.99999 does not equal 1

I don't care how many 9's or how close to 1 it is. Being close, isn't being.

Better question: does it even matter?

Yes. But what does it have to do with recurring decimals?

Yes it is.  No it doesn't, yes I do, and nothing.

All .9999...=1 means is that the limit, as the number of 9's increases without bound, equals 1.

Fancy drawings don't make you right.

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

nope.
There's still a very very very small number between .999999 and 1.

nope.
There's still a very very very small number between .999999 and 1.

Yes. But there isnt between 0.9999....... and 1.

Yes there is! It's just even smaller. T'would be 0.0000...1.

Is this just trolling?

Is this just trolling?

Either way, it's dumb.

All .9999...=1 means is that the limit, as the number of 9's increases without bound, equals 1.

what are you finding out when you find a limit? what direction does the equation head towards but then you must also determine if there is an asymptote at that number ie. graph the equation f(x)=1/(1-x) and you will find that there is an asymptote at 1 because x can never=1

Quote from: jdoe on August 09, 2008, 06:26:12 PM

All .9999...=1 means is that the limit, as the number of 9's increases without bound, equals 1.

what are you finding out when you find a limit? what direction does the equation head towards but then you must also determine if there is an asymptote at that number ie. graph the equation f(x)=1/(1-x) and you will find that there is an asymptote at 1 because x can never=1

What the heck are you talking about? If you wanted to graph the equation in his limit you would use f(x)= 1- (1/10^(x)). Notice now that there is no way for there to be a zero in the denomenator, and no vertical asymptotes exist for f(x)= 1/10^x.  You could use L'Hospitals rule to prove that the limit approaches zero as n approaches infinity if you wanted.

Jdoe's proof works perfectly fine. Its obvious which direction the limit heads, as it was derived from an infinite series starting at k=1, to infinity. He dirived his limit by using the rule that determines the sum of a convergent  geometric series where the sum = term 1/1- common ratio. Since the sum is a limit, finding the limit will find the sum, which he proved was one.

Quote from: cbarnett97 on August 09, 2008, 08:29:06 PM

Quote from: jdoe on August 09, 2008, 06:26:12 PM

All .9999...=1 means is that the limit, as the number of 9's increases without bound, equals 1.

what are you finding out when you find a limit? what direction does the equation head towards but then you must also determine if there is an asymptote at that number ie. graph the equation f(x)=1/(1-x) and you will find that there is an asymptote at 1 because x can never=1

What the heck are you talking about? If you wanted to graph the equation in his limit you would use f(x)= 1- (1/10^(x)). Notice now that there is no way for there to be a zero in the denomenator, and no vertical asymptotes exist for f(x)= 1/10^x.  You could use L'Hospitals rule to prove that the limit approaches zero as n approaches infinity if you wanted.

Jdoe's proof works perfectly fine. Its obvious which direction the limit heads, as it was derived from an infinite series starting at k=1, to infinity. He dirived his limit by using the rule that determines the sum of a convergent  geometric series where the sum = term 1/1- common ratio. Since the sum is a limit, finding the limit will find the sum, which he proved was one.

was I arguing against his equation or was I using a nice simple equation so those who are not strong at math could follow it better. thanks for missing the point try again later

 I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

what does a limit find; an answer or an approximate value?

Quote from: [][][] on August 09, 2008, 09:15:32 PM

I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

what does a limit find; an answer or an approximate value?

The limit gives an exact a sum to an infinite series  (the sum is 1).  Since when the hell is a limit ever an approximation? I think the definition of a limit is obvious to anyone who has ever had a class dealing with them (a calculus class), and you are no substitute for a textbook to those who have not.

Quote from: cbarnett97 on August 09, 2008, 09:33:38 PM

Quote from: [][][] on August 09, 2008, 09:15:32 PM

I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

what does a limit find; an answer or an approximate value?

The limit gives an exact a sum to an infinite series  (the sum is 1).  Since when the hell is a limit ever an approximation? I think the definition of a limit is obvious to anyone who has ever had a class dealing with them (a calculus class), and you are no substitute for a textbook.

In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. Informally, a function assigns an output f(x) to every input x. The function has a limit L at an input p if f(x) is "close" to L whenever x is "close" to p. In other words, f(x) becomes closer and closer to L as x move closer and closer to p. More specifically, when f is applied to each input sufficiently close to p, the result is an output value that is arbitrarily close to L.

Ouch for you!

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

Fuck calculus.  This is all the proof I really need.

