More on 0.999~ vs 1: Comparing Numbers

[BSMSMSTMSPHD]

Here's how I've always though of it (with details left out for the sake of brevity):

If two real numbers are different, then you can find another real number between them (their average, for example). So, assume .999... and 1 are different numbers. Can you find a number between them? Since you cannot, the numbers are therefore the same.

 

[dansydney]

On the side a bit, is 1/3 equal to 0.3333 because the 3's are infinite, the same with 2/3 the 6's are infinite. Are 1/3 and 0.333(repeater) really the exact same number? 1/3 is one third of something whereas 0.333 is 0.333 of something and nomatter how many 3's are added it will not give the same result as 1/3 because you cannot use infinite? Not sure if you will understand what I am trying to say but feel free to make me look stupid. Cheers dansydney

 

[HallsofIvy]

On the side a bit, is 1/3 equal to 0.3333 because the 3's are infinite, the same with 2/3 the 6's are infinite. Are 1/3 and 0.333(repeater) really the exact same number? 1/3 is one third of something whereas 0.333 is 0.333 of something and nomatter how many 3's are added it will not give the same result as 1/3 because you cannot use infinite? Not sure if you will understand what I am trying to say but feel free to make me look stupid. Cheers dansydney

Okay, I'll feel free! "it will not give the same result as 1/3 because you cannot use infinite" is incorrect. You don't have to "use infinite" because an "infinite number of digits" arises only in the base 10 representation of a number and has nothing to do with the number itself.
Once again, the definition of a base 10 representation of a number is that 0.a₁a₂...a_n... is the limit of the sequence 0.a₁, 0.a₁a₂, ..., 0.a₁a₂...a_n,... (not the sequence itself- that's the confusion that leads to such nonsense).
It's easy to prove that the limit of such a sequence exists and is a real number (it's obviously an increasing sequence and has upper bound 0.a₁+ 0.1). It's even easier to prove that 0.333... which, by definition, means the limit of the sequence 0.3, 0.33, 0.333, 0.3333, ... is the limit of the sequence of partial sums of the geometric series
\Sum_{n=0}^\infty 0.3(0.1)^n
and so has sum \frac{0.3}{1-0.1}= \frac{1}{3}.

Similarly, 0.9999... means the sum of the geometric series
\Sum_{n=0}^\infty 0.9(0.1)^n
and so has sum \frac{0.9}{1-0.9}= 1.

 

[HallsofIvy]

Here's how I've always though of it (with details left out for the sake of brevity):

If two real numbers are different, then you can find another real number between them (their average, for example). So, assume .999... and 1 are different numbers. Can you find a number between them? Since you cannot, the numbers are therefore the same.

Can you prove that there is no number between them? The fact that I can't do something hardly proves that it can't be done! (I've proven that many times!)

 

[dansydney]

I sort of get what your saying. I am only a yr11 student so I haven't covered all the limits topics thus far.

 

[Robokapp]

Imagine the fraction...(x^2-x)/(x-1)=0
It can be simplified in x(x-1)/(x-1)=0

Find the limit as x -> 1

lim- = 1
lim+ = 1
lim exists therefore at x=1
What is the value of the function at x=1? There isn't any bc 1 is not a ponit of continuity (i wish i knew how to write fancy the lim as x-> a of f(x)=f(a) proves that point x, f(x) belongs to f(x) and it is a ponit of continuity)

So...0.99999999999 is part of the function and one/infinity of a unit to the right, 1 is NOT a point that belongs to the graph. Sorry to break your heart.

Also think this way. To be equal, in the true equality meaning, it must have the exact same characteristic...like a triangle. Equality in two traingles means all sides, angles, perimeter, area...whatever is identical...basically if you don't label them you can't tell the difference.

Well I can tell the difference between 1 and 0.999999... which no one here can even write fully...I know this is not math but it's 6:30 am lol.

 

[HallsofIvy]

So...0.99999999999 is part of the function and one/infinity of a unit to the right, 1 is NOT a point that belongs to the graph. Sorry to break your heart.

I have absolutely no idea what this means. What does it mean to say that "0.99999999999 is part of the function" and what is "one/infinity of a unit"?

I certainly agree that "1 is NOT a point that belongs to the graph."- 1 is not a point at all! If you mean that there is no point with x-coordinate 1 on the graph of f(x)= (x²-1)(x-1) (I assume that's what you meant. You never did say what f was.), that's true. f(x) = x+1 for x not equal to 1 and is not defined at x=1. But I don't see what that has to do with whether 0.9999...= 1. There is no point with x= 0.999... on the graph.

