Does 0.9999(repeating) = 1?

Yes 705 54.95%
No 578 45.05%
Voters: 1283. You may not vote on this poll

Does 0.9999(repeating) = 1?

First Riot Post
Comment below rating threshold, click here to show it.

Bagel Arms

Member

01-23-2013

Quote:
Originally Posted by Griftrix View Post
Ancient Greece disagrees with your first statement.
The more of your posts I read the more you become my favorite red, Griftrix. <3 The way you are sort of telling people that you find them to be wrong but you're not reaaaaaally telling them that, just that X/Y/Z source of higher credibility disagrees with their arguments is wonderful and funny.


Comment below rating threshold, click here to show it.

Aliath

Senior Member

01-23-2013

Please leave OT. esp when it's to suck up. We are noble here.


Comment below rating threshold, click here to show it.

WhattayaBrian

Engineer

01-23-2013
28 of 32 Riot Posts

Quote:
Originally Posted by David Hume View Post
Your statement is correct, but just want to point out to everyone else that "Infinity * 2" having the same size (cardinality) as regular Infinity does not imply that they are equal.

Taking "Infinity" to mean the first/"smallest" infinite ordinal number, the cardinality of the natural numbers, expressions like "Infinity + 1" and "Infinity * 2" are perfectly well defined.

In set theory, the standard definition of the ordinal numbers are as follows:
0 is the empty set, { }.
1 is the set containing 0, {0} ( = { {} } )
2 is the set containing 0 and 1 {0, 1} ( = { {}, {{}} }
3 is the set containing 0, 1, and 2 {0, 1, 2}
etc.

In general, we define "n + 1" to be "the set containing n, and also every number inside of n." We also define < to mean that "A < B" if A is an element in the set of B (remember, numbers are just special kinds of sets in this system, and 3 really does contain 2, which is our justification for saying 2 < 3. You can think of the < sign as being equivalent to the ∈ or ⊂ sign if that makes you feel better - they're all equivalent on these particular sets.

So the "first infinity" is ω, the set of all the normal counting numbers, {0, 1, 2, 3, 4, ... }. ω is therefore both the set of all finite numbers and itself a number that is "bigger" than all of those numbers (in the sense that 3 < ω, 5 < ω, 100,000,000 < ω).

Since it's a number, we can +1 it to get the next number. Here goes:
ω + 1
= the set containing everything in ω (0, 1, 2, 3, .... ) and ω itself
= {0, 1, 2, 3, ... , ω}

which is a perfectly well defined number in our system. Since ω is a thing in ω + 1, then
ω < ω + 1
is a true statement - BUT ω and ω + 1 have the same cardinality (which is the bijection / bragging contest thing WhattayaBrian brought up earlier.) So in terms of cardinality, ω and ω + 1 have the same size (i.e., |ω| = |ω + 1| ) but at the same time, in terms of less than/greater than in terms of ordinality, ω < ω + 1.

If you extend "+ 1" into "+ n" in the sane way that gives you normal addition, and define multiplication in terms of this addition, it turns out that ω * 2 is still perfectly well defined number - it comes out to ω + ω, which is {0, 1, 2, 3, ... , ω, ω + 1, ω + 2, ω + 3, ... } which is greater than both ω and ω + 1 (and ω +2, etc) but is the same size as both of them.

so
ω < ω + 1 < ω + 500,000 < ω + ω = ω * 2
but
|ω| = |ω + 1| = |ω + 500,000| = |ω + ω| = |ω * 2| (the |x| means "the cardinality of x")

see http://en.wikipedia.org/wiki/Ordinal_number and http://en.wikipedia.org/wiki/Cardinal_number if your brain can still handle more.
This is an interesting discussion. It may be common nomenclature (I'll be the first to admit my Set Theory is lacking), but I have trouble using the word "size" to mean something separate from magnitude. Ordinality is all about ordering (of course). I can say very easily that "a" comes before "b", and that the alphabet is a well-ordered set. But I would not say "a" is "less than" "b"; though we do in fact use that quite often in engineering to do lexicographical sorting.

And also, for sure, we haven't strictly and clearly defined what set of numbers we want to answer the topical question within.

Does .9... == 1 in the natural numbers? A strange question since the decimal representation almost requires at least thinking about fractional numbers, but it seems well formed all the same, so sure.

Does .9... == 1 in every number space? I have no idea. You can come up with your own rules, and as long as they're consistent, you can do whatever you want.



