i saw somone posted this about 6 months ago: Can anyone help me prove that .99999 repeating = 1

[Hanpan]

<< Sorry but I have to throw in one more thing.

Any of you who think you "KNOW" things about math, you might want to be more careful.

There have been times when I was so sure that I was right I would argue it until I was blue in the face, only to come to the realization later that I was dead wrong.

Sometimes your logic isn't the correct logic. >>




This whole argument hinges largely on definition as well as the idea of infinity. In order for this properly to be dicussed it would be very helpful to agree on at least these things. wbwither is correct in saying that in reality nothing afaik expcet ideas have been shown to be infite. That is why there is theorectical math

 

[bizmark]

<<

<< The entire point of the .999 like I said before is to denote that it is NOT 1, otherwise it would be called 1 in the first place! Juast as .2444444444444444444444444 does not equal .25 >>



There are lots of notations that are equivalent in math.

We still have 2/2, and yet I could just call it 1.

We have sin^2 x + cos ^2 x, yet I could just call it 1.

We have 7890437984379843798479204739^0, yet I could just call it 1. >>



Yes, but in those cases, the equation is true. I agree that FOR ALL PRACTIAL INTENTS AND PURPOSES, .9999999 (even just to that resolution, 7 decimal places) is 1. However, technically, it is not. The LIMIT, as I said above, is equal to 1. However, the limit depends on infinity, which is a concept that we use as a tool. Infinity does not exist.



<< Indeed it's the repeating 9s that actually equal the next highest value. >>





<< Any of you who think you "KNOW" things about math, you might want to be more careful. >>



Listen to your own advice, buddy

 

[Cerebus451]

Another interesting way to show it......

0.9999..... is a rational number, meaning it can be represented as a fraction

10/11 = 0.909090.....
1/11 = 0.090909.....

10/11 + 1/11 = 0.99999.....
1 = 0.999999.......

(BTW, you can use 100/101 and 1/101, 1000/1001 and 1/1001, it just changes the number of 9's and 0's grouped together)

 

[Cerebus451]

<< However, the limit depends on infinity, which is a concept that we use as a tool. Infinity does not exist. >>


But we are talking about mathematics here, and as far as math is concerned, not only does infinity exists, but an infinite number of infinities do exist. Whether or not you choose to believe in infinity in practical terms, when dealing with math, get used to 'em.

 

[b0mbrman]

Didn't we come to the conclusion that (0.333... + 0.666... = 0.999...) is equiv. to (1/3 + 2/3 = 3/3 = 1)?

 

[Cerebus451]

<< Didn't we come to the conclusion that (0.333... + 0.666... = 0.999...) is equiv. to (1/3 + 2/3 = 3/3 = 1)? >>


There are several ways to do it. There are several repeating fractions that you can add to get 0.9999.... (when using the decimal equivalent) and they should all add to 1 if you add the fractions.

Here is an interesting postulation:

Given a1/n + a2/n + ... + an/n = 0.9999.... as a decimal, a1/n + a2/n + ... + an/n = 1 as a fraction.
However, it is 11:30PM now, I am still at work, and I refuse to think any more about this. If anyone else wants to offer up the proof, have at it.

 

[bobcpg]

there is a way to prove it, i remember doing it in high school. i'll see if i cant find it....

-bob

 

[bizmark]

<< This whole argument hinges largely on definition as well as the idea of infinity. In order for this properly to be dicussed it would be very helpful to agree on at least these things. >>



Good idea. Suggestions, anyone?

I would refer everybody to the standard delta-epsilon definition of a limit:

LIM(x-->x_0) f(x)=A

if and only if

for every E>0, there exists d>0 s.t. if (x-x_0)<d then (f(x)-A)<E.

In plain English, if we have a function f(x), and we say the limit of f(x) as x approaches x_0 is A, then that means that, given any arbitrarily close distance E that f(x) must be from A, we can find a distance d (probably as a function of E) that x must be within x_0.

For example (this is so hard to write out on a computer w/o latex or something):

The limit of f(x)=2*x, as x approaches 2, is 4. Meaning, that we can pick any small interval around 4 that we want f(x) to be in, and then we'll figure out an appropriate interval for x, such that if x is within that interval, f(x) will be within the chosen interval around 4. Let's take our interval distance E=0.01. That is, we want f(x) to be within [3.99, 4.01].