 

[arildno]

Robokapp:
I have no idea, either!
I thought the real number "1" was an equivalence class of (a subset of) rational sequences with respect to a particular equivalence relation, but obviously, I'm mistaken.

 

[neophysique]

I think there's something illogical about comparing a number that
is not complete (and which can never be complete) like 0.999... to
a number that is completed, like 1. It's no different than someone
claiming to know the biggest number in the infinite set of all integers.
And wouldn't make any less sense if someone also claimed that number
to be 1.

It's like there's a 1 meter track. A runner (the infinite sequencer)
is programmed to keep shrinking and shrinking forever as he runs
that 1 meter track. He by definition can never finish the track,
though it started only as 1 meter long when he was "fully" sized.
For all practical purposes, the track has became infinitely long
to the runner and not 1 meter.

So how does one compare a track that is .999... meters long (one
that can never be completed) with a 1 meter long track ran
by a "normal" runner?

 

[matt grime]

I think there's something illogical about comparing a number that
is not complete

I think there is something illogical about using the word complete without defining what it means for a number to be complete (or incomplete).

 

[StatusX]

The key point about 0.999... is that it is defined as:

0.999... = \lim_{n \rightarrow \infty} \sum_{k=1}^n 9 \frac{1}{10^k}

When people say things like "it can never be complete," I can only imagine that they are trying to visualize a number with many many 9's. This is not what 0.999... means. It is true that for any number n, the above sum is less than 1. It is also true that as n gets larger and larger, the numbers get closer and closer to 1 (and not to any other number), and this is what we mean when we say the limit of this series is 1. In fact, infinity is only really defined in terms of limits. When someone says:

\sum_{n=1}^\infty a_n = L

They are really just using shorthand for:

\lim_{N \rightarrow \infty} \sum_{n=1}^N a_n = L

In turn, an infinite limit like:

\lim_{n \rightarrow \infty} a_n = L

just means that for any e>0, no matter how small, there is some N (which will in general be larger and larger as e gets closer to 0) such that for all n>N, |a_n-L|<e, that is, the terms get arbitrarily close to L. Note that every step in the definition of infinite limits and sums involves normal, finite numbers. You never have to stretch your mind thinking about infinity if you just remember that it is usually just a shorthand for these straighforward definitions.

 

[Robokapp]

\lim_{n \rightarrow\ &quot;1&quot;} \ (x^2-x)/(x-1)

Okay, I'll try to write it the fancy way. Question is...is 0.9999999 = 1
Well f(x)=x(x-1)/(x-1) so the graph looks just like f(x)=x but it's having a hole at x=1 Correct?

Well point x=0.99999 does belong to the graph. Point x=1 does NOT.
Point 0.99999... is where the \lim_{n \rightarrow\ &quot;1-&quot;} occurs. It's a real point, (0.(9), 0.(9)) on the graph. Point for x=1 does not have a y-value. big difference between the two!

 

[Hurkyl]

Point 0.99999... is where the \lim_{n \rightarrow\ &quot;1-&quot;} occurs.

As you've pointed out before,

\lim_{x \rightarrow 1^-} f(x) = 1

Now, for what value of x are you claiming that f(x) = 1?

 

[matt grime]

\lim_{n \rightarrow\ &quot;1&quot;} \ (x^2-x)/(x-1)

Okay, I'll try to write it the fancy way. Question is...is 0.9999999 = 1

no that is not the question. 0.9999999 is not equal to 1. 0.99... is equal to one (as base 10 representations of real numbers).

 

[HallsofIvy]

I think there's something illogical about comparing a number that
is not complete (and which can never be complete) like 0.999... to
a number that is completed, like 1.

I wanted to start this "I think there's something illogical about" your post but Matt Grime beat me to it!

It's no different than someone
claiming to know the biggest number in the infinite set of all integers.
And wouldn't make any less sense if someone also claimed that number
to be 1.

Well, yes there is a difference! There is no such number. There IS such a number as 0.9999... and apparently even you agree to that. Do you insist that there's a "biggest number" lest than 1? If not, what such numbers are larger than 0.9999...?

