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[Squirrel Killer]

Ok, since it's been kicked out of the Blizzard thread, how can the theory that "if there's no number between two other numbers, the two numbers are equal" be true? Since equality is transitive (if A=B and B=C, then A=C, how does this theory not result in everything being equal? Anyone got a non-math nerd explaination of this?

Also, how can .99... = 1.00... given the definition of equal? It would seem to me that while two numbers may be infinitesimally close, and, for practical purposes, roughly equivalent, they would still be, by definition, two seperate, non-equal numbers.

 

[LakeEarth]

They are equal. If you don't believe it go to England and kick Issac Newton's ass.

 

[Mihail]

Yeah, this is a better idea than hijacking those poor dudes' thread.

OK, is there a number between 1 and 2? Yeah, there are lots of them. One of them is 1.1.

Is there a number between 1.0 and 1.1? Yeah, there are lots of them. One of them is 1.01.

Is there a number between 1.0 and 1.01? Yeah, there are lots of them. One of them is 1.001.

Is there a numbe between 1.0 and 1.001? Yeah, there are lots of them. One of them is 1.0001.

No matter how close two different numbers are, there is always an infinite amount of numbers between them.

The only time there are no numbers between them is when there are equal. What are the numbers between them? None, because there is no difference.

There are no numbers between 1 and 1 because they are equal. They are equal because there are no numbers between them. There are no numbers between .999... and 1, thus they are equal.

Also, how can .99... = 1.00... given the definition of equal? It would seem to me that while two numbers may be infinitesimally close, and, for practical purposes, roughly equivalent, they would still be, by definition, two seperate, non-equal numbers.

There is a great fallacy in this argument, which is that it basically boils down to "They can't be equal because they're not equal." You're using the (false) conclusion as an assumption in the argument.

 

[Tortfeasor]

How about this? .99_ and 1 and not equal. They are just sequential.

 

[Mihail]

How about this? .99_ and 1 and not equal. They are just sequential.

What does that mean?

 

[Squirrel Killer]

They are equal. If you don't believe it go to England and kick Issac Newton's ass.

Huh? That's not in non-math nerd terminology

The only time there are no numbers between them is when there are equal. What are the numbers between them? None, because there is no difference.

There are no numbers between 1 and 1 because they are equal. They are equal because there are no numbers between them. There are no numbers between .999... and 1, thus they are equal.

This seems like a very backward definition of "equal." The definition of "equal" is not "no numbers between," it's "the same or identical to in value." "No numbers in between" is just a property of being equal. All automobiles have an engine (by definition), however not everything that has an engine is an automobile. Sure, there are no numbers between .99... and 1.00..., so it seems entirely possible to have two nonequal numbers with no intervening numbers. If two numbers are infinitesimally close to one another are equal, and since equality is transitive, then the two infinitesimally close numbers on either side of the first two numbers are also equal. And so on, and so on, until all numbers are equal to one another.

It really sounds like you're explaination is a begging the question type of fallacy. That is to say, .99... is equal to 1.00... because there are no intervening numbers, and no intervening numbers means numbers are equal because .99... is equal to 1.00...

There is a great fallacy in this argument, which is that it basically boils down to "They can't be equal because they're not equal." You're using the (false) conclusion as an assumption in the argument.

That's not a fallacy, it's a tautology.

 

[LakeEarth]

Anyone who hasn't taken a calculus class should be banned from discussing this stuff. If you haven't learned about limits then there is no point in trying to figure it out.

Huh? That's not in non-math nerd terminology

I dunno what you mean by that. Issac Newton is the 'inventor' of calculus, though its more of a discovery than an invention. Its always been there, in the numbers.

 

[Papi]

1 - .999_ = ?

 

[koam]

.999_ is equal to 1.

The only way of getting 0.999_ is by taking division and multiplication.

1 = 1/3 + 1/3 + 1/3

or

1 = 0.333_ + 0.333_ + 0.333_

You can't actually write 0.999_ since you would never be able to finish writing it, it's infinate.

Basically, 0.999_ is 1 because 0.999_ doesn't exist

 

[LakeEarth]

Zero.

Here's the best, simplest explanation I could find on the net.

http://mathforum.org/dr.math/faq/faq.0.9999.html

Why does 0.9999... = 1 ?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl).
The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum.

So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers

9/10,
9/10 + 9/100,
9/10 + 9/100 + 9/1000,
...."


