0.9 repeating = 1?

[Ian Curtis]

Originally Posted by jmanela

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,



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?

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 )

http://en.wikipedia.org/wiki/Geometric_series

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.

[Andrew.Jensen]

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.

[DonRotolo]

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 for The Answer:

A Vinculum

[Ian Curtis]

Originally Posted by Andrew.Jensen

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.

I'm no math major (though I did make the mistake of taking a proof based Linear Algebra class once, never doing that again! ), 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?

[Mr. Lim]

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

[EricH]

Originally Posted by Mr. Lim

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...

[DjMaddius]

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

[Dad1279]

Originally Posted by DjMaddius

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~

[AndrewJS]

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.

[Molten]

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.