34 replies to this topic

[wretched]

My friend asked me this question... Is one equal or greater than .9999...(that goes on to infinity) and support/prove ur answer somehow. I got my own theories but I wanna hear urs first

[Xilenx]

1 is always greater than .999x by .000x value.

It's similar to the post modern theory, in which basically you cannot achieve the truth (the end), but only work so close to it, yet never achieve it. It's kinda confusing so this might explain it.

Assuming the truth was "10", and you can only achieve the truth by reducing it by half, so from 0, you'll get to 5, then 7.5, then 8.75, then so on, until you are around 9.999x but you would never achieve "10". Hence, you would never be able to achieve "truth", but only gather more and more information about it, never solving the puzzle, only getting closer, yet infinetely away.

[special_angel]

Sorry, Xilenx, you're wrong.
Your error is in assuming that the 9's stop eventually, when, in fact, wretched made it clear they go on forever. Here are a few explainations i have found in case it still doesn't make sense (don't worry, it took me a long time to get it, too).

Why does 0.9999... = 1 ?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright © 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl).
The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum.

So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers

9/10,
9/10 + 9/100,
9/10 + 9/100 + 9/1000,
...."


One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1:

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1


Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
0.9999... = 1

Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.


Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99. How can .999999.... equal 1?
**********************************************************
Date: 03/21/2001 at 15:07:26
From: Emily F. and Jenny B.
Subject: .999999..... I still don't get it

Dr. Math,

In my math class in school, my math teacher always talks about how
whenever she has a problem she goes to your site and finds it or
writes to you. I have a problem.

I know .999999.... is supposed to equal 1. My teacher demonstrated
the subtracted thing and the other stuff you have on your site. I
still don't get it. If .99999999.... goes on forever, wouldn't it be
just a little below one? There would be just a tiny gap between it and
one. Please explain this to me.

Thanks,
Emily and Jenny


--------------------------------------------------------


Date: 03/21/2001 at 16:22:11
From: Doctor Ian
Subject: Re: .999999..... I still don't get it

Hi Emily and Jenny,

There's no doubt that this equality is one of the weirder things in
mathematics, and it _is_ intuitive to think: No matter how many 9's
you add, you'll never get all the way to 1.

But that's how it seems if you think about moving _toward_ 1. What if
you think about moving _away_ from 1?

That is, if you start at 1, and try to move away from 1 and toward
0.99999..., how far do you have to go to get to 0.99999... ? Any step
you try to take will be too far, so you can't really move at all -
which means that to move from 1 to 0.99999..., you have to stay at 1.

Which means they must be the same thing!

Here's another way to think about it. When you write something like

0.35

that's really the same as 35/100,

0.35 = 35 / 100

right? Well, you can turn that into a repeating decimal by dividing by
99 instead of 100:
__
0.35353535... = 0.35 = 35 / 99

Play around with some other fractions, like 2/9, 415/999, and so on,
to convince yourself that this is true. (A calculator would be
helpful.)

In general, when we have N repeating digits, the corresponding
fraction is

(the digits) / (10^N - 1)

Again, some examples can help make this clear:
_
0.1 = 1/9
__
0.12 = 12/99
___
0.123 = 123/999

and so on.

So, here's something to consider: What fraction corresponds to
_
0.9 = ?

It has to be something over 9, right?
_
0.9 = ? / 9

The _only_ thing it could possibly be is
_
0.9 = 9 / 9

right? But that's the same as 1.

Ultimately, though, this probably won't _really_ make sense until you
come to grips with what it means for a decimal to repeat _forever_,
instead of just for a r-e-a-l-l-y l-o-n-g t-i-m-e.

When you think of 0.999... as being 'a little below 1', it's because
in your mind, you've stopped expanding it; that is, instead of

0.999999...

you're _really_ thinking of

0.999...999

which is not the same thing. You're absolutely right that 0.999...999
is a little below 1, but 0.999999... doesn't fall short of 1 _until_
you stop expanding it. But you never stop expanding it, so it never
falls short of 1.

Suppose someone gives you $1000, but says: "Now, don't spend it all,
because I'm going to go off and find the largest integer, and after I
find it I'm going to want you to give me $1 back." How much money has
he really given you?

On the one hand, you might say: "He's given me $999, because he's
going to come back later and get $1."

But on the other hand, you might say: "He's given me $1000, because
he's _never_ going to come back!"

It's only when you realize that in this instance, 'later' is the same
as 'never', that you can see that you get to keep the whole $1000. In
the same way, it's only when you really understand that the expansion
of 0.999999... _never_ ends that you realize that it's not really 'a
little below 1' at all.
*******************************************************
The Infamous .999... = 1

Date: 01/12/2002 at 08:14:36
From: Dan T
Subject: The Infamous .999... = 1

Dear Dr. Math,

I have seen answers to the question that .9999... = 1, but some
people in my 8th grade class STILL don't agree. My geometry teacher
says that it is all about giving infinity a value. A student in the
class says I am trying to prove that 1 = 1/infinity.

