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Dr Unne
05-10-2004, 06:21 AM
Let .999... represent a decimal point followed by infinitely many 9's. Which of the following is true? (And prove it.) Does math make you happy? It should.

.999... < 1
.999... = 1
.999... > 1

crono_logical
05-10-2004, 06:49 AM
Divide by 9 to get 0.111... Notice that this new value is equal to 1/9. Therefore 0.999... = 9 * 1/9 = 1 :D

EDIT: Yes, maths makes me happy, you got a problem with that? :p

The Captain
05-10-2004, 06:54 AM
Math oftentimes makes me happy, just not so late at night!

.999... < 1
.999... = 1
.999... > 1

All three are possible answers, depending on what type of math you use, how you round, what the question pertains to, etc, etc. Circumstances, they make the world go round.

Take care all.

HOOTERS
05-10-2004, 06:54 AM
MathS makes me cry like a baby. A baby that doesn't like maths.

Peegee
05-10-2004, 07:01 AM
I like math theory actually, but I don't like doing pages and pages of factoring or whatever it is you want me to do (like err, calculus)

How does 0.9999... > 1 btw? The other two 'proofs' are simple enough.

Doomgaze
05-10-2004, 07:10 AM
.999... = 1. The other two are both completely incorrect. .999... is no more less than 1 than 3 is less than 4.


oh, yeah, proof:

there exists no number between .999... and 1, therefore .999... = 1. I don't remember the name of the theorem or law or whatever that says that, but it exists.

Rainecloud
05-10-2004, 07:21 AM
Apologies, but I hate and detest maths. I have a head for languages, but not mathematics.

Death Sign
05-10-2004, 07:22 AM
The Lucky Seven Status from FF7 makes me happy!!! =)!

Strider
05-10-2004, 08:16 AM
Oh, GOD, I hate proofs!

I'm so completely rusty, so I wouldn't have to been able to answer that question to save my life. Arche has thoroughly enlightened me.

Peegee
05-10-2004, 08:45 AM
Darn, doom reminded me that 0.99....is an irrational number, and thus I can't use some sort of pseudo logic to state that 1/9 < 1

*shakes fist*

Meat Puppet
05-10-2004, 09:45 AM
I hate maths :(
I'll pick one at random, .999 > 1

BatChao
05-10-2004, 10:07 AM
The limit of 0.9999.... is equal to one, but 0.999... itself does not equal to one and is alwasy less than one. So I'd say 0.999<1

Some math is cool... I like algebra up to simple calculus, but after that it's no fun... and then applying math to stuff like signals, circuits, and the like is the worst.

Mikztsu
05-10-2004, 10:30 AM
No, this is the only correct answer: .999... < 1 What's there to prove while you can see it. :)

Loony BoB
05-10-2004, 12:33 PM
About that 1/9 strategy, how can you prove that 1/9 is 0.111... ? You can't use something that has no proof to prove something else of a similar nature like that.


.999... = 1. The other two are both completely incorrect. .999... is no more less than 1 than 3 is less than 4.


oh, yeah, proof:

there exists no number between .999... and 1, therefore .999... = 1. I don't remember the name of the theorem or law or whatever that says that, but it exists.
I believe, using the ... strategy instead of the other dot thing, the number between .999... and 1 is 0.0...01.

Yay?

crono_logical
05-10-2004, 03:24 PM
I should hope you're capable of working out 1/9 on paper and seeing for yourself, BoB :p You're essentially asking to prove 1/7 is 0.142857142857... with the 142857 pattern repeated infinitely, which is the same principle really. The repeated patterns in 1/9 and 1/7 (and other recurring numbers) occur because of a repeating sequence of remainders to carry over to the next place value when you do the division on each successive place value. Since the remainders repeat, obviously the digits in the answer will have to repeat in exactly the same way once all the place values after a certain point are the same (which is all place values after the decimal point in integers, which will all be 0), and as you've now got a repeating sequence, it's sufficient to show it occurs infinitely :p

I believe, using the ... strategy instead of the other dot thing, the number between .999... and 1 is 0.0...01a) I think you mean 0.9...01
b) 0.9...01 is invalid anyway, since the ... means forever, and you can't stick something on the end after forever, otherwise you imply it has an end, which is contradicting the ... in the first place :p

Levian
05-10-2004, 03:39 PM
Anything below 1 is lower than 1. I don't really see the problem.(.999... < 1)

Dr Unne
05-10-2004, 04:21 PM
If .999... is less than 1, then there's a number between .999... and 1. What number is between them? 0.00...1 isn't anything that makes sense. "A decimal, followed by 0's that never end; and once they end, put a 1 after the last one". No such thing.

MecaKane
05-10-2004, 07:01 PM
Super, just super.
Now I just have to remember this crap for the fall. @_@

Sephex
05-10-2004, 07:05 PM
I am terrible at math so math makes me a sad person.

Steve
05-10-2004, 07:16 PM
Algebra fine let me at it I can work that stuff baby its just all about the X = Y to the power of the Z ok that I can suffer it's kinda like making love to a girl you really hate but she's got the best body on the planet so you'll suffer just so you can say afterwards "See I did that babe there" and watch your mates drop in awe. However other than that get it the hell away from me

Loony BoB
05-10-2004, 07:31 PM
If .999... is less than 1, then there's a number between .999... and 1. What number is between them? 0.00...1 isn't anything that makes sense. "A decimal, followed by 0's that never end; and once they end, put a 1 after the last one". No such thing.
Prove it.

Dr Unne
05-10-2004, 07:56 PM
In .000...1, the 1 comes AFTER the 0's.

Suppose the series of 0's is infinite. Then you can't put a 1 after it. If something is never-ending, you can't have anything after the end of it; it has no end. Infinity is not a number; infinity means never-ending. You can invent a number system where infinity is a number, or where there is a "greatest possible integer", but that number system is not our number system. Do you need a proof that there is no greatest integer in our number system? Suppose n is the greatest possible integer; I give you n+1; therefore n is not the greatest possible integer; this is a contradiction, therefore there is no greatest possible integer.

Suppose you CAN put a 1 after the series of 0's. Then by definition, the series of 0's is finite, because it has an end. If the series of 0's is finite, then it isn't between .999... and 1, no matter how many 0's you have. Suppose you have .0001. Well .0001 + .99999 = 1.00009. No matter how many 0's you have, I can have that many 9's plus one more 9 at the end, and then the sum will be greater than 1.

Yamaneko
05-10-2004, 08:01 PM
I get it... I think.

I detest math. The arts are for me.

Loony BoB
05-10-2004, 08:11 PM
Suppose the series of 0's is infinite. Then you can't put a 1 after it. If something is never-ending, you can't have anything after the end of it; it has no end. Infinity is not a number; infinity means never-ending. You can invent a number system where infinity is a number, or where there is a "greatest possible integer", but that number system is not our number system. Do you need a proof that there is no greatest integer in our number system? Suppose n is the greatest possible integer; I give you n+1; therefore n is not the greatest possible integer; this is a contradiction, therefore there is no greatest possible integer.

Suppose you CAN put a 1 after the series of 0's. Then by definition, the series of 0's is finite, because it has an end. If the series of 0's is finite, then it isn't between .999... and 1, no matter how many 0's you have. Suppose you have .0001. Well .0001 + .99999 = 1.00009. No matter how many 0's you have, I can have that many 9's plus one more 9 at the end, and then the sum will be greater than 1.
If you can't have 0.000....1, then you can't have 0.999...

Your number is just as irrational and infinite as mine is. I can simplify my number by calling it "the second smallest number possible." Or, if I like

1 - 0.999... = x

You're only going by the theory of a computer. It's just a matter of "the next number above zero" - it's a number, sure. It's just impossible to say what it is. Use your imagination... it's a number beyond comprehension. 0.999... is just the same. How many 9's are there? An infinite amount. How many zeros are there? An infinite amount. You just need to think outside the square, so to speak.

I can't see any proof against what I said earlier in what you just said. You're just using definitions. I don't care about definitions. I'm saying that there would be an irrational number between 1 and 0.999.... - it exists, it's just like Pi - you can't write it. It's impossible. So they give it a symbol instead! If nobody ever gave Pi the name of "Pi" and the little symbol, pi would not exist. But we all know it exists, it's just irrational. So give 1 - 0.999... (or, as I prefer to say, 0.000...01) a name. A symbol. And it will then exist. People shouldn't be so narrow minded so as to say that there is nothing inbetween one number and the next - all rational numbers have an infinite amount of numbers between them. All irratoinal numbers have other irrational numbers inbetween them.

Of course, neither of us can really prove the other wrong. It's just like a hyperbole, really. When you continuously half a number, and half it again, and so forth... do you ever end up with zero? Of course not. You just have a new half. That, too, will go on for infinity. There are so many irrational and infinite numbers that nobody can really comprehend it without naming irrational numbers and limiting infinite numbers.

EDIT: Just to play around even more with numbers.

0.9 + 0.9 = 1.8
0.99 + 0.99 = 1.98
0.999 + 0.999 = 1.998
0.999... + 0.999... = 1.999...98

:) I love it when logic is overtaken by imagination. The possibilities become endless.

Laguna
05-10-2004, 08:14 PM
maths makes my head hurt- but each to their own!

ZeZipster
05-10-2004, 08:17 PM
.999 = A screwed up decimal system?

Peegee
05-10-2004, 08:25 PM
hi BoB

the 'second smallest number' is simply the lim (1/x) x->∞

The idea is odd. There is no number between lim (1/x) x->∞ , or that the number is infinitely small.

Loony BoB
05-10-2004, 08:31 PM
or that the number is infinitely small.
Bingo.

Flying Mullet
05-10-2004, 08:31 PM
there exists no number between .999... and 1, therefore .999... = 1. I don't remember the name of the theorem or law or whatever that says that, but it exists.
I don't see how no number between .999... and 1 proves that they are the same number. .999... will always be less than 1. You don't need multiple numbers in between two numbers to prove that they are greater than or less than each other. .999... will always be a fraction less than 1, no matter how small that fraction may be, thus .999... < 1.

Peegee
05-10-2004, 08:32 PM
infinitely small does not mean 0.00000...1 or whatever it is you are thinking. It's smaller than that, but greater than 0.

In fact the 'number' is still lim (1/x) x->∞

I don't see how no number between .999... and 1 proves that they are the same number. .999... will always be less than 1. You don't need multiple numbers in between two numbers to prove that they are greater than or less than each other. .999... will always be a fraction less than 1, no matter how small that fraction may be, thus .999... < 1.

