# 0.9 repeating = 1?

I was googling around today and i came across something rather odd. http://en.wikipedia.org/wiki/0.999…. Now, this wikipidia page claims that .99999 repeating = 1.

Here is a quick algebraic proof to support this,

1/3 = 0.3~ \common knowledge, let (~ stand for repeating)

(3)1/3 = 0.3~(3) \multiply 3 to both sides

3/3 = 0.9~ \evaluate above

1 = 0.9~ \simplify and voila!

Now that is an algebraic point of view. I can’t speak for calculus because i haven’t taken that yet. What is your view on this topic?

What is the difference between .9 repeating and 1?

Hint: a point.

Now whats the dimension of a point?

When you get to calculus they’ll have you prove that .9999=1 using geometric series.

I like this method. between any 2 different real numbers, there is at least one number between them not equal to either of them.

You don’t need calculus for this, just Series and Sequences which is an Algebra II/Precalc class. (You will make extensive use of series and sequences in Calc II and you will learn to hate them :rolleyes:)

Consider .9~ to have a seed value of .9 with a ratio of .1. Use the formula proven on the wiki page. Voila!

They also do an example problem under Repeated Decimals.

I actually figured out that same proof in the Wikipedia article earlier this school year before reading about it. But 0.99 repeating does equal one. There is enough proof, and even common sense, as 1-0.999… =0.000… which is just 0.

You’d be surprised how many people argue that “it can’t be true”. But it is. The key word is “repeating”.

Now for the extra credit: What’s the name of that line that goes over the last digit to indicate that it repeats infinitely?

[spoiler]A Vinculum[/spoiler]

I’m no math major (though I did make the mistake of taking a proof based Linear Algebra class once, never doing that again! :ahh:), but I don’t really see the proof. Wouldn’t accepting that 0.000… is equal to zero hinge on accepting that 0.999… is equal to one? What am I missing? :o

Then I guess this means that (lim x-> 1-) is NOT 0.999~

You mean (lim x-> 1-) of x is not 0.999~, don’t you? As it is, you aren’t taking a limit at all.

By the way, after a certain point, it won’t matter; your machinist won’t go to that tight of a tolerance…

Your wrong. 1/3 = 0.33333333333333333. an infinite amount of 3’s. Not 0.33. So multiple .33333~*3 = .99999~, Not 1.

(1/3) * 3 = 1 = .33333~*3 = .99999~

Just for curiosity, what would happen if you approached the problem backwards, starting at 1, then subtracting:
1 - .1 = .9
1 - .01 = .99
1 -.001 = .999

1 - 10^-n = .9~ as n approaches infinity

You could argue the 10^(-infinity) approaches 0 and thus 1 = .9~
But then again, the equivalent expression, 1/10^n divides the number 1 into smaller and smaller parts. Just because the parts are smaller doesn’t mean they are nonexistent. To me, this expression states that you can always divide 1 into smaller parts, therefore you can never really “reach” infinity, and therefore never “reach” .9~

It’s just like 1/x . What happens as x --> infinity? “The limit is 0.” Yes, but does the function itself every EQUAL 0? It gets closer to it, and closer, and closer, .01 .001 .00001 .00000000000000000000001 But there’s always that 1 at the end, no matter how many zeros you throw in there. And you can’t really put in “an infinite” amount of zeros in between, because then the 1 would be coming after “infinity” which nulls its definition.

I may be wrong, but at least presents an argument different from the standard 1/3 stuff.

I agree that .9~=1 if .3~=1/3. I just think it’d be more accurate to say that both are extremely close approximations. Though most mathemeticians will disagree with me.

between any 2 different real numbers, there exists an uncountably infinite set of different real numbers.

Lets try it this way.

X=.99999…
Thus X10=9.99999…
now 9X = X
10 - X = 9.999… - 0.999… = 9.000
9X = 9
X = 1

And proven.

X=.99999…
Thus X10=9.99999…
now 9X = X
10 - X = 9.999… - 0.999… = 9.000
9X = 9
X = 1

if X=1 then 10X=10 and
9X=10-.999999…

I’m not sure anything was proven!

I’m not quite sure if I completely understand this, but here’s another look at it:

1/9 = 0.111…

91/9 = 90.111…

9/9 = 0.999…

1 = 0.999…

The mathematical meaning of the repeating decimal .999… is a limit.

It is the limit of the sequence of partial sums of the infinite series 9/10 + 9/100 + 9/1000 + …

The sequence of partial sums of the above series is equal to (1-1/10), (1-1/100), (1-1/1000), … (1-1/10^n)

Using the definition of limit of a sequence:

A real number L is said to be the limit of the sequence Xn if and only if for every real number ε > 0, there exists a natural number N such that for every n > N we have | Xn−L | < ε.

It can be shown that the limit of the above sequence is “1”.

Therefore, “1” and “.999…” mean exactly the same thing. They are two different ways of writing the same real number.

The limit of a series of partial sums is not equal to the actual sum at all. This is by definition of a limit. A limit is what the function must approach ever closer without ever reaching it. If it ever actual reaches it at any point, then it is not truly its limit.