FAQ 
Calendar 


#1




Why doesn't .9999~ = 1?
Someone asked this question on another message board I go to. I'm asking it here because the other message board is mostly about video games and I figured someone here would have a better idea.
1/3 = .3333~ 2/3 = .6666~ 3/3 = .9999~ But 3/3 = 1 So why doesn't .9999~ = 1? 
#3





#4







#5




Read your link xray vision...
It does. 
#6




I sent the wrong link and I can't find the one I meant to send, but .999... will always be smaller than 1.
And 1/3 does not equal .333... 
#7




Might want to read your cites a bit more thoroughly, xray vision:
Quote:

#8




Quote:

#9




Re: Why doesn't .9999~ = 1?
Quote:
The first of your statements that I quoted is false, and can only be discussed using the principles of "The Calculus." Your second quoted statement is true, following the basic rule of math that states that any number divided by itself equals unity, or 1. While 1/3 can be said to be .333333..... and 2/3 can be said to be .6666666......, it is silly to simply multiply the infinite decimallyexpressed value of 1/3 by 3 to arrive at a statement like "3/3 = .9999~." 


#10




Quote:

#11




Re: Re: Why doesn't .9999~ = 1?
Quote:
Since 1/3 + 2/3 = 3/3 and .33333... + .66666.... = .99999 Therefore 3/3 = .99999 Q.E.D. 
#12




Put .... after both instances of .99999 in my above post.

#13




Anyone who thinks it is untrue that 0.999... = 1 does not understand limits. (Which is not a terrible thing; lots of Calculus students don't understand limits.)
Imagine you had a blackboard with 0.999 on it. And every second you added another 9 to the end. The number you would have at the "end of time" would not be the same as 0.999.... Also, it is quite possible for every entry in a sequence to be less than the limit of that sequence. That doesn't mean that its limit is not its limit. Q.E.D., please be a little more careful with your ellipses. They're more important when discussing this problem than any other time, in my estimation. 
#14




TBone2 asserted:
Quote:
Using "The Calculus" ( or analysis, if you prefer;specifically, properties about infinite series): 0.333... = S3/10^{i} = 1/3 ( sum of an infinite G.P.) and so 1 = 3/3 = 3*1/3 = 3S3/10^{i} = S9/10^{i} = 0.999... 


#15




Quote:

#16




here is another webpage purporting that .999... doesn't = 1, but check out the last two paragraphs (bolding mine).
Quote:

#17




It's an infinite series. When they talk about it never reaching one, they are referring to an increasing, but still finite, number of 9s being tacked onto the end.

#18




Quote:

#19




Those are just upsidedown 9s.



#20




No, the paragraohs you quote are quite correct. It is an important feature of limits that they do noy have to reach their limiting value after any finite number of steps.
For a ( usually) less controversial example, consider the sequence 1/n. Its terms are 1, 1/2, 1/3, 1/4, 1/5, ... Most people would agree that this sequence tends to 0 as n tends to infinity, and they would be correct. The fact that no term of the sequence is actually 0 is wholly irrelevant. Similarly, when we say that 0.999... = 1 we mean that 1 is the ( unique) limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... 
#21




xray vision...in what way does that web page purport that .999... doesn't equal one? The entire page is devoted to showing that it does equal one! Just like your first cite.
Anyway, your more recent cite contains my favourite argument as to why 0.999...=1, namely: Quote:

#22




We can always bring out the formula for those not bothering with the other threads. We use the basic method for turning a repeating decimal into a fraction.
let x = 0.9999... then 10x = 9.99999... Now we subtract 10x  x = 9.99999...  0.99999... Solving both sides 9x = 9.00000 Reducing x = 9/9 = 1 Therefore 0.9999... = 1 No calculus involved.
__________________
"East is East and West is West and if you take cranberries and stew them like applesauce they taste much more like prunes than rhubarb does." Purveyor of fine science fiction since 1982. 
#23




RC: If you take calculus as the study of limits, then strictly you are using calculus as soon as you write
0.999... and also when you multiply it by 10 to obtain 9.999... 
#24




Quote:



#25




Response to Tbone2
You stated that ...
"following the basic rule of math that states that any number divided by itself equals unity, or 1." Zero divided by zero is not 1 nor is it zero for that matter. I believe mathematicians call the answer either "indeterminate" or "undefined". And Achernar, the proper phrase for "..." is ellipsis and not ellipses. 
#26




Orbifold, I meant to write does, not doesn't. I need some sleep!

#27




The limit of the sequence:
{ 0.9, 0.99, 0.999, 0.9999, ... } is 1. This is really the nomen ludi. You realize that "limit" isn't just a term mathematicians throw around, right? It has a strictlydefined meaning, and it's not ambiguous. It may not be the most intuitive meaning to you, which could be causing the confusion. For instance, it is not correct to think of the limit as the last term in an infinite sequence. The limit does not have to be found in the sequence at all. Anywhere. 
#28




This entire discussion hangs on a pretty thin thread  the limitations of the decimal number system.
Q.E.D. and Jabba, adding the decimal values for 1/3 and 2/3 makes no more sense than multiplying the decimal value of 1/3 by 3. Why strain yourself? 1/3 * 3/1 = 3/3 = 1. (Duh.) 1/3 + 2/3 = 3/3 = 1. (Duh again.) 
#29




Nicely done, RealityChuck!!!