No. You're thinking of 9/n + 9/n≤ + 9/n≥ + 9/n<sup>x</sup> + 9/n<sup>x+1</sup> + ... for { x | x є Z } ; that's a finite series, not the irrational number 0.999... ; finite series != 1. Infinite series = 1.

Loony BoB's stuff

the 1 = 2 proof required some funny factoring that is wrong, if I'm not mistaken.

Your point about 0.999...being irrational (infinite) and 1 being rational (finite) is wrong because 0.999... is not irrational (it's 1!)

xD

Flying Mullet
05-10-2004, 08:37 PM
But as long as we can define a number that is tangeble, it is subtractable from 1. It may take millions of miles of paper to write 3,000,000,000,000,000 9's, but there is still a value that can be obtained by subtacting that from 1.

Loony BoB
05-10-2004, 08:37 PM
infinitely small does not mean 0.00000...1 or whatever it is you are thinking. It's smaller than that, but greater than 0.

In fact the 'number' is still lim (1/x) x->∞
Then that's the number. :) I just didn't know of any way to write it. Maybe they actually do have some kind of symbol out there, and we just don't know about it.

And the easiest way to prove that 0.999... != 1 is to look at it this way:

Equal = The same as.

0.999... = infinite.
1 = finite.
finite != infinite

So maybe it can be bigger, maybe it can be smaller, but it's impossible for it to be equal to. And as any number less than one begins with "0.", you can't say that it is greater than 1. So therefore it is less than 1. Although anyone with a sensible brain knows that. Trying to get around it is useless... although pretty fun. It's similar to this theory I once read that 1 = 2. I can't remember for the life of me how it goes, but it was really confusing and stuff, much like this theory is.

Flying Mullet
05-10-2004, 08:42 PM
And if .999... and 1 are equal to each other, give me a number, other than 1, that when subtracted from 1, will result in an answer of 0.

Peegee
05-10-2004, 08:43 PM
But as long as we can define a number that is tangeble, it is subtractable from 1. It may take millions of miles of paper to write 3,000,000,000,000,000 9's, but there is still a value that can be obtained by subtacting that from 1.

It's silly to find the digits of a repeating irrational number. Now, finding the digits of pi, that's different (don't ask me why)

And your 'smallest number less than one' is still easily written as 1 - lim (1/x) x->∞. Why write 0.11111111111111111111111111111111111111111111111111111111111111111111111111111... when you can write 1/9?

Flying Mullet
05-10-2004, 08:49 PM
'smallest number less than 1'?
When did I say that? :confused:

<b>EDIT:</b> Also, I can see where 1 - .999... = .000...1, and as you never get to the 1 in .000...1, 1 and .999... are the same.

Good stuff. Yeah, math is fun. :)

Peegee
05-10-2004, 08:58 PM
But as long as we can define a number that is tangeble, it is subtractable from 1. It may take millions of miles of paper to write 3,000,000,000,000,000 9's, but there is still a value that can be obtained by subtacting that from 1.

The smallest 'tangible' number subtractable from one, when subtracted from one, is the largest number less than one. (forgiveness pls. I err....thought to say 'number with the smallest difference in value to one, which is less than one)

and it would be 0.999999999999999999999999999999999999999999... according to some sort of primitive way of writing math. Instead I prefer 1 - lim (1/x) x->∞.

Doomgaze
05-10-2004, 08:59 PM
No, no. Everyone who is not Unne or myself are wrong.

First off, repeating decimals ARE rational, because they can be converted into a fraction.

.222... is 2/9, and .999... is 9/9, or 1.

On that note, pi is irrational, because it cannot be made into a fraction.

Now, .999 with ANY number of 9s after it is less than 1. Yes, it approaches 1, but doesn't reach it.

"But as long as we can define a number that is tangeble, it is subtractable from 1. It may take millions of miles of paper to write 3,000,000,000,000,000 9's, but there is still a value that can be obtained by subtacting that from 1."

And that number is less than .999...


But infinity is not a number. The limit approaches inifinity, see? And when you REACH an infinite number of 9s after the decimal point, you get 1.

Flying Mullet
05-10-2004, 08:59 PM
Ahh, okay. I was looking at it as a quote.

Anyway, I am curious to see if any knows:
<i>if .999... and 1 are equal to each other, give me a number, other than 1, that when subtracted from 1, will result in an answer of 0.</i>

It just sounds like an interesting proof.

Peegee
05-10-2004, 09:01 PM
I hate math again :(

Damn you Doomy I am still right about '2nd smallest number' :aimmad:

Flying Mullet
05-10-2004, 09:03 PM
.999... is 9/9, or 1.
No, .999... is 999.../1000..., where the 9's and 0's are always the same amount.

Peegee
05-10-2004, 09:08 PM
Did you fail gr 10 math?

0.<span style="text-decoration: overline">027</span> = 27/999

0.<span style="text-decoration: overline">1</span> = 1/9

therefore

0.<span style="text-decoration: overline">9</span> = 9/9 = 1

I think I was mistaken about the 'irrational number' because I thought that any number with no end of digits is irrational. If that's not the case, then what's irrational? A bunch of digits with no end, where the numbers don't repeat? I also thought a number expressible as a fraction is a radical.

:D

crono_logical
05-10-2004, 09:10 PM
If you can't have 0.000....1, then you can't have 0.999...Yes you can :p Your 0.000....1 doesn't work because of your contradiction of sticking something after the end of something with no end - the 0.999... has no contradiction in it's meaning.



I can't see any proof against what I said earlier in what you just said. You're just using definitions. I don't care about definitions.If you don't define anything, you can't really prove anything either since you'll have no rules to follow for a proof and can change meanings of everything to what you feel like, which, quite literally, is nonsense :p



I'm saying that there would be an irrational number between 1 and 0.999.... - it exists, it's just like Pi - you can't write it. It's impossible. So they give it a symbol instead! If nobody ever gave Pi the name of "Pi" and the little symbol, pi would not exist. But we all know it exists, it's just irrational. So give 1 - 0.999... (or, as I prefer to say, 0.000...01) a name. A symbol. And it will then exist.No, there would only be a number in between 1 and 0.999... if 1 and 0.999... were different numbers. Since they're the same, as shown earlier, it doesn't exist, and you saying "it exists" on its own doesn't prove anything. Pi has nothing to do with anything here, you've just plucked a particular irrational number out of the air that happens to have a name. Since there's infinitely many irrational numbers, I should hope not every single one has a name :p They still all exist nameless though, giving a number a name doesn't magically bring it into logical existance.



People shouldn't be so narrow minded so as to say that there is nothing inbetween one number and the next - all rational numbers have an infinite amount of numbers between them. All irratoinal numbers have other irrational numbers inbetween them.All pairs of numbers have a number in between, yes. Except when both numbers in the pair are identical.



Of course, neither of us can really prove the other wrong. It's just like a hyperbole, really. When you continuously half a number, and half it again, and so forth... do you ever end up with zero? Of course not. You just have a new half. That, too, will go on for infinity. There are so many irrational and infinite numbers that nobody can really comprehend it without naming irrational numbers and limiting infinite numbers.You're straying from the topic here :p




0.999... + 0.999... = 1.999...98
Actually, you'll find the answer turns out to be 1.999... if you did it long hand. Again, you're contradicting yourself by putting something at the end of something with no end.



I don't see how no number between .999... and 1 proves that they are the same number. .999... will always be less than 1. You don't need multiple numbers in between two numbers to prove that they are greater than or less than each other. .999... will always be a fraction less than 1, no matter how small that fraction may be, thus .999... < 1.I showed you in my first post how they're the same :p They're just different ways of writing 1. Would you say 65/65 < 234/234?


For everyone saying 0.999... is smaller than 1, prove it to me, instead of saying over and over that it just is without any backing :p Just because it looks smaller at a glance isn't sufficient, much like how many people mistakingly think square rooting a number will always lead to a smaller number, which doesn't work for numbers between 0 and 1.

Mikztsu
05-10-2004, 09:11 PM
Dammit, I can't believe how complicated many of you can make simple and OBVIOUS thing like this to look like!

crono_logical
05-10-2004, 09:12 PM
Did you fail gr 10 math?

0.<span style="text-decoration: overline">027</span> = 27/999

0.<span style="text-decoration: overline">1</span> = 1/9

therefore

0.<span style="text-decoration: overline">9</span> = 9/9 = 1

I think I was mistaken about the 'irrational number' because I thought that any number with no end of digits is irrational. If that's not the case, then what's irrational? A bunch of digits with no end, where the numbers don't repeat? I also thought a number expressible as a fraction is a radical.

:D
Irrational numbers are numbers that cannot be expressed as fractions.

Doomgaze
05-10-2004, 09:13 PM
PG - if you mean there is no second smallest number, you're right

"No, .999... is 999.../1000..., where the 9's and 0's are always the same amount."

infinity is not an amount

"what's irrational? A bunch of digits with no end, where the numbers don't repeat? I also thought a number expressible as a fraction is a radical."


Irrational numbers cannot be written as whole numbers or as fractions. My mind is blanking on what exactly a radical is, but it has something to do with roots.

Peegee
05-10-2004, 09:17 PM
We need more of these threads.

Lord Xehanort
05-10-2004, 09:17 PM
*is just a sad Algebra II student*
*tries to solve problem*
*head explodes*

Flying Mullet
05-10-2004, 09:18 PM
0.<span style="text-decoration: overline">027</span> = 27/999

0.<span style="text-decoration: overline">1</span> = 1/9

therefore

0.<span style="text-decoration: overline">9</span> = 9/9 = 1

I agree, but your proof is wrong. You need to use the same number of nines as length of number you are dividing into (27/99, not 27/999).

0.<span style="text-decoration: overline">27</span> = 27/99
0.<span style="text-decoration: overline">1</span> = 1/9
therefore
0.<span style="text-decoration: overline">9</span> = 9/9 = 1

Yeah, nit-pick detail. :rolleyes2

crono_logical
05-10-2004, 09:19 PM
We need more of these threads.I agree :D

bennator
05-10-2004, 09:19 PM
my math teacher had an interesting proof based upon sums of infinite series... I'll see if I can reproduce it.

.999999999..... = 9/10 + 9/100 + 9/1000 +.... = .9 * (1/10)^(n-1)</super>

agree so far?

If you see that so far, you can take the infinite sum as X of that series. The infinite sum of any geometric series [a<sub>n</sub> = a<sub>1</sub>*r^(n-1)] is a<sub>1</sub>/(1-r).

