A simple, easy to understand way of proving the equality is to think of it in terms of a number line. If you have two separate numbers, you can be sure that there are numbers between them.

Take 2 and 3 for example. Draw 2 and 3 on a number line, between those two numbers you can find 2.1, 2.2, 2.3, 2.31, 2.32, and so forth.

On the other hand, if you draw 4 and 4 on a number line, you won't be able to find any numbers between them.

There is so much evidence as to why .9...=1, it's not even funny. But people like me will continue to believe otherwise just because we simply refuse to accept it.

Why would you point out that there is so much evidence, but continue to believe otherwise?

I didn't watch that video, so I don't know if it says this but here is another example (I also like dednoob6's):

Say .999... is equal to x.
Multiply each side by, say, 10.
You now have 9.999... = 10x
Now subtract from both sides...
You now have 9 = 9x
This can be simplified to 1 = x, or 1 = .999...

.999... = x
9.999... = 10x
9 = 9x
1 = x

If you refuse to accept it, well... you should be open to new ideas. Some of them are true, like this one.

If a number goes on INFINITELY, it is essentially the same as the whole number. I COULD say that 15 is 14.9999...., but it makes it a lot easier to say 15.

That violates the change of base rule set by mathematics.

A simple, easy to understand way of proving the equality is to think of it in terms of a number line. If you have two separate numbers, you can be sure that there are numbers between them.

Take 2 and 3 for example. Draw 2 and 3 on a number line, between those two numbers you can find 2.1, 2.2, 2.3, 2.31, 2.32, and so forth.

On the other hand, if you draw 4 and 4 on a number line, you won't be able to find any numbers between them.

What numbers can you find between 0.999... and 1?

Why would you point out that there is so much evidence, but continue to believe otherwise?

Another excellent way of putting it. Didn't think of it.

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I always wanted to argue with a brick wall, I suppose the internet is the second best option.

I like MrDenco. But I can not be afraid to point out when he's wrong,

Your correct that the difference between the magnitudes expressed by 0.9~ and 1 is infinitesimal, and that for certain fields of mathematics infinitesimal and zero are functionally equivalent.

However the key mathematic field of Interest in the process of numbering is Set Theory.

And Set theory leaves room for an infinite amount of numbers between any different two expressible values...

There certainly different values despite the fact that they sit extremely close together on the number line.

Heres a very simple example i just thought of to illustrate the difference between the numbers which will always work on any calculator or computer forever...

Heres a very simple example i just thought of to illustrate the difference between the numbers which will always work on any calculator or computer forever...

And Set theory leaves room for an infinite amount of numbers between any different two expressible values...

Firstly, nice to see you again, and I am glad you like me!!! Haha

So, everything you said is correct. The part where you are actually wrong is that .999 repeating isn't a rational number, therefore not an expressible one. There is now way to represent .9 repeating with a fraction.

A simple, easy to understand way of proving the equality is to think of it in terms of a number line. If you have two separate numbers, you can be sure that there are numbers between them.

Take 2 and 3 for example. Draw 2 and 3 on a number line, between those two numbers you can find 2.1, 2.2, 2.3, 2.31, 2.32, and so forth.

On the other hand, if you draw 4 and 4 on a number line, you won't be able to find any numbers between them.

What numbers can you find between 0.999... and 1?

Why would you point out that there is so much evidence, but continue to believe otherwise?

This proof involves a logical fallacy, so it is incorrect.

If you put the same number on a numberline twice, then there would be no values between them. This is a true statement. Let's call this P -> Q.

"If there is no space between two numbers, then they are the same number". Q -> P.

Also, a simple proof that 0.9 repeating does NOT equal 1:

Let A = 0.999… , where A has as many 9s in it as possible. Suppose A = 1.
The greatest number of 9s possible is P, the penultimate number. So A consists of a decimal followed by P nines.
Now let B = 0.00000…1 \neq 0 , where B has P zeros in it (including the one before the decimal). Then:
A + B = 1
1-(A+B ) = 1-(1)
1-(A+B ) = 0
(1-A) + B = 0 (Associative property)
(1-1) + B = 0
B = 0 *contradiction*
Therefore, A \neq 1. QED.