One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1:

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1


Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
0.9999... = 1

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


Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99.

References
R.V. Churchill and J.W. Brown. Complex Variables and Applications. 0.9999... = 1 ed., McGraw-Hill, 1990.

E. Hewitt and K. Stromberg. Real and Abstract Analysis. Springer-Verlag, Berlin, 1965.

W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.

L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.

So in short, until you get a paper published in a math journal, just accept it.

 

[Mihail]

Huh? That's not in non-math nerd terminology


This seems like a very backward definition of "equal." The definition of "equal" is not "no numbers between," it's "the same or identical to in value." "No numbers in between" is just a property of being equal. All automobiles have an engine (by definition), however not everything that has an engine is an automobile. Sure, there are no numbers between .99... and 1.00..., so it seems entirely possible to have two nonequal numbers with no intervening numbers. If two numbers are infinitesimally close to one another are equal, and since equality is transitive, then the two infinitesimally close numbers on either side of the first two numbers are also equal. And so on, and so on, until all numbers are equal to one another.

It really sounds like you're explaination is a begging the question type of fallacy. That is to say, .99... is equal to 1.00... because there are no intervening numbers, and no intervening numbers means numbers are equal because .99... is equal to 1.00...


That's not a fallacy, it's a tautology.

Yes, but it can be logically proven that statement p (there are no numbers in between A and B) implies statement q (A and B are equal). And you don't have to link me to math terms I know what they mean.

You used tautology incorrectly. You used it to say Q <-> Q, but a tautology needs two different statements, such as P <-> Q.

As far as it seeming perfectly possible to have 2 nonequal numbers with no intervening numbers -- that's simply not true. Can you give me an example of 2 such numbers without using the number (.999...) in question?

 

[Trident]

HAY GUYS, I'm no MATH WIZARD, which is important since math is comprised of ELEMENTAL MAGICKS, but I don't see how a SERIES can be INFINITE. I mean, JEEZ, if you can't explain it to me without using some sort of MATH NERDERY, then clearly you are wrong!!!

 

[Alcibiades]

The answer does not exist in algebra, but it does in calculus under a
concept known as limit theory.

Here's how it works. Assume that .9 forever is not equal to one. If it is
not equal, then you could suggest a number (say .00000001) that represents
the difference. But no matter what number you suggest, I can run a string of
9's that proves you wrong. Therefore, .9 forever must be equal to 1.

In proper math terminology, for every delta (the proposed difference), there
exists an epsilon (a smaller actual difference). Therefore the ultimate
difference is zero, so the two numbers must be equal.

 

[cvxfreak]

The answer does not exist in algebra, but it does in calculus under a
concept known as limit theory.

Here's how it works. Assume that .9 forever is not equal to one. If it is
not equal, then you could suggest a number (say .00000001) that represents
the difference. But no matter what number you suggest, I can run a string of
9's that proves you wrong. Therefore, .9 forever must be equal to 1.

In proper math terminology, for every delta (the proposed difference), there
exists an epsilon (a smaller actual difference). Therefore the ultimate
difference is zero, so the two numbers must be equal.

All I can say is: Damn.

 

[WHOAguitarninja]

As others have mentioned, if you've taken calculus, limits would have been the first thing that popped into your head as an answer to your question. That's what it's there for (well....sortof)

 

[Error Macro]

In proper math terminology, for every delta (the proposed difference), there exists an epsilon (a smaller actual difference). Therefore the ultimate
difference is zero, so the two numbers must be equal.

So basically, it's like throwing up your hands and saying "whatever."

 

[borghe]

show me a way to derive .999... from an equation WITHOUT adding together repeating decimals.

The answer does not exist in algebra, but it does in calculus under a
concept known as limit theory.

Here's how it works. Assume that .9 forever is not equal to one. If it is
not equal, then you could suggest a number (say .00000001) that represents
the difference. But no matter what number you suggest, I can run a string of
9's that proves you wrong. Therefore, .9 forever must be equal to 1.

In proper math terminology, for every delta (the proposed difference), there
exists an epsilon (a smaller actual difference). Therefore the ultimate
difference is zero, so the two numbers must be equal.

I meet your .999... bullshit number with .000...1 bullshit number.

