1=.999999...?

[gib]

@ladyjane

at infinity there is the 7

At infinity??? You’re talking about infinity as if it were an actual place. Get this concept through your head: INFINITY MEANS “NO END” - ergo, there’s no “place” where a 7 suddenly shows up!

Now you’re going to say:

@ladyjanes

if there is no last number
same applies to .9999…
so .99999… can only = .99999…
and thus .9999… =/1

To which I will say:

@gib

Do you understand the conceptual difference between .999[bar] and 1 that I was talking about earlier? That ‘.999[bar]’ is a way of imagining 1 by approaching it from below and ‘1’ is a way of imagining it by directly being there already. They’re not different quantities, they’re the same quantity conceptualized in different ways.

And at this point, you’ll start your whole argument over from the beginning… as if repeating it over and over again will eventually make it right.

I think I’ll carry on with this for a little while longer… just to see what else you come up with… but discussions like this in which no progress is made tends to get dull and I eventually leave out of shear bordom.

[Wonderer]

@gib

@Wonderer

Think of it this way: there are two words you can use to describe your rear end. There’s “glutious maximus” and there’s “your ass”. But just because the word “glutious maximus” is bigger and more complex doesn’t make your butt bigger and more complex when you use that word. The bigness and complexity is just a property of the wording, just as the infinite length of the decimal expansion in .333[bar] and .999[bar] is a property of the notations themselves.

to each his own in the end right?

but on a more rational note why would you ruin a perfectly good number? we say .333[bar] because thats the only way we can plug them into calculators.

its use is in fact avoided due to its obvious lack of precision. as it can never truly equal one third.

i understand the misconception you understand, but it is irrelevant in light of the fact that .333[bar] is imprecise to begin with. it in fact does not equal one even if you had "never ending 9’s, because they would never tally up to 1, cause theres only 9’s.

i just thought of a really good mental example.

imagine a singularity… a grain of sand you could say, except it is infinitely divisible… (this represents 1).

the singularity has a value of 1.

now imagine i give you a pick to pick away at this grain of sand, but you can only chip away .9 of what is there at a time…

your first you collect stroke… .9

.1 remains

you collect on the second stroke… .09

.01 remains

its easy to say that no matter how long he chips away, there will always be a remainder…

what is easy to misconstrue however is this.

when you chip away at the stone infinitely you will not consume the whole thing, rather you will be left with an infinitely small remainder

cheers

[ladyjane]

They’re not different quantities, they’re the same quantity conceptualized in different ways.

sorry
.99999… is not the same as 1
you can keep saying they are but -that does not make it so
if the 6s in .66666. never end
same applies to the 9s in .999999…

and
to repeat
0-7 =3
so at infinity the last number must be 7
if it is not then
1.00000… -.333333… must be indeterminate by your argument as you cant take
a number that does not end away from a number that does not end -by your argument
ie there is no last numbers to subtract anything from
therefore
1.00000… -.333333… must be indeterminate

[gib]

@ladyjane

and
to repeat

Yeah, I know you’re repeating.

[ladyjane]

to give support to the claim that 1.0000[R] - .333333[R] = .666666…7
GO TO SCI LOGIC

talkaboutscience.com/group/s … 99419.html

ctually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

A sequence has more to do with an ordinal
than wie a cardinal, as it orders its elements.

Omega denotes the standard ordinal, which
we normally connect with infinite decimal
numbers. What elsiemelsi does, is simple he
puts forward omega+1.

But there are much much more orders and ordinals,
we could even have omega+omega, or omega+mirror(omega),
omega+omega+1, omega^2:

12... 12...    = omega + omega
12... ...12    = omega + mirror(omega)
12... 12... 1  = omega + omega + 1

12...
2....
.....          = omega^2
.....
.....

Nothing special. For example Z, the positive and
negative numbers, are easily seen as mirror(omega)+
omega, namely:

... -2 -1 0 1 2 ...

But mirror(omega)+omega is not healthy for numbers,
because the get big!

... 3333.3333 ....

Quite a big number, isn’t it?

talkaboutscience.com/group/s … 99487.html

Actually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

OK. Now how does one handle

…333… x .333… ?

In this New Arithmetic™ we should get something like

…111…11088…889
<w+i> <2w>

where ‘w’ is omega.

Since 1/3 = .333… the product 1/3 x 1/3 is clearly
not represntable by a fraction (1/9 = .111… does not
extend past ), and therefore the rational numbers is
not closed under this new system.

This is a minor problem, to be sure…but it is a problem.
I’m assuming you have a fix.

[MagsJ (..a chic geek)]

Whatever happened to good’ol rounding up to 2 deciaml places, and done - what’s with the debate on it? Why is it such a big issue: it wasn’t before!

This level of math is very 15 year old/GCSE stuff, LJ… I fail to see the big deal with the dilemma you have set-out, and math or the Universe itself, doesn’t care…

[gib]

Wonderer,

Interesting thought experiment, but your misunderstanding lies here:

@Wonderer

when you chip away at the stone infinitely you will not consume the whole thing, rather you will be left with an infinitely small remainder

There is only a remainder when you stop dividing in a finite amount of time.