Also also, 0.33333... does not equal 1/3.

Even a penultimate number of 3s after the decimal won’t reach 1/3. Thus the .333…*3 = .999… analogy doesn’t apply. It only tends to approach 1/3, which is why it saves time in calculations to call them equal. In calculus, though, 0.3333... repeating, by definition, will never equal 1/3. 1/3 would be 0.333...4.

According to that article, in order for you to prove that I used a logical fallacy, you would have to prove that P is not the only sufficient condition for Q.

If you can't find points between any given two points on a line, then they are the same point.
You can't find points between 0.999... and 1
Therefore, 0.999... = 1

So what is the relationship between these numbers that allows them to occupy different points on a line, but not have any points between them?

If you have a few minutes, you should also watch this video:

Also, a simple proof that 0.9 repeating does NOT equal 1:

Let A = 0.999… , where A has as many 9s in it as possible. Suppose A = 1.
The greatest number of 9s possible is P, the penultimate number. So A consists of a decimal followed by P nines.
Now let B = 0.00000…1 \neq 0 , where B has P zeros in it (including the one before the decimal). Then:
A + B = 1
1-(A+B ) = 1-(1)
1-(A+B ) = 0
(1-A) + B = 0 (Associative property)
(1-1) + B = 0
B = 0 *contradiction*
Therefore, A \neq 1. QED.

Also also, 0.33333... does not equal 1/3.

Even a penultimate number of 3s after the decimal won’t reach 1/3. Thus the .333…*3 = .999… analogy doesn’t apply. It only tends to approach 1/3, which is why it saves time in calculations to call them equal. In calculus, though, 0.3333... repeating, by definition, will never equal 1/3. 1/3 would be 0.333...4.

0.333 repeated suggests that the 3 repeats infintely. You can't say the last one will equal 4 eventually, because there is no last one. That is the way people and calculators write it to save time, in the same way that 0.33334 is a way of rounding the number to an estimate, repeating 3 a finite amount of times and putting a 4 at the end does not equal 0.333 with an infintely repeating sequence of 3s.

0.333 with an infinitely repeating sequence of 3s, literally equals 1/3.

1/3 x 3 = 3/3

3/3 = 1

If you have a minute, you should also watch this video:

You're right that there is a fundamental material inexpressibility of these strange numbers...

Infact my arithmetic example was heavily grounded in my knowledge of the IEEE floating point hardware data conversion rules being used inside calculators which include special conditions for among other-things the storage of signed infinitesimal values...

This was actually a total cop-out in many respects, and I knew it, but really I was just wanting to generate some verifiable proof in my favour...

for all natural numbers beyond one - there is no special conversion rules able to preserve the fidelity of infinitesimal values.

( or put simply ) My example only works because there is a special state floating-point-numerical-computer-representations can be in which logically means 'too small to store, but definitely not 0'

so, my example doesn't work for the other cases..
for example, given the correct parameters it would actually return that 14.9~ is equal to 15.

But I'm afraid to say; this is where your wrong señor Denco.

I was trying to hold back on the use of set-theory, but your inexpressibly argument really drives home the reason why set theory really is the only court in which this hearing can be presided.

Set theory is about defining sets, where a set is just a number of numbers...

For example the set of numbers 'between' 0 and 1 includes the number 0.999~ but does not include the number 1.

But what about the fact that you can't even Express ! that number, to begin to ask the question ? ( you ask )

Thats actually not a problem in set-theory : )

The reason is that set theory deals with n'th order logic; both for sets and for numbers.
And there are certainly higher-order perfectly accurate logical definitions for these weird numbers.

