Ending the 0.999~ = 1 debates

[J'Dona]

In case anyone here hasn't noticed, there have been a lot of debates on this online between people on other forums, some with a good grasp of mathematics and some without. The debate is over whether 0.999 does or does not equal 1.

There are many different proofs to prove that it does, but because some people understand some proofs better than others, and some not at all the debates still go on. I've personally been involved in these sort of debates and I can think up at least five different proofs, all based on different sorts of logic, that show that they are equal. The more forms we can find, the better the chances are that someone will understand it.

I thought that perhaps we could do a service to the internet and try and sort this out once and for all, by compiling the most comprehensive list of proofs on the internet. There are already some sites like this, but they only cover one or two forms of the proof and many are based on questionable things (like 0.333 = 1/3 even though it's just an approximation). If this thread eventually appears high enough up on Google, people might find it for themselves and we can stop arguments caused by a poor knowledge of maths across the globe. Anyone think that this is a worthy goal? :cool:

[blike]

I'll post a few as soon as I get dave's mimetex module fixed :-D

[dave]

We did a proof for this in my analysis module. Went something along the lines of this:

Let x = d0.d1d2d3... with di in the set {0,1,...,9}.

From our definition of the decimal, we have that x = sumn=0 -> infinity(dn*10-n). So for 0.999 we can say that for i > 0, di = 9. Hence if we let x = 0.999, then we have:

x = sumn=1 -> infinity(9/(10^n)).

This is a very easy geometric progression, and if you work it out, you get x = 1.

The other proof (which I'm somewhat dubious about) I learnt at GCSE (when I was 16):

x = 0.999, so 10x = 9.999. Now subtract one equation from the other, so you get 9x = 9 => x = 1. As I said, I'm dubious to say the least :-)

[dave]

Also, moving this to the Number Theory forum because I think it's probably a better place for it.

[psi20]

This proof works with infinite geometric sequences.

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

so you have a geometric sequence.
You put it into the formula. a1/(1-r) where a1 is the first term of the sequence and r is the ratio.

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

[dave]

That's what I just proved :-p

[psi20]

Oh, hehe, I haven't ever seen that notation before. :D

[dave]

No problem, it's exactly the same argument, just in a more formal styling.

[Radical Edward]

there are no irrational numbers between 0.99999... and 1.

[MandrakeRoot]

If i remember correctly the decimal representation is defined to have this property.
Thus making 0.4999999999999999etc = 0.5 and 0.9999999999etc = 1.

The geometric mean argument is i think the most easily understood.

A pseudo-intituive argument could be something like :
Let x = 0.99999999999999999etc. It is easily seen that x <= 1 and also that for any n > 1, 9*sum_m=1^n (1/10)^m < x, So for any eps > 0, 1 - eps < x <= 1. It follows that x could only be equal to 1.
Which basicaly comes down to the geometric mean argument.

Hey Radical edward your argument is not finished yet...you have to show why it is impossible for a rational to be in between x and 1, not using the fact they are equal;

Mandrake,

[dave]

I was personally under the impression that most of these proofs go along the same kind of lines; anyone got a different one?

[J'Dona]

I remember that infinite series form of the proof... I accidentally came up with it one lunch and probably annoyed my friend because we were meant to be starting a game of chess. But I wondered whether it counted as I've heard that the formula for the sum of an infinite series (if it's finite) was only an approximation, though I don't see how.

There's another form of the proof that's a bit less mathematical but hopefully understandable by most people. Say that x = 0.999. In that case, 1 - x = 0.0001. There's an infinite number of zeroes before the 1. But infinity can be defined as a number with no endpoint, so you never get to the 1 and therefore 0.0001 = 0.000 = 0, and 0.999 = 1.

Ah, yes they are mostly similar dave. It's just that most people online don't understand the majority of mathematical proofs (at least on the forums where this is argued over) and so a lot of different versions are needed before one might see one that they understand. Mostly it's due to people having only done maths at a level where every number follows the same mathematical rules but who are interested in terms like infinity, hence they assume things like 1/0 = infinity, and don't know what undefined answers are. So they think that you can do things like multiply by infinity and divide by zero (unless you really can and I'm just the wrong one... I've heard that 0/0 = 0 but that's another thing altogether)

[YT2095]

forgive my nievetity here, but after reading all this, it looks an awefull lot like "Rounding Up"?

like to the nearest whole number etc...

[dave]

It's something along those lines, yes, but just defined in a more rigourous way.

[YT2095]

so it`s kinda like you`re demonstrating HOW and WHY we "round up" then?
as opposed to just doing it because the teacher or exam paper tells you to, sorta thing? :)

[dave]

Basically what we're saying is that as you add more and more 9's to the end of 0.9, eventually (i.e. in the limit) you'll get 0.999 = 1. There's been lots of debate about this in the mathematical community (as there usually is with rather small and little things such as notation).

[YT2095]

so effectively 10/3 may be said to be just 3.3 or 3.3333 depending on how accurate you need it to be, but when taken to infinity, 0.999... may as well be just 1 as near as damnit, and for the sake of sanity we just say 1 to avoid it getting ridiculous! :)

[blike]

I remember arguing with fafalone over this years ago.

looks like I win, dear fafalone :)

[felinlasv]

Don't we just say that 0.999..=1 because it doesn't really matters in calculations since you can ignore the small mistake you make? Same thing in chemistry, if something is small enough compared to another value we can neglect it, which doesn't mean it doesn't exist.. I think that 0.999.. is definitely NOT the same thing as 1 (it simply isn't, same thing like 1+1=1), but since the difference is so small you can say 0.999.. is EQUAL to 1 in calculations. It all depends how you look at it or what you have to compare it to..

[YT2095]

well personaly 0.999... is just that, it`s never 1. but for the sake of sanity, it`s often prudent to consider it to be 1 unless a huge degree of accuracy is required, a bit like 3.14 will do just fine when using Pi for many applications. 3.1415926535897932384626 blah blah would be a triffe excessive :)

[blike]

think that 0.999.. is definitely NOT the same thing as 1

Thats the thing though, according to all our calculations, it is.


Here's one:

X = .999~
10X = 9.999~

Subtracting the original equation:

10X = 9.999~
- X = .999~
____________
9X = 9
X = 1

[YT2095]

that`s just because a calc rounds up :)

try this:

10/3 =3.333...

ans X 3 = 10 :)

[blike]

Nooo, ^ thats just because a calc rounds up ;)

3.333~ * 3 = 9.999~

[YT2095]

in all honesty it sounds a bit like that old bell bow and the 27 notes and where`s the missing note type thing, it`s a quirk.
I`m not sure WHAT the quirk is exactly, but I feel sure it`s something quirky/odd like that :)

[J'Dona]

Hmm, maybe we'd better sort it out amongst ourselves first and just give a "presentation" about our conslusions after, or people might get worried about solidarity. ;)

I think the main problem in sorting this out all lies in the fundamental properties of infinity, and what happens when you divide by it. At first glance one would say that 0.999 is certainly less than 1. But if it's less than 1, what is the difference? I'd say it's 0.0001, as that's all that makes sense (although a number after an unending number of zeroes doesn't). Since any number divided by any other number, no matter how large, always leaves a vanishingly small part, it can't be zero. But that only applies to finite numbers. Infinity intrinsically defies normal rules of maths, as in you can't actually multiply by it (or at least in my level of maths that's the dogma we've been fed). But you can divide by it, and when you divide by it you get zero. Since 0.0001 is an infinitely small number, it would be zero, right? Otherwise the infinite series formula would not work, and something like Zeno's paradox would be true, in which case movement would be impossible.

DISCLAIMER: I may be wrong.

[aommaster]

lol YT! That really summed it up! That was just what I was gonna say!

[dave]

Don't we just say that 0.999..=1 because it doesn't really matters in calculations since you can ignore the small mistake you make? Same thing in chemistry, if something is small enough compared to another value we can neglect it, which doesn't mean it doesn't exist.. I think that 0.999.. is definitely NOT the same thing as 1 (it simply isn't, same thing like 1+1=1), but since the difference is so small you can say 0.999.. is EQUAL to 1 in calculations. It all depends how you look at it or what you have to compare it to..