Can you straighten this out? Thanks.

Dan T.


--------------------------------------------------------------------------------


Date: 01/12/2002 at 09:27:23
From: Doctor Tom
Subject: Re: The Infamous .999... = 1

Hi Dan,

I assume you've seen the Dr. Math FAQ:

0.9999... = 1
http://mathforum.org...faq.0.9999.html

If not, take a look.

Rather than saying "giving infinity a value," it's perhaps a bit
clearer to say, "giving the concept of a limit of an infinite sequence
of numbers a value."

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.

There is nothing special about .999... The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.

And it doesn't limit itself to single repeated decimals. When we say:

1/7 = .142857142857142857...

none of the finite parts of the decimal is equal to 1/7; it's just
that the more you add, the closer you get to 1/7, and in addition, 1/7
is the UNIQUE number that they all get closer to.

Finally, you can show for all such examples that doing the arithmetic
on the series produces "reasonable" results:

Since:

1/3 = .333333...
2/3 = .666666...

1/3 + 2/3 = .999999... = 1.

By the way, there is nothing special about 1 as being a non-unique
decimal expansion. Here are a couple of others:

2 = 1.9999...
3.71 = 3.709999999...
2.778 = 2.77799999999999...

...and the student who says you're trying to show that 1 = 1/infinity
is wrong.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/


--------------------------------------------------------------------------------


Date: 01/12/2002 at 16:40:11
From: Dan T
Subject: The Infamous .999...=1

Sorry, the equation that the student meant was:

1 = 1 - 1/infinity


--------------------------------------------------------------------------------


Date: 01/12/2002 at 16:48:09
From: Doctor Tom
Subject: Re: The Infamous .999...=1

Hi Dan,

Then he's basically right. As you add each new 9 to the expansion, the
errors look more and more like:

1/100, 1/1000, 1/10000, ...

Thus, in a sense, the error begins to look like "1/infinity," which
semms as if it should be zero.

The 1/infinity is meaningless, but the concept of limit is not. We can
say that the limit of the sequence above is zero, and can rigorously
prove it.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/
*******************************************
Sorry about this being so long, and thanks to Dr. Math

[Xilenx]

oh wow, lol you told me.

Well, I also presented the post modern theory, which is why my outlook is more philosophical than through mathematical rounding.

First off, everything after the first two solutions is based around rounding and fractionalizing (i doubt thats a word, but basically converting to a fraction, which does not solve the question of whether or not it is equal to 1.)

But with your first two formulas..well mainly your second, the first one still needs to be broken down to me, because i hate math. so this is going outta my way to push a point.

Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

Basically, you've just put a cap on infinity on this formula, and you assuming that infinite would be reach, however, you are still missing the .000...1, however since it is infinity, the 1 in this formula is a paradox. it exist yet it doesn't, and if you accept it doesn't then the infinity continues (which is does), however, when you accept that fact, you also accept the fact that the 1 does exist, yet you dont b/c it continues, so its confusing like that. however, this formula does not work out b/c you cannot logically take 1 off of infinity, because infinity is forever, hence, you cannot just - x as simply as that.
---
some place down further...

0.1 = 1/9
__
0.12 = 12/99
___
0.123 = 123/999

last time i checked 0.1 = 1/10, and 0.12 = 12/100, etc.

thats just furthering the "rounding process" of math, which is self defeating because arithematic is logical, and logic must make sense, in which you cannot cut corners and you cannot chance anything. hence, the entire concept of math is flawed.
---
.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

that explains what i believe it, somewhere in your post. you do get infinity closer, i grant you that, but you never get 1.

[WEBS]

(let S stand for sigma)

infinity
S (1/2) ^ i
i=0


you would have:
(1/2)^0 + (1/2)^1 + (1/2)^2 + ... and so on.

It seems like there is no answer, since the addition lasts infinitely, however the accepted mathematical solution to that is 2

1/(1-r) where r is the constant, is the accepted formula for solving such geometric sequences (and Im sure there is a derivation for it, but i wont go into that)

therefore: 1/(1-1/2) = 2

PS: I hope I did everything correctly here. Please correct me if I didnt.
Also this is just an example where something that doesn't seem solvable because of an infinite quality can in fact be solved.
And lastly, I remember reading an explanation for why .9999... does not equal 1 and that there is a flaw in the general mathematical logic that says it does, however that explantion may be wrong

[BKBuffyGurl]

I originally thought that 1 would be greater than .99999 but special_angel's many explanations seem to hold a lot of truth. Now all I have to say about this is only in Tech. Barely anywhere would one find this the topic of debate. But I appreciate that, I guess it reflects our level of inteligence.