You have to understand this in terms of limits (think calculus). In terms of calculus, we would say that the process of dividing 1 into the parts .9 and .1, and then repeating on the .1 part, approaches 1. Because this process goes on forever, we never do arrive at 1 in a finite amount of time. That’s why in time there will always be a remainder. But given an infinite amount of time, we actually do end up reaching 1.

Of course, the number .999[bar] is not a process. It just has an infinite number of 9s right now. Therefore, it has already reached the number 1, and is therefore equal to it.

I understand what you mean by an infinitely small remainder, but this doesn’t make sense to me. It’s like saying you have an infinitely small number - like there’s a smallest possible quantity. To me, such a concept is incoherent as ALL numbers can be divided further. In effect, if there is a smallest possible quantity, it’s 0.

[gib]

@Magsj

Whatever happened to good’ol rounding up to 2 deciaml places, and done - what’s with the debate on it? Why is it such a big issue: it wasn’t before!

This level of math is very 15 year old/GCSE stuff, LJ… I fail to see the big deal with the dilemma you have set-out, and math or the Universe itself, doesn’t care…

Yeah, but it’s fun to argue

[ladyjane]

This level of math is very 15 year old/GCSE stuff, LJ

sorry you where never taught

ctually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

now colin leslie dean has just destroyed your whole education and mathematics as well

[MagsJ (..a chic geek)]

@gib

Yeah, but it’s fun to argue

I will continue to follow the, er, argument!

[MagsJ (..a chic geek)]

@ladyjane

sorry you where never taught

I was taught very well, thanks! so no need to feel sorry for me, LJ…

ctually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

now colin leslie dean has just destroyed your whole education and mathematics as well

No!

I just jest! we were taught the dilemma of math, but it does not deter the uasge of it’s application!

[gib]

@Magsj

we were taught the dilemma of math, but it does not deter the uasge of it’s application!

Very well put.

[ladyjane]

so you have a proof that
.9999[R] =1
and colin leslie dean proof that
.9999[R] =/ 1

so now maths is in contradiction and as colin leslie dean has shown ends in meaninglessness

he australian philosopher colin leslie dean shows you it does not
another example that maths ends in meaninglessness
let
x=.33333[BAR]
3x = .99999999 [BAR]
but you say
.99999999 [BAR] = 1
then
3x = 1
now 3x-x = 1- .333333[BAR]
2x = .6666666…7

actually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

therefore
x= (.6666666…7)/2

therefore

x= .333333…35
3x=1.000000…05
therefore we have now
.999999[bar] = 3x= 1.0…5 -a contradiction thus maths ends in
meaninglessness

ctually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.

[gib]

@ladyjane

he australian philosopher colin leslie dean shows you it does not

We all know that Colin Leslie Dean is you, LJ. Quoting yourself doesn’t make you any more reputable.

[ladyjane]

support that 1/3=/ .33333[R]
FROM SCI LOGIC
talkaboutscience.com/group/s … 99512.html

you say 1/3 + 1/3 +1/3 = 1

but

33333… + .33333… + .33333… = .9999…
not as you say .9999… = 1
as dean has shown
9999…=/1

another proof 1/3 =/ .3333…
3333… is shown to converge to 1/3
thus 1/3 must be a different number from .33333…
but you say 1/3 = .3333…
thus they are the same number
but that means there has been no convergence
but this contradicts the cauchy convergence test

All right, since 1/3 != .333…, what is it equal to?
(This does solve the rational closure problem, though.)

[MagsJ (..a chic geek)]

@gib

@Magsj

we were taught the dilemma of math, but it does not deter the uasge of it’s application!

Very well put.

Merci, Gib!

@gib

@ladyjane

he australian philosopher colin leslie dean shows you it does not

We all know that Colin Leslie Dean is you, LJ. Quoting yourself doesn’t make you any more reputable.

Haha, you sound like Anthem: when he had had enough of LJ, too! LJ is good for testing one’s patience, so you do get something out of it!

[Wonderer]

@gib

In effect, if there is a smallest possible quantity, it’s 0.

sorry but here is where our notions diverge.

if something is zero than it has no quantity.

when you say “stop dividing” you already ruined our idea of infinite division…

if you never stopped, you would never be able to measure the remainder, it would be infinitely small.

once again i do not believe that something infinitely small equals zero… this assertion seems ludacris to me

[Rouzbeh]

@Wonderer

this assertion seems ludacris to me

I’ve got hoes in different area codes.

[Wonderer]

@Rouzbeh

@Wonderer

this assertion seems ludacris to me

I’ve got hoes in different area codes.

so no objections?

[Gaiaguerrilla]

This problem I think is perfectly reflective of discussions in physics that are still discussed. I don’t think I could disguss either as well as the professionals do, but they are both essentially the same problem. Even in logic- the universe has no finite structure. Yet- we see finite structure all the time. We are slaves of schrodinger’s cat, yet we see a LIVING cat or a DEAD cat every time. Is the problem we struggle with regarding quantum foam the same as this problem? The fact that multiple realities seem to exist? That the greater multiverse is deterministic, but not the universe? (And it’s not hippy talk, that last phrase is strictly out of textbook)

Gib seems to present it in a way that makes me want to consider .999bar = 1. The proof with multiplication is completely sensible. But so is the statement that two fractions cannot have the same numerator or denominator (pick one) , and yet different numerators or denominators (pick the other one) and still be the SAME number.

I’m not convinced of either side.