No, 0.999 is definately 1. It's just that if you tried to write it down, it wouldn't be.

[bloodhound]

yeah. everything i would say has been said. any number with a decimal form containing repeating block for example 0.12345234234234234234234 can be expressed as a fraction. conversly any rational number must have a terminating or a repeating decimal form.

[bloodhound]

Proofs nicked from my pure maths lectures

http://members.lycos.co.uk/bloodhound/26.GIF
http://members.lycos.co.uk/bloodhound/27.GIF

[J'Dona]

Thanks :)

Unfortunately it's still going on. Look at this:
http://home.earthlink.net/~ram1024/

And the resulting 9 page cataclysm on another forum (the World of Warcraft off-topic forum, as it were, one of the only other forums I look at and I don't know why...):
http://www.battle.net/forums/wow/thread.aspx?FN=wow-offtopic&T=105122&P=1&ReplyCount=163#post105122

[dave]

It's basically a very silly argument. It's unfortunate such things happen in mathematics (like the definition of 0^0, and things like that) - but for some reason they persist, and I'm not quite sure why. Life would be much easier if we spent less time arguing about trivial things such as this.

[MandrakeRoot]

Hi,

When you formalyl define numbers, with the Peano axioms at the basis. The irrationals are defined as the limits of a sequence of rational numbers.
So the decimal representation of an irrational number itself is also defined and in this definition 0.999999999etc is equal to one. So the equality holds by definition.
Just like 1 + 1 = 2 holds by definition.

Mandrake

[bloodhound]

i hate definitions

[dave]

Yes, definitions definately suck. Especially in analysis.

[Aeschylus]

Here's nice little proof (well, it's not laid out formally and it uses inequalities that really should be proved first (though they're easy inequalties to prove), but I've gauged it exactly for the sort of person who will argue 0.99.... is not equal to 1. Though are alot better and more comphrehensive proofs, people who've I've shown this one seem to liek it) that I came up with:

(Firstly just about every crank whose claimed that 0.999.. is not equal to 1 claims that there are nbo numebrs inbetween 0.999.. and 1, this proof really sets out to deal with this claim)

if x = 0.999... then there are 3 possible relations that x and sqrt(x) could have

1. sqrt(x) > 1

Howver if this is the case then 0< x < 1 must be UNTRUE as for any 0 < x < 1, 0 < sqrt(x) < 1. If this is the case x > 1

2. sqrt(x) = 1

If this is the case then x = 1 is the only solution, meaning that 0.999 does indeed equal 1.

3. sqrt(x) < 1

If this is the case then there is a number between 1 and x (as if 0 < x < 1 then sqrt(x) > x); what is the decimal representation of this number? Of cousre at this point someone could argue that in this case sqrt(x) = x, but this means that x msut equal 1 or zero).

[T_Scagel]

This just looks like Achilles and the turtle again (What can exist realistically vs. what can exist mathematically).
Does .9999999 = 1 ? The answer depends on what application of the value is in question.

[bloodhound]

Does .9999999 = 1 ?

can't believe that this question is still being asked, with all of the posts and the proofs posted surely everyone can see that it is indeed true.

[dave]

This just looks like Achilles and the turtle again (What can exist realistically vs. what can exist mathematically).
Does .9999999 = 1 ? The answer depends on what application of the value is in question.

Well from a mathematical sense it's certainly true.

[T_Scagel]

Well from a mathematical sense it's certainly true.

That's exactly what I said!

[dryan]

I don't know if this helps at all, as Blike (sort of) said it.

1 = 1/3 + 1/3 + 1/3
In decimal form:
1.0000 = .33333 + .33333 + .33333

If you line those numbers up to add vertically,
.33333
.33333
+ .33333
________
.99999
Adding vertically, it never rounds up. So .9 = 1

I realize this is hard to understand. When I was younger, my math teacher didn't understand me and I tried forever to prove it to him.

[dave]

That's exactly what I said!

I was trying to clarify your point, sorry if it came out otherwise :-)

[jordan]

I understand exactly what everyone has said and can agree. I guess the biggest problem is that it is ingrained in our minds for so long that 1>.9 and so anything after the .9 brings you closer to 1 but you'll never reach it, just get closer. Until you get more and more advanced, you are presented no reason to change that view. Then you get some more complex stuff like you guys all explained and you see why it was wrong.

[superstorm]

Think of it in this kindergarten point of view. If .999... isnt equal to 1, then what is the outcome or difference? 1 - .999... = ?

[aommaster]

A small number that ends with something like 0.0001 for the answer to 1-0.9999. The numbers just get smaller as the number of nines increases

[jordan]

So then .00\bar01 is equal to zero also? Assumably by all the same proofs above.

[aommaster]

Nope. No matter how many zeros come before the one, it will never equal to zero, it will just get closer

[jordan]

So you're taking the stance that .999\bar9 doesn't equal 1?

[superstorm]

Im guessing so.

[J'Dona]

If you're assuming that 1 - 0.999 = 0.0001, then of course it will equal zero. Here's the simplest reasoning I can think of why:

0.0001 is a decimal point followed by an infinite number of zeroes, with a 1 at the end.

But infinity has no endpoint, hence you never, ever get to the 1. So it doesn't exist.

[JaKiri]

Nope. No matter how many zeros come before the one, it will never equal to zero, it will just get closer

You're completely misinterpreting how recursion works. It's not '0.33.. that's not it, lets add another 3!' it's infinity from the off. It doesn't change over time, so there can be no 'getting closer'.

[dave]

I always find it helps to look at these things in terms of series; it makes it a lot easier to understand.

[aommaster]

This is really confusing! And yes, I am tking the stance that 0.99999 does NOT equal to 1

[JaKiri]

This is really confusing! And yes, I am tking the stance that 0.99999 does NOT equal to 1

You're wrong, and find a flaw in any of the proofs posted in this thread if you don't want to be labled as a big big fool!

[Sayonara³]

This thread advertised an end to the debates.

I want my money back.

[Cap'n Refsmmat]

0.9 is 0.01 less than one.

edit:
wait until I figure out how to use the denominator tags.
Edit 2:
Did it!
Edit 3: made it 0.01

[Sayonara³]

Isn't it 01 (etc) less?

[Cap'n Refsmmat]

oh... yeah.

[J'Dona]

This thread advertised an end to the debates.

I want my money back."Ending" suggests present tense, i.e. it is currently ending, which holds in any point in time. Therefore it didn't advertise, but rather is currently advertising that end, and it wouldn't be doing that if we'd already reached that point, and no refunds are available unless the product is faulty, and it hasn't been delivered yet so that's impossible to determine. Corporate loopholes ahoy. I keep your money. :cool:

Erm, actually, I'd just hoped that everyone here would agree on the answer and work out the clearest proof of it, to end the debates about it elsewhere online. Maybe we need to work on our own solidarity before trying to help 9-year-olds on gaming forums...

[JaKiri]

0.9 is 0.01 less than one.

edit:
wait until I figure out how to use the denominator tags.
Edit 2:
Did it!
Edit 3: made it 0.01

No, it's 0.01, which is equivilent to 0.

By the way, that's the mathematical definiton of 'equivilent', which is 'Under all circumstances identical'.

1 - 0 = 1, if you didn't know that already.

[bloodhound]

No, it's 0.01, which is equivilent to 0.

.
I am not sure if thats true, but i am too lazy to find out. can u give us a proof.

[dave]

It's pretty obvious really.

0.\overline{0}1 = \lim_{n\to\infty} 10^{-n} \sum_{i=1}^{n} 0\cdot 10^{-i} = 0

[matt grime]

0.99.. is equal to 1, that is provable from the definition of what the real numbers are. i'd be interested to see if any of the people who are so sure they are not equal can actually define the real numbers.

[JaKiri]

0.99.. is equal to 1, that is provable from the definition of what the real numbers are. i'd be interested to see if any of the people who are so sure they are not equal can actually define the real numbers.

0.99..i is equal to i as well!

[joml88]

I just wanted to note that for the ppl who believe that .9999.... is not equal to one then they would be asserting that Zeno's Paradoxes were true. Which in turn would disprove motion. And that would probably screw up a lot of stuff.

