Is 1 = 0.9999......

[MadRat]

Originally posted by: josphII

Originally posted by: silverpig

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.
The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

the sum you stated above is EXACTLY what 0.333... is, they are both exactly 1/3.

You've been misguided. There is an infinitely small difference between the sum of (3 * .333...) and (3 * 1/3) that makes them different. It is the same broken logic used to prove .999...=1, trying to use objective measures to define subjective values. There is a difference that you cannot fathom but that doesn't mean it doesn't still exist. You cannot divide or multiply the infinite number of decimal places in either case, which is the premise of your logic.

 

[GtPrOjEcTX]

n = .9999....

This means that 10 x n = 9.9999....

Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

 

[Skyclad1uhm1]

Originally posted by: GtPrOjEcTX
n = .9999....

This means that 10 x n = 9.9999....

Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

You've not read a single post in the thread, have you?

 

[Haircut]

Originally posted by: MadRat

Originally posted by: josphII

Originally posted by: silverpig

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.
The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

the sum you stated above is EXACTLY what 0.333... is, they are both exactly 1/3.

You've been misguided. There is an infinitely small difference between the sum of (3 * .333...) and (3 * 1/3) that makes them different. It is the same broken logic used to prove .999...=1, trying to use objective measures to define subjective values. There is a difference that you cannot fathom but that doesn't mean it doesn't still exist. You cannot divide or multiply the infinite number of decimal places in either case, which is the premise of your logic.

You may choose to use a form of mathematics where 0.333... != 1/3 and 0.999... != 1, but using the math that everyone else uses, the math that has been built up from a number of axioms they are equal. The set of real numbers has been defined from these axioms and using these definitions 0.999... is the same number as 1. It is because it has been defined this way.

 

[spidey07]

This is truely amazing.

Even more proof showing that both are the same number, and yet some continue to not believe it.

"well in my number system and math they are different".

That may be so, but not for the rest of the world.

 

[silverpig]

Originally posted by: josphII

Originally posted by: silverpig

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.

The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

the sum you stated above is EXACTLY what 0.333... is, they are both exactly 1/3.

First, I say 0.999 = 1.

S dne 0.333... because it is a progression. The LIMIT of that sum, however, is 0.333...

 

[silverpig]

Originally posted by: Skyclad1uhm1

Originally posted by: Krk3561

Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

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

What do you think?

9 x .99999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1. Any calculator (except for a very accurate computer) you put that in will give you 9 because it cant process that many digits, so it will round up

The point is that infinity cannot be defined.

x = 0.999...
10x= 9.999...
10x - x =9.999... - 0.999...
9x = 9

This requires you to see that infinity = infinity, that there is no infinity -1 or infinity +1. It's an abstract number. (infinity - infinity can basically be anything)

9.999... (infinity -1 decimals) - 0.999... (infinity decimals) is hard to imagine.

Try this:
Start walking in a circle. As you keep starting the same road again it is infinite. Now walk 10 feet less. Does that mean the circle is less infinite? 'infinite' does not point at the distance covered, but instead it points at the circle.

OMG you are NOT adding or subtracting infinity. You are adding and subtracting an infinite number of decimals, but you are doing nothing with infinity itself.

 

[silverpig]

Originally posted by: MadRat

Originally posted by: josphII

Originally posted by: silverpig

Originally posted by: MadRat
.333... <> 1/3 but it is AN APPROXIMATION that is accepted as being one and the same.

It is not an approximation.
The sum S = (3/10 + 3/100 + 3/1000 + ...) is an approxmation, but this is not what 0.333... is.

the sum you stated above is EXACTLY what 0.333... is, they are both exactly 1/3.

You've been misguided. There is an infinitely small difference between the sum of (3 * .333...) and (3 * 1/3) that makes them different. It is the same broken logic used to prove .999...=1, trying to use objective measures to define subjective values. There is a difference that you cannot fathom but that doesn't mean it doesn't still exist. You cannot divide or multiply the infinite number of decimal places in either case, which is the premise of your logic.

That x = 0.999...
......
x = 1

proof is not the only one.


ALL numbers are subjective. In case you missed my post earlier:

0 = {}
1 = {0}
2 = {0,1}
3 = {0,1,2}
etc...

0 exists subject to the idea of emptiness. 1 exists subject to the existence of 0...

 

[silverpig]

1/3 = 0.333... = lim (n -> inf) [Sum(i= 0..n)(3/10^i)]

In all cases you are already at the value we think of as 1/3. 1/3 just is 1/3. 0.333... is the representation of a 0 with an infinite string of 3s after it. The limit of S is 1/3 and is the value that sum approaches.

note: Sum(i = 0..inf) [3/10^i] is bad form mathematically because you can't actually write 3/10^inf as it doesn't really mean anything. You must deal with limits.

Sum(i = 0..n)[3/10^n] is always less than 1/3 for any finite n.

 

[DoNotDisturb]

Originally posted by: Einz
Dunno if this has been posted yet, but this is how I was taught to think about it:

1/3 = 0.3333....
1/3*3=0.3333...*3
3/3=0.9999...

3/3 = 1, 0.9999... = 1

yeah, i'm just wondering if its correct to actually multiply with an infinite amount by a number. i dont think it'd be infinite if you were to multiply them all by 3s.

 

[silverpig]

You do it all the time.

5 x 3 = 15 right?


You're actually doing

...0005.000... x ...0003.000... = ...00015.000...

Nobody seems to have any problem with it then.

 

[BruceLee]

Calculus uses limits to compute EXACT areas. Riemann sums are approximations. The integral of a function is the LIMIT of the riemann summation of the rectangles that approximate that area. The limit is the exact area, calculus allows us to find this exact area.

