#1




My geometry teacher used to always tell my class that .99999999 with a repeating bar over it is equivalent to 1. He also said that he could prove it mathematically, but he never gave the proof. Could someone out there give the proof that .99999999 with a repeating bar is exactly equal to 1?

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#2




All I know is that fractions with a denominator of 9 are shown as decimals with the numerator repeating. (I know, that was poorly worded. Allow me to demonstrate.)
1/9 is the same as 0.1111... 2/9 is the same as 0.2222... 3/9 is the same as 0.3333... ...and so on. 9/9 would be the same as 0.9999... Since a fraction with the same number as the numerator and denominator equals one, we can conclude that 0.9999... = 9/9 = 1. I hope that was understandable. 
#3




Yup, that's the easiest proof. Well, actually that's a common sense shorthand. The mathematical proof (in thirds for brevity) goes more like:
1/3 * 3 = 1 1/3 = 0.333... 0.333 * 3 = 0.999... Therefore, 0.999... = 1 
#4




Here's another proof:
Given that: x = 0.9999999999 etc. Then: 10 x = 9.999999999 etc. So, then 10x  x = 9.999999999 etc.  0.999999999 etc. = 9 Which implies... => 9x = 9 Which implies... => x = 1 but x = 0.9999999999 etc. so 0.999999 etc. = 1 QED 
#5




Another proof, the one that I like best...
We want to prove that .9rep = 1.
First, let's assign a symbolic name for .9rep so it's easier to deal with. Let's use "x"  it's as convenient as any other letter. So, x = .9rep Now, multiply both sides of the above equation by 10: 10x = 9.9rep Now, subtract x from each side: 9x = 9.9rep  x But since "x" is just .9rep: 9x = 9.9rep  .9rep Or 9x = 9 Divide both sides by 9, and... x = 1 Q(uickly).E(nds).D(at). Slick, eh? Ben 
#6




This is .9 repeating ad infinitum, but never reaching 1.0 since it would take infinity to get there. I believe, in calculus, that that is considered 1.0.
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There are 10 kinds of people in this world: those who understand binary numbers and those who don't. 
#7




Thanks, JS Princeton . . . I was trying to think of that one. I always prefer an algebraic proof to an arithmetic one.

#8




By the by... by using the method I posted you can convert all repeating decimals into whole number fraction. Just instead of 10x use whatever power of 10 is required so that your repetition is cancelled upon subtraction of (10^n)x  x
As a boss of mine at a science museum used to say whenever a visitor was exposed to something educational (even if it was something like a magnet and they had clearly seen it before): "Pretty cool, huh?" 
#9




You, too Ben. #@$@ simulposts.

#10




So does this mean that it's impossible to have a true .9999..., that is, having the 9 repeat itself all the way into infinity? Or, mathematically, do we always refer to .9999... as 1?
In other words... is .9999... ALWAYS equal to 1? 
#11




Well . . . yeah, SPOOFE. I mean, if you could manipulate an infinitely repeating number, 0.999... would always behave as though it were equal to 1.
It would be different if we were talking about a converging series, in which something approaches 1, but never can be said to reach it. But 0.999... is a constant. 
#12




As long as you're certain the 9s go on forever (iterative function, I believe it's called) it's equal to one.
If it stops or changes numbers ANYWHERE, it's not. So, this is a semantics question: how do we know that it repeats forever? Well... I could give you an inductive proof of that (which isn't inductive at all, but deductive... just utilizes a theorem in mathematics that certain things always happen the same for any number on up to infinity)... but we'd have to start with some assumptions. What would those assumptions be? Well, the number would have to be zero followed by an infinite number of nines. But there you see, we really already have the proof (though not rigorous). 
#13




of course, don't forget the all important decimal point that I blatantly left out of my assumption.

