Math problem

Southrend · **Posted:** January 10th, 2012, 12:12 pm

I've been introduced with a math problem today and I don't know what to think of it. My world is slowly falling apart.

Imagine the number 0.9999... followed by an infinite amount of 9's. Does that equal one?

Before you view the 2 solutions I have, just think about it for a second.

Spoiler for solution1:

Spoiler for solution2:

I'm not sure with what solution I should go.

Adbot · **Posted:** January 10th, 2012, 12:12 pm

Evelyn · **Posted:** January 10th, 2012, 12:19 pm

0.9-repeating is 1. There was a formula converting repeating decimals to their fraction equivalents, and when used on 0.9-repeating it produces 1. If you think about it, if 0.3-repeating is 1/3, and 0.6-repeating is 2/3, then 0.9-repeating must be 1.

Also for all intents and purposes, 1/infinity is zero.

Edit: Wolfram|Alpha to the rescue!
http://www.wolframalpha.com/input/?i=0.999999...
http://www.wolframalpha.com/input/?i=1%2Finfinity

Iron Maiden · **Posted:** January 10th, 2012, 2:58 pm

0.999[insert infinity here] = 1

Why?
Let's say you take 1
Divide ''1'' by 3
you get 0.333(infinity)

take your 0.3333... and remultiply it by 3
You get 0.9999(infinity)

1 = 0.99999999999(infinity)
Mind = blown

zant · **Posted:** January 10th, 2012, 3:37 pm

No, it does not equal 1. It is close to 1 by an infinitely tiny amount. If you round, yes it is equat to 1

Jamie · **Posted:** January 10th, 2012, 6:33 pm

.999 repeating to infinity is 1. While an algebraic explanation for this has been offered above, I will offer two other mathematical explanations for why this is true.

.999 repeating to infinity is an infinite series (.9 + .09 + .009 + .0009 ... etc). Notice that this is also a geometric series (meaning that each term in the series can be found by multiplying the previous term by a common ratio). In this infinite series, the common ratio is 0.1. There is a formula to find the sum of an infinite geometric series when the common ratio is between -1 and 1: S=a/(1-r), where a is the the first term of the series and r is the common ratio. In this case, S=0.9/(1-0.1). What does that equal?

If you know calculus, you can also solve this problem using a limit. .999 repeating to infinity is also known as the limit as x->infinity of 1-(1/10^x), which looks fairly similar to the solution2 picture up there. Since 1/10^x will go to zero as x->infinity, the limit is equal to 1.

Evelyn · **Posted:** January 10th, 2012, 7:59 pm

Jamie wrote:

S=a/(1-r)

That's the formula I was talking about.

The calculus approach is the easiest way to figure out 0.9-repeating = 1.

Adbot · **Posted:** January 10th, 2012, 7:59 pm

Duke Juker · RSBandB Donor since 07/01/2010

It's like calling a tomato a fruit. Technically it is a fruit...but it sure as **** doesn't look right or seem right.

So in that regard, I still think 0.999-reapeating is not equal to 1. I think your math problems are committing a certain type of fallacy in logic, but I can't quite put my finger on it.

Earth · **Posted:** February 11th, 2012, 12:42 pm

I think there's a difference between .9inf and .9999999999999999999999999999999999999999999999999999999999
One is infinite, and the other is just a ton of 9s, if that makes any sense at all.

ryan1 · **Posted:** February 13th, 2012, 1:15 pm

We discussed this in my Calculus II class. Very interesting indeed. At first I was reluctant to accept it, but the professor showed it to us in so many ways it was basically undeniable.