Thus, looking at the series, we get .9 / (1 - 1/10) or .9/.9 or 1.

If you want to know where that infinite sum formula comes from, I can post that, too....

Flying Mullet
05-10-2004, 09:22 PM
For everyone saying 0.999... is smaller than 1, prove it to me, instead of saying over and over that it just is without any backing :p
I already did. Give me a number other than 1 that when subtracted from 1 will result in 0. Unless someone can answer this than .999... is equal to and less than 1. :eek:

EDIT: there, that looks better

bennator
05-10-2004, 09:24 PM
The problem is.... .999999 <i>isn't</i> a number in it's own right.... it's another way of saying 1. We claim that 100/100 is one.... 1 - (100/100) = 1... so why can't .99999.... also be 1?

Beside the fact that the proof is pretty watertight

Lord Xehanort
05-10-2004, 09:28 PM
1 - (100/100) = 1

Doesn't 1-(100/100)=0?

bennator
05-10-2004, 09:29 PM
whoops... that too.... that's what I get for posting too quickly

Doomgaze
05-10-2004, 09:30 PM
"Give me a number other than 1 that when subtracted from 1 will result in 0. Unless someone can answer this than .999... is equal to and less than 1. "

Why? I might as well say "unless 2 is green, 2 does not exist"

Skogs
05-10-2004, 09:31 PM
Is it safe to say that .999... + lim (1/x), x --> infinty = 1 ?

Yeah, I'd find some maths quite refreshing given that my degree now consists of absorbing vast quantities of facts, with the occaisional stoichiometrical calculation thrown in.

bennator
05-10-2004, 09:33 PM
oh, and by the way, now that I've started to do calculus, math is quite possibly the most amazing thing ever.

Peegee
05-10-2004, 09:35 PM
no, but I think the answer to your question is "the smallest number greater than one" :D

Skogs
05-10-2004, 09:40 PM
But I think that if we apply the same logic that gives us that .999... =1, then lim (1/x), x--> infinity = 0. I think, anyway. :mog:

Flying Mullet
05-10-2004, 10:17 PM
"Give me a number other than 1 that when subtracted from 1 will result in 0. Unless someone can answer this than .999... is equal to and less than 1. "

Why? I might as well say "unless 2 is green, 2 does not exist"

No.

My formula is 1 - x = 0, where x = any number other than 1. Your answer so far has been .999..., but 1 - .999... != 0.

Basically I'm asking you to prove that 1 - .999... = 0, because if 1 - .999... is equal to zero, then 1 is equal to .999...

<b>EDIT:</b> Bah, even my fiance's against me, I hate arguing math with someone that has a Master's in math. Anyway, just to add to the ".999... = 1" agrument, what is 1/3 + 1/3 + 1/3? Yeah, I think you can follow this train of thought out. it's an easier proof than most of what I've seen.

Dr Unne
05-10-2004, 10:41 PM
Math is only fun once it becomes non-intuitive. :)

.999... is another way of expressing the infinite series:

.999... = 9/10 + 9/100 + 9/1000 + ...

Now even though the series is infinite, so long as it converges, it has a finite sum. So let's find it. (I hope. I haven't done this in a while.)

2) SUM = 9/10 + 9/100 + 9/1000 + 9/10000 ...
3) SUM / 10 = 9/100 + 9/1000 + 9/10000 + ...

Subtract 3) from 2):

(9 * SUM) / 10 = 9/10
9 * SUM = 9
SUM = 1

Put another way:

SUM = 9/10 + 9/100 + 9/1000 + 9/10000 ...
SUM = 9/10 * (1 + 1/10 + 1/100 + ...)
SUM = 9/10 * (1 + (1/10) + (1/10)^2 + (1/10)^3 + ...)

The equation for solving a geometric series of the form

s = 1 + q + q^2 + q^3 + ...

is

s = 1 / (1 - q)

In this case q is 1/10, so

s = (9/10) * ( 1 / (1 - 1/10) )
s = (9/10) * ( 1 / (9 / 10) )
s = (9/10) * (10/9)
s = 1

1.000... and .999... are just two ways of writing the same number.

Flying Mullet: Since we've already proven that .999... = 1, it follows that 1 - .999... = 0, yes. This is another way of saying "There is no real number between .999... and 1", like I said above. Indeed there is no real number between them. If 1 and .999... weren't equal, there would in fact be an INFINITE number of real numbers between them.

Flying Mullet
05-10-2004, 10:42 PM
Flying Mullet: Since we've already proven that .999... = 1, it follows that 1 - .999... = 0, yes. This is another way of saying "There is no real number between .999... and 1", like I said above. Indeed there is no real number between them. If 1 and .999... weren't equal, there would in fact be an INFINITE number of real numbers between them.

Yeah, I know, I've been pretty much convinced for a while now. I'm just playing Devil's Advocate and trying to prove people wrong, without much luck. Somebody's gotta stick up for the ".999... < 1" team. *shrug*

And for my last log to throw on the flame, .999... will always appear behind 1 on the number line. Granted, the space/amount between .999... and 1 will get infinitely smaller, but that space is always there. :p

<b>EDIT:</b> Here's another thread in another forum that my fiance linked me to pretty much summing up that .999... = 1 in a few easy proofs: http://www.physicsforums.com/showthread.php?t=22866

Loony BoB
05-10-2004, 10:51 PM
I've actually been convinced for a while now, too, but arguing against Unne and Doomy is so much fun, and in this thread I get to argue with both at once. In fact, the thing that bennator/Unne posted (the long equation) has actually been shown to me by one of my teachers back in high school. xD Although I do still think that 0.000...001 should be a legitimate number on some level. Not to say that it'll have anything to do with this stuff, but yeah. Maybe I'm just too creative for my own good.

EDIT: That thread you linked to has some interesting thoughts on the 0.0...01 thing, too. :)

frr_vegeta
05-10-2004, 11:02 PM
.999 = .999 *Nods*

Flying Mullet
05-10-2004, 11:34 PM
Okay, I have another "problem" that I think is fun to think about:
2/3 = 0.<span style="text-decoration: overline">6</span>

2/3 + 2/3 + 2/3 = 6/3 = 2

0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = ???

I just find this interesting because the more sets of 3 2/3's(2/3 + 2/3 + 2/3) that we try and add, whole numbers in essence, the more we start to move away from 0.<span style="text-decoration: overline">9</span> and 1 if we try to calculate it by means that we understand, i.e.:

2/3 + 2/3 + 2/3 = 6/3 = 2
0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = 0.666... with an 8 at the end

2/3 + 2/3 + 2/3 + 2/3 + 2/3 + 2/3 = 12/3 = 4
0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = 0.666... with a 6 at the end

and so forth. Yet by our proofs from earlier we should always equal one. Just something to ponder.

It's almost like the more we try to break down infinity, the more it deteriorates.

Loony BoB
05-10-2004, 11:47 PM
How did you get the 8 at the end? x_x

Logan
05-10-2004, 11:56 PM
I have a headache now from TRYING TO FIGURE IT OUT. *sux*

Del Murder
05-11-2004, 02:46 AM
This is the greatest thread in the history of Eyes on Final Fantasy. I love you guys.

Dragonflame
05-11-2004, 02:50 AM
Working by the rules of the decimal system, I am forced to say that .999...=1. However, this is actually an inherent flaw in the decimal system, because the concept of infinity is impossible.There is no way it can possibly be expressed in a way that humans can comprehend. Thus, the whole debate is meaningless because we are working with the rules humans have created for math, when at this point the whole concept is eluding us. We are trying to compress the workings of the universe into a system of numbers when such a system is inadequate for the task. When you really think about it, we have yet to define 1. What is 1? Can anyone explain that?

I have a better question than the one we are currently discussing. Does any of this matter. Is there anyone who will ever need to know whether or not .999...=1? Why don't we accept that there is no answer to the question and move on?

Shlup
05-11-2004, 02:52 AM
That's it! I can't be on staff with this guy anymore! I resign as administrator!

...

...I hate math. :aimmad:

Del Murder
05-11-2004, 02:56 AM
Unne already solved it, you silly goose. All he really needs to do now is prove the convergence of a geometric series is s = 1 / (1 - q), which isn't hard and can already be found in many textbooks.

Dragonflame
05-11-2004, 03:10 AM
But are those textbooks right or not? They were written by people who could not possibly have a complete understanding of the subject, because it is beyond the scope of human experience and thus is impossible to comprehend fully.

Yamaneko
05-11-2004, 03:17 AM
This is the greatest thread in the history of Eyes on Final Fantasy. I love you guys.
Worst, but yeah.

SeeDRankLou
05-11-2004, 03:27 AM
It should be noted that an irrational number is any real number that cannot be expressed as a ratio between two integers. .999.... is a number divisible by 1, 3 and 9. If divided by 9, we get .111...., which is 1/9. So .999....=9*(1/9) (dividing by 3 will give the same results). But numbers must always be looked at in their simplest terms when using theory, so 9*(1/9)=1. The number 1 is not a ratio between two integers (in simplest terms), and therefore .999.... cannot be expressed as a ratio between two integers and is thusly an irrational number. A number cannot exist simultaneously as a rational and irrational number, thus it is not possible for .999.... to equal 1.

HOOTERS
05-11-2004, 03:42 AM
I don't get how there has to be number in between two numbers for the first number to be less than the second number. Won't that mean that every number is equal to every other number? Sooner or later you have to run out of numbers to stick in between. :smash:

escobert
05-11-2004, 03:43 AM
Uhhhh 1+1=2 that's enough math for me...

HOOTERS
05-11-2004, 03:45 AM
Uhhhh 1+1=2 that's enough math for me...

No it doesn't stupid, it equals a window.

Proof: 1 and 1 are the side parts of the window, + is the cross bit that goes in the middle, and = is the top and bottom parts of the window frame. Therefore 1 + 1 = a window :smash:

escobert
05-11-2004, 03:47 AM
Damn HOOT you're too smart for me. Can you teach em the ways of your mathness?

Peegee
05-11-2004, 03:49 AM
I don't get how there has to be number in between two numbers for the first number to be less than the second number. Won't that mean that every number is equal to every other number? Sooner or later you have to run out of numbers to stick in between. :smash:

I asked my friend to prove 2 > 1. He said 2 - 1 > 0 therefore 2 > 1

:D

Dr Unne
05-11-2004, 03:51 AM
<i>However, this is actually an inherent flaw in the decimal system, because the concept of infinity is impossible.</i> --Dragonflame

It's non-intuitive, but things still behave in a certain way even when you throw undefined values into things, like infinity. Infinity is undefined, in that it has no value; but introducing it into an equation doesn't blow the doors off the hinges and make everything else become completely random. Infinity still has "properties", I guess you can say. There are some branches of mathematics which reject the notion of infinity entirely, but it's not necessary to do so.

