0.999999[r] = 1

[Matthias99]

Originally posted by: Cogman
I just thought about it, and an infinity .000 with a one tagged on the end really does not work, because the zeros are infinit there will never actually be and end for the one to be tagged on to. Anyone following what im saying?

Yes; your intuition is correct, and you can actually work out a proof of .999999... == 1.00000... along these lines. There was a HUGE discussion on this topic while back (maybe ~9-10 months ago?) in HT (and OT), and this was presented as a more mathematical proof (the one where you multiply by 10, subtract, and then divide by 10 is not a "proof", just a demonstration). In the real number system, you cannot have "an infinite number of zeroes followed by a 1", since by definition an infinite number of zeroes cannot have anything after it -- if it did, it wouldn't be infinite.

Long story short: .99999... == 1. It's proved, it's over and done with. Arguing against this is like arguing that sin(pi/2) != 1.0 because you don't think it "should" be.

Long story slightly longer:

When someone says ".99999... repeating", what they *really* mean is:

"The sum as x goes from one to infinity of (9 * 10^-x)" -- that is, .9 + .09 + .009 + .0009 + ...

However, you *cannot* define the ".000000...1" number like this. Try to take 1.0 minus the sum above:

1.0 - (.9 + .09 + .009 + .0009 + ...) = 0.1 - (.09 + .009 + .0009 + ...) = 0.01 - (.009 + .0009 + .00009 + ...)

You can see that this will, as you add an infinite number of terms, become the infamous "infinite number of zeroes followed by a one". However, you can also see here that "1.0 - (sum as x goes from 1 to y of (9 * 10^-x))" = (10^-y). Thus, you can find the value of 1.0 - .99999... by taking the limit of the right-hand side of that equation as y goes to infinity. "The limit as x goes to infinity of (10^-x)" is provably zero, which means 1.0 minus the sum defined above (".99999...") must also equal zero, and therefore ".99999..." must equal one.

 

[AnthraX101]

Originally posted by: Matthias99
Long story short: .99999... == 1. It's proved, it's over and done with. Arguing against this is like arguing that sin(pi/2) != 1.0 because you don't think it "should" be.

As fun a ad hominem attacks are (not that both sides aren't participating), what it boils down to is decimal arithmetic is not accurate. You can say the sum (0.3 * (.1*n)) for n=1 -> infin = 1/3. It's a series of approximations that eventualy approach an infinitely repeating number. 0.999[r] = 1 is just an offshoot of the inaccuracy of decimal mathematics.

AnthraX101

 

[Matthias99]

Originally posted by: AnthraX101

Originally posted by: Matthias99
Long story short: .99999... == 1. It's proved, it's over and done with. Arguing against this is like arguing that sin(pi/2) != 1.0 because you don't think it "should" be.

As fun a ad hominem attacks are (not that both sides aren't participating),

I didn't really mean that as an ad hominem attack, nor was it directed at anyone in particular. But, in all seriousness, there is no debate to be had here. Proofs are proofs.

what it boils down to is decimal arithmetic is not accurate. You can say the sum (0.3 * (.1*n)) for n=1 -> infin = 1/3. It's a series of approximations that eventualy approach an infinitely repeating number. 0.999[r] = 1 is just an offshoot of the inaccuracy of decimal mathematics.

AnthraX101

Nope. sum(.3 * 10^-x) as x goes from 1 to infinity (ie, ".33333...") EQUALS 1/3. Exactly. It's not an "approximation", nor does it "approach" it. That's the whole point of infinite sums, series, and limits. There's nothing inaccurate about it.

 

[jman19]

Originally posted by: Amorphus
Wait, so is
0.0000.....1 = 0? assuming an infinite number of zeroes.

You can "prove" things mathematically (like the researchers "proved" that bees could not fly in the oft-cited story), but the issue of .999...=1 is a philosophical question. Can something be so insignificantly small such that it is actually nothing?

Such a number could not exist, as far as I understand.

 

[ss284]

Originally posted by: Matthias99

Originally posted by: AnthraX101

Originally posted by: Matthias99
Long story short: .99999... == 1. It's proved, it's over and done with. Arguing against this is like arguing that sin(pi/2) != 1.0 because you don't think it "should" be.

As fun a ad hominem attacks are (not that both sides aren't participating),

I didn't really mean that as an ad hominem attack, nor was it directed at anyone in particular. But, in all seriousness, there is no debate to be had here. Proofs are proofs.

what it boils down to is decimal arithmetic is not accurate. You can say the sum (0.3 * (.1*n)) for n=1 -> infin = 1/3. It's a series of approximations that eventualy approach an infinitely repeating number. 0.999[r] = 1 is just an offshoot of the inaccuracy of decimal mathematics.

