Thread: Concept of Quantity

whiteflags: "I'm very surprised people are saying things aren't numbers, but they are quantities, and quantities are numbers."
They are not only quantities, that is what I mean. They are quantities with a condition of "being close to another quantity". They are not exact quantities if you like that term. Which makes calling them numbers kind of wrong at my point of view. Isn't exactly the point of calling a quantity a number to give it a specific value? If you have ten apples you can give that quantity a number. If you have "a lot of apples", yes there is a quantity of apples, but you wouldn't give them a number.
If you argue "true, but you have a quantity of apples close to 1000" then still you are not giving them a number, you are comparing them with an imaginary quantity of apples that you could give a number (1000).

But I am not correct with the above unless we clarify the question: is infinity a number? I would say it is fair to call it a quantity but not a number, something like this

So if infinity is not a number, I would say that 0.999... is certainly not a number as it is a sum of infinite numbers.

EVOEx: Despite if I am wrong or right you have to fully finish your equations, because for example:

Code:

1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
= lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

is true, but you imply that it is equal to 0.

Or using

Code:

 
1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = 0

is fine if

Code:

 lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

is equal to 1, which is the exact point of our disagreement.

As far as I can tell both are wrong assumptions. So lets start again more thoroughly

1.

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

We first solve the "lim" part as

Code:

0.9999... =  9*10^-1 + 9*10^-2 + 9*10^-3 + ...

which is correct.

2.

Code:

1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

again solve the lim part

Code:

1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

then using the above

Code:

1 - 0.9999..

all of this are simple and clear. But I fail to see how the above actually equals to zero.
The whole point is why does 0.99999.... or 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... or any other form to point it using either "..." or "Σ" or any other symbol meaning a sum of infinite parts is equal to 1.
The code

Code:

0.9 + 0.09 + 0.009 + 0.0009 + ... = 0.999...

is again perfectly correct.
For me

Code:

 0.999.... -> 1

but not equal. And I believe that limits exactly point that out. Compare

Code:

x = oo
limx(x->oo)(x)

the whole point is that the first doesn't make sense in algebra that is why you have to use a limit to make it more clear. What you are doing with

Code:

0.999.... = 1

defines that purpose**. If that is correct why would you have the symbol "->" at all???

My point is always that using "lim" or "->" to prove the classical "0.999.... = 1" is not the correct method. There can sure be other methods, but using limits is more to solve the problem by adding the symbols "lim" and "->" exactly because the symbol "=" is not enough....

Lasersight: "Ten is another symbol for the number we call ten"
And calling ten is again another symbol, no difference if your write or say it. In mathematics there is not necessarily a practical or physical meaning on symbols...

** I use a lot of that term, but I am now doubting it is correct. I think I am actually saying the opposite of what I mean so please give me the correct phrase

Although I've found this thread entertaining and educational, I'm not really trying to get involved. I'm just going to throw out what I find wrong with the proof and see what happens.

Now, I'm good with some mathematics, but not great with calculus (where I know what I know about limits from), so I may have missed something you are doing with the algebra or the limits.

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = 1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

Now, all of these involved the limit of many (`x' minus one assuming starting at zero) constant functions (which would be a constant) plus the limit of an infinitesimal (having the limit of zero).

I'm fine with those bits even if the equation is justified by the implication of the "..." bits; I'm just stating where I'm coming from.

[Code]1 - 0.9999... = lim(x -> inf) ( 1*10^-x )[/Quote]

Yea, this doesn't work for me. Because you've lost the constants the "equation" doesn't hold because the limit of the function `1*10^-x' as x approaches infinity is simply zero and not some unknown quantity as implied by the "..." bits.

Now, I agree that the limit of the function, for lack of a better word, `1-0.9999...' is zero. That said, the limit of a function doesn't imply anything about the results of the function at the limit (*). The result of a function applied to a given value doesn't imply anything about the limit. You seem to be saying that they are the same with the above "equation". If you were using `lim' on both sides, I'd be right with you.

Anyway, since this doesn't work for me, what follows from the argument also doesn't work for me.

