Curse

around · Jun 05, 2008, 04:07 PM // 16:07

Quote:

Originally Posted by Katsumi

He means that the calculator treats it as .9 infinity, even though it doesn't show it to you as that.

No it doesn't, because calculators only do floating point arithmetic, and can't handle real numbers, of which there are an uncountable infinity of uncomputable ones.

Turtle222 · Jun 05, 2008, 04:09 PM // 16:09

Quote:

Originally Posted by King Symeon

You can't say that unless you have proven that 0.999∞ =/= 1.

Why not? the statement i made proves that it does not equal 1. If 0.999 = 1, then a positive value subtracting itself should equal 0

ie:

.2-.2 = 0

35-35 = 0

1-1 = 0

1- 0.999∞ = ???

But my area of expertise is not in mathematics so i could be wrong, but using logic this seems clear.

Buzzer · Jun 05, 2008, 04:19 PM // 16:19

lol you guys... waiting to pounce on the wrong answers and say 'fail'

There's not really much to discuss.
0.999∞=1 because we cannot express 1/3 in decimal form without the use of limits.

Kattar · Jun 05, 2008, 05:41 PM // 17:41

They're not wrong, buzzer.

As I said before, check the wiki.

It may not seem logical, but mathematically, it's correct.

Lihinel · Jun 05, 2008, 06:01 PM // 18:01

Well technicaly youre wrong in the most chasses (1 << x ; x e N , x > 10), the equatiion only works in base 10.

In any base x ; x > 10:

0.9∞ < 0.A < 0.y∞ = 1 ; y = (x-1)/x

...

+1

...
*insert more trolling here*

awesome sauce · Jun 05, 2008, 06:01 PM // 18:01

Limit as x approaches 1 of (1-x) is 0

1- 0.999∞ is undefined.

Kanyatta · Jun 05, 2008, 06:04 PM // 18:04

Quote:

Originally Posted by t00115577

cant really multiply by infinity as it doesnt exist, but could look either way at it i guess

Infinity exists, its just impossible to reach.

This all has to do with limits, like, if a ball falls 16m and bounces up 8m, then lands and bounces back up 4m, and the bounce distance keeps halving constantly, will it ever stop bouncing?

And 1-0.999∞ = 0.0∞1

Somehow? Infinity zeros followed by a 1.

around · Jun 05, 2008, 06:11 PM // 18:11

Quote:

Originally Posted by spawnofebil

Infinitesimals don't exist.

I hate to quote myself, but there's no other way of saying it.

Unless you weaken the second-order axioms of the real numbers, in which case you're dealing with non-standard analysis.

Also-new challenge for you:

Dissect a ball into two balls with the same volume as the first one. Alternatively, dissect the first ball into another ball with twice the volume.

Dante the Warlord · Jun 05, 2008, 07:14 PM // 19:14

I remember this one... makes sense and we were offered extra credit to find out why it did this..

Also, how did you get the infinity sign? Is there a key for that?!

Kanyatta · Jun 05, 2008, 09:32 PM // 21:32

Quote:

Originally Posted by spawnofebil

Infinitesimals don't exist.

So, small numbers beyond belief don't exist, but 0."infinity 9's" does exist? That seems very contradictory.

I pwnd U · Jun 05, 2008, 10:35 PM // 22:35

Math=/=Double Posting. Stop it, learn to edit.

As it is .999∞=/=1 because if you take a graphing calculator and graph both of those as slopes, they would never touch. Anyways, just my thoughts...

Lord Sojar · Jun 05, 2008, 11:52 PM // 23:52

It doesn't equal 1... it equals .9999 repeating. 0 =/= 0 though. Now that theorem is fun. Regrettably, I don't care enough about any of you to explain it. Have a lovely evening.

awesome sauce · Jun 06, 2008, 06:55 AM // 06:55

Quote:

Originally Posted by I pwnd U

Math=/=Double Posting. Stop it, learn to edit.

As it is .999∞=/=1 because if you take a graphing calculator and graph both of those as slopes, they would never touch. Anyways, just my thoughts...

Neither of those are equations, so you can't graph them.