For example 0.999~ can validly-meaningfully be defined as " 0 followed by an infinite series of 9s after a decimal place "

According to set theory - 1 and 0.9~ belong to different sets and therefor would fail the equality test.

All other fields of mathematics ( besides perhaps the lambda calculus ) attempts to approximately model set theory; which is exactly why the IEEE included the special case for infinitesimal values around 0 and 1 ( ie. because it improves the approximation )

-- So there you have it. Our purest mathematic models say 0.9~ != 1 ---

On a side note, whether 0.9~ = 1 in Lambda is another story all together !!!
And even as a relative expert on λ-calculus, I would certainly-not want to tackle that !
( because in λ all numbers must be defined as sets of recursive operations over the functions λfx.x )

I hope that wasn't overly technical / boring.. But your a smart guy and you deserve a smart answer : )

According to that article, in order for you to prove that I used a logical fallacy, you would have to prove that P is not the only sufficient condition for Q.

If you can't find points between any given two points on a line, then they are the same point.
You can't find points between 0.999... and 1
Therefore, 0.999... = 1

So what is the relationship between these numbers that allows them to occupy different points on a line, but not have any points between them?

No, it doesn't. I can show that your reasoning is faulty without having to provide an alternative or counterexample. I can't claim that your cause is incorrect because you made a fallacy, but I can claim that your arguement is incorrect.

Also, the nature of repeating decimals make your example somewhat faulty. Wouldn't 0.999... repeating get closer to 1 on the number line as the amount of 9's after the decimal place gets larger and larger? And, of course, the number of 9's is infinite. So the mark gets closer and closer to 1 infinitely as you let the 9's generate, but never actually reaches it. Therefore, there are values between 0.999... repeating and 1.

0.333 repeated suggests that the 3 repeats infintely. You can't say the last one will equal 4 eventually, because there is no last one. That is the way people and calculators write it to save time, in the same way that 0.33334 is a way of rounding the number to an estimate, repeating 3 a finite amount of times and putting a 4 at the end does not equal 0.333 with an infintely repeating sequence of 3s.

0.333 with an infinitely repeating sequence of 3s, literally equals 1/3.

1/3 x 3 = 3/3

3/3 = 1

That's the point. 0.3333... repeating will NEVER equal 1/3. It only approaches it. See my numberline example before. For it to equal 1/3, it has to be rounded at some point, because a number that continues infinitely cannot be definite.

0.333... repeating can be better represented as the series

or {0.3 + 0.03 + 0.003 + 0.0003...} and so on.

Notice the infinity sign above the sigma. That means it continues on indefinitely. You cannot create a definite fraction out of an indefinite number.

Heres a very simple example i just thought of to illustrate the difference between the numbers which will always work on any calculator or computer forever...

There are infinitely many values that exist between them, because the 9's repeat infinitely, and with each 9, the number gets slightly closer to 1.

But that's not the number we're talking about. 0.999 repeating already has all the nines, all of them. If there's any 9s left to add, then it's not the number we're talking about.

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I don't get anything about this, so I'm just going to put my view forward, so y'all else can shoot it down and tell me why my reasoning is flawed.

Instead of considering 0.99999... , let's consider the difference between 1 and 0.99999... . (Damn, I have to put a space between the ...'s and punctuation. ;D)

Anyways, as we very well all know:
1- 0.9= 0.1
1- 0.99= 0.01
1- 0.999= 0.001
...
1- 0.99999...=0.00000...1

And so on. Thus, if we consider the reasoning that 0.99999... has all the nines, as MrQuizzles said above, we are also assuming that the difference between 0.99999 and 1.0 is very small, but still existent quantity: 0.00000...1. Correct? Using this logic, this is why I believe that 0.99999...=/= 1.0, but is instead used as merely a very close approximation. Again, feel free to correct me, as I'm rather stupid with mathematics, logic as well (not so smart in general ;D).

I'd like to think of myself as intelligent, but that, given the sheer variety and vastness of the cosmos that surrounds me, it is readily apparent that this is not the case.