Soma

[Edit]
(*) I feel that sounded weird, but the results had done been seen and may be quoted before I get done with the edit.

With that in mind, I'm getting at the fact that a given application of a function may or may not be defined at the limit, may be a value relative to the value associated with the limit, or may be any other value.
[/Edit]

Well I don't see why we couldn't compute 0.999... with limits but it's probably better stated as a sum anyway:

0.999... = Σ(x = 0, x->infinity) 9*10^-x
1 - Σ(x = 0, x->infinity) 9*10^-x = 0
1 = 0.999...

Interestingly if you did the subtraction for any finite length of 0.999... you would eventually have to write that last 1, but for forever places, things are different. No last digit.

EVOEx there is not disagreement yet, it is just what you presented wasn't fully clear.

So basically the "unclear point" was

Code:

lim(x->inf) (1 - (9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x)) = lim(x->inf)(1*10^-x) = 0

but now you clear it, sorry if this wasn't obvious for me.


The key point here is:

Code:

lim(x->inf) (1 - (9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x)) = lim(x -> inf) ( 1*10^-(x-1) - 9*10^-x )

since it is the point you get rid of the "..."

Thinking out loud:

Code:

1*10^0 - 9*10^-1 = 1*10^-1
...
1*10-(n-2) - 9*10^-(n-1) = 1*10^-(n-1)
1*10-(n-1) - 9*10^-n = 1*10^-n

But to get what you are you need to add up all these equations.
The blue will be canceled out so you will end up with

Code:

1 -  (9*10^-1 + ... + 9*10^-n) = 1*10^-n

which is what you want.

So the only thing left actually is

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

since this is where the 0.999... come in place.

To prove this is true I would say you can claim

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

then add

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ... +  lim(x -> inf)(9*10^-x)

but then you add a "lim(x -> inf)" to everything and get what you want.
EDIT: Sorry, scratch that and I think that is where I got confused. I don't think you can add a limit to everything and get what you want. I believe the limit has to be defined to do this which is not really clear here.

Code:

lim(x -> inf)(9*10^-x) + lim(x -> inf)(9*10^-(x-1)) + .... + lim(x -> inf)(9*10^-(x-2)) + ....

so everything is zero. In the end you have

Code:

0.999... = Σ(i=0...inf)(lim(x -> inf)(9*10^-(x-i)))

where it is actually:

Code:

0.999... =Σ(i=0...inf)(9*10^-i)

you see the difference? The thing is that if you use a limit you cannot claim you have

Code:

9*10^-x + 9*10^-(x-1) + ... + 9*10^-1

think of the "x-i" part. You have "i->inf" and "x->inf". You assume that "x is getting smaller and smaller that it gets to 1. Right? But if x is near infinite it will never get to 1 no matter how many times you subtract from it. So concluding:

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

is NOT true.

Well, that is not a "passionate" NOT, please feel free to prove it, I am just thinking out loud here, maybe I am talking crazy and after 5min re-edit...

EDIT2: To clarify I would say that

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ... +  9*10^-i where i->inf

is the proper way to right it, but it is not a limit. The same error as you would do if you had

Code:

lim(1/x) = 0 where x->inf
is equal to
1/x = 0 where x->inf

where this is not true. The limit means "it approaches" a number as is already stated, not that it is equal, so using despite the math in the end putting the "limit" to "0.999..." is not right. Since you can say that the "limit of 0.999... is 1" but that doesn't mean that the equal 1. I believe there is the confusion?

Yes, I would disagree on that part even though I agreed at first

Originally Posted by EVOEx

So we add more and more 9s, and we get closer and closer, but never actually get there. Or: as the number of 9s goes to infinity we get closer and closer to "0.9999...". That's exactly what I'm saying here:

Code:

0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

The "..." is a bit tricky symbol and confuses things. Since you put "x" you assume something like this:

Code:

0.999..... = Σ(x=1...inf)(9*10^-x)

which is fine, but what you do with the above representation is that somehow you use only the last "x". That is how I see it. Where there is a huge difference

x->inf means that x goes to infinity
x=1...inf means that you have a put one by one numbers from 1 to infinity

If you want to use limits you would do:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))

You see that there is a separation on the symbols, mixing x with Z here is what makes this assumption wrong.
The symbols "lim" and "->" follow certain rules and those should be used exactly as they are and not just because they seem right. So the starting point is

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

and you should add limits if you want by starting there.

