sorry
think for yourself and stop repeating orthodoxity
0-7 = 3
so there must be a 7 at infinity
to say the numbers are not countable but there must be a final 7 at
infinity
is just as counter intuitive as cantors idea of different infinities
having different numbers of members or sizes
if cantor can be correct so can the idea of 7 at infinity
Who says you can’t subtract something from something? You’re not subtracting a final 3 from a final 0, but you are subtracting all the other 3s from all the other 0s. No problem there.
You’re the one questioning it. You have to prove to us why it isn’t true. As far as we’re concerned, the proof given in the OP for why .999[bar] = 1 is sound.
But infinity doesn’t exist. Infinity means “no end”. So there is no end to the 6s. There is no 7.
hahaha you have just agreed with me that
.99999[R] =/1
as if what you say about the 6s is true must be true about the 9s -having no end to them
thus
.99999[R] =/1
if you subtract all the .33333333… from all the 00000000000… you must have a 7
as the first 3 you subtract is the last 3 at infinity from the last 0 at infinity
if you say no
then you can get
1.0000…- .333333… = .66666666
for you can never subtract anything as the 0s and 3s never end by your argument
thus
1.0000…- .333333… must always be indeterminate
^^ I’m so sorry for anyone who has to deal with you in real life.
I don’t get why you keep saying this. Why does the infinit expansion of 9s mean that .999[bar] =/ 1? I’ve already said in an earlier post that you’ve got this reverse, remember?
Nope, you just get .666[bar]
I have no idea what this means. The first 3 is the last 3?
It’s not indeterminate. It just equals .666[bar]. When a decimal expansion doesn’t end, that isn’t a problem for doing arithmatic, it’s just a problem for writing them out. The numbers themselves are still there and can be subracted from each other like any other number.
sorry
you only say that because you where taught that
but
1.0000…- .333333… = /.66666666
for you can never subtract anything as the 0s and 3s never end by your argument
thus
1.0000…- .333333… must always be indeterminate
when you do subtraction you always start with the last number of each sequence
ie
.000* -.333* you start with 3*
thus the last 0* at infinity you subtract the last 3*
thus
1.000000… - .3333333… = .666666… 7 it must have a 7 at infinity
sorry YOU SAID
SO THAT MUST ALSO APPLY TO to the 9s in
.999999 [R]
thus because the 9s never end
,99999[R] cant never = 1
you can’t never learn English, can you?
nevertheless from gib argument
SO THAT MUST ALSO APPLY TO to the 9s in
.999999 [R]
thus because the 9s never end
.99999[R] =/ 1
and even though wonderer gave us a proof that
.99999[R] =/ 1
only proves colin leslie dean claim that maths end in meaninglessness
as we have a proof that
.99999[R] = 1
but
we have seen that
.99999[R] =/ 1
a contradiction
as godels invalid proof that maths is inconsistent shows
yOU
DO
NOT
UNDERSTAND
CONVERGENCE.
hey gib says the 6s in
.6666666… never end
so that must mean the same for the 9s in
.999999999…
which means
.99999… =/ 1
because if the 9s never end
thus when u prove
.99999… = 1
you are generating a contradiction in maths
which godels invalid proof of the inconsitentcy of maths shows
and points out the truth of deans claim maths end in meaninglessness
all we can get is
.999999… = 99999999…
[/quote]
You obviously don’t understand what I said. When an infinit series of 9s is there, it must equal 1. 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.
that is just backtracking
you clearly said for .66666[BAR]
SO THAT MUST ALSO APPLY TO to the 9s in
.999999 [R]
thus because the 9s never end
,99999[R] cant never = 1
what applies to one must apply to all the rest
what people don’t seem to understand is that .333[bar] is an imprecise method of manipulating irrational numbers resulting from various fractions.
.333[bar], as i said before… does not equal 1 third because you can never have infinite 3’s. it is however acceptably close assuming a rational mind can look past the imprecision.
on that same note .999[bar] might seem like simply an imprecise way of writing down 1 but because of that imprecision it in fact does not equal one nor ever will.
seriously people… learn fractions
Wait a minute! light bulb goes off
You think that saying .999[bar] = 1 is a way of ROUNDING UP to 1. That’s why you’re using the 7 to show how .666[bar] can be rounded up to .666…7. So either the last digit in .999[bar] gets rounded up “at infinity” so that the whole number becomes 1 (just as the last digit of .666[bar] gets rounded up “at infinity” to 7 - or - .999[bar] goes on forever and never gets rounded up to 1… so it can’t be 1.
Well, here’s where we differ:
I say that .999[bar] already is 1 - there’s no “rounding up”. It’s just a matter of notation and conceptualization, but the quantity so noted or conceptualized is really the same.
You seem to be suffering from the same misconception. What you’re arguing about is a matter of notation, not the actual quantity represented by those notation. It just takes us an eternity to write out .333[bar]. Look: there is a quantity we call “one third”. By convention, we write out that quantity by a variety of symbols. One of them is “1/3”, another is “.333[bar]”, and another is the words “one third” themselves. They all mean the same thing!!! “.333[bar]” does NOT refer to something different than 1/3 or “one third” or “one minues two thirds” or whatever.
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 infinit length of the decimal expansion in .333[bar] and .999[bar] is a property of the notations themselves.
1/3 is not an irrational number. Neither is .333…
there is no rounding up 1.00000… - .33333… = .666…7.
because as i have said
0-3=7
no rounding up
when you take the last 3 from the last 0 you get a 7
no rounding up
and as you said the 9s
in
.9999999… go on forever
so can only .99999… = .999999999…
not .99999… =1
But there is no last number. Therefore you never get a 7. You’re not smart enough to understand this, I know.
at infinity there is the 7
has to be for 0-3=7
just as at infinity there are infinities with different sizes
and again hahaha if there is no last number
same applies to .9999…
so .99999… can only = .99999…
and thus .9999… =/1 as colin leslie dean has proven