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#1
04-24-2003, 06:03 AM
 Dolomite21 Guest Join Date: Mar 2003
Does .99(repeating) = 1

Given that .111(repeating) = 1/9

.222(repeating) = 2/9
.333(repeating) = 3/9
.444(repeating) = 4/9

and on and on

would 9/9 equal 1 and .99(repeating)?
#2
04-24-2003, 06:07 AM
 Jabba Guest Join Date: Mar 2002
Yes
#3
04-24-2003, 06:19 AM
 Dolomite21 Guest Join Date: Mar 2003
So if infinitely speaking a number can be "rounded" up...does 2 = 4
#4
04-24-2003, 06:20 AM
 phraser Guest Join Date: Sep 2002
Try graphing it and you'll have one answer.
Using some math I learned in grade 12 would also give you an answer but I've forgotten how to do that proof :embarrassed:
#5
04-24-2003, 06:21 AM
 Mort Furd Guest Join Date: Apr 2001
No.
9/9=1 - this is second grade mathematics, my niece is just going through this.
.999 repeating does not equal 1 nor does it equal 9/9.
#6
04-24-2003, 06:21 AM
 Jabba Guest Join Date: Mar 2002
Quote:
 So if infinitely speaking a number can be "rounded" up...does 2 = 4
Can you provide a chain of reasoning similar to that in your first post to suggest that it might?
#7
04-24-2003, 06:23 AM
 Liberal Guest Join Date: Nov 1999
Let x = .99999...

10x = 9.99999...

10x - x = 9.99999... - .99999...

9x = 9

x = 1

QED
#8
04-24-2003, 06:26 AM
 Mort Furd Guest Join Date: Apr 2001
That wasn't nice, Lib. Shame on you for playing nasty little games like that.
#9
04-24-2003, 06:26 AM
 Phage Guest Join Date: Mar 2003
no, x = .99999.....

Almost the same, but not. It will be forever under 1. Infinitely small as the gap may be, it still exists. For instance, does the matter in a black hole dissappear when it is formed?
#10
04-24-2003, 06:28 AM
 Mort Furd Guest Join Date: Apr 2001
Quote:
 Originally posted by Libertarian Let x = .99999... 10x = 9.99999... 10x - x = 9.99999... - .99999... 9x = 9 x = 1
Bad Lib, not supported anywhere, and not true besides.
#11
04-24-2003, 06:30 AM
 Jabba Guest Join Date: Mar 2002
There was nothing nasty about Libertarian's proof. It is perfectly valid and 0.999... does equal 1. I wish people who know nothing about mathematics would content themselves with reading these threads rather than posting in them.
#12
04-24-2003, 06:31 AM
 Derleth Guest Join Date: Apr 2000
Give me a number between 0.999... and 1.0.

(We've been through this before. Do a search.)
__________________
"Ridicule is the only weapon that can be used against unintelligible propositions. Ideas must be distinct before reason can act upon them."
If you don't stop to analyze the snot spray, you are missing that which is best in life. - Miller
I'm not sure why this is, but I actually find this idea grosser than cannibalism. - Excalibre, after reading one of my surefire million-seller business plans.
#13
04-24-2003, 06:33 AM
 Liberal Guest Join Date: Nov 1999
Quote:
 Originally posted by Phage no, x = .99999..... Almost the same, but not. It will be forever under 1. Infinitely small as the gap may be, it still exists. For instance, does the matter in a black hole dissappear when it is formed?
There's no gap. To invalidate the proof I gave, you have to find a flaw in the chain of deduction. Stories about black holes, however romantic and mysterious they may be, do not contradict the proof.
#14
04-24-2003, 06:46 AM
 GSV Consolation of Dreams Guest Join Date: Dec 2000

