Discussion:
sqrt2 and 2 equally mysterious, or not mysterious at all, and proof that 1 = 0.999...
(too old to reply)
Franz Gnaedinger
2017-04-21 07:02:27 UTC
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Irrationals not being numbers, and 1 not being 0.999..., are everlasting
topics in sci.math, popping up all the time. Very simple yet forgotten
methods of early mathematics can dissolve the problem.

Approximating the square root of 2 by a number column which I reconstructed
in 1979

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on

99/70 is already a fine value for the square root of 2. (Analogous number
columns approximate the square roots of 3 and 5 and the cube root of 2.)
Now let us look at the square root of 4 by drawing up the analogous number
column

1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
and so on

The ratios, for example 41/20 or 121/61, get ever closer to 2. This means
that 2 as the square root of 4 is equally mysterious as the square root of 2.
Or that the square root of 2 is no less mysterious than 2.

As for 1 = 0.999... let me begin with the Horus eye series of Ancient Egypt,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 or simply '2 '4 '8 '16 '32 '64

1 = '1
1 = '2 '2
1 = '2 '4 '4
1 = '2 '4 '8 '8
1 = '2 '4 '8 '16 '16
1 = '2 '4 '8 '16 '32 '32
1 = '2 '4 '8 '16 '32 '64 '64

From this you get

'2 '4 '8 '16 '32 '64 ... = 1

(The Horus eye series had an astronomical function, since a month of 30
days multiplied by '2 '4 '8 '16 '32 '64 yields 29 '2 '32 days or 29 days
12 hours 45 minutes for one lunation of 29 days 12 hours 44 minutes 2.9
seconds, modern average from 1989 AD - mistake of the old value less than
one minute per lunation, or half a day in a lifetime. One eye of the Horus
falcon was the sun, standing for a month of 30 days, his other eye was
the moon, standing for one lunation or synodic month.)

Another infinite series

1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 '9x10 '10x11 '11x12 ...

'1x2 '2x3 '5x6 '6x7 '9x10 '10x11 ... = pi/4

Now for 0.999... being 1

1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...

Mathematical education should begin with really simple methods, not on
the level of Euclid. Would prevent some people from wasting their life
on kooky notions.
John Gabriel
2017-04-22 03:33:55 UTC
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Post by Franz Gnaedinger
Irrationals not being numbers, and 1 not being 0.999..., are everlasting
topics in sci.math, popping up all the time.
For good reason. Intelligent people are not fooled by such bullshit, only morons like you!
Post by Franz Gnaedinger
Very simple yet forgotten methods of early mathematics can dissolve the problem.
Chuckle.
Post by Franz Gnaedinger
Approximating the square root of 2 by a number column which I reconstructed
in 1979
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on
Idiot. **** A number is the measure of a magnitude ****

Not an approximation or a partial measurement.
Post by Franz Gnaedinger
99/70 is already a fine value for the square root of 2. (Analogous number
columns approximate the square roots of 3 and 5 and the cube root of 2.)
Now let us look at the square root of 4 by drawing up the analogous number
column
1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
and so on
The ratios, for example 41/20 or 121/61, get ever closer to 2. This means
that 2 as the square root of 4 is equally mysterious as the square root of 2.
Or that the square root of 2 is no less mysterious than 2.
As for 1 = 0.999... let me begin with the Horus eye series of Ancient Egypt,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 or simply '2 '4 '8 '16 '32 '64
1 = '1
1 = '2 '2
1 = '2 '4 '4
1 = '2 '4 '8 '8
1 = '2 '4 '8 '16 '16
1 = '2 '4 '8 '16 '32 '32
1 = '2 '4 '8 '16 '32 '64 '64
From this you get
'2 '4 '8 '16 '32 '64 ... = 1
(The Horus eye series had an astronomical function, since a month of 30
days multiplied by '2 '4 '8 '16 '32 '64 yields 29 '2 '32 days or 29 days
12 hours 45 minutes for one lunation of 29 days 12 hours 44 minutes 2.9
seconds, modern average from 1989 AD - mistake of the old value less than
one minute per lunation, or half a day in a lifetime. One eye of the Horus
falcon was the sun, standing for a month of 30 days, his other eye was
the moon, standing for one lunation or synodic month.)
Another infinite series
1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 '9x10 '10x11 '11x12 ...
'1x2 '2x3 '5x6 '6x7 '9x10 '10x11 ... = pi/4
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Bwaaa haaa haaaa.

