2 * 0.999... = 2?

[Canai]

I vote 6

 

[Matthias99]

Originally posted by: JeffreyLebowski
I hate this "new math" 1=1
0.9999~ = 0.9999~
Just as the exponent suggests it is a fraction however close to 1 as it can get it isn't 1. By stating that 0.9999~ = 1, then you are saying that 1.9999~ = 2 and that 1.4999999~ = 1.5 It's stupid stoner math.

If by "stupid stoner math" you mean "calculus", then yes.

From an earlier post I had in one of these threads:

edit: link to Dr. Math, which has more/different explanations:

http://mathforum.org/dr.math/faq/faq.0.9999.html

Ah, these are always fun.

I'd check out that Dr. Math link above for more information, but here's the proof. An honest-to-god actual proof that the number defined by "0.999..." is exactly equal to the number defined by "1.000...".

First off, what you really mean by saying "0.999..." is "the limit as x goes towards infinity of (sum from 1 to x of (9 * 10^(-x)))" -- that is, "0.9 + 0.09 + 0.009 + ...". Start from there.

(Those of you who have taken calc or even pre-calc should recognize that this is a geometric series, and it is possible to prove that all geometric series involving real numbers obey certain rules, and the sum of this series must be [9/10 + 1/10 = 10/10 = 1]. But proving that is more involved, and so I am going to stay along these lines.)

LEMMA: "0.9 + 0.09 + 0.009 + ..." is a real number.
Since each term in the series is a real number (they are actually rational numbers, of the form 9/10, 9/100, 9/1000, ..., but rationals are a subset of the reals), and any two real numbers added together also produces a real number (this is a basic property of addition on reals), therefore the sum of this series must also be a real number. QED.

"1.000..." (AKA 1/1) is also a rational/real number.

Now let's talk about equality of real numbers. There are a number of ways to define equality on real numbers, but one of the more useful ones is to say that "for two real numbers a and b, a = b if and only if a - b = 0" (I'm not gonna delve further into this, but basically, these things are just defined as part of the properties of real numbers). So our question then becomes:

"Does '1.000...' - 'the sequence referred to as "0.999..." ' = 0?"

Taking the sequence above, you can see that the difference between "1" and "0.999..." for a finite number of terms is (sum from 1 to x of (1 - (9 * 10^-x))). Let's look at the values of this sum:

x = 1, value = 1 - .9 = 0.1 = 1/10 = 1 / (10^1)
x = 2, value = 1 - .9 - .09 = 0.01 = 1/100 = 1 / (10^2)
x = 3, value = 1 - .9 - .09 - .009 = 0.001 = 1/1000 = 1 / (10^3)
x = 4, value = 1 - .9 - .09 - .009 - .0009 = 0.0001 = 1/10000 = 1 / (10^4)
x = 5, value = 1 - .9 - .09 - .009 - .0009 - .00009 = 0.00001 = 1/100000 = 1 / (10^5)
...
x = n, value = 1 / (10^n)

The actual difference we are looking for here is whatever the limit of this sum is as x goes to infinity.

The limit as x goes to infinity of (1 / (10^x)) is exactly zero. Thus, "1.000..." - "0.999..." is also exactly zero.

The typical challenge back to this is that "the limit's not zero, it's an infinitely small positive number -- 'an infinite number of zeroes followed by a one' ". No such number can exist in the real number system; the only number that can be the answer to this limit is zero. Attempting to define such a number via sequences, series, and limits -- or any other mathematical method -- is impossible. Intuitively, "an infinite number of zeroes followed by x" (where x is anything that's not "0") cannot exist, since then the sequence of zeroes would not be infinite.

Basically, "0.999..." = "1.000...", no qualms about "logic" or other things like that. When you're working with real numbers, they are identical. The sum of a converging infinite sequence can be exactly equal to a real number -- not just approximately equal. Frankly, that's the entire point of calculus.

 

[BrownTown]

@ thraashman: You really find those hooters girls to be very hot, here they are only sorta "meh". The boobies I am talking about are imo far superior, we are talkign about southern college girls in bikinis and low cut shirts walking around by the hundreds. Also I would note (with considerable pride) that my college was named in the last ranking of hotness to have girls as "A+", one of only three colleges to get this ranking. Why jsut last week playboy was here trying to get girls to audition for their magazine!!

in conclusion: hot southern girls > girls from everywhere else

 

[f4phantom2500]

Originally posted by: Phokus

Originally posted by: f4phantom2500
http://upload.wikimedia.org/math/3/3/1/33190b5d742626c840d4653650ccb472.png

http://upload.wikimedia.org/math/5/b/7/5b764601548173e8e894926fdda91dd7.png

http://upload.wikimedia.org/math/a/a/7/aa7c4f22b682312e877215f27a0ac3c7.png

Take your pick. Basically, yes, .999... * 2 = 2.

can you link the wikientry? that's interesting

yeah, here:

http://en.wikipedia.org/wiki/.999...

 

[thraashman]

Originally posted by: BrownTown
@ thraashman: You really find those hooters girls to be very hot, here they are only sorta "meh". The boobies I am talking about are imo far superior, we are talkign about southern college girls in bikinis and low cut shirts walking around by the hundreds. Also I would note (with considerable pride) that my college was named in the last ranking of hotness to have girls as "A+", one of only three colleges to get this ranking. Why jsut last week playboy was here trying to get girls to audition for their magazine!!

in conclusion: hot southern girls > girls from everywhere else

What college is this and do they have a good graduate program? I could always go back for my masters

 

[Unheard]

I voted no just to piss people off.

 

[thraashman]

I'd just like to point out there is a fallacy in the wikipedia article regarding this. At least in the algebraic proof, that's the easiest one to point out.

1.9999... - 0.9999..... != 1

instead it is undefined because you cannot subtract an infinite from and infinite.

infinity-infinity !=0

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

under indeterminate operators. As far as this argument goes both sides are right. Because it depends on whether you're talking conceptually or literally.

Conceptually

1=0.999... because of calculus and limits and all that crap

literally
0.9999 isn't definable

 

[BrownTown]

IT does point out in the article that many of the "proofs" used are not mathmatially rigorous, but they do serve to help people understand the basic points.

 

[destrekor]

Originally posted by: Zolty
well 0.999... = 1 so yes 2*1=2

you can't round in intermediate steps.. it produces an inaccurate answer

 

[Matthias99]

Originally posted by: thraashman
I'd just like to point out there is a fallacy in the wikipedia article regarding this. At least in the algebraic proof, that's the easiest one to point out.

1.9999... - 0.9999..... != 1

instead it is undefined because you cannot subtract an infinite from and infinite.

"Numbers" like "0.999..." are not "infinite", they are "infinite sequences" -- what you really are saying is (0.9 + 0.09 + 0.009 + 0.0009 + ...), an infinite arithmetic series. You can add and subtract such series, even if you can't add and subtract "infinity" (since "infinity" is not a number).

1 + (0.9 + 0.09 + 0.009 + 0.0009 + ...) - (0.9 + 0.09 + 0.009 + 0.0009 + ...) = 1, since the two infinite sequences are identical and therefore equal (and x - x = 0 by definition; this still applies, since any number of real numbers added together must still be a real number). It's possible to prove this more rigorously (and many many other things related to such sequences), but it gets more complicated.

 

[DrPizza]

I'd vote, but I suspect that after seeing the poll results, I'd end up with even less hope for a future where people are getting smarter, not more stupid.