Originally Posted by C_ntua

Yes, I would disagree on that part even though I agreed at first

The "..." is a bit tricky symbol and confuses things. Since you put "x" you assume something like this:

Code:

0.999..... = Σ(x=1...inf)(9*10^-x)

which is fine, but what you do with the above representation is that somehow you use only the last "x". That is how I see it. Where there is a huge difference

x->inf means that x goes to infinity
x=1...inf means that you have a put one by one numbers from 1 to infinity

If you want to use limits you would do:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))

You see that there is a separation on the symbols, mixing x with Z here is what makes this assumption wrong.
The symbols "lim" and "->" follow certain rules and those should be used exactly as they are and not just because they seem right. So the starting point is

Code:

0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...

and you should add limits if you want by starting there.

Actually, the two are identical:

Code:

lim(Z->inf)(Σ(x=1...Z)(9*10^-x))
0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )

Except that I was too lazy to get the sigma symbol here, so I simply wrote it out with "+ ... +". Let me proof it again for you using the sigma notation.

But first let me discuss this part of what you said:

so you are not getting closer and closer since you will be adding zero. True?
In other words the difference will always be "0.00..001". The more 9s you add the more 0s you can add and you are always on the same point.

But:

Code:

lim(x->inf)(9*10^-x)

Is not a very small number, you have to understand. It's actually 0, per definition of the limit. If x grows towards infinity "9*10^-x" grows smaller. The "9*10^-x" term grows to 0 but never actually becomes 0, meaning the LIMIT is 0.

Now let me proof it using your notation (which is the same except for notational differences). If you still disagree with something please show me exactly what, because I am curious about all this.

Code:

0.9999... = lim[Z->inf]( Σ[x=1...Z](9*10^-x) )
1 - 0.9999... = 1 - lim[Z->inf]( Σ[x=1...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1 - Σ[x=1...Z](9*10^-x) )

There, same three steps as before. Now we start taking terms from the sigma and combining it with the "1" (notice the change in the "1" in the sum).

Code:

1 - 0.9999... = lim[Z->inf]( (1 - 9*10^-1) - Σ[x=2...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-1 - Σ[x=2...Z](9*10^-x) )

And then we take another term:

Code:

1 - 0.9999... = lim[Z->inf]( (1*10^-1 - 9*10^-2) - Σ[x=3...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-2 - Σ[x=3...Z](9*10^-x) )

We repeat this until we get to:

Code:

1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - Σ[x=Z...Z](9*10^-x) )
1 - 0.9999... = lim[Z->inf]( 1*10^-(Z-1) - 9*10^-Z )
1 - 0.9999... = lim[Z->inf]( 1*10^-Z )

As said before, that limit is 0! So:

Code:

1 - 0.9999... = 0
1 = 0.9999...

If you still disagree with something, let me know.


Edit: Okay so you already agreed on it by the time I finished writing it... See that it's the same thing I did before except being too lazy to find out how to use the sigma symbol in here.

Originally Posted by EVOEx

But at the very least, even if you take another definition, to me this seems to prove that there is no number between "0.9999..." and 1: if there were, the limit there would be less than 1!
And one can always find a number between two numbers that aren't equal: their average, for example.

Going on this logic, as said in an earlier post, then you have to consider that

Code:

lim(x->inf)(9*10^-x) = 0

so you are not getting closer and closer since you will be adding zero. True?
In other words the difference will always be "0.00..001". The more 9s you add the more 0s you can add and you are always on the same point.
And the logic you are getting closer and closer is controversial to infinity. If you have infinite space, will you ever reach the end? That is the point, you would never reach the end.
Even worse, if you can reach 1, then if you add another 9s what happens? If you add another and another? Makes no real sense.
Wouldn't you agree that if there is no limit then it will grow bigger and bigger? If there is a limit that would be one, but by definition it will never reach it.