Please shut up. Do not talk about anything which you do not understand.
Post by Franz Gnaedinger
Mathematical education should begin with really simple methods, not on
the level of Euclid. Would prevent some people from wasting their life
on kooky notions.
Mathematics begins with Euclid you infinitely stupid moron. Chuckle.
netzweltler
2017-04-22 06:04:54 UTC
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Post by Franz Gnaedinger
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Doesn't one have to prove that 0.999... is a point on the number line which is located at point 1 exactly? How can that be?

If 0.999… stands for infinitely many commands

Add 0.9 + 0.09
Add 0.99 + 0.009
Add 0.999 + 0.0009


then following all of those infinitely many commands won’t get you to point 1. If you reached point 1 you have disobeyed those commands, because every single of those infinitely many commands tells you to get closer to 1 but NOT reach 1.

Therefore, if you want to define the position of a “point” 0.999… on the number line, it cannot be at point 1 – and for the same reason (“disobeying those commands”) it cannot be short of 1 nor can it be past 1.

So, where is "point" 0.999... located on the number line?
Franz Gnaedinger
2017-04-22 07:12:34 UTC
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Post by netzweltler
Post by Franz Gnaedinger
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Doesn't one have to prove that 0.999... is a point on the number line which is located at point 1 exactly? How can that be?
If 0.999… stands for infinitely many commands
Add 0.9 + 0.09
Add 0.99 + 0.009
Add 0.999 + 0.0009

then following all of those infinitely many commands won’t get you to point 1. If you reached point 1 you have disobeyed those commands, because every single of those infinitely many commands tells you to get closer to 1 but NOT reach 1.
Therefore, if you want to define the position of a “point” 0.999… on the number line, it cannot be at point 1 – and for the same reason (“disobeying those commands”) it cannot be short of 1 nor can it be past 1.
So, where is "point" 0.999... located on the number line?
Each line of the stairway number is 1, so also the infinitely long number
is 1, exactly one. The problem is that you start with modern concepts and
then look backward in time, instead of beginning with very simple pre-Greek
methods, like the stairway development of series.
Jim Burns
2017-04-22 16:06:35 UTC
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Am Freitag, 21. April 2017 09:02:41 UTC+2
Post by Franz Gnaedinger
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Doesn't one have to prove that 0.999... is a point on
the number line which is located at point 1 exactly?
How can that be?
The exact way one proves that depends upon the exact meaning
that is given to "and so on".

If, by 0.999... , we mean the least upper bound of
{ 0.9, 0.99, 0.999, ... } =
{ x e R | x = ( 10^(n+1) - 1)/( 10^(n+1) ) & n e N }
then the least upper bound of that set is exactly 1
and 0.999... = 1 exactly.
If 0.999… stands for infinitely many commands
Add 0.9 + 0.09
Add 0.99 + 0.009
Add 0.999 + 0.0009

then following all of those infinitely many commands won’t
get you to point 1.
What does it mean to follow all of those infinitely many
commands?

None of those results, 0.9, 0.99, 0.999, ... , are from
following all of those infinitely many commands, so it
shouldn't be a problem that none of them are exactly 1 .

I've given you one definition of following all of those
infinitely many commands, and it has many desirable consequences.
Under that definition 0.999... = 1 exactly.
If you reached point 1 you have
disobeyed those commands, because every single of those
infinitely many commands tells you to get closer to 1
but NOT reach 1.
Since my definition of the result of "following infinitely many
commands" does not require it to be the result of any of the
many partial (finite) results, this doesn't look relevant.
Therefore, if you want to define the position of a “point”
0.999… on the number line, it cannot be at point 1 – and
for the same reason (“disobeying those commands”) it cannot
be short of 1 nor can it be past 1.
So, where is "point" 0.999... located on the number line?
Well, you need to define what you mean by number line
for this question to mean anything.