Mathematics in general is about concepts, not about real tangible things. You get to make up the rules when you form your own branch of mathematics. It just so happens that some branches fit with reality better. If you demand that a branch of math fit reality PERFECTLY, then you probably want physics, not pure mathematics. That's my take on it anyways.

<i>But numbers must always be looked at in their simplest terms when using theory, so 9*(1/9)=1. The number 1 is not a ratio between two integers (in simplest terms), and therefore .999.... cannot be expressed as a ratio between two integers and is thusly an irrational number.</i> --SeedRankLou

Huh?

To quote <a href="http://en.wikipedia.org/wiki/Rational_number">wikipedia</a>:

Each rational number can be written in many forms, for example 3/6 = 2/4 = 1/2. The simplest form is when a and b have no common factors, and every rational number has a simplest form of this type. The decimal expansion of a rational number is either finite or eventually periodic, and this property characterises rational numbers. A real number that is not rational is called an irrational number.

1/9 is clearly rational, and .999... is clearly the (periodic) decimal expansion of 1/9. 1/1 is also clearly rational.

<i>Won't that mean that every number is equal to every other number?</i> --HOOTERS

Huh?

<i>Sooner or later you have to run out of numbers to stick in between.</i>

No. For example between 1 and 2, you have 1.1, 1.01, 1.001, 1.0001, etc. etc. That pattern alone produces infinitely many numbers, all of which are between 1 and 2.

SeeDRankLou
05-11-2004, 04:06 AM
To quote <a href="http://en.wikipedia.org/wiki/Rational_number">wikipedia</a>:

Each rational number can be written in many forms, for example 3/6 = 2/4 = 1/2. The simplest form is when a and b have no common factors, and every rational number has a simplest form of this type. The decimal expansion of a rational number is either finite or eventually periodic, and this property characterises rational numbers. A real number that is not rational is called an irrational number.

1/9 is clearly rational, and .999... is clearly the (periodic) decimal expansion of 1/9. 1/1 is also clearly rational.



Incorrect. Yes, 1/9 is rational, but the decimal expansion of 1/9 ends with 8/9, as 9/9=1 not .999.... And even if you were right, and .999....=9/9, that would equal 1, which is not (in simplest terms) a ratio between two integers. Yes 1/1 is rational, but 1/1 is not in simplest terms, and when comparing numbers in theory you must use simplest terms. Either way, .999.... is an irrational number.

Dr Unne
05-11-2004, 04:16 AM
I was really wrong with my last post. I don't know what I'm saying any longer. Enough math for me for one day.

Del Murder
05-11-2004, 05:24 AM
:): The integers are rational numbers. Don't get bogged down by 'simplest terms', they aren't really important. If it helps you can express 1/1 and 2/1 as the 'simplest terms' (a/b with a and b having no <i>common factors</i>) for 1 and 2, since 1 is not considered a common factor. Think of writing 1 as shorthand for writing 1/1, the denominator is always there you just don't write it.

Peegee
05-11-2004, 05:41 AM
now that is pure maths. :D

Doomgaze
05-11-2004, 06:29 AM
Yeah, simplest terms is a completely artificial concept to make math easier to understand. 1 = 1/1 = .999..., and all three are rational (because, of course, they are the same number).

SeeDRankLou
05-11-2004, 09:26 AM
1) In mathematical theory (which I believe is what were are dealing with), simplest terms are always important. You don't say 2A=2(pi)r^2. You don't say .77e=.77mc^2. You don't say a^2 + b^2 + 10=c^2 +10. You can, but you don't. Simplest terms are always used in mathematical theory.

2) You are askewing definition. 2 and 2/1 are interchangable. 2 is an integer, is that to say 2/1 is not also an integer, even though 2/1=2? 1/1 is a fraction, but since 1/1=1 it is also an integer. I'm fairly certain that the definition is specifying a ratio of two integers that is not also an integer itself, or the definition could also go: any real number that cannot be expressed as an integer (making fractions irrational number). Example, integer is defined as any real number with no fractional component. Is that to say 2u0/5 (two and zero fifths) is a compound fraction and not an integer? Even though you can see a fractional component, 0/5=0, thus there is no fractional component.

P.S.: I love math too. This thread is awesome. I had to think about my rebuttle for a rather long time because you guys did give some good evidence to support this notion.

Ran Ciel
05-11-2004, 09:38 AM
Heh, you guys are nuts. This whole thread makes my head hurt.

Flying Mullet
05-11-2004, 02:45 PM
Did you fail gr 10 math?

0.<span style="text-decoration: overline">027</span> = 27/999

0.<span style="text-decoration: overline">1</span> = 1/9

therefore

0.<span style="text-decoration: overline">9</span> = 9/9 = 1


Okay, this proof's been buggin me since yesterday and I know why now.

For a proof to be a proof, there can be no counter example. In your proof 9/9 = 1 IS the counter example unless you can already prove to me that 0.<span style="text-decoration: overline">9</span> = 1. If you don't ALREADY accept and have already proved that 0.<span style="text-decoration: overline">9</span> = 1 this theory is no good.

You're in an endless loop, as you have to accept that 0.<span style="text-decoration: overline">9</span> = 1 to prove that 0.<span style="text-decoration: overline">9</span> = 1, which mean that you have to accept that 0.<span style="text-decoration: overline">9</span> = 1 to prove that 0.<span style="text-decoration: overline">9</span> = 1, etc...

BoB: Huh, good point, I don't know where I got the 8 from.

Anyway, now I'm really intrigued because, as proved before:
1/3 + 1/3 = 2/3 + 1/3 = 3/3 = 1

0.<span style="text-decoration: overline">3</span> + 0.<span style="text-decoration: overline">3</span> = 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">3</span> = 0.<span style="text-decoration: overline">9</span> = 1

Okay, so this shows that we can add infinitely repeating numbers together.

But look at this:

2/3 + 2/3 + 2/3 = 6/3 = 2

but

0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = 0.<span style="text-decoration: overline">18</span>

So why the discrepancy?

I'm sure that this post has some mathematical flaws in it but it's fun to throw around some numbers and try to explain what's happening.

<b>EDIT:</b> Okay, this is a much better post.

crono_logical
05-11-2004, 03:29 PM
I think you mean:

0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = 1.<span style="text-decoration: overline">8</span>

and what gives there :p

The discrepancy is the inaccurateness of how you're adding the decimal representation of the fractions. You're only looking at what's visible and adding that, but not taking into account the carry-overs from the next decimal place which is implied by the bar, then the carry overs from the one after that and so on, hence why you appear to have such a big error.

What you're reading and adding is 0.6 + 0.6 + 0.6, which will give you 1.8. But since the true number you're adding is 0.<span style="text-decoration: overline">6</span>, each item you're adding is 0.0<span style="text-decoration: overline">6</span> away from what you should be adding, and these errors themselves add up to give a 0.2 error in your final answer. Then you're sticking the bar back in again, which yet again changes the answer you got, by adding 0.0<span style="text-decoration: overline">8</span>. Which doesn't actually mean anything, it's only adding that particular value because that happens to be what's after the decimal point in that particular calculation. You can't just ignore the bar in your calculations then stick it back in at the end and hope it's fine, you need to take it into account all the way through :p

Flying Mullet
05-11-2004, 03:35 PM
I think you mean:

0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span> = 1.<span style="text-decoration: overline">8</span>

and what gives there :p
Maybe I do, and maybe I don't. :p

And yes, you are right. And this also shows two things:
Why 1.<span style="text-decoration: overline">9</span> != 2
Why "1/3 + 1/3 + 1/3 = 1" and/or "0.<span style="text-decoration: overline">3</span> + 0.<span style="text-decoration: overline">3</span> + 0.<span style="text-decoration: overline">3</span> = 1" are not accepted proofs for 0.<span style="text-decoration: overline">9</span> = 1

crono_logical
05-11-2004, 03:37 PM
How does it show why 1.<span style="text-decoration: overline">9</span> != 2? You can show it algebraically.

1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>

Let 0.<span style="text-decoration: overline">9</span> = X, then

10X = 9.<span style="text-decoration: overline">9</span>
9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span> = 9
X = 1

We know 1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>, so

1.<span style="text-decoration: overline">9</span> = 1 + X = 1 + 1 = 2

Flying Mullet
05-11-2004, 03:49 PM
As stated before, an easy way to visualize 0.<span style="text-decoration: overline">9</span> = 1 is

1/3 + 1/3 + 1/3 = 1
1/3 = 0.<span style="text-decoration: overline">3</span>
0.<span style="text-decoration: overline">3</span> + 0.<span style="text-decoration: overline">3</span> + 0.<span style="text-decoration: overline">3</span> = 0.<span style="text-decoration: overline">9</span>

Thus 1 = 0.<span style="text-decoration: overline">9</span>

So if we do the same for 2:
2/3 + 2/3 + 2/3 = 2
2/3 = 0.<span style="text-decoration: overline">6</span>
0.<span style="text-decoration: overline">6</span> + 0.<span style="text-decoration: overline">6</span>+ 0.<span style="text-decoration: overline">6</span> = 1.<span style="text-decoration: overline">8</span>

1.<span style="text-decoration: overline">8</span> = 2, but you can see that 1.<span style="text-decoration: overline">8</span> isn't anywhere close to two.

Then again this is why the above 1/3's and 0.<span style="text-decoration: overline">3</span>'s proof is a simplistic way of looking at 0.<span style="text-decoration: overline">9</span> = 1, as it's not an accepted proof, just an easy way to visualize it. The same is true for 1.<span style="text-decoration: overline">8</span> != 2. It's proven in a geometric proof, and the above is a simplistic way of looking at it, even if it isn't 100% concrete.

crono_logical
05-11-2004, 03:52 PM
That's what algebraic proofs are for :p

1.<span style="text-decoration: overline">8</span> != 2 however, no matter how you look at it. It's a result of adding the 0.<span style="text-decoration: overline">6</span>'s incorrectly. 1.<span style="text-decoration: overline">8</span> is 1 + 8/9, which you can clearly see is not 2, so you must have assumed or done something wrong in your previous addition.