d&r, sniggering
#15
04-24-2003, 06:57 AM
 smiling bandit Guest Join Date: Nov 2001
Quote:
 no, x = .99999..... Almost the same, but not. It will be forever under 1. Infinitely small as the gap may be, it still exists. For instance, does the matter in a black hole dissappear when it is formed?
Nope. An infintely small value does not exist, it seems. Perhaps in theory, but not in practice.
#16
04-24-2003, 06:58 AM
 Mort Furd Guest Join Date: Apr 2001
Quote:
 Originally posted by Jabba There was nothing nasty about Libertarian's proof. It is perfectly valid and 0.999... does equal 1. I wish people who know nothing about mathematics would content themselves with reading these threads rather than posting in them.
Following the same "logic" will also prove that .333 repeating =1
Care to argue that one?
#17
04-24-2003, 07:00 AM
 Mort Furd Guest Join Date: Apr 2001
Also, if you allow changing the base of the numbers from ten, then you can prove that everything equals one. So much for a valid proof.
#18
04-24-2003, 07:06 AM
 Derleth Guest Join Date: Apr 2000
Quote:
 Originally posted by Mort Furd Following the same "logic" will also prove that .333 repeating =1 Care to argue that one?
Sure. The number between 0.333... and 1.0 is 0.666.... Therefore, 0.333... does not equal 1.0. As there is no number between 0.999... and 1.0, 1.0 is equal to 0.999....
__________________
"Ridicule is the only weapon that can be used against unintelligible propositions. Ideas must be distinct before reason can act upon them."
If you don't stop to analyze the snot spray, you are missing that which is best in life. - Miller
I'm not sure why this is, but I actually find this idea grosser than cannibalism. - Excalibre, after reading one of my surefire million-seller business plans.
#19
04-24-2003, 07:09 AM
 Dolomite21 Guest Join Date: Mar 2003
so using that logic....would 7 also be the middle number
#20
04-24-2003, 07:14 AM
 Mort Furd Guest Join Date: Apr 2001
Following Libertarians logic:
Let x = .33333...

10x = 3.33333...

10x - x = 3.33333... - .33333...

3x = 3

x = 1

QED?
Something is fishy here.

Look here for a better proof.

This is why I said Lib's proof was nasty. It shows the intended result, but also shows the same result in other conditions where it doesn't apply.
#21
04-24-2003, 07:15 AM
 GSV Consolation of Dreams Guest Join Date: Dec 2000
Mort Furd: Huh?

Let x=.333...

10x=3.333...

10x-x=3.333... - .333...

9x=3

x = 3/9 = 1/3
#22
04-24-2003, 07:18 AM
 Mort Furd Guest Join Date: Apr 2001
Quote:
 Originally posted by atarian Mort Furd: Huh? Let x=.333... 10x=3.333... 10x-x=3.333... - .333... 9x=3 x = 3/9 = 1/3
Erp.
Shuts face and slinks off to hide in a cave.
#23
04-24-2003, 07:19 AM
 Derleth Guest Join Date: Apr 2000
atarian did with mathematical rigor what I alluded to in prose: There is a nonzero difference between 0.333... and 1.0, therefore the two numbers are not equal. No such difference exists between 0.999... and 1.0, as Libertarian proved.
#24
04-24-2003, 07:19 AM
 GSV Consolation of Dreams Guest Join Date: Dec 2000
10x - x = 3x ????
#25
04-24-2003, 07:23 AM
 GSV Consolation of Dreams Guest Join Date: Dec 2000
Thanks Derleth. Don't worry about it Mort, for some real mathematicl ignorance, do a google groups search for "Spaceman Driscoll" whom i referenced previously.

He argues; apparently seriously, that -4 x -4 = -16.
#26
04-24-2003, 07:31 AM
 Mort Furd Guest Join Date: Apr 2001
Not so much ignorance at work here as asleep at the switch. Stupidity is a horrible thing to see onself commit.
#27
04-24-2003, 08:04 AM
 lucwarm BANNED Join Date: Sep 2000 Posts: 2,789
Let x = 1 + 2 + 4 + 8 +16 . . .

Then x-1 = 2 + 4 + 8 + 16 . . .

and (x-1)/2 = 1 + 2 + 4 + 8 . . .

so (x-1)/2 = x

so x - 1 = 2x

so -x = 1

and x = -1

so

1 + 2 + 4 + 8 + 16 . . . = -1

#28
04-24-2003, 08:13 AM
 GSV Consolation of Dreams Guest Join Date: Dec 2000
Neat. Tricky but neat.

I'm thinking the flaw lies in the difference between an infinite sequence tending to a finite vs. an infinity.

