Why does .99999999... = 1?

Page 3 - Seeking answers? Join the AnandTech community: where nearly half-a-million members share solutions and discuss the latest tech.

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
The real line would be flawed if .9... did not equal 1 that would mean that there is a hole in the line. To mathematicians this means that the real line is dense, there are NO HOLES in the real line.

Again because this is an INIFINTE string of 9s there can be NOT be a 1 at the END of it. Do not think of an "infinite" string followed by something. Any digit you put in will NOT be "at" infinity it will be at some finite location followed by an infinite string of digits.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
wildwildwes:


<< Subtract x from 10x and solve for x.

10x = 9.99999...
-x = -0.99999...
-----------------
9x = 9 <--------- 9x is not 9, it is 8.99991
x = 9/9
x = 1
>>



So if 9.9999... - .999.... = 8.9999....1 then what is 9.999.... - 1? Does it equal 8.99999999....?
Does that mean .999.... > 1?

I think, as we discovered in the HT thread, .999... = 1. If you don't accept it, then you don't accept our system of math the way it was created. Its just as easy to say that I don't think an apple represents that spherical fruit that grows on trees. You can disagree with the language all you want, but in terms of the english language apple = a fruit.
 

GoldenBear

Banned
Mar 2, 2000
6,843
2
0
People really don't know how to read a thread before replying. There have been at least 10 posts linking to the HT forum, and 10 stating that it's been posted in HT.

And also someone posted my own solution to the problem :(
 

LaVoS

Senior member
Aug 1, 2001
224
0
0
WEll i do belive that .99999... = 1... but my friends asks " How can .99999... = 1? .999... = .999... and 1 = 1 so how can .999... =1)
lol see what i mean? my math teacher says the same
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Whoa, just found this thread. The cavalry has arrived RossGr :)

Okay, a number of proofs have been given as to why 0.9r = 1. If anyone wants to say that 0.9r < 1, they must show that either:

A. All of the proofs given thus far are flawed (and this doesn't include "they're wrong" as an acceptable counter) or

B. That there exists an explicitly definable number c that is between a (0.9r) and b (1).

Explicitly definable means that you can't use 0.9r or 1 in your description of this number. ie 0.9r + d or 1 - d are not allowed.
 

Noirish

Diamond Member
May 2, 2000
3,959
0
0
i don't think you can assign something like 0.9999999..... whatever to a variable.
if you can express in fraction, then it's okay.
 

GoldenBear

Banned
Mar 2, 2000
6,843
2
0


<< i don't think you can assign something like 0.9999999..... whatever to a variable. >>

0.22222... = 2/9
0.33333.... = 3/9
0.44444.... = 4/9
0.55555.... = 5.9
0.66666.... = 6/9
0.77777.... = 7/9
0.88888.... = 8/9

Hence, 0.99999 = 9/9?
 

littlebig

Member
Jan 31, 2001
175
0
0
Am I in Wonderland??

Man. Grab any advanced calculus or abstract algebra book before explain anything.
0.999.... is not a number. It's just an infinite geometric series representation.
0.9999... means Sum[from n = 1 to infinity] (9/power(10,n))
(sorry, lack of proper tool to show a formula)
For example, 0.999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ... and keep going. And if you know infinite sum of geometric
series, you can fiqure out, [initial term] = 9/10, [common ratio] = 1/10
So, Sum = (9/10) / (1 - 1/10) = (9/10) / (9/10) = ***1***.

It's something like asking WHY "2+1 = 3"? because "2+1" is not "3" (literally! 2+1 has THREE characters, 3 has ONE character).
By asking 0.9999.... = ?, what you are really seeking is an ANSWER for that series, which is ONE.

Any *infinitely repeating* decimal point representation is just a series representation, which could have an answer (like 0.999...) or not. (like Phi = 3.1415.....???, e = 2.71...??? we just assign symbol Phi or e to REPRESENT them.) When I say "answer", I mean in "RATIONAL" numbers (at least in this decimal point case).

