

Does 0.999.... = 1?m0u53m4t
Just a passing thought... Do you think 0.9 recurring = 1 or not?
Dwyer17
Why does it matter. I don't understand the point of this question. The answer is no by the way.
NooBix
Lets say... 0.99999 (x number of recurring 9) 0.9999(recurring) = 10.000(x1 number of recurring 0) and a 1 at the end... so... 0.9 = 10.1 0.99 = 10.01 0.999 = 10.001 etc etc. 1 will not be 0.99999999999999999999999 charseips
Philosophically and mathematically I do not believe that 0.999 recurring will ever equal 1 and to attempt to comprehend it is flawed and too much for our minds. An infinite number of 9s try to visualise it you'll probably end up scaring yourself, feeling very small and insignificant or hurting your brain. I know I do or maybe I think to deeply about these things.
On the other hand thinking pay wise and making money round it up I'm fed up with my interest rate of 3.2145551 where they round down the pennies. TomS
Let's expand:
1/3 = 0.33333....... 3*(1/3) = 1 3 * 0.33333.... = 1 This is because you can't split a whole into 3 party that have the same size. So the mathematical definition ist that 3*0.3333 is 1. What's the definition for 0.9999... ??? drag0n
0.999.. < 1
owned charseips
in the poll does
bigger mean greater than > smaller mean lesser than < same mean equal to = I think it best to be accurate given the question. indeedwrestling
Actually, they are the same.
Let is call this mysterious value .9999999... as X. Then 10x=9.9999999..... So, 9x=9. Thus, x=1. So, point nine repeating is the same as one. Basically, it's a way of discussing infinity. Consider this: there are an infinite number of numbers above one. However, there are also an infinite number of numbers between 0 and 1. Which infinity is bigger? jwellsy
x = 0.9999...
10x = 9.9999... 10x  x = 9.9999...  0.9999... 9x = 9 x = 1. markosaurus
I didn't understand most of the previous posts as they are all a little complicated for my tiny little brain to comprehend, but I have to agree with jwellsy, that seems to work, although I would say I don't think it should, I simply don't see how 0.999999 can ever be equal to 1.
So I have just contradicted myself. Can there ever actually be an answer to this or is it one of those ridiculous hypothetical things that mathematicians like to discuss? indeedwrestling
Isn't really not that complicated.
Here's a simpler example: 1/9 = .111111111111.... Therefore .9999999 is nine times 1/9. (1/9) x (9) = 1. It's just a property of the fact we're in base 10. okullar
Wouw answers are amazing...
You know somethings are only "assumed" not real... 0.999... is so close to 1 that we say they are equal. By the way, how could we explain that there is a number called: 0.9999... if there is no limit, the 9s will go on forever... , can this number be 1? If there is a limit, then it wouldnt be 0.9999... it would be the exact number. With infinity definition, it is hard to say an answer. Well, simply, 1 must be greater than all 0.x we may assume that they are equal, but 0.999... can never be bigger than 1. KD8CPP
NO! you will have .999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 for all I care and it will not equal 1 it will equal .9x whereas x is the number of 9s. the only time plain 1 will equal plain one is if 1=1 is used!
73s Tyler Lewis arecanut
If you're willing to accept the theorem that, between any two real numbers, there must be another real number, then what's a real number between 0.999... and 1?
That's usually the reason I state, only because it's easiest to explain. Of course, if they refuse to believe in the theorem, they probably won't believe any formal mathematical proof you'd try to present, either. carlospro7
while this may be a useless question, and an attempt at getting fri points, .99999 does not equal zero.
1.000000000000.......... > 0.999999999999999999999................ I don't think this is even questionable. osbits
They stand for the same thing.Just like you have a name and nick name.They are the same value.We may think that 1>0.999... because the always a 0.0000... space between them.That not ture.
circuits
non zero numbers have a habbit of being absolute... So 1 > 0.99999999 no matter how many significant digits the latter has. Bummer.
douzy
It's not 1, but can only be rounded up to 1 (approximation)
AftershockVibe
0.9 recurring is equal to one as more than one person has proved above by showing that if;
x = 0.99999999 recurring then 10x = 9.999999 recurring. 9x = 10x  x therefore 9x = 9.9999999999 recurring  0.99999999 recurring = 9. 9x = 9 therefore x = 1. Or to think about it another way... what is 1  0.999999999 recurring? 0.0000.... recurring THEN 1. Recursion will never cease therefore the 1 will never occur. You have to remember that the way we write numbers is only symbolic which is why there is no simple way to fully express recursion and why will will never be able to completely write out an irrational number. benjmd
0.9999.... is only equal to 1 in a discrete number set. It is less than 1 in an infinite number set.
Discrete set: Many have already shown math that proves this, but also consider, conceptually, the complete series of discrete values around "1.". Any discrete number you pick close to 1, but less than 1, will be also less than 0.9999.... Say you pick 0.99999...(23)...999999. You must concede that 0.99999...(24)...999999 is greater. And so on until you have an inconceivable number of 9s, approaching 0.9 repeating. Since 0.9 repeating cannot be *greater* than 1, it must be equal to 1 on the discrete number set.  In short: 0.9 repeating is greater than the greatest discrete number less than 1 but it is not greater than 1. Therefore its discrete value must be equal to 1. Ahh, but here is where it gets tricky. Infinite set: .9 repeating = (the sum [from n=1 to n=infinity] of 9/(10^n)  Note here that in each step of the summation for any upper limit on n, 1/(10^n) is lost from the whole, therefore when we consider the value of 1, we can represent it as: 1 = (the sum [n=1 to n=x] of (9/(10^n)))+(10/(10^(x+1))) for any value x There is therefore, at least conceptually, a value  10/(10^(x+1)), where x approaches infinity, for which 0.9 repeating is separated from 1. Now, for anyone who has even taken precalculus, you know that in our mathematics system as x approaches infinity that term becomes 0 for all intents and purposes. However, it is important to remember that mathematics is just a representation of the Universe. The challenge posited by 0.9 repeating = 1 is one that really ends up being a proof of the existence of infinity. Em_de_Tech
I know this trick. I have no idea how to explain it but i've seen a couple variations of it and they are all written about above. (Actually my friend showed me this whole .999999.... = 1 like 8 days ago) But this is a hard concept to grasp. I don't even think that i am grasping it.
roberpro
Just no thats my answer for that question.Thats have common sense
the1991
.999..... is infinitesimally smaller than 1. That is a true statement. Infer from it what you will.
the_mariska
Damn, this topic really brought me down. OK, I understand that some people have doubts whether God exists or not, the others have doubts whether ghosts exist, I can even imagine having doubts about the evolution, but how the hell can anyone bring the laws of mathematics into discuss??
No matter if you like it or not, 1 = 0.999999... because it's easily provable. [I won't add the proof here, because it's been already posted here three times]. All of the philosophical discussions, whether this could or couldn't be true seem completely senseless to me. To those, who don't think so, I suggest starting a discussion "Is there a square root of (1)?". I bet some of them would never agree that complex numbers could exist fedex220
.99999999999999 Does equal 1. Because you round the .99 up because the second nine is higher than 5 so it will round up causeing the number to be 1
James007
I could kick users who create such topics.
Pointless discussion. close Related topics