#14




Related thread
"Stupid math question" http://boards.straightdope.com/sdmb/...threadid=71810 " How can 1/3+1/3+1/3=1 but .3333...+.3333...+.3333... never truly equal 1? I believe a recent thread tried to explain this, but truthfully I didn't understand the answer. Can someone explain this in plain English. Pretend I'm your six year old. " "Why Daddy, Why?" 
#15




It's a big of a boggler at first...
To think that a number you've known so well may not be what it seems.
The big difficulty that most people encounter with this, I think, is that their internal assumption that every number can only be represented with one syumbol has been violated. But that's is not so strange if you think about it. What is the fraction 2/2? Isn't that just another symbol for "1"? Same with 5/5 and 32,973/32,973. They're all just different symbols for the same number. As is the symbol "1". As is the symbol ".9999...". Etc. Maybe it'll help if you think of numbers like people. I'm Ben. On the boards I am represented by the symbol "ModernRonin." To the people who maintain the computers at my work, I am represented by my login, "bcantric". Some of my friends use my last name because they know more than one Ben. To these people, I am simply "Cantrick." These are all different symbols, but they mean the same person. Similiarly, "1" and ".999..." and etc are just different symbols for the same number. It's a little easier to accept if you can think about it that way. Ben 
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I could tell you why .99999... is not equal to one, but then I'd have to bore you to death. 
#17




what on earth does "any conceivable real world application" mean?
I love it when mathematicians talk dirty ;s) 
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#19




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A slightly more sophisticated answer to the OP: .9999... can be represented as nine times the sum of (1/10)^{n} for n = 1 to infinity. This is a convergent geometric series, and the formula for the limit of such a series (9 * r/(1r), where r is the common ratio) gives us 1. This is not the same formula you get if you start the sum from n = 0; in that case, you'd get 9/(1r), which gives you 10. But it's not the right formula, so we don't care if it gives the right answer. While this is more abstract, it would satisfy pretty much anyone. I've always felt that the argument that starts with x = .999.... is a little lacking in style, so I prefer this one. 
#20




So .999.... is the same as .999999999999..... ?
Or .9..... ? Wierd, but I think I get it. It's just a convention. Except when your .9's go into a black hole. Peace, mangeorge
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#21




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Really, our two statements (yours of teh series, mine with the decimal) are ENTIRELY equivalent. Honestly. If I define a repitend to be infinitely repeating digits, by God, that's what it is. I like your proof, in any case. This has become sorta fun, looking at all the ways you can come up with one. By the by, your "starting with zero" series is the subject of another thread that's been zipping around tonight. I'd parse a URL, but I'm lazy. Look for it yourself all ye with curiousity. 
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#23




I hope that these proofs have been enough to convince everyone, but I'm gonna add one anyway. Not really anything rigorous, but just a logical argument.
In order for two real numbers (however they may be represented) to be different, you should be able to describe some number that would exist between them. Stated another way, there should be some value you can add to one of the numbers that will increase it, but not push it higher than the other number. You should also be able to subtract something from one number without going below the other. Any number you add to .9rep makes it larger than 1. Any number subtracted from 1 makes it smaller than .9rep. 
#24




so I'm curious  mathmatics is supposed to be an absolute. 1=1  therefor wouldnt 0.9999999... = 0.9999999... ?
in any pratical application of the numbers it would be so close to 1 that there would be no point to call it anything but 1, but, it still is not absolutely completely equal to one?
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#25




Here's the way I first wrapped my head around the concept when it was presented to me:
What's 1  0.9999999...? It's 0.000000....1, where the one is preceded by an infinite number of zeros. Effectively, you never reach the 10^{n} that is the difference between 1 and 0.9999999999. So the difference is an infinite decimal expansion of 0.00000..., which is zero. If x  y = 0, then x = y. 
#26




This has been beaten to death, I don't know why I'm jumping in but I always want to...
Consider the axiom of completeness for the real numbers (that is, look it up if you don't know what it means). Any convergent sequence of real numbers converges to something that is a real number. .9rep is just a convienient way to represent a series, which has a sequence of partial sums, that converges. So, to what real number does it converge? Use the axiom that says that for all real numbers A and B, one and only one is true: A>B, A<B, or A=B. Since there's no number in between .9rep and 1, then the first two can't be true. That leaves the last, .9rep=1. 
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#28