It's merely the way we represent Numbers. No significant meaning. PLEASE DO NOT SPEND your life on such questions. I've seen many so called "amature" mathematicians (yes, I'm a professonal mathematician), who are spending so much time in meaningless questions. (like "how to divide arbitrary angle into three equal angles, using only rulers, etc, etc... which have been solved (disproved) hundreds years ago, but hard to explain to somebody who has no professional math knowledge. Have to use Galois Theory, which is very last part of abstract algebra courses). Do not confuse "disproof" with "misunderstanding".

Actually, whole confussion is from the concept of "limit" which most of ppl do not understand properly. Even mathematician confuses sometimes. If you ever learn advanced calculus II or Real Analysis in college, you would remember epsilon-delta definition of limit. Which is defined by double negations (not-not) which does not have any easier way to understand. 0.999.... is not just bunch of numbers. It's a "processing" which keeps going into 1.

Wanna go further? Yes. When people *invented* number system, the first one was Whole (natural) Numbers (obviously).
0, 1, 2, 3, 4.... (ok. Zero is not in the first place but...)

Then when we were trying to solve x + 3 = 0. Oops, no answer. So they invented whole system of negative numbers.
-2, -1, 0, 1, 2,... which make INTEGERs together with whole numbers.

Now, we have bunch of them but still we cannot solve some a*x - b = 0 (1st degree equation) when a, b is interger,
i.e. 2x + 3 = 0. So there goes RATIONAL NUMBERs. (so that x = b / a, obviously a != 0), which means "RATIO" between two integer numbers. (So, "RATIO"nal)

But we still may not have any answer in higher degree polynomials. For example, x^2 = 3... what's x?? no such rational number.
So, we considered *REAL NUMBERs*.

Real Number is not easy thing. There are several way to define it, and one thing is that real numbers are a family of limit points of rational sequences. 0.9999.... is one of such expression (representation).

Mathematically, integers are called "group" ('cause they have an identity (0) and inverse for plus), rational numbers and real numbers are "field" (commutative division ring, which means have two operations (+,*), two identities (0,1), and inverse for both operations and commutative.)

Even though Real Numbers form a field, it's not closed one in the sense that we cannot solve arbitrary polynomial whose coefficients are in real numbers. For instance, there's no answer for this simple quadratic x^2 + 1 = 0 in real field. That's why we are using whole concept of "COMPLEX" numbers. (I saw somebody's mentioning "NON-REAL Numbers". I've studied a math like 20 years and got a PhD in Math, but never seen such term as "NON-Real". Maybe "complex"?)

Complex is the first closed field (algebraic closure) in number system (oh well at least for infinite field). Any degree of polynomials in complex has an answer in it. But still we could have series of complex numbers, in which point we could introduce new number system... but it's beyond scope, I guess.
 

LaVoS

Senior member
Aug 1, 2001
224
0
0
hmmmm... like i said before my firned is stupis so he keeps sayshting twhat i postedearlyer... (.999... is .999... not 1) so if there any way u can put that in eazyer terms for him... OH byt hte way his SN on here is Viper Magic so yeah just so u know
 

spidey07

No Lifer
Aug 4, 2000
65,469
5
76
littlebig,

thanks for the reminder on number systems. it bears much merit on this discussion. while I forget most of what I learned in college the geometric series (calculus) and linear algebra/diff eq, you and ROSS seemed to jar it from my memory. thanks!

this question is so utterly basic I have a hard time regressing to this kind of math...seriously :)
 

littlebig

Member
Jan 31, 2001
175
0
0
spidey07, no problem.
I agree that this is a *real*:) simple question. I just couldn't help myself responding the question.

I've been teaching a math in colleges/consulting HS teachers around 5 years, and believe me, US educational system has a serious problem, especially in mathematics. This is all due to that evil TI CALCULATORS!! :) :)
 

Noriaki

Lifer
Jun 3, 2000
13,640
1
71


<< Thast not what i meant... i emant that .999999999... does not equal one because well does .9 look like 1? no it doenst... >>

*shrugs*

Does 2/2 look like 1? Not really, but it is.
does .5+.5 ? Nope.
How about 3-2?

There are lots ways to write a number.

.999... is one way to write 1.