[JaKiri]

I just wanted to note that for the ppl who believe that .9999.... is not equal to one then they would be asserting that Zeno's Paradoxes were true. Which in turn would disprove motion. And that would probably screw up a lot of stuff.

Not quite; Zeno's paradox works on the assumption that a decreasing geometric series has an infinite sum; the 0.999 1 argument is merely about WHICH finite sum it is.

[aeroguy]

If you're boarding an airliner going across an ocean and you're given the choice between the one with fuel to go 1 * the way there or 0.999 * the way there, which aircraft would you pick?

I rest my case! Hehehe...

[Sayonara³]

The one with enough fuel to do 5 trips.

[aeroguy]

Ehhh?

[aommaster]

Sayonara's humour! We just have to live with it :)

[Sayonara³]

Oi

Like a plane is going to have the exact amount of fuel for 1 trip, never mind less.

Anyway, since a molecule of fuel is logically indivisible (otherwise it is no longer fuel), the analogy breaks down pretty fast.

[JaKiri]

If you're boarding an airliner going across an ocean and you're given the choice between the one with fuel to go 1 * the way there or 0.999 * the way there, which aircraft would you pick?

I rest my case! Hehehe...

What exactly has this got to do with Zeno's paradox?

[aeroguy]

I don't know what Zeno's paradox is, but I do know that 0.999 of the fuel load needed to fly the full distance will make you drop in the drink...
Still convinced that 1 and 0.999 is the same thing?

[Sayonara³]

I'm convinced you can't read.

[J'Dona]

Just to spoil the fun... it wasn't 0.999, aeroguy. It was 0.999, or 0.999~. That means 0.9999999999999999... and so on forever.

Theoretically, that does equal one.

[aeroguy]

Only in mathematics then...

I suppose you could come up with ways to prove just about anything...

Why do you put that _ below the last 9?

[jordan]

Why do you put that _ below the last 9?
It signifies that whichever digit or digits are underlined repeat in that order forever. It can also be a line above the digit(s).

[JaKiri]

I suppose you could come up with ways to prove just about anything...

If it's true, then you're likely (but not certain) to be able to prove it.

If it's not true, then by the definition of true it will not have a valid proof.

If you don't believe it, disprove it please.

[Gauss]

Just to spoil the fun... it wasn't 0.999, aeroguy. It was 0.999, or 0.999~. That means 0.9999999999999999... and so on forever.

Theoretically, that does equal one.

1.\bar0 does not equal 0.\bar9 they are two distinct different numbers.

[Gauss]

1.\bar0 does not equal 0.\bar9 they are two distinct different numbers.

1.\bar 0 does not equal 0.\bar 9 they are two distinct different numbers.

[Gauss]

It's pretty obvious really.

0.\overline{0}1 = \lim_{n\to\infty} 10^{-n} \sum_{i=1}^{n} 0\cdot 10^{-i} = 0

This equation is not correct. All you have done is constructed a statement and stated the LHS equals the RHS.

[Kedas]

Just because we can represent them/it different in mathematics doesn't mean they aren't equal for using it in the real world.

But when 'designing' mathematics I would suggest not swap representations just to be on the safe side.

I guess it comes down to believing how much you think nature is linked to mathematics or the other way around.
mathematics did/can help find somethings we didn't know were there but that doesn't make it the bible.

[Dapthar]

It's pretty obvious really.

0.\overline{0}1 = \lim_{n\to\infty} 10^{-n} \sum_{i=1}^{n} 0\cdot 10^{-i} = 0Well, you're supporting the correct result, but with incorrect arguments. To illustrate, let's calculate a couple of values of your series.

At n=2 we get 10^{-2} \sum_{i=1}^{2} 0\cdot 10^{-i} = 10^{-2}(0 \cdot 10^{-1}+0 \cdot 10^{-2})=0.

At n=3 we get 10^{-3} \sum_{i=1}^{3} 0\cdot 10^{-i} = 10^{-3}(0 \cdot 10^{-1}+0 \cdot 10^{-2}+0 \cdot 10^{-3})=0.
Continuing on in this manner, (I omit the inductive argument, for the sake of brevity), one can see that your series always produces 0, and not 0.\overline{0}1.

I believe what you meant to write was 0.\overline{0}1=\lim_{n\to\infty}10^{-n}.

For this expression, at n=1 we get 10^{-1}=.1.
At n=2 we get 10^{-2}=.01.
At n=3 we get 10^{-3}=.001, and so on.

Using Calculus, it can be shown that 0.\overline{0}1=\lim_{n\to\infty}10^{-n}=0, as desired.

As a final note, I would simply like to mention that 0.\overline{0}1 is not correct mathematical notation, but the explanation of why would most likely spawn some side debate about \infty, so I'll leave that point unaddressed for now.

[jordan]

Please do adress why it's not correct notation. I can't guarentee I'll understand what you're saying, but now I'm interested.

[Dapthar]

Please do adress why it's not correct notation.Sure.

The "bar notation", as I call it, indicates that the sequence under the bar repeats forever, like 1/3=.\overline{3} or 1/7=.\overline_{142857}. So, technically, writing 0.\overline{0}1 is meaningless, since one cannot write something after the zeros.

The whole point of writing .\overline{0}1 is because some people, in an attempt to refute the fact that 1=.\overline{9}, state that 1-.\overline{9}=.\overline{0}1, and what dave was trying to show was that .\overline{0}1=0, thus showing that 1=.\overline{9}.

In fact, the main reason that some people have a problem grasping this equality is that the proof relies heavily upon limits, a tool regularly used in Calculus. In this case, one's intuition runs counter to reality, and thus, a problem arises when people attempt to "intuitively" understand it.

[jordan]

So nothing can ever come after the bar and still be considered correct? Saying something like ".\overline{9}8 is the closest you can come to 1" is nonsense?

[AL]

So nothing can ever come after the bar and still be considered correct? Saying something like "http://www.blike.com/tex/lateximg/pictures/6f558089f6fff0a589c46c4508bf5785.gif (javascript:do_texpopup('.overline{9}8');) is the closest you can come to 1" is nonsense?
It is nonsense because there is no such thing as a "real number that is closest to 1" and not equal to it. It has to do with a property of the real numbers called Completeness.

Suppose there is a number that is "closest" to 1 but not equal to it. We'll call it x. Since x is "closest" to 1, this implies that there is no such number y that is between x and 1 so that x < y < 1.

Now set y = (x+1)/2. Then we get that x < (x + 1)/2 < 1, which implies there IS a y such that x < y < 1. Contradiction.

Thus there is no such thing as a "closest" number in the set of reals.

[Gauss]

I believe what you meant to write was 0.\overline{0}1=\lim_{n\to\infty}10^{-n}.



0.\overline{0}1 is another way of stating or trying to represent \lim_{n\to\infty}10^{-n}. Which is just an identity equation or equivalence statement.

Dave's statement, stated 0.\overline{0}1 = 0

[dave]

I just used the definition of decimal notation.

[Gauss]

x = 0.999... Eq (1)

10 * x = 10 * 0.999...

x+x+x+x+x+x+x+x+x+x = 0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... Eq(2)

Eq (2) - Eq (1)

x+x+x+x+x+x+x+x+x = 0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999... +
0.999...

Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right.

9*x = 8.999...


x = \frac {8.999...}{9}



x = 0.999...

[rbp6]

Gauss, your, proof fails because all you did was prove that .99999....=.9999999....

Unless I'm missing something. All you need to do is use infinite summation with partial fractions and you'll get the right answer. I don't even understand where the debate is, anyone who has taken calc 2 should know the answer instantaneously. Or anyone who understands limits for that matter. Its a fairly simply idea I don't really see why anyone would have trouble grasping it unless they have very limited mathematical and conceptual capabilities.

[Gauss]

Four points to consider

First point:

The proof simply states that x = 0.999... and the method used is smiliar to the one that is used to prove otherwise. The calculations are exact.

Second point:

Take another example:

x = 3 Eq(1)

10*x = 10 * 3 Eq(2)

Eq(2) - Eq(1)

9*x = 27

x = \frac {27}{9}

x = 3

Third Point:

There is no reason whatsoever to use the formula for the sum of an infinite sequence in this case. I know the exact sum of:

S = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + ...