I'll leave it someone else to put it in purty math talk.
#29
04-24-2003, 08:15 AM
 Phage Guest Join Date: Mar 2003
The difference lies in theory, and how things get odd when you talk about infinity. The number .9999.... approaches 1, but would never actually reach it. However, because it is infinitely close it is ok to call it 1 because you will probably never need to be infinitely accurate. It does not matter that there is no number between .9999.... and 1, because .9999.... is as close to 1 as a number less than 1 can be.
#30
04-24-2003, 08:35 AM
 CalMeacham Guest Join Date: May 2000
Quote:
 The difference lies in theory, and how things get odd when you talk about infinity. The number .9999.... approaches 1, but would never actually reach it. However, because it is infinitely close it is ok to call it 1 because you will probably never need to be infinitely accurate. It does not matter that there is no number between .9999.... and 1, because .9999.... is as close to 1 as a number less than 1 can be.
I don't think this is true. The proofs given above are correct, and the weird fact that 0.99999.... = 1 is a quirk of the decimal system. Another way to look at it is that 1/3 = 0.333333.... unambiguously -- there's no other way to represent 1/3 in our decimal system. And 3* (1/3) = 1, so 3 * (0.3333....) = 0.999...., so 0.99999.... = 1.0 .
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#31
04-24-2003, 08:52 AM
 nogginhead Guest Join Date: Dec 2002
Quote:
 Originally posted by CalMeacham I don't think this is true. The proofs given above are correct, and the weird fact that 0.99999.... = 1 is a quirk of the decimal system. Another way to look at it is that 1/3 = 0.333333.... unambiguously -- there's no other way to represent 1/3 in our decimal system. And 3* (1/3) = 1, so 3 * (0.3333....) = 0.999...., so 0.99999.... = 1.0 .

I don't think it's a quirk of the decimal system. It has to do with the implied infinite series that we casually denote with the '...' after the last 9. If we stopped at any number fo decimal places, we'd have a number less than 1, and we could find a number (actually an infinite number fo numbers) between that fixed-decimal-place doohickey and 1, just by adding decimal places.

But the infinite series denoted by '...' means that we never stop at a given number of decimal places. I would say that loosely this means that .999... is as close to 1 as it is possible to get, and that in any real application (if such a thing exists) it is close enought to 1 that you can use 1 in its place.
#32
04-24-2003, 08:58 AM
 sailor Guest Join Date: Mar 2000
Re: Does .99(repeating) = 1

Quote:
 Originally posted by Dolomite21 Given that .111(repeating) = 1/9 .222(repeating) = 2/9 .333(repeating) = 3/9 .444(repeating) = 4/9 and on and on would 9/9 equal 1 and .99(repeating)?
That is begging the question. IF (big operative term IF) for you .22222(repeating) =2/9 then yes, in the same measure, .9999(repeating) = 9/9 . But you have not proven that .22222(repeating) = 2/9 so it is not a given. And if it is a given then it is also a give that .9999 = 9/9 and it requires no proof. The whole premise and OP is flawed but if you want to play games you could say .99999~ = .44444~ + .55555~ and use that as proof. As I say, it does not stand scrutiny.
#33
04-24-2003, 08:58 AM
 Spiritus Mundi Guest Join Date: Nov 1999
Nasty lucwarm. Pretending that the sum of one divergent series can be subtracted from the sum of another divergent series and leave a number behind.

I bet you learned that in "Mean Math Tricks 101".
#34
04-24-2003, 09:03 AM
 Spiritus Mundi Guest Join Date: Nov 1999
Oh, and Lib, never let it be said that you do not learn from your mistakes, eh?
#35
04-24-2003, 09:20 AM
 Orbifold Guest Join Date: Oct 2000
Quote:
 Originally posted by lucwarm Let x = 1 + 2 + 4 + 8 +16 . . . [stuff omitted] so 1 + 2 + 4 + 8 + 16 . . . = -1
I probably shouldn't mention the 2-adic system of numbers where is this in fact a perfectly valid proof, and the limit of the infinite series 1+2+4+... is indeed -1, should I?