It's like there's a 1 meter track. A runner (the infinite sequencer)
is programmed to keep shrinking and shrinking forever as he runs
that 1 meter track. He by definition can never finish the track,
though it started only as 1 meter long when he was "fully" sized.
For all practical purposes, the track has became infinitely long
to the runner and not 1 meter.
So how does one compare a track that is .999... meters long (one
that can never be completed) with a 1 meter long track ran
by a "normal" runner?

No, it's not the same. You are making the same mistake I mentioned before: 0.9999... is not the sequence 0.9, 0.99, 0.999, ..., (which you could think of as "never ending"), it is the limit of that sequence and as far as 'never ending' is concerned, no different from the numbers 2 or 3 or any other number.

 

[Robokapp]

As you've pointed out before,

\lim_{x \rightarrow 1^-} f(x) = 1

Now, for what value of x are you claiming that f(x) = 1?

I'm so confused I don't even know if I'm lost or not but...What i am claiming is that my equation is a y=x scenario with a hole for x=1. basically i created a straight line with the domain and range of {R|x or y not = 1}

so the 0.999 is there, fine and well but 1 is not. I'm not really sure what the question of the topic is since now I see some stuff I didn't read...but if you're talking about approximation, my approach would be that since you can't write all those 9s in there, you got to round down or up. rounding is closer to up into 0 than down into a zero...so one 9 somewhere down the lnie turns into a 0...cascading the others into 0s until the units turn into a 1.


Edit: hurkyl you're post 6999 :D write something special for 7000 and mention my name for noticing you hehe.

 

[Integral]

I'm so confused I don't even know if I'm lost or not but...What i am claiming is that my equation is a y=x scenario with a hole for x=1. basically i created a straight line with the domain and range of {R|x or y not = 1}

so the 0.999 is there, fine and well but 1 is not. I'm not really sure what the question of the topic is since now I see some stuff I didn't read...but if you're talking about approximation, my approach would be that since you can't write all those 9s in there, you got to round down or up. rounding is closer to up into 0 than down into a zero...so one 9 somewhere down the lnie turns into a 0...cascading the others into 0s until the units turn into a 1.


Edit: hurkyl you're post 6999 :D write something special for 7000 and mention my name for noticing you hehe.

Unfortunately your function says NOTHING about the equaltiy of 1 and .999... . You have stated but not proven that .999... is "there"... What ever that means.

edit:
OK I guess I will post this...Again...http://home.comcast.net/~Integral50/Math/proof2a.pdf"

 

[MalayInd]

about the equaltiy of 1 and .999... .

I will say hat 0.999... is equal to 1 if the the trailing dots indicate that the number of repeating nines is greater than any natural number.
1=9/9
Now divide 9 by 9 by taking the quotient 0 rather than 1
You will get 9 as the remainder, then we follow it by zero and thus our quotient builds to 0.999... (a never ending repitition)
I would like to have your comments on this.


Keep Smiling
Malay

 

[Integral]

I will say hat 0.999... is equal to 1 if the the trailing dots indicate that the number of repeating nines is greater than any natural number.
1=9/9
Now divide 9 by 9 by taking the quotient 0 rather than 1
You will get 9 as the remainder, then we follow it by zero and thus our quotient builds to 0.999... (a never ending repitition)
I would like to have your comments on this.


Keep Smiling
Malay

Yes ( ... ) is called an ellipsis, it denotes the repetition of a pattern endlessly. So .999... means repete the pattern 999 endlessly.

I have seen the 9/9 division before, it is yet another Demonstration of the equality.

 

[shmoe]

I have seen the 9/9 division before, it is yet another Demonstration of the equality.

Once you've proven that doing division like this will end up giving you a decimal representation of your number of course.

The same will go for the proof using 3*(1/3)=3*(0.333...) etc. While swell that this seems to convince a lot of people, it's often overlooked why we are justified performing operations on non-terminating decimals like this.

 

[Integral]

Notice that I did not use the word proof! I specifically said DEMENSTRATION. This would imply something less then a proof. I have never seen anybody, suddenly accept that 1= .999... simply because someone showed them a different infinitely repeating number.

 

[Hurkyl]

You have a process that never ends.

There's your problem. 0.999~ is a number, not a "process".

 

[neophysique]

There's your problem. 0.999~ is a number, not a "process".

So how many nines does 0.999... has, if it is a number?

 

[arildno]

So how many nines does 0.999... has, if it is a number?

A completely irrelevant question.