Quote from: [][][] on August 09, 2008, 09:35:28 PM

Quote from: cbarnett97 on August 09, 2008, 09:33:38 PM

Quote from: [][][] on August 09, 2008, 09:15:32 PM

I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

what does a limit find; an answer or an approximate value?

The limit gives an exact a sum to an infinite series  (the sum is 1).  Since when the hell is a limit ever an approximation? I think the definition of a limit is obvious to anyone who has ever had a class dealing with them (a calculus class), and you are no substitute for a textbook.

In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. Informally, a function assigns an output f(x) to every input x. The function has a limit L at an input p if f(x) is "close" to L whenever x is "close" to p. In other words, f(x) becomes closer and closer to L as x move closer and closer to p. More specifically, when f is applied to each input sufficiently close to p, the result is an output value that is arbitrarily close to L.

Ouch for you!

Right, the intuitive definition of a limit, please explain where I am wrong, if you feel that I am wrong in my conception of a limit. (So far the only thing you have proved to me is your copy/paste ability)

Quote from: cbarnett97 on August 09, 2008, 09:49:13 PM

Quote from: [][][] on August 09, 2008, 09:35:28 PM

Quote from: cbarnett97 on August 09, 2008, 09:33:38 PM

Quote from: [][][] on August 09, 2008, 09:15:32 PM

I just don't understand what your equation, and your subsequent explaination of asymptotes has to do with jdoe's proof. What did you  clarify for people? Please explain yourself.

Is this your clumsy way of explaining that the limit of (1/10^x) is 0 as x approaches infinity?

what does a limit find; an answer or an approximate value?

The limit gives an exact a sum to an infinite series  (the sum is 1).  Since when the hell is a limit ever an approximation? I think the definition of a limit is obvious to anyone who has ever had a class dealing with them (a calculus class), and you are no substitute for a textbook.

In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. Informally, a function assigns an output f(x) to every input x. The function has a limit L at an input p if f(x) is "close" to L whenever x is "close" to p. In other words, f(x) becomes closer and closer to L as x move closer and closer to p. More specifically, when f is applied to each input sufficiently close to p, the result is an output value that is arbitrarily close to L.

Ouch for you!

Right, the intuitive definition of a limit, please explain where I am wrong, if you feel that I am wrong in my conception of a limit. (So far the only thing you have proved to me is your copy/paste ability)

well you wanted something out of a textbook since you did not believe me earlier when I essentially stated the same thing so just remember, in math there is a big difference between an approximate value and an exact answer

I was criticising the clumsy way you tried to explain the concept of limits earlier in the thread, I am not saying you don't know what a limit is. Granted my first response mistook your response for a criticism of jdoe's proof.  A limit is just that, a limit, not an approximation.

I was criticising the clumsy way you tried to explain the concept of limits earlier in the thread, I am not saying you don't know what a limit is. Granted my first response mistook your response for a criticism of jdoe's proof.

well then that is that

Quote from: Hara Taiki on August 09, 2008, 07:18:17 PM

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

Fuck calculus.  This is all the proof I really need.

But how do you know .33333... = 1/3?

Quote from: Roundy the Truthinessist on August 09, 2008, 09:51:36 PM

Quote from: Hara Taiki on August 09, 2008, 07:18:17 PM

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

Fuck calculus.  This is all the proof I really need.

But how do you know .33333... = 1/3?

you can round it off, but they are not equivalent terms

Quote from: Roundy the Truthinessist on August 09, 2008, 09:51:36 PM

Quote from: Hara Taiki on August 09, 2008, 07:18:17 PM

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

Fuck calculus.  This is all the proof I really need.

But how do you know .33333... = 1/3?

Okay, fine.  I guess there's calculus there, in the proof that .333... = 1/3.  But I know that from elementary school.  I think it was 5th or 6th grade that I figured out that .999... = 1.

you can round it off, but they are not equivalent terms

I believe you're mistaken here.

unless you want to write 3's forever(as far as we know right now) then you are indeed rounding off

You don't need to put 3s forever.  That's what the ellipses represent.  Or putting a line over the number, if you prefer.  But it's a fact that .333..., if you understand the ellipses to represent an infinite number of 3s, is equal to 1/3.  And .999... is equal to 1.

You don't need to put 3s forever.  That's what the ellipses represent.  Or putting a line over the number, if you prefer.  But it's a fact that .333..., if you understand the ellipses to represent an infinite number of 3s, is equal to 1/3.  And .999... is equal to 1.

my bad did not see the line so I went into literal mode, so you are almost correct but .9999 repeating does not equal 1, that is one of the proofs of rounding errors