Its S = 0.\bar9

However I would use the sum of an infinite sequence in the following case:

S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n}

Obviously, as n \rightarrow \infty \mbox{\, } S_n \rightarrow 1

Fourth point:

This is the case for 2/9

x = 0.222... Eq (1)

10 * x = 10 * 0.222...

x+x+x+x+x+x+x+x+x+x = 0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... Eq(2)

Eq (2) - Eq (1)

x+x+x+x+x+x+x+x+x = 0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... +
0.222... Eq(2)

Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right.

9*x = 1.999...

x = \frac {1.999...}{9}

x = 0.222...}

[rbp6]

I'm sorry, I don't know how to use that whole forum math thing so bear with me.

.99999......=9/10+9/100+9/1000+9/10,000.....

9/10+9/100+9/1000+9/10,000.....=infinite summation(9/10^n)where n goes from 0-infinity

infinite summation(9/10^n)=(9/10)/(1-1/10) (geometric series summation)
(9/10)/(1-1/10)=1

Proven

[dave]

I don't really want to get drawn into this debate because as far as I can see, it's just a matter of definition, and I really, really hate to argue over such minute and tedious things.

Please don't let this stop you though :-)

[matt grime]

Since addition or multiplication is normally carried out from Right to Left and since this is an area of contention of what occurs or happens after the repeating dots (ellipses) then addition or multiplication in this instance is carried out from Left to Right.

There is no contention there at all. Arithmetic operations can and are easily defined on infinitely long decimal representations.

[paganinio]

has this been posted before?
0.999...=0.333...+0.666...=1/3+2/3=1

[Gauss]

Note the following pattern:

n = 0, \mbox { }then \mbox { } 1 - \frac {1} {10^0} = 0

n = 1, \mbox { }then \mbox { } 1 - \frac {1} {10^1} = 0.9

n = 2, \mbox { }then \mbox { } 1 - \frac {1} {10^2} = 0.99

n = 3, \mbox { }then \mbox { } 1 - \frac {1} {10^3} = 0.999

n = 4, \mbox { }then \mbox { } 1 - \frac {1} {10^4} = 0.9999

\mbox{Then }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n

Proof

Let P(n) be the statement that 1 - \frac {1} {10^n} has n 9's after the decimal place.

\forall_n \mbox { where } n \in \mathbb{N}

n = 0, P(0) = 1 - \frac {1} {10^0} = 0 \mbox {, thus P(0) is True.}

n = 1, P(1) = 1 - \frac {1} {10^1} = 0.9 \mbox {, thus P(1) is True.}

n = 2, P(2) = 1 - \frac {1} {10^2} = 0.99 \mbox {, thus P(2) is True.}

n = 3, P(3) = 1 - \frac {1} {10^3} = 0.999 \mbox {, thus P(3) is True.}

n = 4, P(4) = 1 - \frac {1} {10^4} = 0.9999 \mbox {, thus P(4) is True.}

Assume that P(n) is true that 1 - \frac {1} {10^n} (Basis of Induction) has n 9's after the decimal place. We shall show that P(n+1) is true. (Inductive Step)

P(n+1) = 1 - \frac {1} {10^{n+1}}

P(n+1) = P(n) + a_{n+1}

P(n+1) = 1 - \frac {1} {10^n} + \frac {10^1 - 1}{10^{n+1}}

P(n+1) = 1 - \frac {(10^n - 1)(10^{n+1})} {(10^n)(10^{n+1})} + \frac {(10^1 - 1)(10^n)}{(10^{n+1})(10^n)}

P(n+1) = 1 - \frac {(10^{2n+1} - 10^n)}{10^{2n+1}}

P(n+1) = 1 - \frac {1} {10^{n+1}}

\mbox{Therefore P(n+1) is True}

\mbox{Therefore P(n) is True}

\mbox{Therefore }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n

\mbox{Therefore }1 = 0.9_19_29_39_4,...9_{n-1}9_n + \frac {1} {10^n}\mbox{ }\forall_n \mbox { where } n \in \mathbb{N}

[matt grime]

So what has an inductive statement about a finite decimal expansion (hence only apllicable to the rationals) got to do with the real numbers in generality. Oh, wait you're not about to conclude the "infinite" case follows from the finite cases inductively....

and 0.00...01, an infinite number of 0s then a 1 makes no sense as a decimal expansion for anyone still under the impression it was meaningful.

there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions.

[dave]

I'm sorry for not living up to your exacting standards.

I have not yet completed my degree - indeed, I am only just about to start my second year. Therefore I'm beginning to understand what a role definitions have to play in the world of mathematics, but as hard as I try, I find them tedious and boring - although necessary, I do not wish to spend my free time talking about them on an internet chat forum.

I also know that I'll never become anything of any great significance to the mathematical community, basically because I'm not intelligent enough. I do (moderate) this because I enjoy doing mathematics in general and discussing it with other people. Unlike you, I am not a research mathematician and do not have the experience of such.

I guess what I'm trying to say is: we're not all research mathematicians, so cut me a little slack, will you?

[matt grime]

sorry, but being a research mathematician has nothing to do with it, being a mathematician does (i don't use real numbers in my research, none of my colleagues uses them in research either). if you don't actually pay attention to the definition of any of the terms in such a question as "why is 0.999.. equal to 1" then you've not answered the question since the truth of the statement *follows* from the definitions and the principal reason people do not understand this particular question is because they do not understand the definition of the terms. Those two objects are certainly different *representations* of the same real number but they are definitely the same *real* number. The reals are a basic object that we all use, though often without knowing what's going on. asking this question is a good sign of curiosity, understanding the (correct) answer a good sign of aptitude.

www.dpmms.cam.ac.uk/~wtg10/decimals.html

you do not need to discuss them, they are not up for discussion, they simply *are* if you will.

[dave]

sorry, but being a research mathematician has nothing to do with it, being a mathematician does.

I'm not going to go into it here because I don't want to get off-topic, but I don't believe this statement to be correct.

The reals are a basic object that we all use, though often without knowing what's going on. asking this question is a good sign of curiosity, understanding the (correct) answer a good sign of aptitude.

If you have a problem with me (which, considering the general language you used in your post), then take it up with me in PM instead of trying to insult me.

[matt grime]

Sorry, apologies for sounding insulting. I have no problem with you, though i have a problem with answers that don't use the properties of the objects in the question.

[matt grime]

An olive branch:

yes, having to just deal in basic definitions is tedious and dull, no argument there, and often it is unnecessary to make explicit reference to them all. However *this* particular question exactly boils down to people not knowing what the definitions are (complete totally ordered field, or completion of the rationals in the euclidean norm, or the set of all dedekind cuts, whatever tickles your fancy).

[dave]

Fair enough - I'll admit that I haven't really studied the matter of decimals in any significant form, and I don't understand them and their notation as much as I should. I also apologise for my outburst - I'm a little tired, been working for about 11 hours today and have driven ~200 miles so I'm a bit tetchy :-)

[Gauss]

So what has an inductive statement about a finite decimal expansion .

The statement is not about a "finite decimal expansion". What is it that you do not understand about "for all of n in \mathbb{N}. The set \mathbb {N} has never been finite and will never be finite.

(hence only apllicable to the rationals) got to do with the real numbers in generality.

Rationals are a subset of the reals. Therefore any number that is rational is also a real number.

Oh, wait you're not about to conclude the "infinite" case follows from the finite cases inductively....

You may do that, but I will never deduce that. The proof states what it states.

and 0.00...01, an infinite number of 0s then a 1 makes no sense as a decimal expansion for anyone still under the impression it was meaningful.

I agree with you unreservedly on this. It is very bad language to be used by any mathematician. I would also state that 0.999... expressed as such is also very bad language (but thats my opinion). It is not rigorous enough, but everybody uses it that way, so by general agreement I also use it. Doing operations on this form leads to problems.

there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions.

Mathematics is only valid according to the definitions (and proofs) used and hence is only valid in that particular context in which it is used.