Thanks, kinoons
Well I thought I had it, but; How would you notate the difference between .9999... and one? In other words, The difference between .9999 and 1 is .00001, right? How would one write the difference between .9999... (to infinity)and 1? .0000....1 ? Peace, mangeorge 
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#32




guys, your applying a mathematical proof to a philosophical debate  the number line is not only a mathematical concept, but a philosophical one as well. Every number on the number line has its own position on the number line. No matter how you manipulate the math to try and prove that two different values can occupy the same location on the number line cannot be true. It is akin to saying that two different particles of matter can occupy the same space. It is simply not possible. The two pieces of matter can get so absolutely close to one another that the difference cannot be measured, but the two particles cannot be at the same place at the same time.
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#33




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please, please do keep adding the 9 to the end of 0.999... and I'll keep adding the 0 to 1.000  the two numbers will never be the same, they can not be. It is fundamentally not possible for any two numbers to occupy the same location on the number line.
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#38




Hmm...
I'm not a math person, but I play one on T.V. And I have a question: If .99999999..... = 1, then does .11111111..... = 0? 
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#40




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Another proof: If .99999... is not exactly 1, then there is some number epsilon such that 1  epsilon = .9999.... But, no nonzero epsilon can be demonstrated to satisfy this. Kinoons would say epsilon is 0.000...01, but that is only do to a bad definition of "..." Quote:

#41




humor me.

#42




okay, you want to prove to me that 0.999... is equal to 1.000... then do this for me...
draw a line. Label one point on the line 0.9. one inch later on that line mark a point 1.000... (add as many 0's as you would like. now, draw a line for 0.99. Now draw a line for 0.999. Now draw a line for 0.9999. Now one for 0.99999. Continue to keep drawing lines please. Call me when you draw an additional line on 1.000
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#43




The inability for the thickness of the ball in my pen that applies the ink to create thinner and thinner lines as I draw them does not justify your point.

#44




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not convinced  okay 100 feet apart...1000 feet? How much space do you need to be convinced that a line will never be placed on the exact same locaton?
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Kinooning it up for over 1,500 posts and counting. 
#45




wait wait wait.......I reread your first post. I'm on your side here, I thought you said "Okay you want me to prove that 0.99999 = 1?" Sorry. Pretend that's what you were trying to say, then reread my post, then I'll make sense.
0.99999.....(to infinity) does not = 1. If you say they are the same, then you are just saying "infinity is = 1". And that's just not true. Infinity is a concept ONLY, and not to be considered a number. "1" is just a number. Apples are fruits. Oranges are round, like apples. Does that mean Oranges are Apples? Saying an infinite number (like .99999...) is like 1 is comparing apples to oranges. They are two different figures in a huge way. 
#46




kinoons, you're being silly. Obviously nobody has the time to draw the infinite number of lines that your demonstration would require.
Look at the proofs in the 2nd, 3rd, 4th and 5th posts in this thread. They are very simple and should erase all doubt. I mean, this is not a point of controversy in mathematics. It's gradeschool stuff. 
#47




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Present us with a list of 10 numbers between 0.9999... and 1. There should be an infinite number there for you to choose from, if they are really different numbers. Or, to go back to you number line  if they occupy different places, then there must be some space between them. Draw 10 lines in that space that are on neither 0.999... nor 1. _{Yeah, Saltire brought this one up first, I know...} 
#48




I would just like to mention that, by defition, "0.000....01" does not go on for infinity. It has an end (the "...01"), and thusly, you cannot say that .999... has a difference of .000...01 from 1.
Infinity cannot be confined between two points. 
#49




But wait a minute, if 0.999...= 1 then 0.999...8= 0.999... then 0.999...7= 0.999...8 then 0.999...6= 0.999...7 etc. etc. Then if you did it infinite times, you'd soon come to 0 so are you saying that 1=0?
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#50




Burnt Toast Read SPOOFE's post above. You can't put infinity between those two numbers. If you're putting it between those two numbers, that means that it has two ends and you've restricted an infinite number.
kinoons  You can put two numbers on the same point on the number line if they are equivalent. For example, I could put 100/100 on the same point on the number line as 1, because they are the same number. Because .9rep = 1, (the proofs shown are the ones I was taught), then .9rep CAN go on the same spot on the number line as 1. 
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