(NOTE TO ANY LURKING WEB KOOKS: the 2-adic numbers are not the real numbers. They are an entirely separate number field used pretty much exclusively by number theorists. Don't be a web kook! Thank you.)
#36
04-24-2003, 09:44 AM
 CookingWithGas Charter Member Join Date: Mar 1999 Location: Tysons Corner VA Posts: 8,995
Re: Re: Does .99(repeating) = 1

Quote:
 Originally posted by sailor That is begging the question. IF (big operative term IF) for you .22222(repeating) =2/9 then yes, in the same measure, .9999(repeating) = 9/9 . But you have not proven that .22222(repeating) = 2/9 so it is not a given.
It is not difficult to show by induction (IMHO, IANAM) that 2/9 = .22222 even though it was presented as a given in the referenced post. I think anyone that can do long division can be convinced of this.

When I started reading this thread I was highly skeptical . However, now I agree and I do think it's a quirk of the decimal system. Suppose we do this in base 9, then the apparent paradox disappears.

1/9 = 0.1111...10 = 0.19
9/9 = 0.9999...10 = 1.09
1 = 1.010 = 1.09

Further, using base 9 you can create the same paradox for a different number:

1/8 = 0.1111...9
3/8 = 0.3333...9
4/8 = 0.4444...9
8/8 = 0.8888...9 = 1.09

In fact, this works in any base where the divisor is one less than the radix.
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#37
04-24-2003, 09:45 AM
 CookingWithGas Charter Member Join Date: Mar 1999 Location: Tysons Corner VA Posts: 8,995
Re: Re: Re: Does .99(repeating) = 1

Quote:
 Originally posted by CookingWithGas It is not difficult to show by induction (IMHO, IANAM) that 2/9 = .22222
Make that .2222....
#38
04-24-2003, 09:49 AM
 ultrafilter Guest Join Date: May 2001
For those who think that .9... is not equal to 1, please reconcile your thoughts with the axiom of completeness. It would be nice if you could use this version: "A monotonically increasing sequence converges to its least upper bound".
#39
04-24-2003, 09:51 AM
 Wendell Wagner Charter Member Join Date: Jul 1999 Location: Greenbelt, Maryland Posts: 10,607
Here are two of the many threads where we discussed this subject to death:

#40
04-24-2003, 09:57 AM
 Dolomite21 Guest Join Date: Mar 2003
Great...but is it equal
#41
04-24-2003, 10:01 AM
 ultrafilter Guest Join Date: May 2001
Quote:
 Originally posted by Dolomite21 Great...but is it equal
Yes.
#42
04-24-2003, 10:52 AM
 Achernar Guest Join Date: Aug 1999
I just want to say that I do not like the proof that Libertarian gave. It's the classic proof, but it's informal, and you can use a similar-looking "proof" to "prove" things that are wrong. You shouldn't be subtracting two series when you haven't even proved they converge. But I won't argue with the result. That's all.
#43
04-24-2003, 11:15 AM
 Bippy the Beardless Guest Join Date: Mar 2002
Since the OP has been pretty well answered, can I expand the question a little...

0.9... = 1.0...
where ... indicates the infinite repetition of the digit to its left.
Has been shown above.

But in considering the real number line, you can consider that 0.9... aproaches 1.0... as the repitition of the disgits tends to infinity, but can not become 1.0...

Is there any value in equating 0.9... and 1.0... to two adjacent real numbers? That is two real numbers that have no real numbers between them.

I find this a way to consider the real number continuum, and I wonder if this is terribly flawed.

1.0... = 0.9...9 = 0.9...8 = 0.9...7
where 0.9...8 would be the imagined number that begins 0.9... but whose 'final' ot infinitieth digit is an 8.

it can be seen that this set could have a countable infinite number of entities (0.9...7... would be a valid entity) each precisely equal to 1.0... Which at least to me gives me a sense for the difference between a continuum (the reals) and a countable infinity.

Please comment or point me to a web site that exposes the problems with this way of thinking.

Cheers, Bippy
#44
04-24-2003, 11:21 AM
 ultrafilter Guest Join Date: May 2001
There is no final digit in the standard theory of decimal representations.
#45
04-24-2003, 11:23 AM
 panamajack Guest Join Date: Apr 2000
Another one of the things for the web kooks to stay away from : surreal numbers. These are discussed in the first thread Wendall Wagner gave the link for; on the second page MrDeath gives a good explanation of what they are. Though 1.0 and 0.999... would be unique surreal numbers, again these are NOT real numbers (hence the name). As has been shown, these forms do represent the same number in the reals.

