Is 1 = 0.9999......

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DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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Originally posted by: pipster
Fun to reflect on, but not as interesting as pondering what 0/0 equals (zero divided by zero):

is it equal to zero?
is it equal to one?
is it equal to infinity?

Perhaps deserving its own poll/thread, i'm sure some unsuspecting lady or gentlemen will open that pandora's box someday...

(excuse the NooBness if this has already been done before...)

Doesn't need it's own poll/thread.... it's not a pandora's box. Simply, 0 divided by 0 is undefined, and makes absolutely no sense to ask.

However, to further confuse you, if the limit of a function as x approaches some value, such as 0 results in 0/0, then following the application of L'Hopital's rule, you can end up in a much wider variety of answers than just 0, 1, and infinity. For example, what is the value of Sin(3x) divided by x, as x approaches zero? At x = zero, it results in 0 over 0, which is undefined. But, as x approaches 0, the value of that function actually approaches 3. (This is where from mathematics, I believe, the above posters used the term "indeterminate")
 

Yax

Platinum Member
Feb 11, 2003
2,866
0
0
Originally posted by: Jeff7181
Originally posted by: Muzzan
Originally posted by: Jeff7181
0.9999... = 1

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

what if I do this...


x=0.9r
10x = 9.9r
10x-x = 9.9r - 0.9r
now to simplify, we ad x to each side of the equation
(10x - x) + x = 9.9r - 0.9r + x
9x = 9 + x

I do not wish to be rude, but you really need to work on your arithmetic.

10x - x + x = 10x.


I know, I'm typing fast and making typos

10x != 9.9r when x=0.9r
10x = 9.r instead.
 

Simoncifer

Junior Member
Dec 16, 2002
9
0
0
Really? magna cum laude? Me, too, I missed summa cum laude by like 0.03 or something like that. But mine is not applied math, it's abstract math.

P.S. are you sure about you learning this in class?

Originally posted by: DrPizza
Originally posted by: Simoncifer
No, I am stating that for all purpose of Applied Math, 0.99999..... = 1. But strictly speaking, like in number theory or abstract algebra, 0.99999.... != 1.

Sorry, Simon.... but I graduated magna cum laude with a bachelor's degree in applied mathematics. For applied mathematics, .999 repeating does = 1. But I learned this fact first in an abstract algebra course. .9999repeating = 1 in abstract algebra as well.
 

Simoncifer

Junior Member
Dec 16, 2002
9
0
0
ah.... I think I see what's the big deal is, no one has ever found the fraction that define/create 0.99999......., huh? oh well, mankind would eventually find it..... someday. Just like Fermat's Last Theorem.... took humanity more than 300 years to prove it, while this guy Fermat just say it's trival.....
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: Simoncifer
ah.... I think I see what's the big deal is, no one has ever found the fraction that define/create 0.99999......., huh? oh well, mankind would eventually find it..... someday. Just like Fermat's Last Theorem.... took humanity more than 300 years to prove it, while this guy Fermat just say it's trival.....

Ha, d00d, I found this fraction, it's very very easy to see. It's posted about 10 times in this thread.

That fraction is 1/1. I'm not going to post the entire thing again, but I'll point you in the right direction.

1 = 0.9 + 0.1

Do a long division of 1 into 1.00000... but instead of saying 1 goes into 1 once, say it goes 0.9 times with a remainder of 1... Write it out and I'm sure you'll see what I mean.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
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Originally posted by: Simoncifer
Really? magna cum laude? Me, too, I missed summa cum laude by like 0.03 or something like that. But mine is not applied math, it's abstract math.

P.S. are you sure about you learning this in class?

Originally posted by: DrPizza
Originally posted by: Simoncifer
No, I am stating that for all purpose of Applied Math, 0.99999..... = 1. But strictly speaking, like in number theory or abstract algebra, 0.99999.... != 1.

Sorry, Simon.... but I graduated magna cum laude with a bachelor's degree in applied mathematics. For applied mathematics, .999 repeating does = 1. But I learned this fact first in an abstract algebra course. .9999repeating = 1 in abstract algebra as well.

Summa's the higher one? I have them backwards.. 3.85 overall, 4.0 in major. And, I guess it was summa then, not magma (my university only does one or the other, whichever it is)
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
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Originally posted by: Simoncifer
ah.... I think I see what's the big deal is, no one has ever found the fraction that define/create 0.99999......., huh? oh well, mankind would eventually find it..... someday. Just like Fermat's Last Theorem.... took humanity more than 300 years to prove it, while this guy Fermat just say it's trival.....