But that's not the number we're talking about. 0.999 repeating already has all the nines, all of them. If there's any 9s left to add, then it's not the number we're talking about.

Infinity isn't a countable number. You can't have "all of them". It's part of the concept.

If there aren't any 9's left to add, then the decimal wouldn't be repeating anymore. It would only be 0.99999999...9. Ending in a nine rather than repeating more. That, of course, does not equal 1, not any more than 0.99 would.

And so on. Thus, if we consider the reasoning that 0.99999... has all the nines, as MrQuizzles said above, we are also assuming that the difference between 0.99999 and 1.0 is very small, but still existent quantity: 0.00000...1. Correct?

Here's the problem: You can't put a number at the end of an infinite sequence. It's an infinite sequence. The 0s will never end. That 1 will never be there. You'll get just 0.

Infinity isn't a countable number. You can't have "all of them". It's part of the concept.

If there aren't any 9's left to add, then the decimal wouldn't be repeating anymore. It would only be 0.99999999...9. Ending in a nine rather than repeating more. That, of course, does not equal 1, not any more than 0.99 would.

My point was that you can't keep adding 9s because they're already all there. It is impossible to make it closer to 1 by adding more 9s. It's so close to 1 that it IS 1.

You're right that there is a fundamental material inexpressibility of these strange numbers...

Infact my arithmetic example was heavily grounded in my knowledge of the IEEE floating point hardware data conversion rules being used inside calculators which include special conditions for among other-things the storage of signed infinitesimal values...

This was actually a total cop-out in many respects, and I knew it, but really I was just wanting to generate some verifiable proof in my favour...

for all natural numbers beyond one - there is no special conversion rules able to preserve the fidelity of infinitesimal values.

( or put simply ) My example only works because there is a special state floating-point-numerical-computer-representations can be in which logically means 'too small to store, but definitely not 0'

so, my example doesn't work for the other cases..
for example, given the correct parameters it would actually return that 14.9~ is equal to 15.

But I'm afraid to say; this is where your wrong señor Denco.

I was trying to hold back on the use of set-theory, but your inexpressibly argument really drives home the reason why set theory really is the only court in which this hearing can be presided.

Set theory is about defining sets, where a set is just a number of numbers...

For example the set of numbers 'between' 0 and 1 includes the number 0.999~ but does not include the number 1.

But what about the fact that you can't even Express ! that number, to begin to ask the question ? ( you ask )

Thats actually not a problem in set-theory : )

The reason is that set theory deals with n'th order logic; both for sets and for numbers.
And there are certainly higher-order perfectly accurate logical definitions for these weird numbers.

For example 0.999~ can validly-meaningfully be defined as " 0 followed by an infinite series of 9s after a decimal place "

According to set theory - 1 and 0.9~ belong to different sets and therefor would fail the equality test.

All other fields of mathematics ( besides perhaps the lambda calculus ) attempts to approximately model set theory; which is exactly why the IEEE included the special case for infinitesimal values around 0 and 1 ( ie. because it improves the approximation )

-- So there you have it. Our purest mathematic models say 0.9~ != 1 ---

On a side note, whether 0.9~ = 1 in Lambda is another story all together !!!
And even as a relative expert on λ-calculus, I would certainly-not want to tackle that !
( because in λ all numbers must be defined as sets of recursive operations over the functions λfx.x )

I hope that wasn't overly technical / boring.. But your a smart guy and you deserve a smart answer : )

I don't get anything about this, so I'm just going to put my view forward, so y'all else can shoot it down and tell me why my reasoning is flawed.