Flying Mullet
05-11-2004, 03:54 PM
That's what algebraic proofs are for :p

1.<span style="text-decoration: overline">8</span> != 2 however, no matter how you look at it. It's a result of adding the 0.<span style="text-decoration: overline">6</span>'s incorrectly. 1.<span style="text-decoration: overline">8</span> is 1 + 8/9, which you can clearly see is not 2, so you must have assumed or done something wrong in your previous addition.
Which is why using the 0.<span style="text-decoration: overline">3</span>'s and 1/3's approach for 0.<span style="text-decoration: overline">9</span> =1 is not an accepted proof.

I didn't say that it proves that 1.<span style="text-decoration: overline">9</span> != 2, I just said that it's a simplistic way is displaying it, much like the 0.<span style="text-decoration: overline">3</span>'s and 1/3's proof, although neither are accepted proofs.

crono_logical
05-11-2004, 03:56 PM
Oh, guess I misunderstood you then :p

Flying Mullet
05-11-2004, 03:56 PM
How does it show why 1.<span style="text-decoration: overline">9</span> != 2? You can show it algebraically.

1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>

Let 0.<span style="text-decoration: overline">9</span> = X, then

10X = 9.<span style="text-decoration: overline">9</span>
9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span> = 9
X = 1

We know 1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>, so

1.<span style="text-decoration: overline">9</span> = 1 + X = 1 + 1 = 2

Doesn't this algebraic proof rely on multiplying the infinite decimals, which can't really be done?

crono_logical
05-11-2004, 03:59 PM
Why can't you do it though? They might be infinite, but they're also all defined and known and known to be the same at every single place value, so multiplying by whatever base the numbers are in at the very least (so multiples of 10 in this case) would be safe, since all you'll be doing then is shifting the location of the decimal point. I'm not sure if multiplying by other numbers would be safe though, not without more careful thought being put into the process to make sure you don't introduce errors by mistake.

Flying Mullet
05-11-2004, 04:08 PM
Also, you can't assume that 0.<span style="text-decoration: overline">9</span> = 1 in this proof. I don't remember the exact reason (maybe someone here does), but proving that 1.<span style="text-decoration: overline">9</span> = 2 is like proving that 0.<span style="text-decoration: overline">9</span> = 1 so you can't use the 0.<span style="text-decoration: overline">9</span> = 1 assumption.

<b>EDIT:</b> Heh, as I was searching the web for 1.<span style="text-decoration: overline">9</span> =/!= 2 info, I found the proof that you are using above. :)

crono_logical
05-11-2004, 04:14 PM
I didn't assume 0.<span style="text-decoration: overline">9</span> = 1 though, I proved it first before sticking it in the last equation :p

If you want though, we can do it another, but similar, way.

Let X = 1.<span style="text-decoration: overline">9</span>
10X = 19.<span style="text-decoration: overline">9</span>
9X = 19.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span>
9X = 18
X = 18/9 = 2

Flying Mullet
05-11-2004, 04:17 PM
I didn't assume 0.<span style="text-decoration: overline">9</span> = 1 though, I proved it first before sticking it in the last equation :p
Okay, but here's a counter argument/proof:

1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>

Let 0.<span style="text-decoration: overline">9</span> = X, then

10X = 9.<span style="text-decoration: overline">9</span>
9X = 9.<span style="text-decoration: overline">9</span> - 1 = 8.<span style="text-decoration: overline">9</span>

X = 8.<span style="text-decoration: overline">9</span>/9 = ??

Either way once you hit this point dividing into an infinite number is not easy to do.

See, in your proof you are inter-changing 0.<span style="text-decoration: overline">9</span> and 1. All if takes for me to prove this proof wrong is to interchange the 0.<span style="text-decoration: overline">9</span> and 1 in a couple of different places. If your proof were true then it would have the same answer no matter where 0.<span style="text-decoration: overline">9</span>'s or 1's are used.

crono_logical
05-11-2004, 04:22 PM
Then I'd just go on to show you that 8.<span style="text-decoration: overline">9</span> = 9. You haven't actually given a counter example :p

EDIT: Also there's a flaw in your counter-argument. How can you subtract 1 from 10X to get 9X when you haven't proven X = 1 yet? I can interchange them freely in mine though, but only after I've proven 0.<span style="text-decoration: overline">9</span> = X = 1 already.

Flying Mullet
05-11-2004, 04:25 PM
I didn't assume 0.<span style="text-decoration: overline">9</span> = 1 though, I proved it first before sticking it in the last equation :p

If you want though, we can do it another, but similar, way.

Let X = 1.<span style="text-decoration: overline">9</span>
10X = 19.<span style="text-decoration: overline">9</span>
9X = 19.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span>
9X = 18
X = 18/9 = 2


Nope, that won't work either.
10X = 19.<span style="text-decoration: overline">9</span>
<i>9X = 19.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span></i>

You subtracted 1 from the left side and 1.<span style="text-decoration: overline">9</span> from the right side.

crono_logical
05-11-2004, 04:26 PM
Nope, that won't work either.
10X = 19.<span style="text-decoration: overline">9</span>
<i>9X = 19.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span></i>

You subtracted 1 from the left side and 1.<span style="text-decoration: overline">9</span> from the right side.Er, I subtracted X from the LHS, which I said was 1.<span style="text-decoration: overline">9</span> in an earlier line, so there's nothing wrong there :p

Flying Mullet
05-11-2004, 04:27 PM
Also there's a flaw in your counter-argument. How can you subtract 1 from 10X to get 9X when you haven't proven X = 1 yet? I can interchange them freely in mine though, but only after I've proven 0.<span style="text-decoration: overline">9</span> = X = 1 already.
Then you have to subtract 1X from both sides, not 1, when going from 10X to 9X, because you haven't proven that X = 1 until a couple of lines after that statement.

crono_logical
05-11-2004, 04:28 PM
10X = 9.<span style="text-decoration: overline">9</span>
<b>9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span> = 9</b>
X = 1

I did subtract 0.<span style="text-decoration: overline">9</span> from the right though :p

Flying Mullet
05-11-2004, 04:38 PM
How does it show why 1.<span style="text-decoration: overline">9</span> != 2? You can show it algebraically.

1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>

Let 0.<span style="text-decoration: overline">9</span> = X, then

10X = 9.<span style="text-decoration: overline">9</span>
9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span> = 9
X = 1

We know 1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>, so

1.<span style="text-decoration: overline">9</span> = 1 + X = 1 + 1 = 2

Alright, I'm going to start over on this proof and look at it two ways:
<b>The first says that we haven't proved that 0.<span style="text-decoration: overline">9</span> = 1</b>

So 0.<span style="text-decoration: overline">9</span> = X

We don't prove that X = 1 until after these series of statements so we have to use 0.<span style="text-decoration: overline">9</span> for all instances of X
10X = 9.<span style="text-decoration: overline">9</span>
9X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>
9X - 0.<span style="text-decoration: overline">9</span> = 9
9X = 9 + 0.<span style="text-decoration: overline">9</span>

9X = 9.<span style="text-decoration: overline">9</span>
X = 9.<span style="text-decoration: overline">9</span> / 9

Okay, now we're dividing into an infinite number so this is a lost cause.

Now, I will take a second approach:
<b>Let's assume that 0.<span style="text-decoration: overline">9</span> = 1 from a previous proof.</b>
So 0.<span style="text-decoration: overline">9</span> = 1 = X (we can use these interchangably.
10X = 9.<span style="text-decoration: overline">9</span>
9X - 1 = 9.<span style="text-decoration: overline">9</span> - 1
9X = 8.<span style="text-decoration: overline">9</span>

X = 8.<span style="text-decoration: overline">9</span> / 9

And we're back to where we started, dividing into an infinite number.

It doesn't matter which way you look at it, you can counter-proof the proof.

Remember, the burden of the proof lies on the proof itself. It has to be able to stand up to all counter-proofs/attacks. If even one causes it to fail it's not a proof.

crono_logical
05-11-2004, 04:49 PM
Alright, I'm going to start over on this proof and look at it two ways:
<b>The first says that we haven't proved that 0.<span style="text-decoration: overline">9</span> = 1</b>

So 0.<span style="text-decoration: overline">9</span> = X

We don't prove that X = 1 until after these series of statements so we have to use 0.<span style="text-decoration: overline">9</span> for all instances of X
10X = 9.<span style="text-decoration: overline">9</span>
9X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>This step's incorrect, so this counter proof is invalid. YOu've taken 0.<span style="text-decoration: overline">9</span> from the right, but you've taken 0.<span style="text-decoration: overline">9</span> <b>and</b> X from the left.



Now, I will take a second approach:
<b>Let's assume that 0.<span style="text-decoration: overline">9</span> = 1 from a previous proof.</b>
So 0.<span style="text-decoration: overline">9</span> = 1 = X (we can use these interchangably.
10X = 9.<span style="text-decoration: overline">9</span>This step's also invalid, though I think it's a typo judging from your next line, so I'll continue on this one :p



9X - 1 = 9.<span style="text-decoration: overline">9</span> - 1
9X = 8.<span style="text-decoration: overline">9</span>

X = 8.<span style="text-decoration: overline">9</span> / 9

And we're back to where we started, dividing into an infinite number.Yes, we're back to dividing by an infinite number, but you've stopped midway. You can then see 8.<span style="text-decoration: overline">9</span> = 8 + 0.<span style="text-decoration: overline">9</span> = 8 + 1 = 9, so that resolves the dividing by an infinite number, if you don't like dividing such things. Stopping somewhere just because you can't see a way to continue doesn't provide a counter-proof.



Let's try writing this proof out even longer so it's easier to see what's going on :p

We want to proove 0.<span style="text-decoration: overline">9</span> = 1.

Let's start with 0.<span style="text-decoration: overline">9</span>.
Let's define X = 0.<span style="text-decoration: overline">9</span>.

Now 10X = 9.<span style="text-decoration: overline">9</span>, because we've shifted the decimal place as we're multiplying by the base.

Next we subtract X from both sides.

10X - X = 9.<span style="text-decoration: overline">9</span> - X

For the LHS, 10X - X = 9X, as you should know from algebra.
For the RHS, the only value of X we know is 0.<span style="text-decoration: overline">9</span> since that's what we defined earlier, so the RHS becomes 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>, so we now have

9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>

You should be able to see the RHS simplifies to 9, giving us

9X = 9

Divide both sides by 9, and we get

X = 1.

Nowhere earlier have I used X = 1 in obtaining this value.