[BrainMan]

The problem is that if you think .999... is not the same as 1, then any proof using a limit is question begging. The very concept being questioned by those that think the numbers are not equal is that a number getting closer and closer to some other number,

x = .9
x = .99
x = .999
...

never actually reaches that number, even after an infinite number of iterations. An infinite limit is presumed to ignore minute differences, so using it as a proof is not valid if there might be a miniscule difference between .999... and 1. In other words, the definition of a limit is really what is at question here (along with all other operations with infinite iterations to be completed). [Nonstandard analysis, in fact, defines limits as operations that ignore infintesimal differences by mapping them onto the closest real numbers, which conforms with the intuitions of those that think .999... is not equal to 1.]

[BrainMan]

there have been several worrying expressions of dislike of definitions (esp by a maths moderator). worrying because maths IS its definitions.

The question is, COULD math have developed differently (remaining consistent) such that .999... does not equal 1, but is only very close to it?

[matt grime]

The statement is not about a "finite decimal expansion". What is it that you do not understand about "for all of n in \mathbb{N}. The set \mathbb {N} has never been finite and will never be finite.



Rationals are a subset of the reals. Therefore any number that is rational is also a real number.



You may do that, but I will never deduce that. The proof states what it states.



I agree with you unreservedly on this. It is very bad language to be used by any mathematician. I would also state that 0.999... expressed as such is also very bad language (but thats my opinion). It is not rigorous enough, but everybody uses it that way, so by general agreement I also use it. Doing operations on this form leads to problems.



Mathematics is only valid according to the definitions (and proofs) used and hence is only valid in that particular context in which it is used.


the point was, what the hell did your post have to do with anything? it dealt with an infinte set of finitely long decimal expansions. so?

[matt grime]

The question is, COULD math have developed differently (remaining consistent) such that .999... does not equal 1, but is only very close to it?

of course it could, since these are just notations for objects with a formal set of properties. we could have given them any meaning we wanted. in fact in base 11, 0.999... doesn't equal 1.

the thing about being very close to it indicates that you want a system where the archimidean axiom is not true, that isn't a very useful system really but you may define it, and use it. note that in that system you cannot prove almost any theorem that you take for granted in analysis (even limits are no longer unique).

[jcarlson]

Yes, but in base 11, assuming a is the 11th digit, .aaa... does equal 1.

[Newtonian]

Thats exactly what i was trying to convey earlier,but my math is crap!

[Johnny5]

In case anyone here hasn't noticed, there have been a lot of debates on this online between people on other forums, some with a good grasp of mathematics and some without. The debate is over whether 0.999 does or does not equal 1.

There are many different proofs to prove that it does, but because some people understand some proofs better than others, and some not at all the debates still go on. I've personally been involved in these sort of debates and I can think up at least five different proofs, all based on different sorts of logic, that show that they are equal. The more forms we can find, the better the chances are that someone will understand it.


Well I would enjoy seeing just one of your five proofs, because they aren't equal. In fact, 1 is a number, and .99999... isn't a number, so your proof is guaranteed to contain an error, you will equate apples to oranges, which is similiar to adding apples and oranges, but not identical. I'm not playing semantical games either. In this particular problem, semantics happens to matter.

Regards

[jcarlson]

.\overline{9} is most certainly a number. Not only that, its a real number, and also a rational number. It is merely a different representation of the same real number that 1 represents.

[lamefif]

I think u guys r trying to make a hole in water. The real world is not precision perfect but fuzzy & mathenatics is one of the language the mortal humans r trying to make sense of it. But if u get to close to the edge u wont c it, is like going to the north pole and asking where exactly is it.
Good luck

[matt grime]

Yeah, and some of us think that "u" isn't perfect for precision for "you"....

[mcoy]

ok...just consider the fact that 0.99999...99 cannot be expressed by a fraction, in which case i propose that 0.9999....999 is an irrational number....i might be wrong of course. and one more thing, where the heck will you use 0.999...999 for anyway? it has no use other than just being argued at.....

[J'Dona]

Well I would enjoy seeing just one of your five proofs, because they aren't equal. In fact, 1 is a number, and .99999... isn't a number, so your proof is guaranteed to contain an error, you will equate apples to oranges, which is similiar to adding apples and oranges, but not identical. I'm not playing semantical games either. In this particular problem, semantics happens to matter.

RegardsOkay, I'll post one of the proofs for you (although it's probably already been said two or three times in this thread in much better form than I could reproduce it).

Let's say that 0.999... recurring is the sum of a geometric progression. This is pretty easy to demonstrate: if the first term, a, is 0.9 and the ratio, r, is 0.1, then the progression will be 0.9, 0.09, 0.009, and so on, and the sum will be 0.999 (which is just my way of writing 0.999... recurring). This has already been written mathematically in other posts by people like dave, but I can't figure out the LaTeX well enough to reproduce it without quoting and I don't fully understand the terminology anyway.

Here's the formula for the sum of a geometric series:

http://www.blike.com/tex/lateximg/pictures/cf00569d26ae2e6e2ff5bf0e06050e00.gif (http://javascript%3Cb%3E%3C/b%3E:do_texpopup%28%27S_%7Bn%7D%20=%20frac%7Ba%281 %20-%20r%5E%7Bn%7D%29%7D%7B1%20-%20r%7D%27%29;)

In the case of 0.999, n equals infinity, so the term rn is actually 0.1infinity, which is zero, and (1 - rn) simplifies to 1. So the formula simplifies to this:

http://www.blike.com/tex/lateximg/pictures/54b97c9f498489620ff5dd84e669b7fa.gif (http://javascript%3Cb%3E%3C/b%3E:do_texpopup%28%27S_%7Binf%7D%20=%20frac%7Ba%7 D%7B1%20-%20r%7D%27%29;)

Substituting the values of a and r in (0.9 and 0.1 respectively), we get this:

http://www.blike.com/tex/lateximg/pictures/92d12f02fbb8ac6665c71a30765bbcb6.gif (http://javascript%3Cb%3E%3C/b%3E:do_texpopup%28%27S_%7Binf%7D%20=%20frac%7B0.9 %7D%7B1%20-%200.1%7D%27%29;)

http://www.blike.com/tex/lateximg/pictures/b0fd1065e2e0a1a57327c9d37bc12908.gif (http://javascript%3Cb%3E%3C/b%3E:do_texpopup%28%27S_%7Binf%7D%20=%201%27%29;)

and so if the math is all right it looks like the sum to infinity of the series, 0.999, is equal to 1.

The only thing I can see wrong with this proof is the assumption that 0.1infinity does equal zero and not just some ridiculously small number, though I'm not sure what else it might be, and falls into the "infinite zeroes means no endpoint" scenario again.

I can't believe my first post after 2 months was in this stupid thread again...

[dave]

Welcome back, J'Dona :-)

Yes, this thread has rather degraded; I don't know whether there's actually much discussion left in it anymore. That's why I stopped posting in it a while back.

[Daymare17]

I think I might have found a way to debunk the simplest proof that 0.9 = 1. The proof goes like this:

N = 0.9
10N = 9.9
10N - N = 9
N = 1

The debunking goes like this:

Infinity is not a number but a concept. This is apparent from the fact that you cannot divide by infinity but only by successive approximations, i.e. continually increasing numbers towards a limit that equals infinity, yet never reaching infinity itself. 0.9 "in practice" has to be 0 followed by a set amount of decimal nines. It can be any amount but has to be a certain amount. It can be followed by a million nines, or a googolplex nines, or three nines, or five nines. Let's try five nines and see what we get.

N = 0.99999

10N = 9.9999
- N = 8.99991

And then the whole proof falls to pieces. The same thing would happen no matter how many decimal nines we had.

As we see, in order for the proof to be true we have to proceed from the supposition that N is two different numbers, namely 0.9 (followed by x decimal nines) and 0.9 (followed by x+1 decimal nines)

When you proceed from contradictory premises it's no wonder the result is contradictory.

I haven't found a way to debunk the proof that uses the infinite geometrical progression (let alone dave's first proof, which I don't understand at all :D) but I imagine that it may have a similar flaw.

[ed84c]

Hmm, this means

0.99999999999... is infinitley close to 10. And we know 1/infinity= 0 and surely hence this means 9.9....=10??

[ed84c]

After making a cup of tea i have concluded on the following:

If a mathematical proof requires real numbers then 10 != 9.999........