[Xilenx]

I dont know of thats a tech issue, more or less just a boredom issue..

[z2z007]

heh, I learned 1/2 of these proofs when I was in 6th grade since I went to Mark Twain and had Mr. Herman for my Math/Computer Talent teacher.

He used go like BONUS CARD! whenever sum1 figured out sumthing hehehe...

Aw now I know why math class is sooo boring at Tech! lol

[Neo]

Seeing how all the solutions to this questions are being presented purely in numbers, I thought I'd take a different approach...the abstract approach. When thinking about numbers decreasing "exponentially" in a downwards curve on a graph, it occurred to me that there is a physical structure in the universe that follows the rules of its mathematical counterpart: a black hole. So then, if anything falling into a black hole is reduced in size exponentially to infinity, I came up with two possible theories.

1. If a black hole does in fact have a corresponding inverted funnel, anything that decreases on one side would logically increase on the other. Likewise, the exponential reduction of a value into infinity would eventually return to its original value. This behavior is somewhat similar to a universe that is "curved," thereby making it that traveling in a single direction will only lead back to the starting point. Thank you Einstein.



2. A black hole does not have a corresponding inverted funnel and everything continues to fall inwards. Because time slows down as you fall deeper into the hole, from the standpoint of anything falling deeper and deeper, the universe (everything outside the black hole) will accellerate in its development. In light of the prediction that the universe will collapse in on itself (the "Big Crunch") and once more begin to expand (the "Big Bang"), one can only assume that time will stop/begin with it. With this concept in mind, anything would that occurs in our timeline before the crunch itself would be inverted in the next expansion. This behavior is similar to the very old television sets where the image is projected onto the screen in a way that reverses it in appearance. Now imagine that reversing effect happening to all matter and energies in the universe.

Basically, 0.999n comes out to equal itself and only itself.

[Xilenx]

That is one of the most awkward answers I've ever heard of, but it agrees with what I said, so its all good.

---

However, one point about time slowing down as you pass through the funnel of a black hole, that kind a goes with the post modern theory, and that the other side of the funnel is never reached, b/c you can infinitely slow down, but you can never actually stop.

[Neo]

Quote

However, one point about time slowing down as you pass through the funnel of a black hole, that kind a goes with the post modern theory, and that the other side of the funnel is never reached, b/c you can infinitely slow down, but  you can never actually stop.

Come to think of it, that is a very valid point. And yet, taking that point into consideration, you still cannot rule out theory #1, but you can merge them. Observe...

Lets imagine that we have 2 people. Person A is falling into a black hole. Person B is stationary in a position just outside the reach of the gravitational field. Based on the nature of time distortion inside/outside the black hole, Person B will see Person A as slowing down, and Person A will see Person B as accelerating. Now, keep in mind that the fact of the matter is Person B is not the only thing that will be accelerating in the eyes of Person A; the entire universe will appear to be accelerating. At the same time, the universe is also approaching the Big Crunch, during which everything will converge into a single point. After that will come a Big Bang, where the universe will expand in a way that we would consider to be "inverted," including time itself and everything that occurred in chronological order. What ends up happening is that Person A, who was going INTO the black hole during the previous iteration of the universe will actually be coming OUT of it in the next. The same would go for 0.999n. If you think about it, the diagram I made applies to both theories. Essentially, the two funnels are connected by the inversion of the universe, an operation involving a Big Crunch followed by a Big Bang.

[z2z007]

There are alot of interesting things being brought up, but isn't this supposed 2 be a humor thread?

[Neo]

Perhaps you might find it funny that I'm sitting here discussing matters of theoretical physics without ever having studied anything beyond SQ2 math.

[Xilenx]

Hmm, if person A has stepped into the black hole, and B is right outside of its gravitational pull, then in the beginning, won't both subjects see each other as accelerating? Since A is being pulled, and B is stationary, they inversely see one another accelerating, along with the entire universe through person A's eyes. Which is kind of confusing, because when is the exact time that the person will begin to decelerate and slow down? However, let's assume that once person A begins his entrance into the funnel, they already begin to slow down.

This is where i don't agree with you; and that's when person A reaches the crunch point, because person A will only infinitely slow down, yet never reach crunch point, hence will never invert and come out the other end.

The Black Hole is such a vague concept that it breaks down the laws of physics, and onto the laws of philosphy. To me, I don't see how scientists can assume that the gravitational pull exists, since we've never been that close to one (i might be wrong, a website would prove me so), and that there is this crunch point, as well as an inversion of the universe. There is no law of physics when we consider the Black Hole, because it breaks down numbers and physical boundaries, into subjective concepts. We can continue to theorize about it, (which is probably what they do with the crunch, etc), but even then, that is very specific especially for a concept way too broad that we should not even assume that much.