If what you mean by number line only includes rational numbers,
then you have problems.

Consider _hypothetically_ a right isosceles triangle ABC.

d(AB) = d(BC) = p/q for some rational (of course,
since there are only rationals, by assumption).

angle(ABC) is a right angle.

Apparently, the line through both points A and C can't exist.
(Because no line can have both a rational point and
a rational point sqrt(2)*p/q away from that point.
There are only rational numbers, remember?)

If line AC can't exist, how can the hypothetical right
isosceles triangle ABC exist?

It might be that you have some very strange idea of what the
number line is, but for myself and for most people it would be
very strange that we would not be able to have any right
isosceles triangles.

On the other hand, we could say that the numbers on the
number line (i) follow all the familiar rules of rationals
for +, -, *, /, <, and also (ii) any two lines (or curves)
which _cross_ also _meet_ at least once.

I haven't worked it out in detail, but I suspect that with (ii)
we could prove right isosceles triangle ABC exists by virtue
of lines AB, BC, and AC intersecting at points A, B, and C.

If what we mean by number line includes rule (ii), then I
think that we need to have a definition of "following
infinitely many commands" for finite but unbounded many
commands that approaches but never reaches point C along
line AC -- never reaches C _exactly_ when finite. By that
definition, 0.999... = 1 exactly.

(There is a proof that
"Two curves which cross also meet"
is logically equivalent to
"Any bounded non-empty set has a least upper bound".)
John Gabriel
2017-04-22 19:08:49 UTC
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Post by Jim Burns
Am Freitag, 21. April 2017 09:02:41 UTC+2
Post by Franz Gnaedinger
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Doesn't one have to prove that 0.999... is a point on
the number line which is located at point 1 exactly?
How can that be?
The exact way one proves that depends upon the exact meaning
that is given to "and so on".
Nonsense. The definition was given by Euler as S = Lim S.

There is nothing left to do. It is neither a theorem nor does it require proof.

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel
Post by Jim Burns
If, by 0.999... , we mean the least upper bound of
{ 0.9, 0.99, 0.999, ... } =
{ x e R | x = ( 10^(n+1) - 1)/( 10^(n+1) ) & n e N }
then the least upper bound of that set is exactly 1
and 0.999... = 1 exactly.
Nonsense again because you can't have 0.999... unless you perform a super task.

<No time for too much bullshit>
netzweltler
2017-04-22 21:26:19 UTC
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Post by Jim Burns
Am Freitag, 21. April 2017 09:02:41 UTC+2
Post by Franz Gnaedinger
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Doesn't one have to prove that 0.999... is a point on
the number line which is located at point 1 exactly?
How can that be?
The exact way one proves that depends upon the exact meaning
that is given to "and so on".
If, by 0.999... , we mean the least upper bound of
{ 0.9, 0.99, 0.999, ... } =
{ x e R | x = ( 10^(n+1) - 1)/( 10^(n+1) ) & n e N }
then the least upper bound of that set is exactly 1
and 0.999... = 1 exactly.
If 0.999… stands for infinitely many commands
Add 0.9 + 0.09
Add 0.99 + 0.009
Add 0.999 + 0.0009

then following all of those infinitely many commands won’t
get you to point 1.
What does it mean to follow all of those infinitely many
commands?
None of those results, 0.9, 0.99, 0.999, ... , are from
following all of those infinitely many commands, so it
shouldn't be a problem that none of them are exactly 1 .
"following infinitely many commands" can be easily defined:

t = 0: moving along the number line from 0 to 0.9
t = 0.9: moving along the number line from 0.9 to 0.99
t = 0.99: moving along the number line from 0.99 to 0.999
...