However

If a mathematical proof can be conceptual, using infinity (not defining it, just using it)
Then 10=9.9999......

[AndrewMech]

Try this,

A frog has 2 meters to travel,
On its first leap it can jump 1 meter,
Every leap after its first it can only leap
half the distance of the previous leap.

How many leaps before it travels 2 meters?

[timo]

http://redwing.hutman.net/~mreed/warriorshtm/necromancer.htm

[dave]

Infinity is not a number but a concept.

We're not using it as a number anywhere in the thread; certainly not in my posts :-)

This is apparent from the fact that you cannot divide by infinity but only by successive approximations, i.e. continually increasing numbers towards a limit that equals infinity, yet never reaching infinity itself. 0.9 "in practice" has to be 0 followed by a set amount of decimal nines. It can be any amount but has to be a certain amount. It can be followed by a million nines, or a googolplex nines, or three nines, or five nines. Let's try five nines and see what we get.

But we're not concerned with this. "0.999..." doesn't mean "have a finite number of 9's in the expansion". We're only concerned with the limit, so none of this counts.

I haven't found a way to debunk the proof that uses the infinite geometrical progression (let alone dave's first proof, which I don't understand at all :D) but I imagine that it may have a similar flaw.

I'm betting you won't find one either.

After considering this thread for a bit, I thought of something a little more interesting. If you have a problem accepting 0.999... = 1, then it follows that you must have a problem with things like exponentials, and more specifically, exp(). After all, we define e using an infinite sum (or equivalent limit):

e = \sum_{n=0}^{\infty} \frac{1}{n!}

Really guys, what's so difficult to accept about this? Whilst I accept that it may not be completely intuitive, that's pretty much true of a lot of mathematics out there.

[myrkle]

Well I'm gonna try to do this like my calculus teacher told me:
1/9 = 0.111111
2/9 = 0.222222
3/9 = 0.333333
etc...
so 9/9 = 0.999999

therefore 0.99999 is equal to 1

[Zanthra]

I think of it by the fact that any number with a repeating decimal is a rational number.

In the case of .333 it is equal to 1/3.

In the case of .131313 it is equal to 13/99.

For such numbers there is an easy algebraic way to turn it into a fraction.

Take X to be the repeating decimal:

x = 3.723969696

Break X into two parts the repeating portion and the nonrepeating portion:

x = y + z

y = 3.723
z = .000969696

First find the fraction of z:

z = .000969696 : Multiply both sides untill you have one complete repetion to the left of the previous repetition, equal to 10 times the number of digits in the repetition, in this case 100.

100z = .0969696 : Subtract one of the variable from both sides. On the left use the representative variable, and on the right use the repetitive value.

99z = .096 : Now that you have gotten rid of the repetition, you can find a fraction by simplyfying.

z = .096 / 99
z = 96 / 99000

Now go back and simplify the same fraction for the origonal nonrepetitive portion:

y = 3.723
y = 3.723 / 1
y = 3723 / 1000

Multiply them to have the same denominator:

z = 96/99000
y = 368577/99000

x = y + z : Put them back together and simplify.
x = 368673/99000
x = 122891/33000

122891/33000 = 3.723969696


This works for any number with a repeating decimal, including .999:

x = .999

x = y + z

y = 0
z = .999

10z = 9.999

9z = 9
z = 1

x = y + z
x = 0 + 1

x = 1
.999 = 1


There is no number with a repetative decimal that does not work in this method. There are also no numbers with repetative decimals that do not exqual an integer ratio. Meaning that .999, 1.999 etc. all are integer ratios 1/1, 2/1 etc...

[MattC]

Regarding the quirk in the proof
x = .99999
10x = 9.9999

The quirk is this: If you define a set number of 9's, it doesn't work. When you say it goes on forever, you side-step this.

Use 4 nines, for instance:

.9999
9.999

9.999 - .9999 = 8.9991, not 9
to make that work, you have to pretend that x has two different numbers of 9's

when you multiply by 10 you add a nine

It's all moot, though, because the difference between .999 repeat and 1 is infinitely small, by definition.

however, in real terms, there is a difference. It is just so small that it is impossible to measure OR prove mathematically, because we are using a finite type of math to describe an infinite (non-real?) quantity

So technically that is not a proof of anything but the limitations in our math system, in my opinion.

At the same time, I'd say that .999 repeat is equal to one. It's a moot point to say otherwise.

Of course, math is not my specialty. I agree with YT though.

[Zanthra]

I have to disagree with the fact that, repeting decimals are not a finite number. I think they are well defined by the integer ratios they turn into through the method I described. It works for simple ratios such as 1/3 = .333, and even more and more complex ones, what is wrong with it working or .999

You are saying if you have X nines after the decimal point, then there will be X - 1 nines if you multiply by ten. That leaves 1 nine behind when you are done with the subtraction.

The problem here is that you are saying infinity - 1. For all practical purposes infinity * 50 = infinity.

For example there is the problem of the infinately large hotel. There are rooms labeled from 0 ... infinity. One person wants a room but all the rooms are full. So in solution to this, the hotel calls up everyone in the hotel and tells them each to move to their room + 1. This makes room 0 vacant and the new arival sleeps there. If 2 people arrive at the hotel, you can have everyone move to their room + 2. Now what about the case where an infinate number of people want rooms in the hotel. In that case the hotel can once again call up everyone in the hotel and tell them each to move to their room * 2. This frees every odd numbered room in the hotel, and there is room for an infinate number of people. This shows how finite numbers and infinite numbers can have relationships, and how infinate variables can do strange things in word problems.

Another problem that was described to me is if you have two buckets. You do the following steps an infinate number of times.

Have 2 variables, X and Y both at zero.

Place X and X + 1 in the first bucket, and increment X by 1.
Take Y from the first bucket and place it in the second bucket and increment Y by 1.
Repeat.

When you are done, what numbers are in each bucket? Since you do it an infinate number of times, there are no numbers that can be in the first bucket, and all numbers will be in the second bucket, dispite the fact that logic will state that there should be the same number of numbers in the first bucket as the second. This shows how logical assupmtions and resonable truths can be bent by the rules of infinity.

In the case of the repeating decimal, we are not talking about a finite number. Instead think of it as saying that after each repetition there is another repetition.

For example if you take 9999 and you take off a nine from the front, if there is always another repetition it is still 9999 you take off 100 million, or more and you still get 9999

All fractions are exactly equal to some number with repeating decimals. Many of them have repeating zeroes however, and it that case the zeroes are simply not written.


Infinity is a fickle thing and hard to understand. When I was first posed with the bucket example, I tried to find some way to get an equal size of each of the buckets, when I did so I got it wrong. When the answer was revealed however it made perfect sense. I think that is where I really began to understand infinity.

[bascule]

"debates"? What debate? That makes the issue sound as if it were debatable.

Sadly, idiots v. sensible, intelligent people is a debate that will never end.

[Zanthra]

I do not believe there are any idiots in the world. Just different forms of Inteligence.

Since the real question about the difference between 1 and .999 is what infinity means, I would like to show you this page that I found a few minutes ago. http://diveintomark.org/archives/2003/12/04/infinite-hotel It shows the significant difference between thinking of infinity and thinking of real numbers is that infinity is the lack of a bound. Infinity means there is no number larger than it, even itself, and even itself * itself, or itself to the power of itself.

[MattC]

I agree that, in conceptual terms, .999 repeat is 1. And who can say that .999 repeat exists in anything outside of conceptual terms, anyway? Of course, if such a thing can exist, so too can an infinitely small number that is the difference between 1 and .9999. If infinity can exist, then so too can a difference, that is infinitly small, between .999 repeat and 1. And if there exists such a difference, then we can say that the two are not equal. In one sense, at least.

I also think that people who go around calling others idiots, simply because they attempt to understand things in a different (and in this case, unpopular) way, are most often the idiots. But then, I am being a bit abrasive here, too. I suppose I am just trying to say that I took offense to being called an idiot, indirectly, for adding my thoughts to this thread.

[Zanthra]

Do you consider infinity a number. Because for any number X, X != X + 1 infinity does.

Anything that is small but has any size at all has a finite size. Infinity is the lack of a bound, so 1 / infinity would equal zero.

