Author Topic: .9 repeating equals 1?  (Read 21319 times)

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Offline aeTIos

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Re: .9 repeating equals 1?
« Reply #15 on: December 11, 2013, 01:49:25 pm »
Quote
an infinitely small gap is not a gap at all
Well, it is. It's just infinitely small.
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Offline TIfanx1999

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Re: .9 repeating equals 1?
« Reply #16 on: December 11, 2013, 02:14:40 pm »
@Shmibs: No, I haven't taken calculus. A decimal can extend as far as it wants in one direction, but that still will not make it equal to one from what I can see.

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Re: .9 repeating equals 1?
« Reply #17 on: December 11, 2013, 02:17:37 pm »
Art_of_Camelot:

If you want to learn calculus: I'm making a series of lessons here:

http://ourl.ca/20293

It does not cover convergence yet, but it has covered limits (to an extent) :D
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Offline AngelFish

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Re: .9 repeating equals 1?
« Reply #18 on: December 11, 2013, 02:21:54 pm »
To me, it equals 1.

Simply accept that as long as 41.9999...99958=42, I'm right.


That "notation" is heavily misleading and, somewhat unfortunately, crops up whenever this debate does. What does the use of the ellipsis denote? An infinite string of characters extending to the right. The tricky part here is the word "infinite". An infinite string has very different behavior from a finite one. In particular, the notion of an "end" does not exist. You can't append to the end of an infinite string because there isn't an end to append to. By placing numbers after the ellipsis, you're attempting to denote this. It's not a valid operation and hence what you get is wrong. 41.999...=42 would be the only way to write that statement and thankfully, it's true.

To see why, let's take a quick look at what the reals are:

Reals intuitively consist of all the numbers along a number line. This can be formalized with something called a dedekind cut, which is essentially as follows: For every point X along a line, define sets A and B such that A consists of all the points below X and B consists of all points not in A. We can define the reals as every point that we can do this for. Note that each A has the property that while there is an upper bound on the set (X), there is no greatest member of A. In the discussion 0.9999... = 1, what we're really asking is whether there's a a distinct real number from 1 such that there is an infinitesimal difference between them. In other words, is there a greatest member in set A? As I stated earlier, this is a property of the way the set is defined that there is not. Therefore, if 0.9999... is distinct from 1, it cannot be in A. It must therefore be a member of B, as the smallest member. Unfortunately, this is exactly where 1 is located. Thus 0.999... = 1.

If this confuses you, that's fine. It's simply a more intuitive explanation of the formalization of reals. Note that, unlike what shmibs, says, I am an uninteresting person (err, i mean you can have systems where the equivalence is not true), but it's not possible with the *reals*.


EDIT: Thanks to shmibs for edits and *ahem* punctuation.
« Last Edit: December 11, 2013, 04:44:29 pm by AngelFish »
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline harold

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Re: .9 repeating equals 1?
« Reply #19 on: December 11, 2013, 03:30:53 pm »
Dedekind cuts probably aren't going to convince the deniers, though. Let's be honest, they don't really make intuitive sense, and the biggest reason the deniers have is "but it doesn't make sense"..
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Offline shmibs

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Re: .9 repeating equals 1?
« Reply #20 on: December 11, 2013, 03:37:30 pm »
@Shmibs: No, I haven't taken calculus. A decimal can extend as far as it wants in one direction, but that still will not make it equal to one from what I can see.
that's true. no matter how many 9s you care to add to the end, you will still have a finite number of them. adding 9s to the end can never help you to reach infinitely many 9s any more than it can help you to reach 1. thus, both 1 and .9 repeating are greater than any number you can reach by simply adding more and more 9s. as .9 repeating is also, obviously, not greater than 1, the two must be equal.

>inb4 that's not exactly true from fish :P
« Last Edit: December 11, 2013, 03:43:55 pm by shmibs »

Offline AngelFish

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Re: .9 repeating equals 1?
« Reply #21 on: December 11, 2013, 04:46:36 pm »
* AngelFish acknowledges the wisdom of the shmibs and that it's a much better way of saying what he tried to get across with dedekind cuts
* AngelFish also acknowledges that he is in fact a banana
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline Legimet

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Re: .9 repeating equals 1?
« Reply #22 on: December 11, 2013, 04:49:57 pm »
0.9999... is just 9 * (1/10 + (1/10)^2 +(1/10)^3 +...) = 9 * (1/10) / (1- 1/10) = 1. To do this formally, you'll need calculus and infinite series.