 

[matt grime]

I didn't agree that 0.9999... is a number

Well, it is a representation of a real number, so you're reasoning from a faulty premise in the first place.

I think the statement 0.9999... = 1 then is like comparing
a movie that is ever in the making with one that is out
in the theaters. Sure, if the movie in the first case is
ever completed (but it can't because of a director who
is the utimate perfectionist) it might be identical to
the one in the theaters now but since it will never
be completed, it's just as logical, maybe even more so
to say that it will be nothing like the one in the theaters
now.

I think that it is a mathematical question, so reasoning mathematically will get you the answer, not using spurious analogies.

So The argument that the limit of the function 0.9999...

it is not a function, it is a representation of a real number.

approaches
one by no means proves they are equivalent. In fact, one can
argue just as validly that since 0.9 < 1, 0.99 <1 , 0.999 <1 , 0.9999 < 1,
regardless of how many 9's we add to the sequence,

Why does some property about a finite number of nines after the decimal place tell you a priori anything about the string with an infinite number of 9s after the decimal place?

the function 0.999... will always generate a number less than 1.

it is not a function.

So the expression 0.9999... = 1 makes no sense to me in two ways.

what it means in your opinion is not relevant since you have not related your opinion to any of the mathematics of what these symbols represent.

First, it seems to be comparing a function process with the value
of a function.

this opinion is wrong.

Second, the limiting process leads to the conclusion
that 0.9999... approaches 1 but is always less than 1.

0.999... does not approach anything, just as 0.5 does not approach 1/2.

Logically one can see it as this:

You have a process that never ends.
In order to assign a finite value to this function,
you have to end the process, contradicting
the premise that you have a process that never ends.

That is not at all what is going on, logically or otherwise.

 

[HallsofIvy]

I'm so confused I don't even know if I'm lost or not but...What i am claiming is that my equation is a y=x scenario with a hole for x=1. basically i created a straight line with the domain and range of {R|x or y not = 1}

so the 0.999 is there, fine and well but 1 is not. I'm not really sure what the question of the topic is since now I see some stuff I didn't read...but if you're talking about approximation, my approach would be that since you can't write all those 9s in there, you got to round down or up. rounding is closer to up into 0 than down into a zero...so one 9 somewhere down the lnie turns into a 0...cascading the others into 0s until the units turn into a 1.

If you really mean "0.999" then, yes, it is on the graph (I'm giving up repeatedly pointing out that what you really mean is "a point with x coordinate 0.999 is on the graph") but irrelevant to the discussion of
0.999... (infinitely repeating 9s).

If you really mean 0.999... is on the graph, then you have repeatedly asserted that without giving any reason. Asserting that 0.999... is on the graph but 1 is not is just asserting that 0.999... is not equal to 1 which is the whole question. Do you have a proof of that?

 

[shmoe]

Notice that I did not use the word proof! I specifically said DEMENSTRATION. This would imply something less then a proof.

I would never have imagined that implication, but alright (I also didn't think you would have believed it was a rigorous proof/demonstrartion/arm-waving/etc, and just wanted to point it out to the viewers)

 

[Robokapp]

Well...if you want a number in between 0.(9) and 1.(0)... just do the following:

(1-0.99999.)*1/2 :lol:

 

[shmoe]

Well...if you want a number in between 0.(9) and 1.(0)... just do the following:

(1-0.99999.)*1/2 :lol:

I assume you mean (1+0.99999.)*1/2?

Your next task is to prove this is strictly great than 0.999... and strictly less than 1 *without* assuming that 0.999... is strictly less than 1.

 

[Robokapp]

yeah. +. sry about that. well...prove that 0.(9) is < than 1.(0) is simple...let's assume you take infinitely many sig figs in both numbers.

The largest spot sig fig place in 1.(0) is the units. The first sig fig place for
0.(9) is the tenths of units. If you consider that 0 a sig fig...0<1.

I don't really know how else to prove that 0.(9) < 1.(0) because as my grandfather used to say..."it just is". so the average mean has to be... 0.(9)<x<1.(0) assuming x is the wrid number we're looking for.

I am very curious to see the application of this number that is half of an infinity of a unit less than 1 but...Math doesn't always reflect real life.

 

[HallsofIvy]

yeah. +. sry about that. well...prove that 0.(9) is < than 1.(0) is simple...let's assume you take infinitely many sig figs in both numbers.

The largest spot sig fig place in 1.(0) is the units. The first sig fig place for
0.(9) is the tenths of units. If you consider that 0 a sig fig...0<1.