A brief quote may help explain a bit without having to pull up that thread :
Originally posted by MrDeath in another thread
Quote:
 But clearly we're missing something if we try this out with .999999... = 1 - 1/omega. The problem is, we can get at 9.99999.... two different ways - we can multiply .999999..... by 10, or we can add 9 to it instead. (This is the trick that makes the proof work.) In the world of surreal numbers, the two methods are not compatible and give you different answers: 9 + (1 - 1/omega) = 10 - 1/omega, but 10(1 - 1/omega) = 10 - 10/omega, which is a distinct surreal number, although the two are only infinitesimally different.
#46
04-24-2003, 12:20 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Springfield, IL Posts: 15,587
Quote:
 Originally posted by Bippy the Beardless But in considering the real number line, you can consider that 0.9... aproaches 1.0... as the repitition of the disgits tends to infinity, but can not become 1.0... Is there any value in equating 0.9... and 1.0... to two adjacent real numbers? That is two real numbers that have no real numbers between them.
Well, strictly speaking, a single number can't "approach" anything; it just sits there having whatever value it has. A series or sequence of numbers (like .9, .99, .999, ...) can approach a limit, and a repeating decimal like .9999... can be defined/understood to mean the limit of this sequence, which is 1.

Under the standard conception of the real numbers, there is no such thing as "two real numbers that have no real numbers between them." If two numbers are not the same, then there are always infinitely many other numbers between them (like (a+b)/2, for instance).

You may be wanting to use "infinitessimals": "infinitely small" numbers that are greater than 0 yet smaller than any other number. In the very early days of Calculus, mathematicians used this idea to explain and develop what they were doing, but it became apparent that this didn't really make a whole lot of sense and didn't hold water logically, so eventually the ideas of Calculus were reformulated in terms of the modern delta-epsilon definition of a limit, which put things on firmer ground.

Fairly recently, as I understand it, something called "nonstandard analysis" has been developed, which is an attempt to actually define and use infinitessimals in a logically consistent way.
#47
04-24-2003, 12:44 PM
 Bippy the Beardless Guest Join Date: Mar 2002
Sorry about being unclear I meant by
you can consider that 0.9... aproaches 1.0... as the repitition of the disgits tends to infinity
The series 0.90..., 0.990...., 0.9990... continuing as the number of 9s tends to infinity.
Is there a form of logic interpretation where we can say that
1.0... = 0.9... AND 1.0... > 0.9... AND NOT 1.0... < 0.9... in a consistant fashion?

Clearly this doesn't map all real numbers (as pi for minstance cannot be expressed in this way) but is there a sence in considering a last digit, or last set of digits after an infinite repitition of digits, even though they must have a value equal to zero or infinity.

( 1 integer as a real in decimal notation could be considered 0... 1. 0..., any number not starting 0... is numerically infinite, so we drop the 0... whn writing a finite number )

Cheers, Bippy

p.s. does anyone have a method to overline a number on this board to allow for the more usual symbol for repeating digits.
#48
04-24-2003, 12:49 PM
 andymurph64 Guest Join Date: Sep 2002
Quote:
 by ultrafilter For those who think that .9... is not equal to 1, please reconcile your thoughts with the axiom of completeness. It would be nice if you could use this version: "A monotonically increasing sequence converges to its least upper bound".
GAH!!! Must flush memories of Real Analysis from brain.

GAH!

I hate you ultrfilter!
#49
04-24-2003, 01:28 PM
 Mangetout Charter Member Join Date: May 2001 Location: Kingdom of Butter Posts: 47,520
10 / 3 = 3.333...

3.333... + 3.333... + 3.333... = 9.999...
#50
04-24-2003, 02:19 PM
 Liberal Guest Join Date: Nov 1999
Quote:
 Originally posted by Spiritus Mundi Oh, and Lib, never let it be said that you do not learn from your mistakes, eh?
Yep. This exact topic is one of the things I often cite when asked whether I've changed my mind about anything due to my activities here at Straight Dope. The proof is both valid and sound. After seeing it, there was nothing else to consider. I had to change my mind — much like some atheists who have seen the modal ontological proof of God's existence.

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