.999... is not a number. there is no way to arrive at it aside from the theoretical bullshit of adding together repeating decimals. you can't divide one number by another number and arrive at .999... thbe fact that some jokers insist you can add two repeating numbers together, even though I proved in the blizzard thread that you really can't logically perform mathematics on repeating numbers without resorting to performing math on the actual fractions the repeating numbers are derived from originally, doesn't make it fact or reality.

subtract .888... from 1.444.... And don't do the math on the fractions but on the decimal representations of the fractions. the only way to arrive at the correct answer is by doing the math on the fractions which ends up being .555...

as someone else said, repeating decimals are just decimal approximations of fractions. doing math on a repeating number assuming out to infinity is theoretical at best.

 

[Trident]

The Real number line is continuous. IE, for every two numbers, there is an infinite number of numbers between them. That's the difference between a continuous number line and a discrete number line. Since there are NO numbers between .999_ and 1, they must be the same number. This has been said several times... I don't see what the problem is.

What the hell do you mean, not a number? Any sequence of digits is a number...

 

[borghe]

The Real number line is continuous. IE, for every two numbers, there is an infinite number of numbers between them. That's the difference between a continuous number line and a discrete number line. Since there are NO numbers between .999_ and 1, they must be the same number. This has been said several times... I don't see what the problem is.

What the hell do you mean, not a number? Any sequence of digits is a number...

so by this rule any number that ends in 9's repeating is equal to the next number up.

so .8999_ = .9
.567849999_ = .5675

I call bullshit on this with no proof other than people are saying so. there is no proof in math aside from a rule that says numbers have to exist between other numbers.

 

[Mihail]

so basically .000_1 is between .999_ and 1

For the record, you never proved that you can't add 2 repeating decimals together. You most certainly CAN add repeating decimals together.

.000_1 is an impossible number. That would be like me writing 5.cow -- it makes no sense. You can't put a 1 at the end of an infinite number of 0s.

.999... is not a BS number. It is a real, rational number. And we can treat it as such because we know exactly what happens when you extend it to infinity... the 9s keep repeating. The same thing is true for .333... and .666...

We can add them because we know exactly what happens to them... their digits repeat forever. I don't know where you get the crazy idea that you can't add repeating decimals. Repeating decimals are RATIONAL numbers. And, hell, we can add irrational numbers like Pi anyway. If we couldn't, all this circumference and other circle stuff you've been doing in math falls apart.

 

[Red Scarlet]

Ahhh I hate this argument.

I'm not a calculus person, it's all blahblah example/theory vs another one. They aren't equal to me, sure call me dumb, but so what.

Sure you can do some trick on the calculator to make it 1, but at the same time, add or subtract the numbers and you won't get 0 or 2, either. ITS A MAGIC TRICK! yeah

 

[Mihail]

Ahhh I hate this argument.

I'm not a calculus person, it's all blahblah example/theory vs another one. They aren't equal to me, sure call me dumb, but so what.

There's no need for calculus or any kind of math... just use logic. If there is no number between .999... and 1, they must be equal, no?

 

[borghe]

.999... is not a BS number.

to get to 15 I can multiply 3x5, divide 45/3, add 10+5, etc.

how do you get to .999... without resorting to bullshit mathematics on repeating numbers?

and why can't you have a theoretical .000_1? any sequence of numbers is real, right? that is a theoretical sequence.

 

[Mihail]

so by this rule any number that ends in 9's repeating is equal to the next number up.

so .8999_ = .9
.567849999_ = .5675

I call bullshit on this with no proof other than people are saying so. there is no proof in math aside from a rule that says numbers have to exist between other numbers.

WTF? You can't pick and choose proofs. That IS a proof. And it's not the only proof.

x = 0.999....
10x = 9.999...
10x - x = 9.999... - 0.999
9x = 9
x = 1

is another proof. And there are more!

 

[borghe]

There's no need for calculus or any kind of math... just use logic. If there is no number between .999... and 1, they must be equal, no?

there is a .000_1 difference between the two. so there would be an infinity times .5 number that existed between them.

 

[Red Scarlet]

They aren't on the same place on the number line, as minutely apart, they are still not at the exact same spot. This topic needs a burning, searing, iodine thrown on it painful death! :lol

All it is is 'in theory vs in theory' wordplays it seems, over and over, ad nauseum.

x = 0.999....
10x = 9.999...
10x - x = 9.999... - 0.999
9x = 9
x = 1

Yeah, that calculator trick. You can do it to make it equal 1, but you can do one to also not equal one; its an endless argument.

 

[Mihail]

to get to 15 I can multiply 3x5, divide 45/3, add 10+5, etc.

how do you get to .999... without resorting to bullshit mathematics on repeating numbers?