#30




Quote:

#31




Quote:

#32




Quote:

#33




None of the terms of the sequence is 0, but the limit of the sequence is. Sometimes the limit of a sequence is one of the terms. For example,
1, 1, 1, ... certainly tends to 1 and so does 1, 1/2, 1, 2/3, 1, 3/4, 1, ... But it is by no means necessary that this be so. Perhaps a look at the formal definition of a limit will help. For a sequence of real numbers, we say that a_{n} tends to the limit l if: Given any real e>0, we can choose a natural number N such that a_{n}  l < e whenever n>N. In other words, if you give me any positive margin of error, I guarantee that all the terms of the sequence from some point on will approximate the limit number l to within that margin of error. Let us apply this to the two sequences we have been discussing. I have claimed that 1, 1/2, 1/3, 1/4, ... converges to 0 and that 0.9, 0.99, 0.999, 0.9999, ... converges to 1. As an example take e, the magin of error, to be 0.001 For the first sequence, any term after the 1000th is less than 0.001 and so is within 0.001 of the proposed limit. For the second, every term after the 3rd is within 0.001 of 1. If instead we take e = 0.000001, then we just have to go further along the two sequences: the 1,000,000th term will do for the first sequence; the 6th term for the second. In general, whatever positive e you choose, I can guarantee that all terms from some point on will be within e of my claimed limit. That is what we mean by convergence, and in particular by the value of an infinite sum. 
#34




.000~1



#35




Quote:

#36




Now now, I'm with you Q.E.D., but performing arithmetic on infinity isn't going to strengthen our cause.

#37




Q.E.D., what does 2  .9999... = x. What would you have x be?

#38




1

#39




Scratch the "what does".



#40




1  .9 = 0.1
1 .99 = 0.01 1  .999 = 0.001 I can go on forever and never get 0, however, 1 1 =0. My conclusion is 1 cannot be the same as .999... 
#41




Quote:
Given any two distinct real numbers, there is always a third real number lying strictly between the first two. Between 0 and 1, there is 1/2. Between 1 and 1/2, there is 3/4. Between x and y, there is (x+y)/2. By the same token, if there is no such third number then the first two numbers can't be distinct. There's no number x such that 1 < x < 1, which is why 1 and 1 are the same number. Now, what about 0.999... and 1? I claim there is no number between them. Indeed, suppose 0.999... < x < 1; I claim this leads to a contradiction. For starters, it must be true that 0 < x0.999... < 10.999... . But 10.999... is clearly less than 10.9=0.1. It's also less than 10.99=0.01, 10.999=0.001, 10.9999=0.0001, and so on. In fact, given any positive number y I can show that 10.999... is less than y by writing out enough 9's. But somehow 10.999... isn't less than x0.999..., even though x0.999... is a positive number. This is the contradiction. So there does not exist any such number x. The same thing happens with the sequence 0.1, 0.01, 0.001, ... . No term in the sequence is 0. But the limit must be zero. After all, the limit certainly isn't negative. And the limit can't be postive either, for if it were then it would be smaller than 0.000......00001 for some number of zeroes, and how can the limit of a decreasing sequence be smaller than any of the terms in the sequence? Zero is the only possibility remaining. 
#42




XRay Vision, the whole point behind infinite arguments is that is you go on forever you do get 0.
These threads always come down to this one basic factor. Either you get the idea of infinity and understand this equality or you don't. What's ironic is that your argument  that we keep getting smaller and smaller but never reach zero  is arguing for an infinitesimal. But infinitesimals and limits are incompatible ideas in our arithmetic. The limit as .999999... goes to infinity is 1. 
#43




Quote:
0.999... has an infinite number of 9s. Using .9, or .99, or .999 is an approximation, which is why you don't get exactly zero. But the approximation gets closer by a factor of 10 for each digit you add. So, what happens when you use an infinite number of digits? 
#44




Quote:



#45




Let me expand on my previous response.
.9 differs from 1 by (1.9)/1, which is .1 or 1/10. .99 differs from 1 by (1.99)/1, which is .01 or 1/100. .999 differs from 1 by (1.999)/1, which is .001 or 1/1000. Let n be the number of 9s you are using in the approximation. Thus, the difference between 1 and .9999.... is 1/(10^{n}). For n=infinity, the difference is 1/(10^{infinity}) or 1/infinity. Since 1/infinity equals zero*, the difference between .999... and 1 is zero. Saying that the difference between two numbers is zero is equivalent to saying that they are the same number. QED. *Yes, I know that it isn't technically correct to say that 1/inf equals 0, but rather lim_{x>inf} 1/x = 0. Small technicality. 
#46




Quote:
Quote:

#47




Let's leave infinite arithmetic out of this, and not cloud the issue with technicalities.
Others have already said most of what I would, and I've been there and done that on this topic, so let me just add that the fact that .9... = 1 is a direct consequence of the most important property of the real numbers, and if they're not equal, everything falls apart. 
#48




Q.E.D. your wrong again!!
If you went .999 of the distance to 1.0, then guess what? You never got there!! Terribly obviously wrong again!! If you went .99999999999~~ way around the world you never reached your starting point so .99999~ will never equal 1!! I don't care who tells you it does!! WRONG!! got your back again xray!! 
#49




Quote:



#50




To all the mathematicians out there give it up; those who don't understand limits and calculus and choose not to study will never understand. To all of the math challenged out there let's discuss something a bit more on your level: How many angels can dance on the head of a pin?

Closed Thread 
Thread Tools  
Display Modes  