AnthraX101

Nope. sum(.3 * 10^-x) as x goes from 1 to infinity (ie, ".33333...") EQUALS 1/3. Exactly. It's not an "approximation", nor does it "approach" it. That's the whole point of infinite sums, series, and limits. There's nothing inaccurate about it.

Agreed. There is no inaccuracy in decimal mathmatics when limits are used. This is assuming of course, that an infinite amount of sums are taken of the series. But anyways, mathematically, .99[r] = 1/1 just like .33[r] = 1/3. Anything beyond that is left up to philosophy. Mathematically, the statement is sound.

-Steve

 

[unlockthesource]

What did they expect was going to happen when an unlimited "value" is introduced to a system of definite limits?

 

[silverpig]

Originally posted by: Matthias99

Originally posted by: AnthraX101

Originally posted by: Matthias99
Long story short: .99999... == 1. It's proved, it's over and done with. Arguing against this is like arguing that sin(pi/2) != 1.0 because you don't think it "should" be.

As fun a ad hominem attacks are (not that both sides aren't participating),

I didn't really mean that as an ad hominem attack, nor was it directed at anyone in particular. But, in all seriousness, there is no debate to be had here. Proofs are proofs.

what it boils down to is decimal arithmetic is not accurate. You can say the sum (0.3 * (.1*n)) for n=1 -> infin = 1/3. It's a series of approximations that eventualy approach an infinitely repeating number. 0.999[r] = 1 is just an offshoot of the inaccuracy of decimal mathematics.

AnthraX101

Nope. sum(.3 * 10^-x) as x goes from 1 to infinity (ie, ".33333...") EQUALS 1/3. Exactly. It's not an "approximation", nor does it "approach" it. That's the whole point of infinite sums, series, and limits. There's nothing inaccurate about it.

Um no. That series approaches 1/3. Look at your language:

"as x goes from 1 to infinity"

This implies a progression.

lim n->inf Sum (.3 * 10^-x) for x=1..n is 1/3 though.

The series is a progression that doesn't actually equal anything until you put a limiting condition on it. Fortunately though, this is exactly how we define 0.999... as it is a decimal point followed by an infinite string of nines. This is the limiting value of the sum .9 + .09 + .009 + ...

 

[Matthias99]

Originally posted by: silverpig

Originally posted by: Matthias99
Nope. sum(.3 * 10^-x) as x goes from 1 to infinity (ie, ".33333...") EQUALS 1/3. Exactly. It's not an "approximation", nor does it "approach" it. That's the whole point of infinite sums, series, and limits. There's nothing inaccurate about it.

Um no. That series approaches 1/3. Look at your language:

"as x goes from 1 to infinity"

This implies a progression.

lim n->inf Sum (.3 * 10^-x) for x=1..n is 1/3 though.

Nitpicking. I said:

sum(.3 * 10^-x) as x goes from 1 to infinity

(emphasis on the sum, not the "going")

You know, like, the sum of the infinite series. If you want to get really technical, yes, I should have taken the limit of the sum as you wrote, but I felt this was a clearer way of stating it. This is probably why I wasn't a math major.

My point is that the sum of the (infinite) series is exactly 1/3. Not an "approximation" of 1/3. Obviously, the sum of a finite number of terms approaches 1/3, but will never reach it.

 

[unipidity]

Surely arguing that 1/3rd = 0.333r is exactly the same problem as 0.99r = 1? In that an infinte series of the same nature equals 1?

Anyway. Im a believer, but thats the nature of infinity...non-physical. Nothing that is anything less than 1 is 1, but 0.999r is not less than 1... Stupid freakni 'number'.

 

[imported_Nail]

Mathematics is a human construct

Mathcematics is an observation, numbers are an invention.

 

[imported_adamj]

Thats why these numbers are called irrational and not infinite.

 

[imported_Nail]

1 is irrational?

 

[imported_adamj]

Originally posted by: Nail
1 is irrational?

No. But i was refering to the .3(r) .6(r) and .9(r) being irrational. But i was incorrect because these numbers can be writen as fractions with real integers and still be accurate. Real irrational numbers cannot be writen as fractions.

 

[mobobuff]

Originally posted by: Matthias99

Originally posted by: silverpig

Originally posted by: Matthias99
Nope. sum(.3 * 10^-x) as x goes from 1 to infinity (ie, ".33333...") EQUALS 1/3. Exactly. It's not an "approximation", nor does it "approach" it. That's the whole point of infinite sums, series, and limits. There's nothing inaccurate about it.

Um no. That series approaches 1/3. Look at your language:

"as x goes from 1 to infinity"

This implies a progression.

lim n->inf Sum (.3 * 10^-x) for x=1..n is 1/3 though.