EDIT: Yes, a proof has to stand up to potential counter-proofs and attaks, but they have to be valid counter-proofs or attacks and those cannot have flaws in themselves :p

Flying Mullet
05-11-2004, 04:55 PM
We're not proving that 0.<span style="text-decoration: overline">9</span> = 1, we're proving whether or not 1.<span style="text-decoration: overline">9</span> = 2.

Flying Mullet
05-11-2004, 05:06 PM
Fine, how's this?
<b>The first says that we haven't proved that 0.<span style="text-decoration: overline">9</span> = 1</b>

So 0.<span style="text-decoration: overline">9</span> = X

10X = 9.<span style="text-decoration: overline">9</span>
10X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>
10X - 0.<span style="text-decoration: overline">9</span> = 9
10X/10 - 0.<span style="text-decoration: overline">9</span>/10 = 9/10
X - 0.0<span style="text-decoration: overline">9</span> = .9
X = .9 + 0.0<span style="text-decoration: overline">9</span>
X = .9<span style="text-decoration: overline">9</span>

That just proved that 0.<span style="text-decoration: overline">9</span> is equal to itself, not that 0.<span style="text-decoration: overline">9</span> is equal to 1.

Dr Unne
05-11-2004, 05:07 PM
Neat, overlines.

<i>Okay, now we're dividing into an infinite number so this is a lost cause.</i> --Flying Mullet

No, you can divide an infinite series by a number, or multiply it by a number, etc. I did so in one of my proofs.

(1/10 + 1/100 + 1/1000 +...) / 10 = (1/100 + 1/1000 + 1/10000+...)


<i>1) In mathematical theory (which I believe is what were are dealing with), simplest terms are always important. You don't say 2A=2(pi)r^2. You don't say .77e=.77mc^2. You don't say a^2 + b^2 + 10=c^2 +10. You can, but you don't. Simplest terms are always used in mathematical theory.</i> --SeedRankLou

I don't see any reason why you have to simplify everything. I've never heard this. Could you explain why this is true?

<i>I'm fairly certain that the definition is specifying a ratio of two integers that is not also an integer itself, or the definition could also go: any real number that cannot be expressed as an integer (making fractions irrational number).</i>

The definition of rational number is a ratio of two integers, where the denominator isn't 0. That's the only restriction. 1/1 is a rational number. 10/2 is a rational number. It doesn't matter if the rational number can be written in another form as an integer. Just a different way of representing the same value.

The list of rational numbers is like this (pretty sure):

<pre>...
... -1/2 -1/1 1/1 1/2 1/3 1/4 1/5 ...
... -2/2 -2/1 2/1 2/2 2/3 2/4 2/5 ...
... -3/2 -3/1 3/1 3/2 3/3 3/4 3/5 ...
...</pre>

If you want a set where no element is repeated, you can go through and eliminate all the repeats. (2/1, 3/1 for example)

1/1 = 2/1 = 3/1 = 1 = 1.<span style="text-decoration:overline">0</span> = 9/9 = .<span style="text-decoration:overline">9</span>, all just different ways of representing the same rational number.

Is .999 repeating a rational number? Well, a number is rational if it can be written as A/B (A over B): .3 = 3/10 and .55555..... = 5/9, so these are both rational numbers. Now look at .99999999..... which is equal to 9/9 = 1. We have just written down 1 and .9999999 in the form A/B where A and B are both 9, so 1 and .9999999 are both rational numbers. In fact all repeating decimals like .575757575757... , all integers like 46, and all finite decimals like .472 are rational. -- http://mathforum.org/dr.math/faq/faq.integers.html

Flying Mullet
05-11-2004, 05:10 PM
<i>No, you can divide an infinite series by a number, or multiply it by a number, etc. I did so in one of my proofs.

(1/10 + 1/100 + 1/1000 +...) / 10 = (1/100 + 1/1000 + 1/10000+...)
</i>
True, but I meant that we aren't going to come up wih X = 1 after the division at that point, not that it's impossible to divide. Plus it was divided by 9, and when there are remainders it makes it a lot more difficult, then when they are divided by powers of 10.

crono_logical
05-11-2004, 05:11 PM
Ok, I'll do this out in full too, and I might do it two ways again :p I'll do it the way you're having a problem with first :p

<b>Method 1)</b>

Start with 1.<span style="text-decoration: overline">9</span> = 1.<span style="text-decoration: overline">9</span>.

You can split the integer and non-integer part one side as follows:

1.<span style="text-decoration: overline">9</span> = 1 + 0.<span style="text-decoration: overline">9</span>

Now we need to do something about 0.<span style="text-decoration: overline">9</span>. Let 0.<span style="text-decoration: overline">9</span> = X. X = 0.<span style="text-decoration: overline">9</span> at this point in time only, and nothing else. We can rewrite that equation as

1.<span style="text-decoration: overline">9</span> = 1 + X

This is Equation A.

Now, moving back to our definition of X, we have

X = 0.<span style="text-decoration: overline">9</span>

Multiply both sides by 10:

10X = 9.<span style="text-decoration: overline">9</span>.

Subtract X from both sides:

10X - X = 9.<span style="text-decoration: overline">9</span> - X

Simplify the LHS:

9X = 9.<span style="text-decoration: overline">9</span> - X

Since X = 0.<span style="text-decoration: overline">9</span> at this point in time only, and nothing else, we substitute this value in the RHS:

9X = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>

Simplify the RHS:

9X = 9
X = 1

Now we're allowed to use X = 1 because we've shown it. We can stick this back in Equation A and get

1.<span style="text-decoration: overline">9</span> = 1 + 1
1.<span style="text-decoration: overline">9</span> = 2 ∎


<b>Method 2)</b>

We've got 1.<span style="text-decoration: overline">9</span>. Let's call this Y. Y = 1.<span style="text-decoration: overline">9</span>. Y is nothing else at this point in time. Y is only 1.<span style="text-decoration: overline">9</span> at the moment.

Y = 1.<span style="text-decoration: overline">9</span>

Multiply both sides by 10:

10Y = 19.<span style="text-decoration: overline">9</span>

since when multiply by the base we're working in (10), you just shift the (decimal) point.

Subtract Y form both sides:

10Y - Y = 19.<span style="text-decoration: overline">9</span> - Y
9Y = 19.<span style="text-decoration: overline">9</span> - Y

Now Y = 1.<span style="text-decoration: overline">9</span>. Y is nothing else at this point in time. Y is only 1.<span style="text-decoration: overline">9</span> at the moment. So we can only substitute this value in the RHS.

9Y = 19.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span>

Simplify the RHS:

9Y = 18

Divide both sides by 9:

Y = 2

Since Y = 1.<span style="text-decoration: overline">9</span> though, then

2 = 1.<span style="text-decoration: overline">9</span> ∎

crono_logical
05-11-2004, 05:12 PM
Fine, how's this?
<b>The first says that we haven't proved that 0.<span style="text-decoration: overline">9</span> = 1</b>

So 0.<span style="text-decoration: overline">9</span> = X

10X = 9.<span style="text-decoration: overline">9</span>
10X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>
10X - 0.<span style="text-decoration: overline">9</span> = 9
10X/10 - 0.<span style="text-decoration: overline">9</span>/10 = 9/10
X - 0.0<span style="text-decoration: overline">9</span> = .9
X = .9 + 0.0<span style="text-decoration: overline">9</span>
X = .9<span style="text-decoration: overline">9</span>

That just proved that 0.<span style="text-decoration: overline">9</span> is equal to itself, not that 0.<span style="text-decoration: overline">9</span> is equal to 1.I should hope you can prove anything equals itself :p Proving something equals itself doesn't rule out the possibility it equals something else at the same time, or has an alternate way of being represented. If you proved 0.5 = 0.5, are you saying 0.5 cannot equal 1/2?

Flying Mullet
05-11-2004, 05:16 PM
Okay, I see how you're solving it, but here's what's bothering me:


Now, moving back to our definition of X, we have

X = 0.<span style="text-decoration: overline">9</span>

Multiply both sides by 10:

10X = 9.<span style="text-decoration: overline">9</span>.

Subtract X from both sides:

10X - X = 9.<span style="text-decoration: overline">9</span> - X
You see, here, because X = 0.9, I can use 0.9 instead of X, so I get
10X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>

And that throws this equation out of whack.



<b>Method 2)</b>

We've got 1.<span style="text-decoration: overline">9</span>. Let's call this Y. Y = 1.<span style="text-decoration: overline">9</span>. Y is nothing else at this point in time. Y is only 1.<span style="text-decoration: overline">9</span> at the moment.

Y = 1.<span style="text-decoration: overline">9</span>

Multiply both sides by 10:

10Y = 19.<span style="text-decoration: overline">9</span>

since when multiply by the base we're working in (10), you just shift the (decimal) point.

Subtract Y form both sides:

10Y - Y = 19.<span style="text-decoration: overline">9</span> - Y
9Y = 19.<span style="text-decoration: overline">9</span> - Y
And again, the same thing here, I can insert 1.<span style="text-decoration: overline">9</span> for Y, so I get
10Y - 1.<span style="text-decoration: overline">9</span> - 10.<span style="text-decoration: overline">9</span> - 1.<span style="text-decoration: overline">9</span>

Flying Mullet
05-11-2004, 05:17 PM
I should hope you can prove anything equals itself :p Proving something equals itself doesn't rule out the possibility it equals something else at the same time, or has an alternate way of being represented. If you proved 0.5 = 0.5, are you saying 0.5 cannot equal 1/2?
That's my point, this proves that it equals itself, not that it equals 1.

<b>EDIT:</b>Nevermind, I remember that you have to use all X's or all it's value, and not a mix of the two.

Man, this has been a good thread. It's been helping to get the cobwebs out.

And here's another good thread debating this: http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_20828798.html
I especially like reading some of the arguments against .999... = 1, there's convincing stuff on both sides.

crono_logical
05-11-2004, 05:29 PM
The other thing you're forgetting when you want to say let's do

10X - 0.<span style="text-decoration: overline">9</span> = 9.<span style="text-decoration: overline">9</span> - 0.<span style="text-decoration: overline">9</span>

instead, is that as valid a step it is, it's not the <i>only</i> valid step you can do there. Which can make some other proofs, especially those involving trig functions and powers of such, rather nasty, since if you do the wrong step (which is still valid) at a certain place, you can make the equation more complicated or end up going round in circles instead (and proving something equals itself :D), instead of actually doing the valid step you need to get to the result you're trying to show :p

Flying Mullet
05-11-2004, 05:36 PM
Yeah, that's it, it's the equation's fault. :) :p

Polaris
05-11-2004, 06:16 PM
Kill me, please!!!