If you have 1 / X, then X (1 / X) = 1 but in that case 2X ( 1 / X ) = 2. That means that since infinity * 2 = infinity then infinity ( 1 / infinity ) = 1 and infinity ( 1 / infinity ) = 2 a contradiction.

Math cannot be done with infinity as a number, however when you use limits you can, in that case 1 / infinity = 0 and 1 / 0 = infinity, meaning that something infinately small is equal to nothing.

Infinity is not a number that math can be done with.

[dave]

Not quite. If you're using limits (as I hoped you were in the last question), then \lim_{x\to 0}\frac{1}{x} is undefined. You need to take either the left or right limit to get plus or minus infinity respectively.

[AL]

I agree that, in conceptual terms, .999 repeat is 1. And who can say that .999 repeat exists in anything outside of conceptual terms, anyway? Of course, if such a thing can exist, so too can an infinitely small number that is the difference between 1 and .9999. If infinity can exist, then so too can a difference, that is infinitly small, between .999 repeat and 1. And if there exists such a difference, then we can say that the two are not equal. In one sense, at least.
An infinitely small number can exist in another system, like the hyperreals (*R), but not in R. Once you agree that we are talking about the real numbers, then you must play by R's rules. There is no such thing as a real non-zero number with an infinitesimal value in R -- it follows directly from the Archimedean property. Further, that 1 and .999_ are equal follows directly from that, since there is no infinitesimal value you can get by subtracting them.

[Bio-Hazard]

In my opinion:

if it isn't a 1, it is not a 1.

If it doesn't look like a 1 it isn't a 1.

[Tom Mattson]

There's no way that you or anyone else could possibly reach any sensible mathematical conclusion without falling back on mathematical proof, and proofs that establish the truth of the statement 0.999...=1 abound both in textbooks and on the internet.

[dave]

In my opinion:

if it isn't a 1, it is not a 1.

If it doesn't look like a 1 it isn't a 1.

That's a bit silly really. What's the difference between saying "2/2 doesn't look like a 1, therefore it isn't a 1"? The two are mathematically equal, and I'm willing to bet that you won't be arguing that fact.

[Bio-Hazard]

you have two parts of something that was divided into two, making it into a whole one.

it may be one.. however.. it isn't exactly one nor is it originally whole. sure it may be together.. but not originally whole since it is divided and then combined. Of course then standard for being whole is an abstract idea of what whole means..

I'm done here. Talking about the earth number system is pointless. It's all an abstract way of putting things into symbols anyways. who cares?

anyways.. .001 + 0.999 = 1

.0 (infinite 0s) 1 + 0.0 (infinite 9s) 9 = 1

i think since humans are analog (i will not believe of infinite as a concept.. some of you already know what I think of infinite anyways).. i believe it is a reality.. and since a feature of analog in the universe is that analog beings can reach any high number achievable infinite is a reality. However digitally infinite is constrained..

[TD]

anyways.. .001 + 0.999 = 1

.0 (infinite 0s) 1 + 0.0 (infinite 9s) 9 = 1

What you're saying isn't possible. You can let the last digit be recurring, but not another one. So we're talking about 0.999 and not 0.999, not that it really matters in this case. It does matter with 0.001, which isn't possible. You cannot have an infinite number of recurring 0's, and then still 'add a 1'.

[Tom Mattson]

it may be one.. however.. it isn't exactly one nor is it originally whole.


No, 2/2 is exactly 1.


anyways.. .001 + 0.999 = 1

.0 (infinite 0s) 1 + 0.0 (infinite 9s) 9 = 1


No, 0.001 is not a real number.

[Archrono]

Every proof used has a problem. You are treating 0.999... as a discrete number and comparing it to 1. 0.999... has no value in terms of discrete numbers such as 1. How much is 0.999...? Infinitely close to 1. It is not a discrete number and is not therefore comparable.

Intuition tells us that 0.999... is less than 1. Arithmetic tells us that 0.999... is equal to 1. However, logic tells us that 0.999... is not arithmetically related to 1 because it is infitecimal. The proofs apply discrete arithmetic to a non-discrete number.

Now you're goin to prove that 1.000.... = 0.999... with the same silly arithmetic. Stop dividing infinitely, stop adding infinitely. You can't make apple pie from florida oranges.

[jcarlson]

Every proof used has a problem. You are treating 0.999... as a discrete number and comparing it to 1. 0.999... has no value in terms of discrete numbers such as 1. How much is 0.999...? Infinitely close to 1. It is not a discrete number and is not therefore comparable.

Intuition tells us that 0.999... is less than 1. Arithmetic tells us that 0.999... is equal to 1. However, logic tells us that 0.999... is not arithmetically related to 1 because it is infitecimal. The proofs apply discrete arithmetic to a non-discrete number.

Now you're goin to prove that 1.000.... = 0.999... with the same silly arithmetic. Stop dividing infinitely, stop adding infinitely. You can't make apple pie from florida oranges.

Logic? Logic tells me that .\=(9) is equal to one. There is a very simple, mathematically rigorous proof for this.

All real numbers are BY DEFINITION defined by the infinate LIMITS of their decimal representations, not the representations by themselves. Many times the limit and the representation are one in the same, however other times it is not. One blatently obvious case is the set of all irrational numbers: \pi, e, \phi, sqrt(2), etc, which, if they weren't defined as limits, couldn't be irrational in the first place, because no matter how many decimals you added to the sequence you would still end up with a rational number.

So with that in mind, let us examine the number .\=(9)

By definition, this decimal is defined as:

lim_{x\to\infty}\sum_{n=1}^{x}\frac{9}{10^n}

[K9-47G]

I noticed that if I type .9999999999 (ten nines) into my TI-83 calculator and press enter, it gives me the answer to be .9999999999 (ten nines), But if I type .99999999999 (eleven nines) into my calculator and press enter it gives the answer to be one. I suppose my TI-83 rounds to the 10th decimal place.

[AL]

Every proof used has a problem. You are treating 0.999... as a discrete number and comparing it to 1. 0.999... has no value in terms of discrete numbers such as 1. How much is 0.999...? Infinitely close to 1. It is not a discrete number and is not therefore comparable.
If by "discrete number" you meant "integer," I fail to see how that makes your argument meaningful. All I see is question begging. "They're different, so they're not equal."


Intuition tells us that 0.999... is less than 1. Arithmetic tells us that 0.999... is equal to 1. However, logic tells us that 0.999... is not arithmetically related to 1 because it is infitecimal. The proofs apply discrete arithmetic to a non-discrete number.

Now you're goin to prove that 1.000.... = 0.999... with the same silly arithmetic. Stop dividing infinitely, stop adding infinitely. You can't make apple pie from florida oranges.

You and all the other proponents of 0.9_ not equaling 1 really need to get over this idea that if some expression is to be evaluated numerically in an infinite number of steps that it must therefore be some sort of phantom number. Consider that the square root of 2 has an infinite number of digits. If you multiplied it by itself, you'd be multiplying two infinitely long numbers, yet even a grade schooler can see that the product is 2 and wouldn't argue to "stop [multiplying] infinitely."

Likewise, the solutions to other infinite operations can be deduced without resorting to numerical/empirical methods that involve actually carrying out an infinite number of steps. As has already been shown a zillion times, 0.9_ is easily written as an infinite geometric series, with a sum that is rigorously deduced to be 1.

[RyanJ]

I noticed that if I type .9999999999 (ten nines) into my TI-83 calculator and press enter, it gives me the answer to be .9999999999 (ten nines), But if I type .99999999999 (eleven nines) into my calculator and press enter it gives the answer to be one. I suppose my TI-83 rounds to the 10th decimal place.

Yup, I think most (If not all) calculators do at some point.

I know you guys have gone through everything in the link but maybe you cna find something there you missed to mull over ;)

http://mathforum.org/dr.math/faq/faq.0.9999.html

Cheers,

Ryan Jones

[cosine]

Every proof used has a problem. You are treating 0.999... as a discrete number and comparing it to 1. 0.999... has no value in terms of discrete numbers such as 1. How much is 0.999...? Infinitely close to 1. It is not a discrete number and is not therefore comparable.