By t = 1 we have followed all of the infinitely many commands, but we didn't reach 1. If we reached 1 by t = 1 we have disobeyed the commands.
Post by Jim Burns
I've given you one definition of following all of those
infinitely many commands, and it has many desirable consequences.
Under that definition 0.999... = 1 exactly.
If you reached point 1 you have
disobeyed those commands, because every single of those
infinitely many commands tells you to get closer to 1
but NOT reach 1.
Since my definition of the result of "following infinitely many
commands" does not require it to be the result of any of the
many partial (finite) results, this doesn't look relevant.
Therefore, if you want to define the position of a “point”
0.999… on the number line, it cannot be at point 1 – and
for the same reason (“disobeying those commands”) it cannot
be short of 1 nor can it be past 1.
So, where is "point" 0.999... located on the number line?
Well, you need to define what you mean by number line
for this question to mean anything.
If what you mean by number line only includes rational numbers,
then you have problems.
First of all, the number line includes number 1. Are you saying, I have problems locating 0.999... on this number line?
Post by Jim Burns
Consider _hypothetically_ a right isosceles triangle ABC.
d(AB) = d(BC) = p/q for some rational (of course,
since there are only rationals, by assumption).
angle(ABC) is a right angle.
Apparently, the line through both points A and C can't exist.
(Because no line can have both a rational point and
a rational point sqrt(2)*p/q away from that point.
There are only rational numbers, remember?)
If line AC can't exist, how can the hypothetical right
isosceles triangle ABC exist?
It might be that you have some very strange idea of what the
number line is, but for myself and for most people it would be
very strange that we would not be able to have any right
isosceles triangles.
On the other hand, we could say that the numbers on the
number line (i) follow all the familiar rules of rationals
for +, -, *, /, <, and also (ii) any two lines (or curves)
which _cross_ also _meet_ at least once.
I haven't worked it out in detail, but I suspect that with (ii)
we could prove right isosceles triangle ABC exists by virtue
of lines AB, BC, and AC intersecting at points A, B, and C.
If what we mean by number line includes rule (ii), then I
think that we need to have a definition of "following
infinitely many commands" for finite but unbounded many
commands that approaches but never reaches point C along
line AC -- never reaches C _exactly_ when finite. By that
definition, 0.999... = 1 exactly.
(There is a proof that
"Two curves which cross also meet"
is logically equivalent to
"Any bounded non-empty set has a least upper bound".)
Jim Burns
2017-04-22 23:30:14 UTC
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Post by netzweltler
Post by Jim Burns
What does it mean to follow all of those infinitely many
commands?
None of those results, 0.9, 0.99, 0.999, ... , are from
following all of those infinitely many commands, so it
shouldn't be a problem that none of them are exactly 1 .
t = 0: moving along the number line from 0 to 0.9
t = 0.9: moving along the number line from 0.9 to 0.99
t = 0.99: moving along the number line from 0.99 to 0.999
...
None of these, no matter how far you continue your '...',
will be the result of following infinitely many commands.

Keep in mind the natural numbers 1, 2, 3, ... .
Those are all finitely many S-operations from 0.
Post by netzweltler
By t = 1 we have followed all of the infinitely many commands,
but we didn't reach 1. If we reached 1 by t = 1 we have
disobeyed the commands.
What is the first command (above) that we would have disobeyed,
if we reached 1 by t = 1 ?

By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's not 1
in this case.

----
I will remind you that least upper bound of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...

No one can force you to use it. But equally, you can't
force anyone else not to use it. The best you can hope
for is to show that using it produces some sort of
inconsistency, or maybe some other non-logical reason
not to use it, more broadly conceived.

That's not what you're doing. You're just saying that,
if we don't define these infinite decimals the way
we have defined them, 0.999... won't be 1. Okay.
But we _do_ define them that way.

Did you have anything to say about why we should or
should not define them that way?

[...]
Post by netzweltler
Post by Jim Burns
Post by netzweltler
So, where is "point" 0.999... located on the number line?
Well, you need to define what you mean by number line
for this question to mean anything.
If what you mean by number line only includes rational numbers,
then you have problems.
First of all, the number line includes number 1.
Are you saying, I have problems locating 0.999... on
this number line?
I am saying that either this number line includes irrational
numbers or it is not a number line in any normal sense.

For example, it would not allow you to draw a right
isosceles triangle. That would not be normal.