I don't really know how else to prove that 0.(9) < 1.(0) because as my grandfather used to say..."it just is". so the average mean has to be... 0.(9)<x<1.(0) assuming x is the wrid number we're looking for.

I am very curious to see the application of this number that is half of an infinity of a unit less than 1 but...Math doesn't always reflect real life.

Your ability to write nonsense is amazing. First, "significant figures" apply to measurements in applications of mathematics and have no significance in abstract mathematics. Second, saying "it just is" means "I have no idea but I can talk loud". Third "half of an infinity of a unit" is meaningless. Yes, "math doesn't always reflect real life". You, on the other hand, reflect neither real life nor mathematics.

 

[shmoe]

The largest spot sig fig place in 1.(0) is the units. The first sig fig place for
0.(9) is the tenths of units. If you consider that 0 a sig fig...0<1.

I have no idea how this is supposed to be relevant to showing x is greater than 0.999... or less than 1.

I don't really know how else to prove that 0.(9) < 1.(0)

First, you haven't proven it. Second, it's not true.

Take the time to learn a proper definition of what 0.999... and other decimals actually is. There's lots of posts on this topic, go find them and read carefully.

I am very curious to see the application of this number that is half of an infinity of a unit less than 1 but...Math doesn't always reflect real life.

"half of an infinity of a unit" is meaningless rubbish.

 

[Robokapp]

:( I'm sorry. I have the bad habit to assume what I think is right and what makes sense to me to actually be right. Well when I was in grade school my grandfather used to help me w/ math and...he'd not take time to expalin stuff so he'd just say "what do you mean why? it just is". Well after a few years of that...you rememeber it more than the reasoning process.

Once again...sorry. I'll try to shut up more.

 

[scott_alexsk]

Am I incorrect in the premise that 0.9... is an irrational number?
-scott

 

[d_leet]

Am I incorrect in the premise that 0.9... is an irrational number?
-scott

Yes you are since it is equal to 1, And 1 is certainly a rational number.

 

[HallsofIvy]

Am I incorrect in the premise that 0.9... is an irrational number?

scott_alexsk: any repeating decimal is rational. Even if you didn't know that 0.9...= 1, the fact that it is repeating tells us that it is rational.

 

[Curious3141]

Your ability to write nonsense is amazing.

Sorry for the OT remark, but I just wanted to say that *that* made me LOL. Literally. I'm still LOL-ing, I don't know why I find it so funny, it just seems so deadpan.

 

[scott_alexsk]

Well my thought was 0.3... and 0.6... are rational since they can be written as fractions, 0.9... cannot since 3/3=1. But I guess I was wrong.

-scott

 

[arildno]

Eeh? Isn't 3/3 a fraction??

 

[Robokapp]

Isn't an irrational number a number where the next decimal does not repeat in a pattern and there are infinitely many? or something like that?

that's not a deffinition, that's my interpretation but...a number that repeats 9s is rational. it's not natural, nor whole, but it belongs to rational numbers...right?

 

[StatusX]

Rational numbers are real numbers which are equal to p/q for some (and in fact infinitely many) integers p and q. That they have repeating or terminating decimal expansions is a secondary fact about them, but is not a good property to define them by, simply because base 10, or any other base, is completely arbitrary. 0.999... is a rational number because 1 and 1 are integers, and 0.999...=1/1. By the way, if you agree 0.999... is rational, what are some integers p,q with p/q=0.999...? Remember, all integers are finite numbers (neither infinity nor infinity-1 is an integer).

 

[d_leet]

Well my thought was 0.3... and 0.6... are rational since they can be written as fractions, 0.9... cannot since 3/3=1. But I guess I was wrong.

-scott

Ok but you do know that the sum of two rational numbers will still be rational right, so then notice that .9... = .6... + .3...

 

[chroot]

Well my thought was 0.3... and 0.6... are rational since they can be written as fractions, 0.9... cannot since 3/3=1. But I guess I was wrong.

-scott

That's sort of the point kiddo. The fraction 3/3 can also be written as 1, and can also be written as 0.999...

- Warren

 

[HallsofIvy]

Isn't an irrational number a number where the next decimal does not repeat in a pattern and there are infinitely many? or something like that?

that's not a deffinition, that's my interpretation but...a number that repeats 9s is rational. it's not natural, nor whole, but it belongs to rational numbers...right?