ANything is better than Math!!!!!!

SeeDRankLou
05-11-2004, 07:35 PM
<i>1) In mathematical theory (which I believe is what were are dealing with), simplest terms are always important. You don't say 2A=2(pi)r^2. You don't say .77e=.77mc^2. You don't say a^2 + b^2 + 10=c^2 +10. You can, but you don't. Simplest terms are always used in mathematical theory.</i> --SeedRankLou

I don't see any reason why you have to simplify everything. I've never heard this. Could you explain why this is true?

<i>I'm fairly certain that the definition is specifying a ratio of two integers that is not also an integer itself, or the definition could also go: any real number that cannot be expressed as an integer (making fractions irrational number).</i>

The definition of rational number is a ratio of two integers, where the denominator isn't 0. That's the only restriction. 1/1 is a rational number. 10/2 is a rational number. It doesn't matter if the rational number can be written in another form as an integer. Just a different way of representing the same value.

It's just math etiquette. Similar to writing etiquette. When you are just writing, you can write however you wish. But if you are writing a formal paper, there are rules you follow, like don't end a sentence with a preposition, use commas appropriately, ect. ect. When dealing with formal math (i.e. theory) there are rules you follow, and one of them is that all terms and equations must end as their simplest form. There are good reasons for this. One is irrelevence. In theory, extra terms are irrelevent. You are only dealing with what is at hand. Is .999....=1, not is .999....=1/1. For example, if you have an equation and you are solving for y and you get 10y=10x+20, in practicality that's fine, but in theory that's not fine. There is a factor 10 in that equation that in no way affect the value of y, and therefore the 10 is irrelevant and shouldn't be there. And to say 10y/10=(10x+20)/10 is also has irrelevent terms for the same reason. If people are going to be overly practical, then people could write something like 1(y+1-1)/1=1(x+2+1-1)/1. Yes, this is correct, but all of those 1s are irrelevent, they do nothing to the equation. Another reason is duplicity. Lets say I have the number 12. If I write 12/1, do I not still have 12? If I write 15-3, do I not still have 12? If I write (square root of 4)*(24/4), do I not still have 12? If I write log (base 2) 4096, do I not still have 12? If I write 24*(cos 45)^2, do I not still have 12? Numbers can be written in hundreds of differents fashions, but all of the above can be simplified to 12. Is that to say the number 12 is an integer, an arithmetic expression, a fraction, a compound fraction, a logarithm, and a trigonometric function, or is 12 just an integer? 12 can be expressed as all of the above things, but all of the above things can be expressed as 12. So which is right? Technically they're all right, but in mathematical theory, 12 is right. When numbers are in their simplest terms, they always lack duplicity, so the number is only what it is and nothing more, making a number concretely and inarguable what it is. If a number isn't in simplest terms, it can be something else, but if it is in simplest terms it is only what it is. That is probably the main reason for simplest terms. That and making equations as short as possible (i.e. laziness).

And your definition isn't entirely correct. A rational number is a number capable of being expressed as an integer or a ratio of two integers, excluding zero as the denominator. That distinction is made for a reason. A ratio of two integers is not itself suppose to represent an integer, or you would simply write integer and not the above phrase. Ratios in theory are in simplest terms, and if a 1 is in the denominator, then it is irrelevent.

omnitarian
05-11-2004, 07:57 PM
Blast from the past :p


Your 0.000....1 doesn't work because of your contradiction of sticking something after the end of something with no end - the 0.999... has no contradiction in it's meaning.

Couldn't you consider .999... sticking a 9 after a series of infinite nines?

It comes down to which logic you use for infinity. And infinity isn't very logical to begin with. So I don't think there's a "real" answer here.

crono_logical
05-11-2004, 08:09 PM
Couldn't you consider .999... sticking a 9 after a series of infinite nines?It'd be part of the infinite string of 9's, so you could simplify it back to 0.999... and not have the contradiction :p


SeedRankLou: Simplifying the final solution makes sense (and makes things more readable too), although I wouldn't say working in simplest terms all the time is necessary during the working out stage of your solution. For example, I might prefer not to simplify certain algebraic expressions in an equation because I might foresee the need to unsimplify it again later on to get some terms to cancel out more easily. Or when working out nth roots of complex numbers in (r,theta) notation, you want to take into account values which are not the simplest or principal value of your number if you want to get all solutions.

Peegee
05-11-2004, 08:17 PM
the 0.<span style="text-decoration: overline">9</span> = 0.999....9 is right, because the same digit is put in the end, so it really doesn't matter. To bring it back to SeedRankLou's point, it's just redundant to say it.

A slightly better example would be 0.<span style="text-decoration: overline">127</span> (127/999). The result would be 0.<span style="text-decoration: overline">127</span>...12, or 0.<span style="text-decoration: overline">127</span>..13, if you were to round it, that is.

Of course that's due to rounding. You never put digits after the repeating digits, so it's logically wrong, not a matter of subjective thinking.

edit: Arche just answered it, so um...factor x<sup>2</sup> + 2x + 2 . Or show me the graph, I don't really care.

omnitarian
05-11-2004, 08:20 PM
So, you're saying that if you wrote a 9 at the end of .999..., you'd get .999... . With the same logic, we could say that 999... + 1 = 999... or infinity + 1 = infinity.

Of course, writing a 9 at the end of an infinite string of nines makes no sense, and neither does the equation x + 1 = x.


If you can't have 0.000....1, then you can't have 0.999...



It comes down to which logic you use for infinity. And infinity isn't very logical to begin with. So I don't think there's a "real" answer here.

:)

Flying Mullet
05-11-2004, 08:22 PM
Yes, just think of the .999... as a line, and as long as there are only 9's, the line continues infinitely.

Now look at 0.000...1, and it's as long as there are only 0's. The line is not infintely long because as soon as we reach a one, it's like reaching the end of the line.

Also, saying that 0.000...1 is infinite is like trying to say that

(.999... + .999...) is greater than .999...

but when dealing with infinity is like a boolean, either it is or it isn't infinite. And as both numbers/equations are infinite, one can't be greater than the other.

Peegee
05-11-2004, 08:29 PM
Of course, writing a 9 at the end of an infinite string of nines makes no sense, and neither does the equation x + 1 = x.

x + 1 = x is true for { x | x є ∞ } and only that. x + 1 != x for { x | x є R }

and as for the concept of 999... , would that just be ∞ ? Or would it be the lim (x<sup>2</sup>) x -> ∞ ?

omnitarian
05-11-2004, 08:30 PM
Now look at 0.000...1, and it's as long as there are only 0's. The line is not infintely long because as soon as we reach a one, it's like reaching the end of the line.

But you never reach the end of the line, because there are an infinite number of zeroes. Just like there is no final 9 in .999... there is no 1 in .000...1.

So, depending on what logic you use, you could say that .000...1 = 0, or that .000...1 = .000...1, or that .000...1 is illogical.

crono_logical
05-11-2004, 08:34 PM
So, you're saying that if you wrote a 9 at the end of .999..., you'd get .999... . With the same logic, we could say that 999... + 1 = 999... or infinity + 1 = infinity.

Of course, writing a 9 at the end of an infinite string of nines makes no sense, and neither does the equation x + 1 = x.





:)Quoting BoB's proofless post doesn't proove or disprove anything though. Is there something you're trying to say? :p

Writing anything at the end of an infinite string doesn't make snese anyway so maybe such a question shouldn't have been asked or answered as such :p The equation makes sense though, it just has no solutions for x.


but when dealing with infinity is like a boolean, either it is or it isn't infinite. And as both numbers/equations are infinite, one can't be greater than the other.
Actually, you'll find that you can get infinites of different "sizes", as it were, so it's not quite boolean :p The infinity you get from counting all rational numbers is a smaller infinity than the infinity gotton from counting all irrational numbers, as one example. This'll require me to go back to my notes from 2 years ago on how to prove this one though :p

Flying Mullet
05-11-2004, 08:34 PM
But you never reach the end of the line, because there are an infinite number of zeroes. Just like there is no final 9 in .999... there is no 1 in .000...1.

So, depending on what logic you use, you could say that .000...1 = 0, or that .000...1 = .000...1, or that .000...1 is illogical.

<i>in∑fi∑nite - adj.
1. Having no boundaries or limits.
bound∑a∑ry n.
1. Something that indicates a border or limit.</i>

If there's a 1 at the end it has a boundary and a limit as it can't proceed beyond the 1.


Actually, you'll find that you can get infinites of different "sizes", as it were, so it's not quite boolean :p The infinity you get from counting all rational numbers is a smaller infinity than the infinity gotton from counting all irrational numbers, as one example. This'll require me to go back to my notes from 2 years ago on how to prove this one though :p[/j]

Yeah, I'm curious to see that proof. I thought that all infinites are equal.

Peegee
05-11-2004, 08:37 PM
0.00...1 is not illogical as much as it is incorrect. What you are thinking of is:

10<sup>-10<sup>n</sup></sup> , where 'n' is a large integer, or whatever.

Flying Mullet
05-11-2004, 08:47 PM
Another way I like to view .000...1 vs. .999... is that .999... is infinite while .000...1 is infintely growing.

Imagine that the big bang is pushing out on some "padding", and that we can never go beyond the boundaries of the big bang's ever expanding borders.

.000...1 is like that border, it is always growing outward, but we will reach the end eventually. .999... is never-ending, we can't reach it's end.

Anyways, that's how I visualize .000...1 and .999...

Dr Unne
05-11-2004, 08:50 PM
<i>Yeah, I'm curious to see that proof. I thought that all infinites are equal.</i> --Flying Mullet

You can look up Cantor, and the diagonal method. http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument for example.

Peegee
05-11-2004, 08:52 PM
Then prepare to think differently because they are both always growing.

.999... is the limit of (1/x + 1) (I think) as x -> ∞

0.000..1 or whatever is the limit of (1/x) as x -> ∞

Flying Mullet
05-11-2004, 08:54 PM
Well yeah, both are infinitely growing, but .000...1 has a limit where .999... doesn't.

I'm just trying to put visuals to it.

omnitarian
05-11-2004, 08:59 PM
<i>in∑fi∑nite - adj.
1. Having no boundaries or limits.
bound∑a∑ry n.
1. Something that indicates a border or limit.</i>

If there's a 1 at the end it has a boundary and a limit as it can't proceed beyond the 1.