So to find d in general for any E, we have the equation d=(1/2)*E. So, since E=0.01, we take d=0.005. Then, look at our intervals around 2: [1.995, 2.005]. Take f(1.995)=2*1.995=3.99 -- that's within E of 4. Take f(2.005)=2*2.005=4.01 -- that's also within E of 4. Take any number x from within the range [1.995, 2.005]. Then f(x) will lie within [3.99, 4.01], guaranteed.

I believe that this entire discussion hinges upon the proper understanding of this definition of limit.

 

[bizmark]

<<

<< However, the limit depends on infinity, which is a concept that we use as a tool. Infinity does not exist. >>


But we are talking about mathematics here, and as far as math is concerned, not only does infinity exists, but an infinite number of infinities do exist. Whether or not you choose to believe in infinity in practical terms, when dealing with math, get used to 'em. >>



No, you're wrong, and your statement that "an infinite number of infinities exists" just shows how bad your understanding of infinity is.

There are two infinities, countable and uncountable.

In this discussion, we're clearly talking about uncountable infinities. The real numbers are an example. Any nonempty, nontrivial interval of the real numbers is also an example, e.g. the interval [0,1] has an uncountably infinite number of points. There exists a one-to-one and onto map from [0,1] to (-infinity, +infinity). These intervals, then, are one and the same. Any nonempty, nontrivial interval of the real numbers has equivalent cardinality (and therefore is equivalent) to the whole of the real numbers.

I am quite comfortable dealing with infinities in terms of limits, as I have stated before. However, I am not prepared to treat it as a real number, because it is not one.



<< Given a1/n + a2/n + ... + an/n = 0.9999.... as a decimal, a1/n + a2/n + ... + an/n = 1 as a fraction.
However, it is 11:30PM now, I am still at work, and I refuse to think any more about this. If anyone else wants to offer up the proof, have at it. >>



This also comes upon something that I haven't studied before, but I have heard about. The whole idea of representing fractions as decimals is something that we all take for granted in our everyday lives, but there's quite a bit of heavy, theoretical math behind it, and I don't think that it's really proven that your equivalences above hold true.

edit Oops, I just realized that even though we're actually dealing with real numbers, we're still talking about a countable infinity (the number of 9's in 0.9999...). My bad.

 

[BuckleDownBen]

wbwither-

i think you are splitting hairs with all your talk of infinity not "existing." you say infinity exists no where in the universe. this may be true, but show me where any finite number exists. you can point to 5 rocks and say that is the number 5, but i can say no, those are just rocks. when you do any math, you must assume some things. you try to make your set of assumptions as small and as discretely defined as possible, but that set of assumptions is always going to be there. i'm a bit rusty on my number theory, but if i remember correctly, if you assume the Natural numbers exist and that every set of numbers will have a smallest element, then you can deduce the concept of infinity.

EDIT:
i was writing my post when you said this

"I am quite comfortable dealing with infinities in terms of limits, as I have stated before. However, I am not prepared to treat it as a real number, because it is not one."

So i withdraw what I said, but I'll leave it in case someone wants to read it.

 

[AkumaX]

PI is 3!!!!!!

*gasp*

 

[Capn]

It is 1, no need for more discussion



** Thus said Capn **

 

[Cattlegod]

to answer this question simply: it does and it doesn't equal 1.

as the number of 9's approachs infinity, delta ( the difference between the two ) gets infinitly small. and thus can be said that it is equal to one because the difference between the two is infinitly small. for all mathematical purposes, this works fine

however, in reality, now matter how many 9's you put there, an infinite number, there still exists a delta of .000...0001 stating that infact, it does not equal one.

 

[BuckleDownBen]

<< however, in reality, now matter how many 9's you put there, an infinite number, there still exists a delta of .000...0001 stating that infact, it does not equal one. >>



this is not true. you have mangled the proof of limits given by wbwither above. there is no delta such that .999... < delta < 1.