Instead of considering 0.99999... , let's consider the difference between 1 and 0.99999... . (Damn, I have to put a space between the ...'s and punctuation. ;D)

Anyways, as we very well all know:
1- 0.9= 0.1
1- 0.99= 0.01
1- 0.999= 0.001
...
1- 0.99999...=0.00000...1

And so on. Thus, if we consider the reasoning that 0.99999... has all the nines, as MrQuizzles said above, we are also assuming that the difference between 0.99999 and 1.0 is very small, but still existent quantity: 0.00000...1. Correct? Using this logic, this is why I believe that 0.99999...=/= 1.0, but is instead used as merely a very close approximation. Again, feel free to correct me, as I'm rather stupid with mathematics, logic as well (not so smart in general ;D).

~Bair

I can understand your reasoning, but I wish to explain why it is wrong. (Still your input is super greatly appreciated!)

This is dealing with a geometric series. A geometric series deals with infinity, and they put real numbers in relation to infinity. the difference goes from .1 to .01, to .001, .0001 .00001 .000001 .0000001 and so on. So essentially where there is a '9' there is a zero, and after the last nine there is a 1 (.999 is .0001, 3 nines, gives 3 zeros, and right after a 1) but infinity has no end, so that there isn't the possibilby of that one coming after the last nine, simply because it doesn't exist

That's the point. 0.3333... repeating will NEVER equal 1/3. It only approaches it. See my numberline example before. For it to equal 1/3, it has to be rounded at some point, because a number that continues infinitely cannot be definite.

0.333... repeating can be better represented as the series

or {0.3 + 0.03 + 0.003 + 0.0003...} and so on.

Notice the infinity sign above the sigma. That means it continues on indefinitely. You cannot create a definite fraction out of an indefinite number.

.

If you watch my video you will find the explanation on why .333... is 1/3. It uses the proven geometric summation theorem. I used the exact example in the video, which is officially recognized by every algebra 2 book in production, college professors, and mathematicians.

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I always wanted to argue with a brick wall, I suppose the internet is the second best option.

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@MrQuizzles, MrDenco

Got it! Thank you for the explanations!

~Bair

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I'd like to think of myself as intelligent, but that, given the sheer variety and vastness of the cosmos that surrounds me, it is readily apparent that this is not the case.

Good times, good times.

<- Please. ._.

Take 2 and 3 for example. Draw 2 and 3 on a number line, between those two numbers you can find 2.1, 2.2, 2.3, 2.31, 2.32, and so forth.

On the other hand, if you draw 4 and 4 on a number line, you won't be able to find any numbers between them.

What numbers can you find between 0.999... and 1?

Why would you point out that there is so much evidence, but continue to believe otherwise?

That violates the change of base rule set by mathematics.

Another excellent way of putting it. Didn't think of it.

I like MrDenco. But I can not be afraid to point out when he's wrong,

Your correct that the difference between the magnitudes expressed by 0.9~ and 1 is infinitesimal, and that for certain fields of mathematics infinitesimal and zero are functionally equivalent.

However the key mathematic field of Interest in the process of numbering is Set Theory.

And Set theory leaves room for an infinite amount of numbers between any different two expressible values...

There certainly different values despite the fact that they sit extremely close together on the number line.

(0-0.999~ + 1)* INF = INF

-However-

(0-1 + 1)* INF = 0

You can run and tell that homeboy. : P

Firstly, nice to see you again, and I am glad you like me!!! Haha

So, everything you said is correct. The part where you are actually wrong is that .999 repeating isn't a rational number, therefore not an expressible one. There is now way to represent .9 repeating with a fraction.

Love the

by the way. Excellent point actually!!!

This proof involves a logical fallacy, so it is incorrect.

If you put the same number on a numberline twice, then there would be no values between them. This is a true statement. Let's call this P -> Q.

"If there is no space between two numbers, then they are the same number". Q -> P.

Your arguement:

P -> Q

Q, therefore P.

This does

notfollow logically.http://en.wikipedia...._the_consequent

Also, a simple proof that 0.9 repeating does NOT equal 1:

Let A = 0.999… , where A has as many 9s in it as possible. Suppose A = 1.

The greatest number of 9s possible is P, the penultimate number. So A consists of a decimal followed by P nines.