I mention this because whatever allows us to make sense
(as we must be able to make sense) of sqrt(2) as
1.41421356237... seems likely to allow us to make sense
of 0.999... as 1. That is certainly the case for the definition
using least upper bound I gave you.
John Gabriel
2017-04-23 02:40:42 UTC
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Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's not 1
in this case.
I will remind you that least upper bound of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
Chuckle.

I don't see what your definition of 0.999... is. All you seem to be saying is that infinitely many elements of the set { 0.9, 0.99, 0.999, ... } are the LUB.

Stupid. Very stupid.
Jim Burns
2017-04-24 08:25:48 UTC
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Post by John Gabriel
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's
not 1 in this case.
I will remind you that *least upper bound* of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
*emphasis added*
Post by John Gabriel
Chuckle.
I don't see what your definition of 0.999... is.
All you seem to be saying is that infinitely many elements
of the set { 0.9, 0.99, 0.999, ... } are the LUB.
If, for all x in C, x =< b ,
then b is an upper bound of a set C,

If b' is an upper bound of C and
if, for any upper bound b of C, b' =< b,
then b' is a least upper bound of C.

So, a least upper bound of a set of real numbers is a real
number. It isn't, for example, a set of real numbers.

----
Interesting facts about least upper bounds:

A set might have a least upper bound and contain it.
X = { x e R | x =< 1 }, LUB(X) = 1
Or, a set might have a least upper bound and not contain it.
Y = { x e R | x < 1 }, LUB(Y) = 1

By the definition of bound, every real number is vacuously a
bound of the empty set, so {} doesn't have a _least_ upper bound.

Some sets of real numbers do not have any bound. For example
{ 2, 3, 5, 7, 11, 13, ... } is not bounded.
These sets do not have least upper bounds, either.

For every other set of real numbers -- that is, the bounded,
non-empty sets of reals -- there is a real number which is
a least upper bound of that set.

For every set of real numbers with a least upper bound --
which is every bounded, non-empty set -- there is one and
only one least upper bound. (Two different least upper bounds
would each be less than the other.)

----
One of my favorite facts about the least upper bound is that
the least upper bound of C is the set union of C
LUB C = Union C
when C is a bounded and non-empty and has only elements
drawn from a certain well-defined collection of subsets of
Q, the rational numbers. For example, if we write,
[0.9] = { x e Q | x < 0.9 }

[0.99] = { x e Q | x < 0.99 }

[0.999] = { x e Q | x < 0.999 }

...

[1] = { x e Q | x < 1 }

then
[0.999...] =def=

LUB{ [0.9], [0.99], [0.999], ... } =

Union{ [0.9], [0.99], [0.999], ... } =

{ x e Q | x < 1 } =

[1]
Post by John Gabriel
Stupid. Very stupid.
John Gabriel
2017-04-24 09:55:45 UTC
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Post by Jim Burns
Post by John Gabriel
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's
not 1 in this case.
I will remind you that *least upper bound* of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
*emphasis added*
Post by John Gabriel
Chuckle.
I don't see what your definition of 0.999... is.
All you seem to be saying is that infinitely many elements
of the set { 0.9, 0.99, 0.999, ... } are the LUB.
If, for all x in C, x =< b ,
then b is an upper bound of a set C,
Monkey! If 0.999... is a LUB, then it must be 1.

There is NO "If, for all x in C, x =< b, ..." because you automatically assume that b is 1.

Whooooooossssshh over your stupid head. Ask yourself this question:

How do you compare all x in C with 0.999... ? Stupid man.
John Gabriel
2017-04-24 10:04:00 UTC
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Post by John Gabriel
Post by Jim Burns
Post by John Gabriel
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's
not 1 in this case.
I will remind you that *least upper bound* of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
*emphasis added*
Post by John Gabriel
Chuckle.
I don't see what your definition of 0.999... is.
All you seem to be saying is that infinitely many elements
of the set { 0.9, 0.99, 0.999, ... } are the LUB.
If, for all x in C, x =< b ,
then b is an upper bound of a set C,
Monkey! If 0.999... is a LUB, then it must be 1.
There is NO "If, for all x in C, x =< b, ..." because you automatically assume that b is 1.
How do you compare all x in C with 0.999... ? Stupid man.
You blithering idiot. If you say that there is always another '9', then by the same token one can tell you that no matter how many 9s you have, 0.999... when compared with 1, will always be less than one.