 

[Capn]

"however, in reality, now matter how many 9's you put there, an infinite number, there still exists a delta of .000...0001 stating that infact, it does not equal one."

I don't agree with you here, if you say the 9's are infinitely repeating then there is no delta. A delta only exists if you stop at some point in time.

This page does a pretty good proof

That being said, I'm an engineer. So .9=1 and Pi = 3.

 

[bizmark]

<< wbwither-

i think you are splitting hairs with all your talk of infinity not "existing." you say infinity exists no where in the universe. this may be true, but show me where any finite number exists. you can point to 5 rocks and say that is the number 5, but i can say no, those are just rocks. when you do any math, you must assume some things. you try to make your set of assumptions as small and as discretely defined as possible, but that set of assumptions is always going to be there. i'm a bit rusty on my number theory, but if i remember correctly, if you assume the Natural numbers exist and that every set of numbers will have a smallest element, then you can deduce the concept of infinity. >>



YES!!! You're exactly right. All numbers are figments of our imaginations, but infinity is even more so. And you stated a part of the Least Upper Bound Property. Fully stated, it is "If A is a set of real numbers, A not empty, and A is bounded above, then A has a least upper bound." (You essentially said greatest lower bound, but they're equivalent statements). The key thing that you left out is the "A is bounded above" part. The set of natural numbers is NOT bounded above.

If it were, that least upper bound would be infinity. Then take infinity minus 1. That number is surely in the set of natural numbers, since infinity was the least of all upper bounds of the real numbers. But then take that number, and add two. It is a property of the natural numbers that if A and B are in N, then A+B is in N as well. So since (infinity-1) is in N, and 2 is in N, then (infinity-1)+2, or infinity+1, is in N. But wait! infinity was supposed to be bigger than all numbers in N! Contradiction. Then we must conclude that the set of natural numbers is not bounded above.

Anyway, you are right in saying that once we assume this property (and 12 other axioms -- commutativity of addition, the distributive property, the Trichotomy Law, etc. -- your basic understanding of how numbers work) we can deduce some concept of infinity. However, this concept of infinity is that which is embodied in limits -- it is one of arbitrary closeness, not arbitrary largeness. From these 13 axioms we can define limits (actually only the first 12 are needed for limits) and integrals and derivatives.



<< So i withdraw what I said, but I'll leave it in case someone wants to read it. >>



Great, thanks for your comments, which I think were good. You've obviously taken a decent math course or two in your life. I'm really trying to be clear here, because teaching math is something that I want to do someday. Is there anything else that I can say to make you believe it? I'm really not trying to be an ass here, but sometimes you have to be forceful or else people won't believe you

 

[bizmark]

<< This page does a pretty good proof >>



Yes, but you'll notice all these annoying little "lim (m->infinity)" things all over the place....

I've never denied that the LIMIT is one. In fact, I stated as much in my first post in this thread. It is very, very, very true. However, the sum itself is not.

 

[nd]

Yet another proof:

1/3 = .333333...
2/3 = .666666...

1/3 + 2/3 = .999999... = 1

If you have trouble visualizing why 0.99999... is actually equal to one, think about it this way. If it's not equal to one, then 1 - 0.9999.... must result in a non-zero value. But for every 9 you add, you have to add a zero before you can place the 1, so there are infinitely many zeros after the decimal place. The result of the subtraction is zero, making the values equal.

Infinitely many 9's after the decimal is just a conceptual thing that we use. It is exactly 1 (not even "only if you take the limit" -- it really means 1), just like 0.33333.... with infinitely many 3's is exactly 1/3. I haven't read all of wbwither's argument, so I'm not sure if he has a valid point. I suspect if he does, it relies on some different assumptions/conventions than what's generally accepted in the mathematics community.

 

[5489]

this is a very interesting thread.

 

[bizmark]

<< Yet another proof:

1/3 = .333333...
2/3 = .666666...

1/3 + 2/3 = .999999... = 1

If you have trouble visualizing why 0.99999... is actually equal to one, think about it this way. If it's not equal to one, then 1 - 0.9999.... must result in a non-zero value. But for every 9 you add, you have to add a zero before you can place the 1, so there are infinitely many zeros after the decimal place. The result of the subtraction is zero, making the values equal.