I won't post a proof or anything since the geometric series proof has been posted several times, and beaten to death in HT>
 

Hanpan

Diamond Member
Aug 17, 2000
4,812
0
0
Finally someone who knows what they are talking about. Thanks to everyone as well. This denial by many people of a simply mathematical equivialnce is astounding. As others have mentioned i think the real problems hinges on both the definition of infinity and the fact that as spacial beings such things as 0.99... and 1 which look totaly different are equivilent.



<< Am I in Wonderland??

Man. Grab any advanced calculus or abstract algebra book before explain anything.
0.999.... is not a number. It's just an infinite geometric series representation.
0.9999... means Sum[from n = 1 to infinity] (9/power(10,n))
(sorry, lack of proper tool to show a formula)
For example, 0.999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ... and keep going. And if you know infinite sum of geometric
series, you can fiqure out, [initial term] = 9/10, [common ratio] = 1/10
So, Sum = (9/10) / (1 - 1/10) = (9/10) / (9/10) = ***1***.

It's something like asking WHY "2+1 = 3"? because "2+1" is not "3" (literally! 2+1 has THREE characters, 3 has ONE character).
By asking 0.9999.... = ?, what you are really seeking is an ANSWER for that series, which is ONE.

Any *infinitely repeating* decimal point representation is just a series representation, which could have an answer (like 0.999...) or not. (like Phi = 3.1415.....???, e = 2.71...??? we just assign simbol Phi or e to REPRESENT them.) When I say "answer", I mean in "RATIONAL" numbers (at least in this decimal point case).

It's merely the way we represent Numbers. No significant meaning. PLEASE DO NOT SPEND your life on such questions. I've seen many so called "amature" mathematicians (yes, I'm a professonal mathematician), who are spending so much time in meaningless questions. (like "how to divide arbitrary angle into three equal angles, using only rulers, etc, etc... which have been solved (disproved) hundreds years ago, but hard to explain to somebody who has no professional math knowledge. Have to use Galois Theory, which is very last part of abstract algebra courses). Do not confuse "disproof" with "misunderstanding".

Actually, whole confussion is from the concept of "limit" which most of ppl do not understand properly. Even mathematician confuses sometimes. If you ever learn advanced calculus II or Real Analysis in college, you would remember epsilon-delta definition of limit. Which is defined by double negations (not-not) which does not have any easier way to understand. 0.999.... is not just bunch of numbers. It's a "processing" which keeps going into 1.

Wanna go further? Yes. When people *invented* number system, the first one was Whole (natural) Numbers (obviously).
0, 1, 2, 3, 4.... (ok. Zero is not in the first place but...)

Then when we were trying to solve x + 3 = 0. Oops, no answer. So they invented whole system of negative numbers.
-2, -1, 0, 1, 2,... which make INTEGERs together with whole numbers.

Now, we have bunch of them but still we cannot solve some a*x - b = 0 (1st degree equation) when a, b is interger,
i.e. 2x + 3 = 0. So there goes RATIONAL NUMBERs. (so that x = b / a, obviously a != 0), which means "RATIO" between two integer numbers. (So, "RATIO"nal)

But we still may not have any answer in higher degree polynomials. For example, x^2 = 3... what's x?? no such rational number.
So, we considered *REAL NUMBERs*.

Real Number is not easy thing. There are several way to define it, and one thing is that real numbers are a family of limit points of rational sequences. 0.9999.... is one of such expression (representation).

Mathematically, integers are called "group" ('cause they have an identity (0) and inverse for plus), rational numbers and real numbers are "field" (commutative division ring, which means have two operations (+,*), two identities (0,1), and inverse for both operations and commutative.)

Even though Real Numbers form a field, it's not closed one in the sense that we cannot solve arbitrary polynomial whose coefficients are in real numbers. For instance, there's no answer for this simple quadratic x^2 + 1 = 0 in real field. That's why we are using whole concept of "COMPLEX" numbers. (I saw somebody's mentioning "NON-REAL Numbers". I've studied a math like 20 years and got a PhD in Math, but never seen such term as "NON-Real". Maybe "complex"?)

Complex is the first closed field (algebraic closure) in number system (oh well at least for infinite field). Any degree of polynomials in complex has an answer in it. But still we could have series of complex numbers, in which point we could introduce new number system... but it's beyond scope, I guess.
>>