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Re: .9 repeating equals 1?
« Reply #23 on: December 11, 2013, 05:43:01 pm »
if you have infinite 9s after the comma then it is equal to 1, if you have a finite number, it will only approche 1 but never equal it.

Also:

0.1111...... = 1/9
0.2222...... = 2/9
0.3333...... = 3/9
0.4444...... = 4/9
0.5555...... = 5/9
0.6666...... = 6/9
0.7777...... = 7/9
0.8888...... = 8/9
0.9999...... = 9/9
« Last Edit: December 11, 2013, 05:44:01 pm by Sorunome »

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Re: .9 repeating equals 1?
« Reply #24 on: December 11, 2013, 07:01:01 pm »
/\ That solves it for me.

Offline aeTIos

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Re: .9 repeating equals 1?
« Reply #25 on: December 12, 2013, 03:40:06 am »
To me, it equals 1.

Simply accept that as long as 41.9999...99958=42, I'm right.


That "notation" is heavily misleading and, somewhat unfortunately, crops up whenever this debate does. What does the use of the ellipsis denote? An infinite string of characters extending to the right. The tricky part here is the word "infinite". An infinite string has very different behavior from a finite one. In particular, the notion of an "end" does not exist. You can't append to the end of an infinite string because there isn't an end to append to. By placing numbers after the ellipsis, you're attempting to denote this. It's not a valid operation and hence what you get is wrong. 41.999...=42 would be the only way to write that statement and thankfully, it's true.

To see why, let's take a quick look at what the reals are:

Reals intuitively consist of all the numbers along a number line. This can be formalized with something called a dedekind cut, which is essentially as follows: For every point X along a line, define sets A and B such that A consists of all the points below X and B consists of all points not in A. We can define the reals as every point that we can do this for. Note that each A has the property that while there is an upper bound on the set (X), there is no greatest member of A. In the discussion 0.9999... = 1, what we're really asking is whether there's a a distinct real number from 1 such that there is an infinitesimal difference between them. In other words, is there a greatest member in set A? As I stated earlier, this is a property of the way the set is defined that there is not. Therefore, if 0.9999... is distinct from 1, it cannot be in A. It must therefore be a member of B, as the smallest member. Unfortunately, this is exactly where 1 is located. Thus 0.999... = 1.

If this confuses you, that's fine. It's simply a more intuitive explanation of the formalization of reals. Note that, unlike what shmibs, says, I am an uninteresting person (err, i mean you can have systems where the equivalence is not true), but it's not possible with the *reals*.


EDIT: Thanks to shmibs for edits and *ahem* punctuation.
I don't really grasp what you're saying here... It seems you assume calculus knowledge which most deniers do not have. I for myself can accept the convention  that 0.99... is 1 but I don't agree with it, mainly because only 1 is 1.
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Offline Jim Bauwens

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Re: .9 repeating equals 1?
« Reply #26 on: December 12, 2013, 04:01:46 am »

Offline Keoni29

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Re: .9 repeating equals 1?
« Reply #27 on: December 12, 2013, 01:54:16 pm »
I love that youtube channel!
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Offline AssemblyBandit

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Re: .9 repeating equals 1?
« Reply #28 on: December 12, 2013, 04:52:16 pm »
If your having math problems, I feel bad for you son, I've got 98.999... problems but .999... is one! Vi's the best!

Offline Roondak

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Re: .9 repeating equals 1?
« Reply #29 on: December 12, 2013, 04:59:31 pm »
In ninth grade my friends and I got so mad about this. We would argue for hours about this, no joke. :D I was always in the camp that said it was equal, but whatever.

As said above, in summation notation it could be expressed as the summation of 9/(10^x) as x goes from 1 to infinity, which of course would be 1.