Intuition tells us that 0.999... is less than 1. Arithmetic tells us that 0.999... is equal to 1. However, logic tells us that 0.999... is not arithmetically related to 1 because it is infitecimal. The proofs apply discrete arithmetic to a non-discrete number.

Now you're goin to prove that 1.000.... = 0.999... with the same silly arithmetic. Stop dividing infinitely, stop adding infinitely. You can't make apple pie from florida oranges.

So .999... isn't a number? Then how can it be less than or equal to 1?

[Curios]

I think there is a slight difference between 0.99 and 1 or 0.999 and 1 etc…
Take an example of a system that succeeds 99.999% of the time to do whatever task and fails the remaining 1*10^-3%. This effectively means that the system will fail once every 100,000 attempts. This sounds a bit faulty since you guys are debating 0.999 and not 0.999 but a system or law of nature that succeeds 99.999% keeps that very slight possibility of failure while a 100% successful system eliminates anything other than success. This may sound stupid but there is a difference because that 0.0001% failure exits in a 99.999% successful system while the 100% leaves no margin for debate! (Failure wise)

An example of this is a quantum fluctuation whereby the possibility of you randomly chosen (by nature) to be tunneled through a concrete wall to another room exits! although very small. The very nature of that small probability existing allows me to debate it, while otherwise I wouldn’t be able to.

Another example I can think of is a trans-planetary missile that is 99.999% accurate. When launching that missile to Pluto for example, a tolerance between where it is aimed and where it will hit exists and is certain.


I have to admit that reading your posts reminded me when I got worried about birth control pills because I thought they were 99% effective (actually 99.99%) and thought the law of large numbers will get me! Hence my 1st post.

[Tom Mattson]

This sounds a bit faulty since you guys are debating 0.999 and not 0.999 but a system or law of nature that succeeds 99.999% keeps that very slight possibility of failure while a 100% successful system eliminates anything other than success.


It is faulty, for that very reason. It is clear that 0.999 \neq 1.

Also, why bother to bring up physical systems or quantum fluctuations? They have nothing whatsoever to do with determining a mathematical truth.

[Curios]

Because we happen to live on earth.

[CanadaAotS]

uh, you say 0.0001% failure exists... well it doesn't. You can't have repeating zero's, then a 1. If it could equal anything it would be 0. Which would just prove the point that it'd be a 0% failure and that 99.9% = 100%.

[Tom Mattson]

Because we happen to live on earth.

That doesn't answer my question.

Mathematics is an a priori discipline. All mathematical theorems follow from mathematical definitions by mathematical rules of inference. The physical universe plays no role in deciding mathematical truths, apart from the fact that material brains are needed to carry out the thought processes.

[Curios]

Quote from Matt Matson: "The physical universe plays no role in deciding mathematical truths"

What if a system is 100% faultless and makes a mistake? It simply can not happen. If it did happen, then all mathematical theorems and laws of nature will fail. The physical universe is the ultimate test to mathematical constructs. They cannot contradict each other and both be correct.

The point I’m trying to explain is that a system with 99.999% certainty still contains a small amount of uncertainty while a system with 100% certainty has 0 uncertainty.

[CanadaAotS]

Curios: thats the entire point, the 99.999% certainty would not contain any uncertainty and be equal to 100%.


Another example I can think of is a trans-planetary missile that is 99.999% accurate. When launching that missile to Pluto for example, a tolerance between where it is aimed and where it will hit exists and is certain.


If it was 99.999% accurate they wouldn't say 99.999%. They'd say 100%. lol

[zyncod]

The point I’m trying to explain is that a system with 99.999% certainty still contains a small amount of uncertainty while a system with 100% certainty has 0 uncertainty

Does it? Give us the +/- number for that level of uncertainty.

[CanadaAotS]

yah, and 0.0001 is not a number that exists, so you'll have a hard time giving an actual number.

funny how this thread is called "Ending the 0.999~ = 1 debates" since this debate will obviously not end no matter how many proofs showing that 0.999~ does = 1 :-p are presented...

[Tom Mattson]

Quote from Matt Matson: "The physical universe plays no role in deciding mathematical truths"

What if a system is 100% faultless and makes a mistake? It simply can not happen. If it did happen, then all mathematical theorems and laws of nature will fail. The physical universe is the ultimate test to mathematical constructs. They cannot contradict each other and both be correct.


Do yourself a huge favor: Pick up an elementary textbook in mathematics and read it.


The point I’m trying to explain is that a system with 99.999% certainty still contains a small amount of uncertainty while a system with 100% certainty has 0 uncertainty.

And the point I am trying to explain is that your goofball notions of mathematics are rejected by those who work in the field.

[Curios]

What's wrong Tom Mattson? Don't tell me you're letting the trivial debate get under your skin.

Please dont give any favours out to anyone but yourself.

Appreciated

[Tom Mattson]

What's wrong Tom Mattson? Don't tell me you're letting the trivial debate get under your skin.


No, not at all. In fact I'm more amused than anything else.


Please dont give any favours out to anyone but yourself.


You are posting on a message board that is devoted to science. That means that you can expect to have your mistakes pointed out, and you can further expect suggestions on how to remedy them.

Mathematics is an academic discipline that requires a great deal of hard work and dedication. There are people here who have done that hard work. When you post things that contradict what every authority in mathematics says about mathematics it sends the message that you think you know the subject better than they do, despite the fact that you clearly have not done the necessary hard work to understand it.

[dave]

Frankly, this thread has gone on long enough. Unless people decide to post some sensible, reasoned arguments then I will simply close the thread.

[Archrono]

The best argument against 0.999.... = 1 is

0.999.... will never equal 1.

You must go to infinity to make a geometric series -or any other proof- prove that 1 = 0.999.... But by infinity's definition it can never be reached, therefore 0.999... never equals 1.

You are never at infinity, you never have infinity in your hands to dividide by and you are never allowed to repeat a geometric series infinitely. You are treating infinity as if it is there for any fool to use.

Reenforcement - As x gets really really big (goes off toward infinity, *eye roll*) 1-(1/x) represents 0.999... but x is never infinity, so it is entirely inappropriate to do arithmetic at infinity.

And this is where we move on to calculus and thus math as a tool rather than a philosopher's play thing.

[TD]

Sorry, but that's nonsense.

An infinite series is per definition a sum of an infinite number of terms. The fact that there's no real infinity in 'nature' doesn't mean it doesn't exist in mathematics (since it does, and it only does so because we defined it somehow).

Infinite series surely exist and are very commonly used.
For example, we can define the number e as \sum\limits_{n = 0}^\infty {\frac{1}{{n!}}} . If it's impossible to determine this infinite sum, then e wouldn't "exist".

You have probably once learned about rational and irrational numbers (which form the real numbers, together). A number was irrational if it has an infinite number of non-repeating decimals. So they never become periodic and they're not finite. It is also impossible to express these number as a fraction of two integers.

Rationals on the other hand, are the numbers which either have a finite number of decimals or which become periodic in their decimals. For example, 22/7 (which is an approximation for pi btw, pi being irrational itself) has the decimal expansion 3.142857142857142857..., or {\rm{3}}{\rm{.}}\overline {{\rm{142857}}} = 3.142857.... Those last two numbers are merely other notations for 22/7 and its the same number 4 times, a rational one.

I assume it's clear that 0.999... has a repeating part as well (being just 9) and therefore, it has to be rational and thus it can be written as a fraction of integers. You tell me how that can be done, if 1 (=1/1=2/2 or whatever) isn't it.

[CanadaAotS]

I think that (one of) the main arguments of the not-equal-to-oners (lol) is that infinity is not a number. Well, in a physical universe you can't have infinite of anything, it just doesn't work. However, like Tom Mattson said, mathematics is not dependent on the physical universe it's just a tool.

But besides that, if .999... does not equal one because it can't progress infinitley then its not a number at all, equal to one, or equal to anything for that matter.

[Tom Mattson]

The best argument against 0.999.... = 1 is

0.999.... will never equal 1.


Can't you see that your premise and your conclusion say exactly the same thing??

I beseech the powers that be to lock this thread. It has descended into sheer crackpot nonsense.

[dave]

I've closed the thread, since it's gone on far too long now.