Now let B = 0.00000…1 \neq 0 , where B has P zeros in it (including the one before the decimal). Then:

A + B = 1

1-(A+B ) = 1-(1)

1-(A+B ) = 0

(1-A) + B = 0 (Associative property)

(1-1) + B = 0

B = 0 *contradiction*

Therefore, A \neq 1. QED.

Also also, 0.33333... does not equal 1/3.

Even a penultimate number of 3s after the decimal won’t reach 1/3. Thus the .333…*3 = .999… analogy doesn’t apply. It only tends to approach 1/3, which is why it saves time in calculations to call them equal. In calculus, though, 0.3333... repeating, by definition, will never equal 1/3. 1/3 would be 0.333...4.

According to that article, in order for you to prove that I used a logical fallacy, you would have to prove that P is not the only sufficient condition for Q.

If you can't find points between any given two points on a line, then they are the same point.

You can't find points between 0.999... and 1

Therefore, 0.999... = 1

So what is the relationship between these numbers that allows them to occupy different points on a line, but not have any points between them?

0.333 repeated suggests that the 3 repeats

infintely.You can't say the last one will equal 4 eventually, because there is no last one. That is the way people and calculators write it to save time, in the same way that 0.33334 is a way of rounding the number to an estimate, repeating 3 a finite amount of times and putting a 4 at the end does not equal 0.333 with an infintely repeating sequence of 3s.0.333 with an

infinitelyrepeating sequence of 3s,equals 1/3.literally1/3 x 3 = 3/3

3/3 = 1

If you have a minute, you should also watch this video:

You're right that there is a fundamental material inexpressibility of these strange numbers...

Infact my arithmetic example was heavily grounded in my knowledge of the IEEE floating point hardware data conversion rules being used inside calculators which include special conditions for among other-things the storage of signed infinitesimal values...

This was actually a total cop-out in many respects, and I knew it, but really I was just wanting to generate some verifiable proof in my favour...

for all natural numbers beyond one - there is no special conversion rules able to preserve the fidelity of infinitesimal values.

( or put simply ) My example only works because there is a special state floating-point-numerical-computer-representations can be in which logically means 'too small to store, but definitely not 0'

so, my example doesn't work for the other cases..

for example, given the correct parameters it would actually return that 14.9~ is equal to 15.

But I'm afraid to say; this is where your wrong señor Denco.

I was trying to hold back on the use of set-theory, but your inexpressibly argument really drives home the reason why set theory really is the only court in which this hearing can be presided.

Set theory is about defining sets, where a set is just a number of numbers...

For example the set of numbers 'between' 0 and 1 includes the number 0.999~ but does not include the number 1.

But what about the fact that you can't even Express ! that number, to begin to ask the question ? ( you ask )

Thats actually not a problem in set-theory : )

The reason is that set theory deals with n'th order logic; both for sets and for numbers.

And there are certainly higher-order perfectly accurate logical definitions for these weird numbers.

For example 0.999~ can validly-meaningfully be defined as " 0 followed by an infinite series of 9s after a decimal place "

According to set theory - 1 and 0.9~ belong to different sets and therefor would fail the equality test.

All other fields of mathematics ( besides perhaps the lambda calculus ) attempts to approximately model set theory; which is exactly why the IEEE included the special case for infinitesimal values around 0 and 1 ( ie. because it improves the approximation )

-- So there you have it. Our purest mathematic models say 0.9~ != 1 ---

On a side note, whether 0.9~ = 1 in Lambda is another story all together !!!

And even as a relative expert on λ-calculus, I would certainly-not want to tackle that !