But you and always will be a moron who can't think for himself. You are deluded and there is no hope for you.
Jim Burns
2017-04-24 12:55:18 UTC
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Post by John Gabriel
Post by Jim Burns
Post by John Gabriel
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's
not 1 in this case.
I will remind you that *least upper bound* of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
*emphasis added*
Post by John Gabriel
Chuckle.
I don't see what your definition of 0.999... is.
All you seem to be saying is that infinitely many elements
of the set { 0.9, 0.99, 0.999, ... } are the LUB.
If, for all x in C, x =< b ,
then b is an upper bound of a set C,
If b' is an upper bound of C and
if, for any upper bound b of C, b' =< b,
then b' is a least upper bound of C.
So, a least upper bound of a set of real numbers is a real
number. It isn't, for example, a set of real numbers.
How do you compare all x in C with 0.999... ?
I've typed out { 0.9, 0.99, 0.999, ... } enough times now.
Hereafter, that is what {0.999...} means, which is an infinite
set of rational numbers.

Let P(k) stand for 1 - 1/10^(k+1) < 1.

P(0) and, for all k in N, P(k) -> P(k+1).
By induction, for all k in N, P(k), that is,
1 - 1/10^(k+1) < 1
1 is a bound of the set {0.999...}

{0.999...} is a non-empty bounded set. Therefore, it has
a least upper bound b.
By trichotomy, b > 1, or b = 1, or b < 1.

~(b > 1). b is the _least_ upper bound and 1 is a bound.

Assume b < 1.

For every d > 0, there is some k' in M such that
b - d < 1 - 1/10^(m+1) < b < 1
otherwise b-d would be a bound. But b is _least_ .

But, for small enough d > 0, b < 1 - 1/10^(m+2)
And b is a _bound_ . Contradiction.

By contradiction, ~(b < 1)

From ~(b > 1) and ~(b < 1), we conclude b = 1.

By proof, LUB{0.999...} = 1.
By definition, 0.999... = LUB{0.999...} = 1.
Post by John Gabriel
Stupid man.
John Gabriel
2017-04-24 14:24:34 UTC
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Post by Jim Burns
Post by John Gabriel
Post by Jim Burns
Post by John Gabriel
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's
not 1 in this case.
I will remind you that *least upper bound* of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
*emphasis added*
Post by John Gabriel
Chuckle.
I don't see what your definition of 0.999... is.
All you seem to be saying is that infinitely many elements
of the set { 0.9, 0.99, 0.999, ... } are the LUB.
If, for all x in C, x =< b ,
then b is an upper bound of a set C,
If b' is an upper bound of C and
if, for any upper bound b of C, b' =< b,
then b' is a least upper bound of C.
So, a least upper bound of a set of real numbers is a real
number. It isn't, for example, a set of real numbers.
How do you compare all x in C with 0.999... ?
I've typed out { 0.9, 0.99, 0.999, ... } enough times now.
You can type it out a thousand more times you moron. It will make your claim no less FALSE than it already is.
Post by Jim Burns
Hereafter, that is what {0.999...} means, which is an infinite
set of rational numbers.
No. That is what YOU say it means. I tried at one point to explain to you what a well-formed concept means but you just didn't have what it takes to understand because you are a moron.

There is NO such thing as an infinite set you twat! You don't get to write an ellipsis and tell me that what you just brain farted is an infinite set! You infinitely stupid BABOOOON!!!!!
Post by Jim Burns
Let P(k) stand for 1 - 1/10^(k+1) < 1.
P(0) and, for all k in N, P(k) -> P(k+1).
By induction, for all k in N, P(k), that is,
1 - 1/10^(k+1) < 1
1 is a bound of the set {0.999...}
Idiot, what are you going on about again? Did I ever disagree with those statements? Pay attention you moron!!!!! You are saying that 0.999... represents an imaginary infinite set. Idiot!!!!!! THINK MAN!! THINK.