Nitpicking. I said:

sum(.3 * 10^-x) as x goes from 1 to infinity

(emphasis on the sum, not the "going")

You know, like, the sum of the infinite series. If you want to get really technical, yes, I should have taken the limit of the sum as you wrote, but I felt this was a clearer way of stating it. This is probably why I wasn't a math major.

My point is that the sum of the (infinite) series is exactly 1/3. Not an "approximation" of 1/3. Obviously, the sum of a finite number of terms approaches 1/3, but will never reach it.

Ah! I knew it would boil down to semantics sooner or later.

What was the name of that model that displayed that balls released simultaneously from the top of a circle and followed chords of different angles to the circumference will meet the edge at the exact same time? I always wanted to construct a mass-scale model of that and take a well-timed photo.

 

[Runamile]

From what I have read, this is what I have come to conclude is as simple terms as I can muster. Think about it, it really does make sence:

First about limits (not because we don't know, but just to be thorough). A limit is a number that something (usually the result of a equations, like in a tan curve, where they are 90 and 270 degrees. But in this case its the 0.9999 repeating) will get infinitly closer to but will never equal. With that being said...

1/3 != 0.3333333...
Yes! Thats what I said!

1/3 is the limit for 0.33333..., which means it will get infinitly closer to 1/3. 1/3 is an ALMOST exact representation, but not PERFECT.

That is the reason why 1/3 * 3 = 1, but 0.333... * 3 = 0.9999...

I'll say it again- 1/3 is the LIMIT for 0.33333..., yet a very acurate representation for mathmatical purposes.

 

[Dinominant]

Originally posted by: Runamile
From what I have read, this is what I have come to conclude is as simple terms as I can muster. Think about it, it really does make sence:

First about limits (not because we don't know, but just to be thorough). A limit is a number that something (usually the result of a equations, like in a tan curve, where they are 90 and 270 degrees. But in this case its the 0.9999 repeating) will get infinitly closer to but will never equal. With that being said...

1/3 != 0.3333333...
Yes! Thats what I said!

1/3 is the limit for 0.33333..., which means it will get infinitly closer to 1/3. 1/3 is an ALMOST exact representation, but not PERFECT.

That is the reason why 1/3 * 3 = 1, but 0.333... * 3 = 0.9999...

I'll say it again- 1/3 is the LIMIT for 0.33333..., yet a very acurate representation for mathmatical purposes.

Wow, hold on there. A limit can have an exact answere:
the limit as 'x' approaches '2' of 1+x is equal to 3

 

[Mojoed]

This problem reminded me of THIS. for some reason.

 

[Matthias99]

Originally posted by: Runamile
From what I have read, this is what I have come to conclude is as simple terms as I can muster. Think about it, it really does make sence:

First about limits (not because we don't know, but just to be thorough). A limit is a number that something (usually the result of a equations, like in a tan curve, where they are 90 and 270 degrees. But in this case its the 0.9999 repeating) will get infinitly closer to but will never equal. With that being said...

1/3 != 0.3333333...
Yes! Thats what I said!

1/3 is the limit for 0.33333..., which means it will get infinitly closer to 1/3. 1/3 is an ALMOST exact representation, but not PERFECT.

That is the reason why 1/3 * 3 = 1, but 0.333... * 3 = 0.9999...

I'll say it again- 1/3 is the LIMIT for 0.33333..., yet a very acurate representation for mathmatical purposes.

Read the thread again. 1/3 == "0.33333..." -- the limit of the sum I gave above (sum(3 * 10^-x)) as x goes to infinity EXACTLY EQUALS 1/3 (similarly, the limit of (9 * 10^-x) as x goes to inifinity exactly equals 1). I suppose this could be stated as saying that "0.33333..." is "infinitely close" to 1/3 -- which is equivalent to saying that they are equal. After all, if there are no points on the number line closer to 1/3 than "0.33333...", they must be equal.

 

[DrPizza]

The biggest problems to understanding this is understanding limits, and at what point you say "is approaching" and at what point you say "is"
And understanding the concept of infinity.

The sum of an infinite number of terms IS the limit. <---- read that 2 or 3 times.
So, since Runamile already agreed that "1/3 is the limit for 0.33333...", then all runamile needs to do is recognize that,
and I repeat myself,
The sum of .3 + .03 + .003 + .0003 +. . . IS the limit -- which he's already stated, is 1/3.

The question isn't "what does the sum approach" but rather, what is the limit as the number of terms approaches infinity.
The sum is this limit.

