself has thousands of practical applications, but first and foremost, math is about numbers, so you can't use it for anything if you don't know what you mean when you speak of numbers. So yes, knowing what numbers are is important, and does have practical applications. For instance, in physics, we describe space and time with a continuum of numbers, which we think is an accurate representation, because space and time both seem smooth. If there was a largest "number" smaller than 1, such as .999999..., we could no longer use numbers to represent a continuum, because numbers themselves would not form a continuum.
-David
E pur si muove!

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Erasmus

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« Reply #143 on: November 11, 2006, 01:44:54 PM »
Quote from: "thedigitalnomad"
I'm just genuinely curious as to what kind of practical application this subject has.


The first one that comes to mind is the proof that there are more real numbers than natural numbers.  The proof goes by showing that there are more real numbers in the interval (0,1) than there are natural numbers (1, 2, 3, ...).

You do this by assuming that there are the same amount, and showing a contradiction ensues.  First, note that any number in the interval (0,1) can be written as an infinite binary string of the form "0 . b1 b2 b3 ..." (it can be done as a decimal string as well, but the proof is easier with binary strings).  Well, if every such number can be written in this fashion, and there are the same number of such numbers as there are natural numbers, then we can write out every such number with a natural number next to it (we can just label the real numbers "x1, x2, x3, ...") so that no natural number is ever repeated and every natural number is used, like so:

x1 = 0 . b11 b12 b13 ...
x2 = 0 . b21 b22 b23 ...
x3 = 0 . b31 b32 b33 ...
...

It is importat to understand that according to our assumptions, every real number in (0,1) occurs exactly once in this list.

Now I will pick a real number y as follows: y = 0 . y1 y2 y3 ... where yk = 0 if bkk = 1 and 1 otherwise.  In other words, you get y by reading off the diagonal digits (starting from the upper-left corner and going down and to the right) of the above list, and then inverting all the digits.

In other words, y = invert_binary_digits(0 . b11 b22 b33 b44 ...).

Clearly 0 < y < 1, so y is in (0,1) by definition.  By our assumptions, every real number in (0,1) is written in that list, so y must be written in that list.  In other words, y = xk for some natural number k.

So which is it?  Is y = x1?  No, because the first digit of x1 is different from the first digit of y, by construction.  Is y = x2?  No, because the second digit of x2 is different fromt he second digit of y, by construction.  Similarly, y differs from xk in the kth digit, by construction, so there does not exist any k such that y = xk: a contradiction.

Therefore our original assumption that we could list all the real numbers in (0,1), one after another, is wrong, which means our assumption that there are just as many real numbers in (0,1) as there are natural numbers is wrong.  Since I have already produced a list of real numbers, one for every natural number, there cannot be less real numbers in (0,1) than natural numbers, so there must be more, q.e.d.

(this proof is dedicated to BOGWarrior89)
Why did the chicken cross the Möbius strip?

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skeptical scientist

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0.9... = 1
« Reply #144 on: November 11, 2006, 02:14:09 PM »
Quote from: "Erasmus"
Quote from: "thedigitalnomad"
I'm just genuinely curious as to what kind of practical application this subject has.


The first one that comes to mind is the proof that there are more real numbers than natural numbers.


Thanks for the proof Erasmus, but nobody but a mathematician would call this an "application". :P

On a more serious note, there's a problem with this proof. It may be that the first number on the list is .011111111..., the next is .01010101..., the next is .101110110111..., the next is .0011100101..., and the diagonal happens to be .011111111.... Then y, as you have defined it, is .10000.... The problem is that this number is on the list: it's .01111111..., which is the first number on the list. Of course, this is easily fixed: just use decimal digits instead of binary, and don't change anything to a 0 or 9, which guarantees that the y created has a unique decimal representation.
-David
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Erasmus

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0.9... = 1
« Reply #145 on: November 11, 2006, 02:17:51 PM »
Quote from: "skeptical_scientist"
Of course, this is easily fixed: just use decimal digits instead of binary, and don't change anything to a 0 or 9, which guarantees that the y created has a unique decimal representation.