GRRRR. You are too stupid. I have wasted my time as in the past. FUCK OFF!
netzweltler
2017-04-24 12:14:12 UTC
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Post by Jim Burns
[0.999...] =def=
LUB{ [0.9], [0.99], [0.999], ... } =
Union{ [0.9], [0.99], [0.999], ... } =
{ x e Q | x < 1 } =
[1]
The Union{ [0.9], [0.99], [0.999], ... } doesn't cover point 1.

How can 0.999... be a point on the number line to the right of the points 0.9, 0.99, 0.999, ..., if [0.999...] doesn't consist of a segment other than the segments

[0,9]
[0.9], [0.09]
[0.9], [0.09], [0.009]
...
netzweltler
2017-04-23 07:13:02 UTC
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Post by Jim Burns
Post by netzweltler
Post by Jim Burns
What does it mean to follow all of those infinitely many
commands?
None of those results, 0.9, 0.99, 0.999, ... , are from
following all of those infinitely many commands, so it
shouldn't be a problem that none of them are exactly 1 .
t = 0: moving along the number line from 0 to 0.9
t = 0.9: moving along the number line from 0.9 to 0.99
t = 0.99: moving along the number line from 0.99 to 0.999
...
None of these, no matter how far you continue your '...',
will be the result of following infinitely many commands.
The list contains a line for each n ∈ ℕ and so is the result of infinitely many commands.
Post by Jim Burns
Keep in mind the natural numbers 1, 2, 3, ... .
Those are all finitely many S-operations from 0.
Post by netzweltler
By t = 1 we have followed all of the infinitely many commands,
but we didn't reach 1. If we reached 1 by t = 1 we have
disobeyed the commands.
What is the first command (above) that we would have disobeyed,
if we reached 1 by t = 1 ?
None of the commands of the list tells us to reach 1.
Post by Jim Burns
By the way, I don't see what your definition of infinitely
many operations is. All you seem to be saying is that it's not 1
in this case.
I'm not just saying that. Which command of the complete list of commands above tells us to reach 1? All we've got is these commands on this list. There is no additional instruction (something like a "last" instruction that tells us to stop right at 1). We don't stop at any point on the number line.
Post by Jim Burns
----
I will remind you that least upper bound of
{ 0.9, 0.99, 0.999, ... } is a _definition_ of 0.999...
No one can force you to use it. But equally, you can't
force anyone else not to use it. The best you can hope
for is to show that using it produces some sort of
inconsistency, or maybe some other non-logical reason
not to use it, more broadly conceived.
That's not what you're doing. You're just saying that,
if we don't define these infinite decimals the way
we have defined them, 0.999... won't be 1. Okay.
But we _do_ define them that way.
Did you have anything to say about why we should or
should not define them that way?
[...]
Post by netzweltler
Post by Jim Burns
Post by netzweltler
So, where is "point" 0.999... located on the number line?
Well, you need to define what you mean by number line
for this question to mean anything.
If what you mean by number line only includes rational numbers,
then you have problems.
First of all, the number line includes number 1.
Are you saying, I have problems locating 0.999... on
this number line?
I am saying that either this number line includes irrational
numbers or it is not a number line in any normal sense.
For example, it would not allow you to draw a right
isosceles triangle. That would not be normal.
I mention this because whatever allows us to make sense
(as we must be able to make sense) of sqrt(2) as
1.41421356237... seems likely to allow us to make sense
of 0.999... as 1. That is certainly the case for the definition
using least upper bound I gave you.
The number line includes sqrt(2) and pi just like it includes 1, but it doesn't include non-terminating representations of sqrt(2), pi or 1.
Franz Gnaedinger
2017-04-22 07:26:05 UTC
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Post by Franz Gnaedinger
Irrationals not being numbers, and 1 not being 0.999..., are everlasting
topics in sci.math, popping up all the time. Very simple yet forgotten
methods of early mathematics can dissolve the problem.
Approximating the square root of 2 by a number column which I reconstructed
in 1979
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on
99/70 is already a fine value for the square root of 2. (Analogous number
columns approximate the square roots of 3 and 5 and the cube root of 2.)
Now let us look at the square root of 4 by drawing up the analogous number
column
1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
and so on
The ratios, for example 41/20 or 121/61, get ever closer to 2. This means
that 2 as the square root of 4 is equally mysterious as the square root of 2.
Or that the square root of 2 is no less mysterious than 2.
As for 1 = 0.999... let me begin with the Horus eye series of Ancient Egypt,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 or simply '2 '4 '8 '16 '32 '64
1 = '1
1 = '2 '2
1 = '2 '4 '4
1 = '2 '4 '8 '8
1 = '2 '4 '8 '16 '16
1 = '2 '4 '8 '16 '32 '32
1 = '2 '4 '8 '16 '32 '64 '64
From this you get
'2 '4 '8 '16 '32 '64 ... = 1
(The Horus eye series had an astronomical function, since a month of 30
days multiplied by '2 '4 '8 '16 '32 '64 yields 29 '2 '32 days or 29 days
12 hours 45 minutes for one lunation of 29 days 12 hours 44 minutes 2.9
seconds, modern average from 1989 AD - mistake of the old value less than
one minute per lunation, or half a day in a lifetime. One eye of the Horus
falcon was the sun, standing for a month of 30 days, his other eye was
the moon, standing for one lunation or synodic month.)
Another infinite series
1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 '9x10 '10x11 '11x12 ...
'1x2 '2x3 '5x6 '6x7 '9x10 '10x11 ... = pi/4
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Mathematical education should begin with really simple methods, not on
the level of Euclid. Would prevent some people from wasting their life
on kooky notions.
Half a century of studies and work led me to the triangle of language
whose corners are