Divide 1 by 1.

and why can't you have a theoretical .000_1? any sequence of numbers is real, right? that is a theoretical sequence.

Just because you call it theoretical doesn't give you license to make up stuff. You can't put an end to infinity -- it's as simple as that. It defies the very concept of infinity.

 

[borghe]

WTF? You can't pick and choose proofs. That IS a proof. And it's not the only proof.

x = 0.999....
10x = 9.999...
10x - x = 9.999... - 0.999
9x = 9
x = 1

is another proof. And there are more!

umm. no, this is the same bullshit "proof" insisting that you can add repeating decimals together.

 

[Trident]

to get to 15 I can multiply 3x5, divide 45/3, add 10+5, etc.

how do you get to .999... without resorting to bullshit mathematics on repeating numbers?

and why can't you have a theoretical .000_1? any sequence of numbers is real, right? that is a theoretical sequence.

.999 is the result of the infinite series 9/(10^n)

I mean, unless you consider infinite series to be bullshit. Now the question becomes, is it a convergent of divergent infinite series? I suppose I could try to remember some convergence tests. Give me a few.

 

[Mihail]

umm. no, this is the same bullshit "proof" insisting that you can add repeating decimals together.

Why is it BS? I don't need to prove I can add 2 numbers together considering they are 2 numbers. Why can't you accept that 1/3 and .333... are the same number? You're the one that has to prove that they're not the same number.

 

[borghe]

Why is it BS? I don't need to prove I can add 2 numbers together considering they are 2 numbers. Why can't you accept that 1/3 and .333... are the same number? You're the one that has to prove that they're not the same number.

because .333... repeating is not a number. it is an approximated decimal notation of a fraction.

 

[Red Scarlet]

Isn't 1/3 an 'estimate' equal to .33333.... since it can't truly be proven, ie an approximation?

.3333333333333333333333 can't be exactly done in fraction form, so 1/3 is used and 'accepted', even though it isn't exactly exact? And that's where the .999999 = 1 comes from.

asldkfhasdlkfasdyraksdra can't read topic anymore, have fun guys!

 

[Mihail]

Let me ask you another question, borghe:

Every repeating decimal can be expressed as the quotient of 2 whole numbers, right? 1/3 = .333..., 45/999 = .045045045...

Right?

So, what you're claiming is that the ONLY repeating decimal not to be expressed as a quotient of 2 whole numbers is .999...?

What is more likely? That .999... has some magical exception properties that makes it different from the infinite other repeating decimals or that your whole premise is wrong?

because .333... repeating is not a number. it is an approximate decimal notation of a fraction.

Speaking of your mistaken premise, there it is. .333... is the EXACT value of 1/3.

Notice that the ... symbolizes that the 3s go on for infinity. This notion of infinity is what makes it equal to EXACTLY 1/3.

Until you see this fact, you can't see the rest of the truth.

 

[Trident]

because .333... repeating is not a number. it is an approximate decimal notation of a fraction.

Infinite series of 3/(10^n)

 

[Arrowgigantic]

The truth of the matter is that nobody knows if .9 repeating = 1 because the Continuum Hypothesis (CH) has yet to be proven, while at the same time it has been proven that the CH cannot be disproven.

Yes, it's very confusing. If you want to really confuse someone, though, confuse them with this CH-related tidbit: there are infinite sets that are demonstrably larger than other infinite sets.

 

[empanada]

Here's what I said in the Blizzard thread:

This the wierd thing with calculus. You have to put your faith in things like infinity and limits for it to work. Just a simple exercise of writing 0.99.... until you reach 1 will never work, even in 10000000000000000000000000000 years non stop. Now I'm not saying that calculus is wrong, but I prefer to call it an approximation.

So I agree that 0.99_ ~ 1.

 

[Trident]

.999_ = infinite series of 9/(10^n)

To see if it converges

Integral test for convergence: http://en.wikipedia.org/wiki/Integral_test_for_convergence

integrate 9/(10^x) from 1 to limit as x approaches infinity

this becomes

-9 /(10^x)

0 + 9/(10) = 9/10

Since this converges, so does the infinite series.

So, unless you
a) don't believe in infinite series
b) don't think i got the infinite series right
c) don't believe in the integration test for convergent infinite series
or
d) don't think i got the integration right

You may have to accept that .999_ is in fact a number. You can do the same proof for .333_ and stop with this approximation bullshit.

