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?
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.
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
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
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
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.
Thanks, JS Princeton . . . I was trying to think of that one. I always prefer an algebraic proof to an arithmetic one.
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?”
You, too Ben. #@$@ simulposts.
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?
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.
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).
of course, don’t forget the all important decimal point that I blatantly left out of my assumption.
Related thread
“Stupid math question”
http://boards.straightdope.com/sdmb/showthread.php?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?”
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
Short answer: No! Long answer: In any conceivable real-world application, yes.
I could tell you why .99999… is not equal to one, but then I’d have to bore you to death.
what on earth does “any conceivable real world application” mean?
I love it when mathematicians talk dirty ;-s)
At the risk of being bored to death, could you give us a hint? Because I find it diffult to imagine .999… being anything but 1.
I’m curious. Or are you talking about the series representation below?
A slightly more sophisticated answer to the OP: .9999… can be represented as nine times the sum of (1/10)[sup]n[/sup] for n = 1 to infinity. This is a convergent geometric series, and the formula for the limit of such a series (9 * r/(1-r), 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/(1-r), 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.
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