Infinitely many 9's after the decimal is just a conceptual thing that we use. It is exactly 1 (not even "only if you take the limit" -- it really means 1), just like 0.33333.... with infinitely 3's is exactly 1/3. >>



Yes, but we've been over this before, there is no such thing as 'infinitely many 9's". Infinity does not exist.

No one should "have trouble visualizing why 0.99999... "is actually equal to one". It's a very intuitive concept: the limit. However, limit equivalences are not normal equivalences.



<< I suspect if he does, it relies on some different assumptions/conventions than what's generally accepted in the mathematics community. >>



Well, I don't know, everything I've learned has been from normal textbooks in normal classes with normal profs. It's somewhat advanced, to be sure, but nothing unreasonable.

 

[Lithium381]

<< this is a very interesting thread. >>




hah yep, i don't think it'll end either, there is no way to to know for sure! i accept both, it is, and is not equal to one.....
when i see .999999999 somewhere, i immediatly round to 1, heh, for simplicity....

 

[nd]

<< Yes, but we've been over this before, there is no such thing as 'infinitely many 9's". Infinity does not exist. >>

Like I said, infinitely many 9's is a conceptual thing that we use. It is generally accepted that 0.3333... is exactly equal to 1/3. Not just a 'limit equivalence', but equal (and it even makes sense if you think about it -- see one of the many 'proofs' above). The "sum" that you're referring to, is also exactly equal to 1 (again, not just 'limit equivilence') if you can imagine there being an infinite amount of operand fractions. For some reason, it seems you're hungup on wanting to take the limit of this sum without considering what it really means conceptually.

 

[silverpig]

There are some no-brainer arguments in this thread, just as there were in the previous one in the HT forum. First of all:


0.9r is NOT a limit people!!! You don't "take" nines, or "add" nines to the end. They are already there. It isn't a number that approaches another number because numbers don't change. Functions change. 2 is 2, 1 is 1, 1 does not approach 2. f(x) = x is 1 when x = 1 and f(x) approaches 2 as x approaches 2.

Second, the arguement that "if 0.9r was equal to 1, it's just be 1 and not 0.9r" is well, just plain ignorant. By that logic, 3/3 wouldn't equal 1 either because if it did, it'd just be 1. 0.9r isn't 1, it equals 1. There is a difference, although it's just a matter of language and not mathematics.

Third, saying 0.9r = 1 and saying the limit of the sum of 0.9 + 0.09 + 0.009 + .... = 1 are two different things. 0.9r isn't a progression, it is a number. I can't stress that enough.

Fourth, there are a large number of proofs to show 0.9r = 1. I came up with one myself back when the other thread was alive that is very very easy to understand provided you know how to do long division. I will show this proof as well as one other one that hasn't been mentioned yet.


PROOF 1:
(long division)

Let's take 1/1 and see what it equals. Doing it the normal way, it's easy to see that 1/1 = 1, but let's just do it via long division for fun. The only difference is, we are going to use the fact that 1 = 0.9 + 0.1 or 1/1 = 0.9 + 0.1. This is like saying 2 goes into 3 one time with a remainder of 1. It's the same here, 1 goes into 1 0.9 times with remainder 0.1. That's true right? 1 = 0.9 + 0.1? Okay...

(this is gonna be hard to type out on here so just write it down if you need to)

1 into 1 0.9 times. Put the 0.9 up top and then multiply 0.9 by the 1 on the left and put the product (0.9) under the 1 inside the division symbol. Subtract as you would normally. You'll get your 0.1 remainder. No biggie right? All we did is show that 1 goes into 1 0.9 times with a remainder of 0.1. Now do the same for this 0.1. Bring down the 0 (you'll have to make the 1 inside the division symbol into a 1.000000000000000...) and you'll have a 10 sitting there. 1 goes into 10 nine times with remainder 1. Put the nine up top (there should now be a 0.99 up there), the product of 9 under the 10, subtract to get a remainder of 1, bring down the next zero and repeat. You can see that 1/1 equals an infinite string of nines after a decimal.