You could try y =1 and y = 1 - 1/x. In that case, you're right, they will never intersect because y= 1-1/x has a horizontal asymptote y = 1. However, the limit as x approaches infinity or negative infinity will be the same for both.

That's the only way that you can deal with this problem, with limits.

RhanoctJocosa · Jun 06, 2008, 08:57 AM // 08:57

cbf reading everyone's posts, but around is correct.

let x= 0.99999∞
10x = 9.99999.... (there is 1 less digit after the . than x, so it doesn't cancel it all out)

around · Jun 06, 2008, 09:04 AM // 09:04

Quote:

Originally Posted by Kanyatta

So, small numbers beyond belief don't exist, but 0."infinity 9's" does exist? That seems very contradictory.

How so? Infinitesimals do not exist (in the standard formulation for real numbers), because there is an axiom which states that if x < z, there exists a y such that x < y and y < z.

The real numbers are continuous, you cannot arbitrarily divide it up into an infinite number of infinitesimals.

Also-Rahja: your 'proof' involves dividing by zero. I can feel it in my water.

I pwnd U · Jun 06, 2008, 10:41 AM // 10:41

Quote:

Originally Posted by awesome sauce

Neither of those are equations, so you can't graph them.

You could try y =1 and y = 1 - 1/x. In that case, you're right, they will never intersect because y= 1-1/x has a horizontal asymptote y = 1. However, the limit as x approaches infinity or negative infinity will be the same for both.

That's the only way that you can deal with this problem, with limits.

Use
Slopes

If you make each into a slope you will slowly see the .9999... one fall away from the 1 slope. y=mx+b, you can instert whatever you want for the other numbers. Just use the slope as each one.

around · Jun 06, 2008, 10:44 AM // 10:44

What are you on about?

Let f(x)=0.9999999... and g(x)=1

If you plot them, you get two STRAIGHT LINES WITH ZERO GRADIENT, one being at y=0.999... and y=1, which differ why the width of a line (ie zero).

Rushin Roulette · Jun 06, 2008, 11:03 AM // 11:03

to prove you wrong as you wanted with pure simple cold logic.

∞ = infinity (by deffinition never ending=

0.999∞ =/= 1

because 1 - 0.999∞ = 0.0000∞1

Machines what so ever are not capable of true ∞ because they are decimal at core which means everything has to end up as a full number at one point or another after the decimal place (in this case when memory runs out if no artificial limits are set)

Fril Estelin · Jun 06, 2008, 11:08 AM // 11:08

The fundamental answer if very simple, as the entity "0.999∞" does not exist (you can't write it semantically, you have to use the syntactic symbol ∞ instead to invoke in a finite step an infinite process). The demonstration on message #2 makes basically a huge math jump after "9x =" when two infinite series are subtracted.

In other words "0.999∞" is a mathematical artefact (based on infinite series) invented for specific purposes, and when you push the limit of the mathematical definition (when subtracting the two series) by an epsilon, you end up with this apparently strange result.

You can see it this way also:
0.999∞ = 1
On the right you have the common understanding of maths, number that are associated with real things (counting stuff with our hands).
On the left you have mathematical constructs that escapes reality (unless you consider quantum mechanic as "the reality" rather than an explanation of microscopic phenomenons).
And you're trying to compare (symbol "=") the two. (the core of the issue is actually here, in this little symbol "=")

A funnier and much more unsettling explanation:
http://mathforum.org/library/drmath/view/55746.html

Zeno's paradoxes is actually much funnier, but based on the same problem (at a logical level, rather than arithmetic):
http://en.wikipedia.org/wiki/Zeno's_paradoxes

And these questions lead naturally to philosophy ...

Rushin Roulette · Jun 06, 2008, 11:19 AM // 11:19

to prove you wrong as you wanted with pure simple cold logic.

∞ = infinity (by deffinition never ending=

0.999∞ =/= 1

because 1 - 0.999∞ = 0.0000∞1

Machines what so ever are not capable of true ∞ because they are decimal at core which means everything has to end up as a full number at one point or another after the decimal place (in this case when memory runs out if no artificial limits are set)