You should have studied number theory a little more! Fermat wrote someplace that he found a proof that wouldn't quite fit in the margin, or something like that. But, he never published or attempted to publish that proof... that came from examinations of his papers some time after his death. However, at a point AFTER he made the comment about discovering a proof, he was making attempts to prove his last theorem for a special case. If he indeed proved it in general, there would have been no point for him to prove the special case... thus, he most likely discovered an error in his former "proof" of this theorem. It's been a while, but I think they're pretty sure what the error was; it's turned out to be a common mistake of people attempting to prove it over the years.
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
And I'm sure that Fermat's Last Theorem is much more than 300 years old. He just made it famous by making that mysterious remark about it.

Odd that it took some Harvard prof guy 10 years of his life to write this many-hundred page proof using a combination of cutting edge math mixed with some math he had to invent in order to prove it...

I managed to prove a while back that the numbers x, y, and z could not be whole multiples of each other in any way, even for n = 2.... I don't remember how I got there as it was a while ago.
 

Simoncifer

Junior Member
Dec 16, 2002
9
0
0
Dude, I knew that he probably found out his mistake, that's why he never published it..... but he didn't publish a lot of other things too..... Nonetheless, it was intimiating for the rest of the mathematicians (a lot of them French) to find in one of his personal notes, after Fermat died, that he wrote in the margin something like "I have discovered a truly remarkable proof which this margin is too small to contain." Something he probably thought would only take a page or two to prove, would actually take hundreds of pages to prove, not to mention modern concepts. This to show that sometimes things that seems simple actually are not.


Originally posted by: DrPizza


You should have studied number theory a little more! Fermat wrote someplace that he found a proof that wouldn't quite fit in the margin, or something like that. But, he never published or attempted to publish that proof... that came from examinations of his papers some time after his death. However, at a point AFTER he made the comment about discovering a proof, he was making attempts to prove his last theorem for a special case. If he indeed proved it in general, there would have been no point for him to prove the special case... thus, he most likely discovered an error in his former "proof" of this theorem. It's been a while, but I think they're pretty sure what the error was; it's turned out to be a common mistake of people attempting to prove it over the years.

 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: Simoncifer
Really? magna cum laude? Me, too, I missed summa cum laude by like 0.03 or something like that. But mine is not applied math, it's abstract math.

you're telling me that you graduated with honors in abstract math and still think that .999r != 1??
I'm sorry, but where did you get your degree?

 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: Simoncifer
Really? magna cum laude? Me, too, I missed summa cum laude by like 0.03 or something like that. But mine is not applied math, it's abstract math.

P.S. are you sure about you learning this in class?

Originally posted by: DrPizza
Originally posted by: Simoncifer
No, I am stating that for all purpose of Applied Math, 0.99999..... = 1. But strictly speaking, like in number theory or abstract algebra, 0.99999.... != 1.

Sorry, Simon.... but I graduated magna cum laude with a bachelor's degree in applied mathematics. For applied mathematics, .999 repeating does = 1. But I learned this fact first in an abstract algebra course. .9999repeating = 1 in abstract algebra as well.

LOL,
Simon, Why do I not believe your claim!

I was not aware that Real Analysis was considered Applied Math, in the programs I participated in (Oregon State and Lehigh U) Real Analysis is required for ALL math majors. And in ALL of the real analysis courses, undergrad, and graduate level, the real number line is dense and .999... =1.

Now as far GPA claims go, I completed my BS in Math in 1984 with Highest Scholarship, that means 4.0.

To bad none of us can prove our claims, makes them pretty worthless dosen't it!
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Now comes that part in every argument when one side relegates the battle to a matter of personal credentials.

It has nothing to do with the argument that .999...<>1 or vice versa.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: MadRat
Now comes that part in every argument when one side relegates the battle to a matter of personal credentials.

It has nothing to do with the argument that .999...<>1 or vice versa.

actually, I am still waiting for the part where you tell us what number is between .999... and 1. It's a simple enough of a question. Why not answer it for once?
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
I'm still waiting for you to prove .999... exists. If its a Real Number then represent the number in fractal form.

If we have a non-terminating sequence of digits, such as .23232323... then what makes it any different? By your simplistic representation the number .23232323... should be equal to .23 or .232 or .2323 etc... The truth is that they are all different values and therefore none of them can represent .23232323...
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: MadRat
I'm still waiting for you to prove .999... exists. If its a Real Number then represent the number in fractal form.

So a real number now needs to have a fractional form?? So I guess pi isn't a number. And either is the square root of 2... Maybe you should lookup the difference between a real and a rational number again. In any case, it is moot for .999... as the rational form is simply 1/1.

Originally posted by: MadRat
If we have a non-terminating sequence of digits, such as .23232323... then what makes it any different? By your simplistic representation the number .23232323... should be equal to .23 or .232 or .2323 etc... The truth is that they are all different values and therefore none of them can represent .23232323...

so in the last quote you claimed a repeating number like .999r doesn't exist, but .23232323... etc does? Which one is it?
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: MadRat
Now comes that part in every argument when one side relegates the battle to a matter of personal credentials.