Quote:
Originally Posted by Kaolla View Post
precisely what i tell people but they don't listen
Saying 9.9... - .9... == 9.0...9 is the logical equivalent of saying 9.9... - .9... == "I want to go to the store to buy some bananas".

9.0...9 is not a number that can exist in the Reals. You can make your own algebra if you'd like where it does exist, as long as you can define rules for it that don't break down.

Math isn't true so much as math is created. There are many systems that model the real world, and those we consider "irrefutable", but these are simply systems like everything else, and can be discarded for another if it suits your purposes.


Comment below rating threshold, click here to show it.

1nnn1a

Junior Member

01-23-2013

I'm not 100% sure if the guy who's playing with the ln(2) expansion has been disproven yet, but if he hasn't I'll just note here:

The fact that ln(2)'s expansion is not absolutely convergent (i.e., the sum of the absolute values of its series diverges (see: http://en.wikipedia.org/wiki/Harmoni...cs)#Divergence )). Since this is the case, we can actually make the series representation of ln(2) converge to ANY real number we desire, and we can also sum it up to infinity or negative infinity by rearranging its terms. Hence, rearranging its terms is actually an invalid way to show that ln(2) = 2ln(2).

If you're mathematically inclined and want to see the proof for non-absolute convergent series you can see here: http://www.math.ucla.edu/~ralston/pub/Rearrange.pdf


Comment below rating threshold, click here to show it.

Zerglinator

Senior Member

01-23-2013

I wonder if someone's going to say "Can I get a Rioter for my 0.999999...th win?"


Comment below rating threshold, click here to show it.

Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
Saying 9.9... - .9... == 9.0...9 is the logical equivalent of saying 9.9... - .9... == "I want to go to the store to buy some bananas".
except that it makes logical sense, unlike whatever you people use.

Attachment 598680

and yes i know the subtraction in that pic is totally wack. it's late and i don't feel like fixing it

the pic just illustrates a point

[edit] nm better edit before **** ppls cry


Comment below rating threshold, click here to show it.

Graffers

This user has referred a friend to League of Legends, click for more information

Senior Member

01-23-2013

Quote:
Originally Posted by Mèdusa View Post
0.9999.... = X
10 X = 9.999....
10X - X = 9
9X = 9
9X / 9 = 1
So, X = 1

That's the standard proof anyway
My only problem with this is in the second line. Who decided that X was equal to .999...? You don't get to solve the equation to solve the equation. I'm gonna give this a shot.

X = 7
10X = 70
9X = 45
X = 5

Did I do that right?

Also, 1/3rd doesn't equal .333(repeating). That's simply the easiest way for people to understand it. Every mathematician understands that there's a remainder that can't be reached. When it's multiplied by 3 that remainder brings 1/3rd to 1, not .999(repeating). Saying that .999(repeating) equals one is far fetched. In fact, I'd go as far to say that .999(repeating) isn't even a number, rational or irrational. There's not a single equation you can come up with that could yield that number.


Comment below rating threshold, click here to show it.

1nnn1a

Junior Member

01-23-2013

Quote:
Originally Posted by Graffers View Post
My only problem with this is in the second line. Who decided that X was equal to .999...? You don't get to solve the equation to solve the equation. I'm gonna give this a shot.

X = 7
10X = 70
9X = 45
X = 5

Did I do that right?
No, you did not do it right. You stated X=7. Then 10X=70 is perfectly valid. However, in the example you cited they simply subtracted the first equation from the second. You did neither, and just made something up at random.


Comment below rating threshold, click here to show it.

Graffers

This user has referred a friend to League of Legends, click for more information

Senior Member

01-23-2013

Quote:
Originally Posted by 1nnn1a View Post
No, you did not do it right. You stated X=7. Then 10X=70 is perfectly valid. However, in the example you cited they simply subtracted the first equation from the second. You did neither, and just made something up at random.
You're absolutely right, I won't try to read math equations when I've been up for so long
I still don't think .999(repeating) is a number at all. Infinity is a cruel mistress that's literally impossible to do math with.


Comment below rating threshold, click here to show it.

Graffers

This user has referred a friend to League of Legends, click for more information

Senior Member

01-23-2013

To demonstrate how out of it I am right now...
X=.999...8
10X=9.999...8
9X=9
X=1
I almost used this equation to debunk the theory, but then I realized .999(repeating) with an 8 on the end could never come up in an actual equation. Then I realized this was mathematics and anything is possible, because screw logic. It's almost like playing with infinity doesn't work with algebra. Hmm...