I don't see how you can construe that calculus is an approximation.

The first part of this is completely true. However, when someone can explain to me how Euler's formula (e^i*pi) is not an approximation I will believe you. Basically what this formula is saying is that one infinite number to another infinite number gives you an integer. This to me makes no sense, and this is where I have always questioned the validity of calculus. I am not really saying I do not believe it, since I have taken it for awhile, but what I am saying is that the whole process of calculus is an approximation. I hope that we can continue this discussion without idiots posting every 5 seconds. Because someone doesn't agree with your so called proof, that doesn't mean you have to go around and call them an idiot. It really just shows how much of an idiot you are for being so myopic. I know the mathematical proofs are there for .999 repeating = 1, however no one here will ever experience "infinity". By this token, it is very difficult for people to sit here and say "I can't believe you won't agree with me, I'm right your wrong" Look, understand that in an "argument" or debate, for your point to be taken seriously, calling someone names will not fly. Also understand that in theory .999 repeating will never = 1. I for one do not believe that .999 repeating = 1, and it will take more proof other than limits (which I belive to be very shady anyways) for me to change my mind. I read over at another forum some very good proof for everyone's case however, showing that they are equal. I will try and dig up the link and post back later.

edit: forgot to add, silverpig you are bringing some great points to the argument, keep them coming.

 

[DoNotDisturb]

yes i already pointed out Bruce Lee about how calculus is based on approximations. I hope people understand that.

 

[BruceLee]

Sorry I forget your stance on this, do you think that .99 repeating = 1?? I am checking out the site in your sig now, and I will see what its about.

 

[silverpig]

Thanks BruceLee

e and pi are not infinite numbers. They are infinitely long in decimal form, but so are all numbers. 1 is really ...0001.000... The only difference is that e and pi don't repeat 0s.

 

[silverpig]

Originally posted by: DoNotDisturb
yes i already pointed out Bruce Lee about how calculus is based on approximations. I hope people understand that.

But calculus extends these approximations to find exact values.

 

[DoNotDisturb]

Calculus teacher so far has taught us that it is based on limits, i'm learning more from you guys (correctly hopefully ) i've been taught to find exact values through riemann sums and indefinite integrals.

 

[BruceLee]

Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

 

[DoNotDisturb]

bruce lee, if you view .999 (with you adding 9s to the end) opposed to .999.... (assuming there is an infinite amount).

there is a big diff.

 

[BruceLee]

Originally posted by: DoNotDisturb
bruce lee, if you view .999 (with you adding 9s to the end) opposed to .999.... (assuming there is an infinite amount).

there is a big diff.

??? I don't understand what you are arguing here.

 

[Haircut]

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

Take a look at the series expansion of Sin (x)

If x = Pi/6 (30 degrees) then the series expansion is

Pi/6 - (Pi^3)/1296 + (Pi^5)/933120 - (Pi^7)/1410877440 + ...
This is an infinite sum, but is equal to a rational value as sin (Pi/6) = 0.5

 

[silverpig]

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

It's just a limitation of how we write numbers down. pi, 3, sqrt(2), 4, 12... they're all single points on the real number line. Looking at them like that, there is no difference in the nature of the numbers.

What if, instead of base 10, we used a number system based on pi?

 

[silverpig]

Originally posted by: BruceLee

Originally posted by: DoNotDisturb
bruce lee, if you view .999 (with you adding 9s to the end) opposed to .999.... (assuming there is an infinite amount).

there is a big diff.

??? I don't understand what you are arguing here.

He's saying there's a difference in 0.999... where you use the ... to represent an already infinite number of 9s on the end.

When he says ".999 (with you adding 9s to the end)" he's referring to a progression where you write 0.9 and then add a 9, and then add another one after that, and then another... The number of nines approaches infinity, but because it's a progression, it never actually gets there.

 

[BruceLee]

Originally posted by: silverpig

Originally posted by: BruceLee
Yes sorry for the notation fault. However, my point still remains: no one knows the exact value of either pi or e, yet with these 2 values in an expression an integer is output. That alone has always bugged me about calculus and is my main reason for not buying into the .999 repeating = 1. I understand that all integers have an infinite amount of zeros behind the decimal point, but the value is in no way approximated.

It's just a limitation of how we write numbers down. pi, 3, sqrt(2), 4, 12... they're all single points on the real number line. Looking at them like that, there is no difference in the nature of the numbers.

What if, instead of base 10, we used a number system based on pi?

Yes I understand all this, however Euler's formula was developed on a base 10 system. These were Euler's own words after stating his equation to someone... "e^i*pi = -1, therefore there is a god". ha I always find that amusing, although somewhat irrelevant to this argument. All I am saying is I have trouble believing all the proofs regarding .999 repeating = 1 for the fact that calculus is an approximation. I do not think that fact can be disputed, although many of you belive otherwise. I respect all the opposition to my points, I just know from all the experience I have had regarding limits and such I think the system is very shady. The idea of infinity in itself is very odd. Extremes are hard to comprehend in math as well as in Physics when approaching the speed of light (although that's a totally different story and argument some very weird things happen when approaching the speed of light, and this is just another reason why infinity is a tough concept to comprehend). Sorry I might be going off on a tangent and not making sense, but I'm trying to watch the Eagles game. Enough math for me, I have bookmarked the thread and will return later to express some points. This is the weekend, and my time off!!!

 

[silverpig]

The point e is the at the same place no matter what number base you use. It will just look different depending on which base you choose. How it looks in decimal form doesn't really matter. It's still just another poing on the line.