It has nothing to do with the argument that .999...<>1 or vice versa.

actually, this argument has everything to do with credentials. If you don't understand the fundamentals of a dense real number line, than your opinions really don't carry much weight.

It is akin to someone arguing that the letter B really comes before the letter A in the alphabet because there is no god-given reason (or "proof") to state otherwise. And in reality, there is no reason why A comes before B, but the way our alphabet is defined, A comes first be definition.

In the same sense, because of the way our number system was defined, .999r = 1. Like I said before, if you don't like the real number system, create your own where .999r and irrational numbers don't exist (apparently it will make things much easier for you to understand). Unfortunately for you, the way our number system has been defined leads to some very "nice" properties and creates a very robust and simple system, so I doubt many people will use your number system (actually, I believe the pythagoreans already failed at trying to convince the world that irrationals don't exist).
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
You've not yet proved any logical reason for the argument that .999...=1 other than its accepted. Then you not only once bring up the part about credentials, but you dwell on it. This is nothing more than a cyclical argument where you fail to prove the point then go on about how anyone that doesn't follow your acceptance is wrong. Seems to me that this is more or less a distraction from the argument. I hope this isn't how all your arguments go. If so then the world must be awfully disturbing once you leave your flat.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: MadRat
You've not yet proved any logical reason for the argument that .999...=1 other than its accepted. Then you not only once bring up the part about credentials, but you dwell on it. This is nothing more than a cyclical argument where you fail to prove the point then go on about how anyone that doesn't follow your acceptance is wrong. Seems to me that this is more or less a distraction from the argument. I hope this isn't how all your arguments go. If so then the world must be awfully disturbing once you leave your flat.


three points:

1) I noticed you conveniently avoided the question, so I'll ask again: are you saying now that irrational numbers and repeating numbers don't exist?

2) Again, you avoided my other question (you seem to be a lot better at ignoring questions then you are at answering them): what number is between .999r and 1?

3) I think you *almost* figured it out. I am offering no "logical" proof that .999r = 1 other than it is accepted. Same way that the letter A comes before B is "accepted". .999r = 1 by definition of our number system, not by some universal law passed down to moses or anything.
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: MadRat
I'm still waiting for you to prove .999... exists. If its a Real Number then represent the number in fractal form.

If we have a non-terminating sequence of digits, such as .23232323... then what makes it any different? By your simplistic representation the number .23232323... should be equal to .23 or .232 or .2323 etc... The truth is that they are all different values and therefore none of them can represent .23232323...


Try 23/99 if you want the fractional represention of .23...

You are the one being simplistic. First of all by settimg up a strawman argument such as this one. Then by refusing to learn of the sublty and depth of the real number system when it is offered up to you on a platter.you prove your close mindedness by blinding yourself to the existance of a theoritcal infinty. Do you really mean to claim that I am not allowed to concieve of an infinitly long string of numbers! What nonesense I just did it! Consider a infnite string of 9s, there I did it again! How can you claim that it does not exist! I just made it exist what are you going to do with it! Not only does it exist because I proposed it, I can represent it. Then I can show that this stirng of 9s exists in every interval surrounding 1. There can be only one number with this property, that is 1.

You claim to be a philosopher, but cannot follow the logic of a rather straight forward mathmaical proof. You are a simple fraud and have virtually no right to claim yourself a logician and therefore have no claim to being a philosopher.
 

Ferocious

Diamond Member
Feb 16, 2000
4,584
2
71
What ever happened to that math "genius" from CAL-TECH that used to post here a lot?

I'd like to see his response.

(damn...what was his name...)
 

Darien

Platinum Member
Feb 27, 2002
2,817
1
0
This thread should make any mathematician/anyone trained in the mathematics adequately to cringe or cry or laugh at the sheer reluctance to accept 0.9r = 1.



this is NOT a debate that anyone other than those trained in math properly (science and engineering majors anyone?) should participate in. If you attempt to take the philosophy road and try and do something to disprove 0.9r = 1, you have yet to prove it in mathematics. Many a philosophers have failed in explaining physics because of the oddities in quantum mechanics and classical physics -- far too much is reliant on MATH and not what people think is a logical argument.
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Originally posted by: Darien
this is NOT a debate that anyone other than those trained in math properly (science and engineering majors anyone?) should participate in. If you attempt to take the philosophy road and try and do something to disprove 0.9r = 1, you have yet to prove it in mathematics. Many a philosophers have failed in explaining physics because of the oddities in quantum mechanics and classical physics -- far too much is reliant on MATH and not what people think is a logical argument.

Excuse me, but OBSERVATION has defined the parameters of science.

Originally posted by: Marauder-
The real question should be does 0.9999 = BUTTER as threadposts aproach infinity...

Was it BUTTER or popcorn? Its buried in the archive too far back to remember.

 
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