PROOF 2:

This is the one you have to disprove in order to convince any mathemetician that 0.9r dne 1.

Two numbers are different so long as you can explicitly define a third number that exists in between the two numbers.


Simple example here: 2 does not equal 4 because there exists a number called 3 that is larger than 2 but is smaller than 4. Therefore 2 dne 4. Now try the same thing with 0.9r and 1.

(note: explicitly defining a number means it can't contain the other two numbers in it's definition, or use a made up symbol. eg [0.9r + 1]/2 is not valid, nor is 0.9r + n)

I'll even get you started here:

let 0.9r = w + x
and let 1 = y + z

w,x,y, and z must be defined numbers though. Now show:

w < a < y
x < b < z
w + x < a + b < y + z

Therefore 0.9r < 1.

All you have to do now, is come up with numbers for w,x,y,z,a, and b. If you can do that, I'll be convinced.

 

[bizmark]

<< 0.9r is NOT a limit people!!! You don't "take" nines, or "add" nines to the end. They are already there. It isn't a number that approaches another number because numbers don't change. Functions change. 2 is 2, 1 is 1, 1 does not approach 2. f(x) = x is 1 when x = 1 and f(x) approaches 2 as x approaches 2. >>



Please see my response to your point #3.



<< Second, the arguement that "if 0.9r was equal to 1, it's just be 1 and not 0.9r" is well, just plain ignorant. By that logic, 3/3 wouldn't equal 1 either because if it did, it'd just be 1. 0.9r isn't 1, it equals 1. There is a difference, although it's just a matter of language and not mathematics. >>



Yes, you're right.



<< Third, saying 0.9r = 1 and saying the limit of the sum of 0.9 + 0.09 + 0.009 + .... = 1 are two different things. 0.9r isn't a progression, it is a number. I can't stress that enough. >>



Riiiiight. By definition, the decimal .8125 is 8/10+1/100+2/1000+5/10000. Similarly, by definition .9999.... is 9/10+9/100+9/1000+9/10000+.... infinitely many times.

It's the same thing as saying that 9,312 is 9*1000+3*100+1*10+2, in the base 10 number system. It's basic number representation. Every number can be written as the sum of numbers (the number of possibilities of which is equal to the base of the system... i.e., in base 10 we have 10 possibilities for the digits) times every conceivable power of the base. In fact,

9312=....+0*10^100+...+9*10^3+3*10^2+1*10^1+2*10^0+0*(10^-1)+....+0*(10^-100)+....

Every number can be uniquely expressed (unique within each base) as an infinite sum as described above. So for all real X,

X=SUM(i=0 to infinity)a_i*10^i+SUM(j=1 to infinity)b_i*10^(-j), for a_i and b_i in the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.



<< Fourth, there are a large number of proofs to show 0.9r = 1. I came up with one myself back when the other thread was alive that is very very easy to understand provided you know how to do long division. I will show this proof as well as one other one that hasn't been mentioned yet. >>



That long division proof is pretty interesting. I actually like it a lot. The only thing that I can think of that is wrong with it is that it relies on an infinite algorithm and for an algorithm to be valid in a proof, it must have a finite number of deterministic steps. The steps are all well-determined (just keep dividing), but you keep going to infinity. I remember this fallacy from one point in my studies. I couldn't find it in any of my books, though. I'll ask a prof about it and report back.



<< Two numbers are different so long as you can explicitly define a third number that exists in between the two numbers. >>



Yes, as I stated at the beginning, the numbers 0.99999.... and 1 are arbitrarily close together. Stated another way, there is no number which lies between them. If you are willing to take this as a definition for equivalence, then by all means, please do so and call them equivalent all you want. However, while doing so, you must realize that you are doing so based on intuitive notions of infinity and not rigorous mathematics, which only allows us to conclude that the two numbers are arbitrarily close together.

edit: made something more rigorous

 

[silverpig]

<<Yes, as I stated at the beginning, the numbers 0.99999.... and 1 are arbitrarily close together.>>


Ummm, sorta. That's not exactly what I was trying to say though. I meant something more along the lines of "if two things are not different from each other, then they are the same." So, if you can show that there exists no difference, you have shown that they are the same...