Of course, now that we established that .999_ is a number, it has to be 1. This is because, as stated before, since real numbers are continuous, every two numbers has an infinite amount of numbers between them. Since .999_ and 1 have NO numbers between them, they must be the same number.

This isn't a bullshit rule. If you took some upper level math classes, they could derive it from the basic axioms of calculus. But so far as internet message boards go, I've already done way too much work.

 

[Trident]

The truth of the matter is that nobody knows if .9 repeating = 1 because the Continuum Hypothesis (CH) has yet to be proven, while at the same time it has been proven that the CH cannot be disproven.

Yes, it's very confusing. If you want to really confuse someone, though, confuse them with this CH-related tidbit: there are infinite sets that are demonstrably larger than other infinite sets.

I'm not too well versed on the continuum hypothesis, but I don't see how it applies to this at all. Doesn't it have to do with cardinality?

 

[El_Victor]

 

[ZombieSupaStar]

...


MY HEAD A'SPLODE

 

[LizardKing]

Ok, Mihail, just as you say that you can never add the .1 to a repeating decimal (of course), wouldn't that mean that .999... could never equal 1?

I understand that there are no numbers between .999... and 1.0 and i also understand that if you subtract 1-.999... it would equal 0.000000... thereby implying that .999... is equal to one. But isn't that more of an error in representation of numbers than proving that .999... is equal to 1? Because even though 1-.999... is written as 0.0000.... (which would be equal to 0.0000... derived from 1-1), when solving 1-.999... isn't it implied that, in order to use this number practically it must end (even if it's at whichever point is chosen)? If we never arrive at a number (using infinity) then wouldn't that number cease to exist?

 

[Shompola]

0.99_ converges to 1... just leave it at that :lol

 

[Mihail]

isn't it implied that, in order to use this number practically it must end (even if it's at whichever point is chosen)? If we never arrive at a number (using infinity) then wouldn't that number cease to exist?

Nope. We can say .333... + .111... = .444...

We can say .111... - .111... = 0

Mathematical operations like addition and subtraction are operations that are valid for all Real numbers. And seeing as .111... is a Real number, you would be the one that has to PROVE that you can't do addition with it. Why would the number cease to exist just because one of its digits goes on forever. It exists. What is a number? It's a conceptual amount, right? It's a FINITE amount. The number itself is not infinite. Again, it is a FINITE amount. All that is infinite is our graphical representation of it. The concept of the number is very real and very finite. It's just the writing system that we adopted that causes its WRITTEN notation to be infinite. The infinite nature of the NOTATION had nothing to do with the finite nature of the actual NUMBER.

 

[borghe]

Ok, Mihail, just as you say that you can never add the .1 to a repeating decimal (of course), wouldn't that mean that .999... could never equal 1?

I understand that there are no numbers between .999... and 1.0 and i also understand that if you subtract 1-.999... it would equal 0.000000... thereby implying that .999... is equal to one. But isn't that more of an error in representation of numbers than proving that .999... is equal to 1? Because even though 1-.999... is written as 0.0000.... (which would be equal to 0.0000... derived from 1-1), when solving 1-.999... isn't it implied that, in order to use this number practically it must end (even if it's at whichever point is chosen)? If we never arrive at a number (using infinity) then wouldn't that number cease to exist?

this is exactly what I am saying. you cannot use infinitely repeating numbers in a practical sense to derive a practical number. going in the other direction what is an _3 + _7? According to your rule you can't say it is 10_ because you couldn't have two ends with an infinite middle (one end being 1 the other being .)

Nope. We can say .333... + .111... = .444...

this relies on the same hyptohesis that you can perform practical math on infinitely repeating numbers. You can definitely say 1/3 + 1/9 = 4/9 which is represented approximately in decimal fashion as .444...

 

[Trident]

0.99_ converges to 1... just leave it at that :lol

Well, numbers don't converge. Series converge.

 

[Squirrel Killer]

Anyone who hasn't taken a calculus class should be banned from discussing this stuff. If you haven't learned about limits then there is no point in trying to figure it out.

So I can't try to learn why .99... = 1? I can't ask questions that might help my understanding and/or hone your explanation? There is no other way to learn other than by taking calculus?

I dunno what you mean by that. Issac Newton is the 'inventor' of calculus, though its more of a discovery than an invention. Its always been there, in the numbers.