No, what just about every one here is saying is that 0.999..., infinitely repeating 9s is a natural number and a whole number (as well as an integer). It is exactly equal to 1.

 

[shmoe]

That they have repeating or terminating decimal expansions is a secondary fact about them, but is not a good property to define them by, simply because base 10, or any other base, is completely arbitrary.

To expand a little- whether a rational number is terminating or repeating depends on base. eg. 1/5=.2 in decimal but .001100110011... in binary.

However, non-terminating and non-repeating in one base, non-terminating and non-repeating in all bases, so you can use this to define irrational/rational, though the p/q version is nicer.

 

[HallsofIvy]

I remember, years ago, a person asking on a math forum, "how to prove that a rational number can be written as a fraction". I stared at that, dumbfounded, for a while before realizing that he must have learned "can be written as a repeating or terminating decimal" as the definition of rational number.

 

[Robokapp]

No, what just about every one here is saying is that 0.999..., infinitely repeating 9s is a natural number and a whole number (as well as an integer). It is exactly equal to 1.

I may lack the capability to explain it...but if it is true, assuming it is...you can't write your number. 0.99999 will eventually be rounded...up or down. if you round it down you end up with a lot of 9s...but not all of them. if you round it up you end up with a lot of zeros but not all of them...and a 1 in front. Out of comodity you don't write 1.00000...and that is correct. you don't have to. but you're not getting away when the 0.999... comes.

The way I see it is like that proof that 1 = 2. all you got to do is divide by (a-b) after you were told that a=b. yeah, 1 does = 2 once you assume 0 divided by 0 = 1.

So until you actaully write the 0.999... number it does not matter what it's equal to. It's not only unreal but inexistent, that's the point. You can write parts of it and talk about the hole, but what you are NOT writing is the tiny difference between many 9s and many 0s.

Well I probably pissed a lot of people off with what I just wrote but that's how i see it. You're comparing apples with photos of oranges and they're almoust the same...only problem is...one of them doesn't really exist.

Well it does, you just can't express it. The difference between 0.999... and 1.000... is so far back in the decimals that you don't get to write it ever. Do you know what it looks like? No. You got no idea. It's like graphing an asymptote. It goes up and ends below the starting point. you can come infinitely close to it, but you're never gonig to reach it...

That's my best approach.

 

[FrogPad]

I may lack the capability to explain it...but if it is true, assuming it is...you can't write your number. 0.99999 will eventually be rounded...up or down. if you round it down you end up with a lot of 9s...but not all of them. if you round it up you end up with a lot of zeros but not all of them...and a 1 in front. Out of comodity you don't write 1.00000...and that is correct. you don't have to. but you're not getting away when the 0.999... comes.

The way I see it is like that proof that 1 = 2. all you got to do is divide by (a-b) after you were told that a=b. yeah, 1 does = 2 once you assume 0 divided by 0 = 1.

So until you actaully write the 0.999... number it does not matter what it's equal to. It's not only unreal but inexistent, that's the point. You can write parts of it and talk about the hole, but what you are NOT writing is the tiny difference between many 9s and many 0s.

Well I probably pissed a lot of people off with what I just wrote but that's how i see it. You're comparing apples with photos of oranges and they're almoust the same...only problem is...one of them doesn't really exist.

Well it does, you just can't express it. The difference between 0.999... and 1.000... is so far back in the decimals that you don't get to write it ever. Do you know what it looks like? No. You got no idea. It's like graphing an asymptote. It goes up and ends below the starting point. you can come infinitely close to it, but you're never gonig to reach it...

That's my best approach.

You have some weird logic my friend.
I am reminded of an event that recently happened. A friend of mine said to me that "a car is a car".
I told him,
"no, a car is not just a car. For examplle, there is a HUGE difference between a Ferarri and a Hyundai".
He said,
"well a car is something that gets you from point A to point B".

We were in the weight room at the time, so I said,
"Well that bench has wheels on it." As I pointed to a weightlifting bench. "I could push you from this spot here, to that spot over there." As I pointed with my finger from one carpet square to another. "I guess you would classify the bench as a car then?"

He said,
"a car is a car"

I just left it at that.

 

[d_leet]

I may lack the capability to explain it...but if it is true, assuming it is...you can't write your number. 0.99999 will eventually be rounded...up or down. if you round it down you end up with a lot of 9s...but not all of them. if you round it up you end up with a lot of zeros but not all of them...and a 1 in front. Out of comodity you don't write 1.00000...and that is correct. you don't have to. but you're not getting away when the 0.999... comes.