In the end .999n = .999n, and not 1.

[Neo]

Quote

This is where i don't agree with you; and that's when person A reaches the crunch point, because person A will only infinitely slow down, yet never reach crunch point, hence will never invert and come out the other end.

I think you got two concepts confused. The "Big Crunch" is where the entire universe (including time) collapses into a single point. The correlation between the Big Crunch and what happens inside a black hole can be thought of in this way:

As it approaches being compressed into a single point, the universe (including time) is a sphere. A black hole would be a spot somewhere on the surface of that sphere (it doesn't matter if it's on the inside or outside, since the black hole is one-dimensional and so is the surface of the sphere). So then, once the sphere shrinks down, imagine it inverting. Not only would space be inverted, but so would time, and as a result, everything that happened. As for the black hole, it too would be inverted, and thus would be on the "opposite" side of the sphere, in the opposite position, and with everything moving the opposite way (outwards). In fact, you can think of time slowing down inside a black hole as being a kind of "buffer" that saves anything that falls into it by preventing it from ever reaching the true center. If it were not for that buffer, and everything that fell in got compressed into a single dimension, it would not be possible for that to be undone when the universe inverts because it would just stay 1-dimensional like it was right before.

Consider the effect in a two-dimensional model (sorry, too lazy to make a fancy illustration):
[b][size=24]>

The top-left and bottom-left (red) lines would be space and time, respectively. As you follow the lines to their joining point (Big Crunch), you will see that beyond the convergence, they will continue to go on, but will be "flipped" in such a way that the top-right (blue) line will be time, and the top-right (blue) line will be time.

[special_angel]

Xilenx, a lot of your argument confuse me; you seem to think that 0.999... can't be worked with mathematically because it contains an infinite number of 9's, per your quote

Quote

however, when you accept that fact, you also accept the fact that the 1 does exist, yet you dont b/c it continues, so its confusing like that. however, this formula does not work out b/c you cannot logically take 1 off of infinity, because infinity is forever, hence, you cannot just - x as simply as that.

0.999... is not infinity, it mearly contains an infinite number of digits. 0.999... is no harder to subtract from than 0.333..., which equals 1/3. One thing which I think was included in my numerous proofs was 1/3 + 2/3= 9/9= 1.
So what does 0.333... + 0.666... equal? 0.999?

[Xilenx]

Neo, so does that mean the black hole is only a universal vacuum for both the first and the second universe? I just think that the theories developed on the black hole seem questionable, although i don't know of any other, because it's suppose to slow down time, and eventually stop, then end up on the other side of the black hole, speed up, and eventually reach normal speed, correct?

Now, since the black hole is a mysterious concept, I apply the post modern idea to it, saying that time and everything in the universe can only slow down, never stop, and it can infinitely slow down, and infinitely approach the other end of the black hole, but never actually make contact with the middle, or the other end.

However, now arguing about the black hole theory, it gets confusing, because it sounds as if it has nothing to do with the .999n thing anyway.
---
Special_angel, I said earlier that when you "fractionalize" or turn decimals + numbers into a fraction, you've changed it, b/c .333n only equals 1/3 when you round or in terms, you get lazy, because the raw value of .333n will never equal a fraction, and will never end. hence, 1/3 + 2/3 = 9/9, but
.3333n + .6666n = .9999n, and never 1.

However, if the number is .333 + .666 then that will equal .999 because you can see the end of that number.

I said it was infinite, in the sense, that you cannot subtract off of an infinite value. It's like saying im going to reduce 1.222222222 and so on, by 1.222222222 and so on, however, with that idea, you will never get 0, because you cannot subtract from an infinite value. It's infinite. You'll never find the end of the entire number.

[Neo]

You can do that kind of operation logically, like a computer, and it would never be completed (operation time-out) because you would have to process each number individually before performing the operation.

Or, you can think of it in abstract terms, like humans, and would reason that ANY value subtracted from itself would result in no value.

[Xilenx]

We could use humans, but there is an objective, clear number of how many humans exists, and it is not infinte. There is a limit on the human race, and if we really want to, we can count each individual, and come to a number.

However with infinity, you cannot even do it on a computer.

What's infinity plus infinity? You cannot logically add these values, because they will never end. Infinity is already on going, and you will never get an answer.

The same goes with other forms of arithmetic, subtraction, mulitplication, division. It is just logically impossible to add or take away value of infinity or add or take away from a nonstop going number order.

Assuming you can, infinity - infinity will somehow get us 0, then by that, you assume that infinity does end somewhere, and that you can calculate it, therefore, contradicting the definition of what infinity is in the first place.

[wretched]

We're not talking about infinity...we're talking about a number with infinite digits. In this case, there are some cases where if we subtract a number with infinite digits by itself, the result is zero.

9.999999...
-9.999999....
0