Since I haven't taken calculus, how would "go to England and kick Issac Newton's ass." help me understand why .99... = 1?

You used tautology incorrectly. You used it to say Q <-> Q, but a tautology needs two different statements, such as P <-> Q.

Actually, a tautology is diagramed A = A (at least in logic, in mathematics it may be different.) The wording may be different but the premises are the same.

As far as it seeming perfectly possible to have 2 nonequal numbers with no intervening numbers -- that's simply not true. Can you give me an example of 2 such numbers without using the number (.999...) in question?

You only need one negative example to disprove a theory. I submit that .9... != 1 is that one example of non-equal numbers that have no intervening numbers. But if that's not enough, how about .99... + .99... = 1.99... != 2.

HAY GUYS, I'm no MATH WIZARD, which is important since math is comprised of ELEMENTAL MAGICKS, but I don't see how a SERIES can be INFINITE. I mean, JEEZ, if you can't explain it to me without using some sort of MATH NERDERY, then clearly you are wrong!!!

...and since everyone has been nice enough to keep Blizzard stuff out of my nice math thread, I've still got a kick to the beanbag saved up. If it's alright with everyone else, I'll just go ahead and give it to this attention whore.

By the way, you all fail in explaining a math topic to the non-math inclined, with LakeEarth taking the Grand Prize in "How to discourage someone from learning." Here's how you should explain it in the future, of course, this explanation understands that .99... is only arbitrarily close to 1.

So here's where I stand on this "issue":

The "no intervening numbers" explanation is bunk. It defines sequentiality, not equivalence.

The "0.99... is 1 because 0.99... doesn't exist" explanation is bunk. .99... exists as surely as .33..., pi, and .25.

The "ultimate difference is zero, so the two numbers must be equal" explanation is bunk. The difference is not zero, it's infinitesimally small.

Once you start explaining convergence or limits, surprisingly enough, non-MATH WIZARDS do understand what you're talking about. And that's where you get to the meat of the explanation, namely, that of course .99... != 1, but for all practical purposes it's close enough that worrying about it is pointless. So call .99... "1" and finish your goddamn homework, you slacker!

 

[Trident]

this is exactly what I am saying. you cannot use infinitely repeating numbers in a practical sense to derive a practical number. going in the other direction what is an _3 + _7? According to your rule you can't say it is 10_ because you couldn't have two ends with an infinite middle (one end being 1 the other being .)

.999_ = infinite series of 9/(10^n)

To see if it converges

Integral test for convergence: http://en.wikipedia.org/wiki/Integr...for_convergence

integrate 9/(10^x) from 1 to limit as x approaches infinity

this becomes

-9 /(10^x)

0 + 9/(10) = 9/10

Since this converges, so does the infinite series.

So, unless you
a) don't believe in infinite series
b) don't think i got the infinite series right
c) don't believe in the integration test for convergent infinite series
or
d) don't think i got the integration right

You may have to accept that .999_ is in fact a number. You can do the same proof for .333_ and stop with this approximation bullshit.

Of course, now that we established that .999_ is a number, it has to be 1. This is because, as stated before, since real numbers are continuous, every two numbers has an infinite amount of numbers between them. Since .999_ and 1 have NO numbers between them, they must be the same number.

This isn't a bullshit rule. If you took some upper level math classes, they could derive it from the basic axioms of calculus. But so far as internet message boards go, I've already done way too much work.

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[Trident]

And that's where you get to the meat of the explanation, namely, that of course .99... != 1, but for all practical purposes it's close enough that worrying about it is pointless. So call .99... "1" and finish your goddamn homework, you slacker!

auuuuuugh nooooo

infinite series... converge... to number... why god... why...

 

[LakeEarth]

and why can't you have a theoretical .000_1? any sequence of numbers is real, right? that is a theoretical sequence.

Dictionary.com, look up the word "infinite". You'll see it means forever, never stopping, never ever EVER end. Going 0.000....01 is impossible because you're ending the number, therefore it's finite.

 

[Mihail]

that of course .99... != 1, but for all practical purposes it's close enough that worrying about it is pointless. So call .99... "1" and finish your goddamn homework, you slacker!

Why is it of course? You have not proven that, so you can't say of course. All your "proofs" are something like:

".999... is not equal to 1 because .999... is not the same number as 1."

That's not a proof.

And what do you mean "for all practical purposes?" This doesn't exist in math. Either something is true, or it's not. It can't be kinda true. There's no "close enough."