And, I'll put it this way as well. Take a near-sighted runner. Place a wall 100 feet to his left. Ask him what he's going to hit when he runs to the left. He can't see far enough ahead, so he runs 50, then 80, then 90, then 95, then 99 feet to his left. He doesn't have to run into the wall to finally realize he's going to get there. He can already answer the question. Likewise, we don't have to add up an infinite number of terms to realize we're going to get to 1/3 (we can't by adding them one at a time, because there's an infinite number of them) But, if we want to add up an infinite number of terms, we use the limit - the sum of the infinite number of terms IS the limit.

 

[zugzoog]

Originally posted by: Dinominant
OK, here is a valid proof aquired from Dr. Math:

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

I disagree with this demonstration with regard to the step

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

Since you have raised 0.9999[r] by an order of magnitude when you complete the subtraction the result will actually be

9x = 9 - 0.000....0009

therefore x != 1.

Ok, the subtracted number is undefinable, but while it is undefinable it is != 0 therefore the logic holds.

The faulty logic may be more easily understood if I follow the logic presented but substitute 0.8888[r] instead of 0.99999[r]

x = 0.8888888...
10x = 8.88888...
10x - x = 8.8888... - 0.88888...
8x = 8
x = 1

Which is clearly false.

 

[Sahakiel]

Originally posted by: zugzoog

I disagree with this demonstration with regard to the step

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

Since you have raised 0.9999[r] by an order of magnitude when you complete the subtraction the result will actually be

9x = 9 - 0.000....0009

in this step, you imply that 0.999... or .999[r] has a tail (terminates), which is clearly a contradiction to the original definition of 0.999... or 0.999[r]. Nevermind the violation of syntax.

therefore x != 1.

Ok, the subtracted number is undefinable, but while it is undefinable it is != 0 therefore the logic holds.

The faulty logic may be more easily understood if I follow the logic presented but substitute 0.8888[r] instead of 0.99999[r]

x = 0.8888888...
10x = 8.88888...
10x - x = 8.8888... - 0.88888...
8x = 8
x = 1

Which is clearly false.

It is clearly false because your subtraction gave in "8x = 8" when it should clearly be "9x = 8", which results in an arguably true conclusion.

 

[blahblah99]

Originally posted by: zugzoog

Originally posted by: Dinominant
OK, here is a valid proof aquired from Dr. Math:

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

I disagree with this demonstration with regard to the step

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

Since you have raised 0.9999[r] by an order of magnitude when you complete the subtraction the result will actually be

9x = 9 - 0.000....0009

therefore x != 1.

Ok, the subtracted number is undefinable, but while it is undefinable it is != 0 therefore the logic holds.

The faulty logic may be more easily understood if I follow the logic presented but substitute 0.8888[r] instead of 0.99999[r]

x = 0.8888888...
10x = 8.88888...
10x - x = 8.8888... - 0.88888...
8x = 8
x = 1

Which is clearly false.

It may just be my ignorance, but where did you get 0.0000..09? 0.9999...repeating times 10 = 9.999999...repeating.


I havn't seen a valid contradicting proof yet in this post, or I'm just stubborn.

 

[zugzoog]

Your stubborness is understandable as the concept of infinities is one of the most difficult concepts in mathematics.

But the point of my post is that 9.999999[r] - 0.999999[r] != 1 when 9.99999[r] is 0.999999[r] * 10

Remember that 10x is 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r]

so that 10x - x is 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r] + 0.99[r]

to then go to the step 10x - x = 9x = 9 assumes that 0.99999[r] = 1 in the attempt to demonstrate that 0.99999[r] = 1

0.0000..09 is my best description of the difference that would result from the subtraction.



Oh and please disregard the logic statement with the substitution of 0.888888[r], I have introduced a false step of my own. Ooops.

 

[Sahakiel]

Originally posted by: zugzoog
Your stubborness is understandable as the concept of infinities is one of the most difficult concepts in mathematics.

Refer to my first post in this specific thread.

 

[sao123]

I follow what zugzoog is trying to say... and it is incorrect...
He is looking for something like this to happen

.99999
x 10
--------
_.00000
9.9999_
---------
9.99990


by definition the repeating decimal .99999999999999999999..... goes on forever...
multiplying such a number by 10 does not produce a number 9.99....990
but it produces 9.99....999 Because there are infinite nines, you never reach the terminal of the number to multiply thus producing the zero you expect.

.9999999999999999999999999999999999999999999999999999999999999999999999999999999999...
x 10
--------------------------------------------------------------------------------------------------------------------
_.000000000000000000000000000000000000000000000000000000000000000000000000000000000...
9.999999999999999999999999999999999999999999999999999999999999999999999999999999999...
----------------------------------------------------------------------------------------------------------------------
9.999999999999999999999999999999999999999999999999999999999999999999999999999999999

Therefore since the last digit of both decimals is a 9, and not a 0, 9-9=0.
therefore 9.99999........... - .99999......... = 9.0