See? You're using the logic that .000...1 is illogical, Or incorrect, or whatever term you want to use. Personally, I don't see it any different.


The equation makes sense though, it just has no solutions for x.

Yes, that's basically what I meant. Ner cow touched it good enough.


Quoting BoB's proofless post doesn't proove or disprove anything though. Is there something you're trying to say? :p

I guess what I'm trying to say is that when you work with numbers that defy logic, you get answers that defy logic. As hard as we try to shackle it in, infinity is still one really crazy number.

That's all I have to say on the issue, really. So it makes no sense to hang around this hornet's nest any longer. *flees*

Peegee
05-11-2004, 09:00 PM
actually 0.999... = 1 :D

I was thinking that the lim (1/x + 1) x-> ∞ was greater than one, but wrote it as less. MY MISTAKE :D

Flying Mullet
05-11-2004, 09:09 PM
See? You're using the logic that .000...1 is illogical, Or incorrect, or whatever term you want to use. Personally, I don't see it any different.
Actually I posted those definitions because you said that .000..1 is infinite, and I was showing you why it isn't. I never said anything about anything being logical or illogical.


I was thinking that the lim (1/x + 1) x-> ∞ was greater than one, but wrote it as less. MY MISTAKE :D
How could you! :surprised

Peegee
05-11-2004, 09:12 PM
You know what? I have no idea!

eestlinc
05-11-2004, 09:58 PM
i'm not going to read this entire thread, but .999... = 1

proof

8/9 + 1/9 = 1

. .88888888888888888...
+ .11111111111111111...
.99999999999999999...

SeeDRankLou
05-11-2004, 10:44 PM
SeedRankLou: Simplifying the final solution makes sense (and makes things more readable too), although I wouldn't say working in simplest terms all the time is necessary during the working out stage of your solution.


and one of them is that all terms and equations must end as their simplest form.

That applies in no way to intermediate steps. I agree with you completely. In fact, there exist many equations that can only be solved by not simplifying terms in intermediate steps.

To Flying Mullet, Moo Moo the New Cow, and omnitarian: The idea of the numbers .999.... and .000....1 being infinite are very fiesible, if written correctly.

(Please forgive these equations, I don't know how to type math symbols)
.999....=(Reiman Sum) 9/(10^n) where n goes from 1 to infinity.
.000....1=1/(10^infinity)

Endless
05-11-2004, 11:00 PM
0.999... = 9*0.111... = 9*1/9 = 1
or
0.999... = 0.333...+0.333...+0.333... = 1/3+1/3+1/3 = 1
or
1-0.999... = 0.000..., but a number with an infinity of 0 is 0, so 1-0.999...=0 and 1=0.999...
or x=0.999...
10x=9.999...
10x-x=9
9x=9
x=1

There are many ways to prove it, but I can't be arsed to compile them all.

Flying Mullet
05-11-2004, 11:04 PM
.000....1=1/(10^infinity)
Is the result of dividing a number by an infinite number an infinite number?

I don't know the answer. Maybe someone else does.

SeeDRankLou
05-11-2004, 11:06 PM
I would image it would be an infinitely small number, give that the denominator is infinity. The closest number to zero without being zero.

Edit: But on that note, I guess the 1 could be replaced with any number greater than zero that isn't infinity, and 10^infinity could just be replaced with infinity.

Peegee
05-11-2004, 11:24 PM
It really doesn't matter since ∞ isn't a number per se, so arithmetic operations involving it typically result in ∞ anyway :D (meaning 1 + ∞ = ∞, ∞ / ∞ = ∞ .. etc)

wait ∞ / ∞ = ∞ ? x / x = 1 only for { x | x є R } ?

:o

SeeDRankLou
05-11-2004, 11:39 PM
Infinity is the idea of a number that nothing is greater than, which does not exist. Still, the concept of infinity can be computed in arithmetic operations. And furthermore, the idea of infinity still fits into the realm of real numbers. Depending on certain factors, infinity/infinity=1 is correctly. Just like 1/infinity would be the smallest fiesible number possible, since you would be having the largest possible denominator.

Dr Unne
05-12-2004, 02:19 AM
<i>A rational number is a number capable of being expressed as an integer or a ratio of two integers, excluding zero as the denominator. That distinction is made for a reason.</i> --SeedRankLou

No reason that I know of. All numbers capable of being expressed as an integer (x) are also capable of being expressed as a ratio of two integers, namely x/1. The part you added to the definition is redundant.

<i>Just like 1/infinity would be the smallest fiesible number possible, since you would be having the largest possible denominator.</i>

Which number system are you using? Infinity is not a real number, and it can't be used as a number. 1/oo is undefined. Limit (x->oo) 1/x is the closest idea to that, that I know of. You might as well ask what 1/BLARJG means.

Peegee
05-12-2004, 02:24 AM
what does 1/BLARJG mean, Dr Unne? :eep: :love: :mad: :greenie: :rolleyes2 :( :p :D

Del Murder
05-12-2004, 03:09 AM
The great thing about math is that it is not subjective.

∞ is not a number, it is a concept. Therefore, 10^∞ is not a number and cannot be used as such.

Lou: You said 'simplest terms' is part of a sort of 'math etiquette' and I could not agree with you more. The important thing to remember here is that etiquette is used for cleanliness and not for substance.

Yamaneko
05-12-2004, 03:16 AM
I always thought whenever we were told to put things in "simplest terms" it was to make us work more and make the teacher's job easier. Of course, I've never used math in real world problems that deemed the use of converting an answer to simplest terms, so I'm guessing that using simplest terms makes everyone's job easier.

Small numbers are better than big ones, I think.

Peegee
05-12-2004, 03:32 AM
btw what *is* the practical use of calculus?

eestlinc
05-12-2004, 04:46 AM
one practical use of calculus is determining rates of change, like acceleration. also area under a curve, which is useful in statistical analysis.

Spuuky
05-12-2004, 10:15 AM
This thread is way too long to read but they are equal.

Flying Mullet
05-12-2004, 01:50 PM
Good job, wanna cookie?

Dixie
05-12-2004, 02:15 PM
.9=tenths
.99=hundreths
.999=thousandths
.9999=ten thousandths
.99999=hundred thousandths
.999999=millionths (or whatever is, I did decimals in the begining of the school year, so I forgot)

It just keeps going.

I like math. 5/6x7/6= 35/36!

SeeDRankLou
05-12-2004, 06:26 PM
Which number system are you using? Infinity is not a real number, and it can't be used as a number. 1/oo is undefined. Limit (x->oo) 1/x is the closest idea to that, that I know of. You might as well ask what 1/BLARJG means.

Oh wow, what was I on when I wrote my last post or two. I guess yesterday was my day for too much math. :D I apologize for the person who temporarily occupied my body and wrote those last one or two very illogical and confusing posts.

And Dr Unne, I think the real question is: is the part I added to the definition redudant or is the /1 in x/1 redundant?

Del Murder, the idea of theory is to minimize possibility so that you can conclude what something is or how something functions. Redundant possibility is just that, redundant. It has no place in the context of something else. It's like giving someone driving directions. If you just stick in some random spot for the driver to take four rights, that wouldn't change anything. After the fourth turn, they would be right back where they started at the first turn (well, technically they'd be a block back from where they started, but you get my drift), and then finish the drive. Dividing by one is like taking four right turns, it does nothing. You can tell the driver to take the four right turns, but why? You can divide by 1, but why? (and when I say 1 I literally mean the number 1, not something equal to 1 to help solve an equation) To make an integer appear as a fraction? What does that do?

Moo Moo the New Cow, another practical use of calculus is a lot of aspects of physics.

crono_logical
05-12-2004, 07:02 PM
And Dr Unne, I think the real question is: is the part I added to the definition redudant or is the /1 in x/1 redundant?You could argue the part you added is redundant, because using the /1 trick is part of the working to show that integers are rational numbers. The final step of the proof would be drawing the conclusion from such representation, and not that representation itself.

The formal definition of the set <i>Q</i> of rational numbers though is:

<i>Q</i> = { x : x = a/b, a, b ∈ <i>I</i>, b != 0 }

(the block is the member-of symbol for you non-unicode users, good thing I couldn't find a not-equals unicode symbol quickly to avoid two blocks :p )

where <i>I</i> is the set of integers, which implies you can have a 1 on the bottom to show integers are rational numbers. It doesn't even mention you have to simplify the fraction either.

blue_midget192
05-12-2004, 08:30 PM
but i know pie to eight didgets with out looking at a calcualtor! :)

SeeDRankLou
05-12-2004, 10:08 PM
Ugh.......*cries*, *gives up*

I hereby refute everything I have posted on this thread before now. I think I have devised a way (and a rather simple way) for me to fathom this possibility.

*Note: RS stands for Riemann Sum

.999.... can be rewritten as RS(i=1->infinity) 9/(10^i) which can further be rewritten:

lim (n->infinity) RS(i=1->n) 9/(10^i)=lim (n->infinity) RS(i=1->n) (9/10)(1/10^(n-1))

which sets up a property of Riemann sums, so this equation becomes:

lim (n->infinity) (9/10)(1-(1/10^n))/(1-(1/10))

As n->infinity, the denominator of 1/10^n become larger and larger, and the quotient gets closer and closer to zero, and thus lim (n->infinity) 1/10^n=0. So we have:

(9/10)(1-0)/(9/10)=1

And that is the jest of it. I guess I have no more arguement left in me to the contrary. You can't fight calculus proving something.

Peegee
05-12-2004, 10:25 PM
Riemann sum symbol = Σ ?

It's in your character map. Here are some other fun symbols:

√ ∞ ∑ ∩ ≠ ≡ ≤ ≥

Flying Mullet
05-12-2004, 10:30 PM
Kids, today on Sesame Street we're going to have fun with math symbols.

SeeDRankLou
05-12-2004, 11:36 PM
Riemann sum symbol = Σ ?

It's in your character map. Here are some other fun symbols:

√ ∞ ∑ ∩ ≠ ≡ ≤ ≥

I tried that. When I posted my last post, it gave me some weird text instead of sigma, so I edited my post.

Peegee
05-12-2004, 11:55 PM
That's odd. But you can read my symbols properly, right?

*moves to the help forum*

SeeDRankLou
05-13-2004, 06:53 PM
I think possibly I was getting the symbol from a different font that the page was using and it couldn't process it. So if I use the correct font:

∑ → √ ∞ ≡ ≠ ∩ ≤ ≥

Edit: yeah, that worked.