( because in λ all numbers must be defined as sets of recursive operations over the functions λfx.x )

I hope that wasn't overly technical / boring.. But your a smart guy and you deserve a smart answer : )

No, it doesn't. I can show that your reasoning is faulty without having to provide an alternative or counterexample. I can't claim that your

causeis incorrect because you made a fallacy, but I can claim that your arguement is incorrect.Also, the nature of repeating decimals make your example somewhat faulty. Wouldn't 0.999... repeating get closer to 1 on the number line as the amount of 9's after the decimal place gets larger and larger? And, of course, the number of 9's is infinite. So the mark gets closer and closer to 1

infinitelyas you let the 9's generate, but never actually reaches it. Therefore, therearevalues between 0.999... repeating and 1.That's the point. 0.3333... repeating will NEVER equal 1/3. It only approaches it. See my numberline example before. For it to equal 1/3, it has to be rounded at some point, because a number that continues infinitely cannot be definite.

0.333... repeating can be better represented as the series

or {0.3 + 0.03 + 0.003 + 0.0003...} and so on.

Notice the infinity sign above the sigma. That means it continues on indefinitely. You cannot create a definite fraction out of an indefinite number.

Then give me one number that exists between 0.999... and 1.

This is, of course, an invalid example because computers and calculators take shortcuts and can't be considered accurate in such circumstances.

There are infinitely many values that exist between them, because the 9's repeat infinitely, and with each 9, the number gets slightly closer to 1.

But that's not the number we're talking about. 0.999 repeating already has all the nines, all of them. If there's any 9s left to add, then it's not the number we're talking about.

Instead of considering 0.99999... , let's consider the difference between 1 and 0.99999... . (Damn, I have to put a space between the ...'s and punctuation. ;D)

Anyways, as we very well all know:

1- 0.9= 0.1

1- 0.99= 0.01

1- 0.999= 0.001

...

1- 0.99999...=0.00000...1

And so on. Thus, if we consider the reasoning that 0.99999... has

allthe nines, as MrQuizzles said above, we are also assuming that the difference between 0.99999 and 1.0 is very small, but still existent quantity: 0.00000...1. Correct? Using this logic, this is why I believe that 0.99999...=/= 1.0, but is instead used as merely a very close approximation. Again, feel free to correct me, as I'm rather stupid with mathematics, logic as well (not so smart in general ;D).~Bair

Infinity isn't a countable number. You can't have "all of them". It's part of the concept.

If there aren't any 9's left to add, then the decimal wouldn't be repeating anymore. It would only be 0.99999999...9. Ending in a nine rather than repeating more. That, of course, does not equal 1, not any more than 0.99 would.

Here's the problem: You can't put a number at the end of an infinite sequence. It's an infinite sequence. The 0s will never end. That 1 will never be there. You'll get just 0.

My point was that you can't keep adding 9s because they're already all there. It is impossible to make it closer to 1 by adding more 9s. It's so close to 1 that it IS 1.

Mhhhmmm it does kind of make sense, but you know, I am 15 so like.... http://bocsupportnetwork.com/wp-content/uploads/2012/07/admin/1427/30/inoneear1.jpg

I can understand your reasoning, but I wish to explain why it is wrong. (Still your input is super greatly appreciated!)

This is dealing with a geometric series. A geometric series deals with infinity, and they put real numbers in relation to infinity. the difference goes from .1 to .01, to .001, .0001 .00001 .000001 .0000001 and so on. So essentially where there is a '9' there is a zero, and after the last nine there is a 1 (.999 is .0001, 3 nines, gives 3 zeros, and right after a 1) but infinity has no end, so that there isn't the possibilby of that one coming after the last nine, simply because it doesn't exist .

If you watch my video you will find the explanation on why .333... is 1/3. It uses the proven geometric summation theorem. I used the exact example in the video, which is officially recognized by every algebra 2 book in production, college professors, and mathematicians.

Got it! Thank you for the explanations!

~Bair

Say we have a function f(x) that gives us the value of the number 0.9... where 'x' is the number of 9s after the decimal place.

f(1) = 0.9

f(2) = 0.99

f(3) = 0.999

and so on.

So now what happens when we do this:

Lim f(x) as X->INF = ??

1, the answer is 1. When we get to infinity 9s, which is what the repeating bit implies, the value of the number is 1.