The fact is that we are not talking about.99999 or even .99999...99 where there are only a finite number of nines, we are talking about the number .999... where there ARE an infinite number of nines, we are not talking about rounding up or down and that really has nothing to do with what this number is EQUAL to. The thing is that we really don't need to write all of the nines because we've come up with notation that represents that this or any pattern will repeat forever and that is the use of ellipses(...).

The way I see it is like that proof that 1 = 2. all you got to do is divide by (a-b) after you were told that a=b. yeah, 1 does = 2 once you assume 0 divided by 0 = 1.

Can you explain in detail how we are doing something similar to this ( dividing by zero or some other mathematical fallacy) when it comes to showing that .999... is indeed equal to 1?

So until you actaully write the 0.999... number it does not matter what it's equal to. It's not only unreal but inexistent, that's the point. You can write parts of it and talk about the hole, but what you are NOT writing is the tiny difference between many 9s and many 0s.

Why do we need to write the whole number when we have something that represents this number perfectly which is .9... or
9\displaystyle\sum_{i=1}^\infty 10^{-n}
These two expressions represent the same thing, the same number, now do you care to explain how this is only part of the number, and not the whole? And how is .999... "unreal" and "inexistent"?

Well I probably pissed a lot of people off with what I just wrote but that's how i see it. You're comparing apples with photos of oranges and they're almoust the same...only problem is...one of them doesn't really exist.

Your analogy makes no sense for several reasons, first of which I'm sure all of the members on this forum would agree that apples are real and just as real as pictures of oranges are. Second can you explain how exactly the difference between 1 and .999... does not exist? Since it has been said and shown it this thread that it does exist and is equal to zero.

Well it does, you just can't express it. The difference between 0.999... and 1.000... is so far back in the decimals that you don't get to write it ever. Do you know what it looks like? No. You got no idea. It's like graphing an asymptote. It goes up and ends below the starting point. you can come infinitely close to it, but you're never gonig to reach it...

Sure we can express it since it's equal to zero. It's not an assymptote because when we write .999... we are not saying that we are continually adding nines to .999 forever and ever but we are saying that THERE ARE AN INFINITE number of them, so we are never trying to reach anything in a sense we're already there.

I don't see you having any complaints in this thread about 1/3 being equal to .333... so why only a problem with .999...

 

[matt grime]

I may lack the capability to explain it...but if it is true, assuming it is...you can't write your number.

Why can't I? 0.999..., there I wrote it, or 1, I just wrote it again. You might mean that it doesn't have a terminating decimal expansion, but then almost no numbers do (only p/q in lowest terms where q is only divisible by 2 and 5 and no other primes, or p/q is an integer). In anycase, this has nothing to do with what 0.99... is.

0.99999

we aren't talking about 0.99999

will eventually be rounded...up or down.

That is not correct.

The way I see it is like that proof that 1 = 2.

No it is not. It is a prefectly valid statement, and easily proved, that as decimal representations of real numbers 0.999... and 1 are equivalent.

It's not only unreal but inexistent, that's the point.

THis is not a debate about platonism, is it?

You can write parts of it and talk about the hole, but what you are NOT writing is the tiny difference between many 9s and many 0s.

what does that mean? (Note: rhetorical question.)

Well I probably pissed a lot of people off with what I just wrote but that's how i see it.

You just demonstrated that you're not going to listen to reason, or read up on people's suggestions, and instead will just continue arguing for something from a poisition of ignorance. PLenty of people have expended time and considerable effort in this thread to explain what is going on and you've steam-rollered over it and ignored it all. If you think that pisses people off you're probably making a safe bet.

You're comparing apples with photos of oranges and they're almoust the same...only problem is...one of them doesn't really exist.

Well it does, you just can't express it.

You've just contradicted your own (bad) analogy.

The difference between 0.999... and 1.000... is so far back in the decimals that you don't get to write it ever.

this demonstrates you've not bothered to consider people's explanations of what decimal representations of real numbers are.

Do you know what it looks like? No. You got no idea. It's like graphing an asymptote. It goes up and ends below the starting point. you can come infinitely close to it, but you're never gonig to reach it...

That's my best approach.

your best approach would be to actually read what other people have read and consider the mathematics behind it: not what you think these things represent but what they actually do represent.