It's every more easily fixed: stick with binary but enforce that no representation ends in an infinite string of 0s.  This is always possible as per the discussion that this thread has revolved around, and was my reason for bringing up this proof in the first place, and I'm just a fool for having forgotten to include it.  Thanks :)

Quote
Thanks for the proof Erasmus, but nobody but a mathematician would call this an "application".


That's just because those people have probably not considered the fact that calculus is based on this stuff and that bridges are based on calculus.
Why did the chicken cross the Möbius strip?

0.9... = 1
« Reply #146 on: November 11, 2006, 05:49:22 PM »
Quote from: "Erasmus"
To Curious: errors?

Quote
when we say "0.333... = 1" we are not saying that some finite truncation of "0.333..." is equal to "1"; we are saying that in fact the infinite sum represented symbolically by "0.333..." is in fact equal to the number represented by "1".


I'll buy the 0.999... = 1, but not the 0.333... = 1

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Erasmus

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« Reply #147 on: November 11, 2006, 05:50:27 PM »
Quote from: "Curious"
I'll buy the 0.999... = 1, but not the 0.333... = 1


Ah, hm, sorry 'bout that.... should have been, in all cases, 0.333... = 1/3.
Why did the chicken cross the Möbius strip?

0.9... = 1
« Reply #148 on: November 11, 2006, 05:56:22 PM »
Quote from: "Erasmus"
Quote from: "thedigitalnomad"
I'm just genuinely curious as to what kind of practical application this subject has.


The first one that comes to mind is the proof that there are more real numbers than natural numbers.  The proof goes by showing that there are more real numbers in the interval (0,1) than there are natural numbers (1, 2, 3, ...).

You do this by assuming that there are the same amount, and showing a contradiction ensues.  First, note that any number in the interval (0,1) can be written as an infinite binary string of the form "0 . b1 b2 b3 ..." (it can be done as a decimal string as well, but the proof is easier with binary strings).  Well, if every such number can be written in this fashion, and there are the same number of such numbers as there are natural numbers, then we can write out every such number with a natural number next to it (we can just label the real numbers "x1, x2, x3, ...") so that no natural number is ever repeated and every natural number is used, like so:

x1 = 0 . b11 b12 b13 ...
x2 = 0 . b21 b22 b23 ...
x3 = 0 . b31 b32 b33 ...
...

It is importat to understand that according to our assumptions, every real number in (0,1) occurs exactly once in this list.

Now I will pick a real number y as follows: y = 0 . y1 y2 y3 ... where yk = 0 if bkk = 1 and 1 otherwise.  In other words, you get y by reading off the diagonal digits (starting from the upper-left corner and going down and to the right) of the above list, and then inverting all the digits.

In other words, y = invert_binary_digits(0 . b11 b22 b33 b44 ...).

Clearly 0 < y < 1, so y is in (0,1) by definition.  By our assumptions, every real number in (0,1) is written in that list, so y must be written in that list.  In other words, y = xk for some natural number k.

So which is it?  Is y = x1?  No, because the first digit of x1 is different from the first digit of y, by construction.  Is y = x2?  No, because the second digit of x2 is different fromt he second digit of y, by construction.  Similarly, y differs from xk in the kth digit, by construction, so there does not exist any k such that y = xk: a contradiction.

Therefore our original assumption that we could list all the real numbers in (0,1), one after another, is wrong, which means our assumption that there are just as many real numbers in (0,1) as there are natural numbers is wrong.  Since I have already produced a list of real numbers, one for every natural number, there cannot be less real numbers in (0,1) than natural numbers, so there must be more, q.e.d.

(this proof is dedicated to BOGWarrior89)


But does that prove there are more? How is infintity less than infinity^2?

0.9... = 1
« Reply #149 on: November 11, 2006, 05:59:26 PM »
dont worry people...i can solve this  :)

1/3 +2/3 = 3/3 = 1/1 = 1

there, all solved  8-)