life with needs and wishes

mathematics, logic of building and maintaining,
based on the formula a = a

art, human measure in a technical world,
based on Goethe's formula all is equal, all unequal ...,
a formula known to artists of all times

Galilei famously wrote that the book of nature is written in the language
of mathematics. Yes, but we can decipher only the first lines on the first
page of the first volume on the first shelf in the first room of the divine
library. God may understand all of nature in mathematical terms, while for us
the equal falls apart into equal unequal. The infinite, equal unequal in
itself, marks the border between the two forms of logic. When we succeed
in taming a part of the infinite on the mathematical side we also increase
the infinite on the other side. For example Georg Cantor's aleph zero of the
rationals and c of the reals revealed infinitely many more and higher alephs
- ever more ever higher infinities. We can't get rid of the other side of
logic. Which is what Goedel proved.

Some people need the illusion of mathematics as a closed and well protected
garden, which might have psychological reasons (holding their life together,
conclusion from long discussions), but they achieve nothing. Progress comes
from taming further parts of the infinite on the mathematical side of logic's
border.
bassam king karzeddin
2017-04-22 12:53:59 UTC
Permalink
Raw Message
Post by Franz Gnaedinger
Post by Franz Gnaedinger
Irrationals not being numbers, and 1 not being 0.999..., are everlasting
topics in sci.math, popping up all the time. Very simple yet forgotten
methods of early mathematics can dissolve the problem.
Approximating the square root of 2 by a number column which I reconstructed
in 1979
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on
99/70 is already a fine value for the square root of 2. (Analogous number
columns approximate the square roots of 3 and 5 and the cube root of 2.)
Now let us look at the square root of 4 by drawing up the analogous number
column
1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
and so on
The ratios, for example 41/20 or 121/61, get ever closer to 2. This means
that 2 as the square root of 4 is equally mysterious as the square root of 2.
Or that the square root of 2 is no less mysterious than 2.
As for 1 = 0.999... let me begin with the Horus eye series of Ancient Egypt,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 or simply '2 '4 '8 '16 '32 '64
1 = '1
1 = '2 '2
1 = '2 '4 '4
1 = '2 '4 '8 '8
1 = '2 '4 '8 '16 '16
1 = '2 '4 '8 '16 '32 '32
1 = '2 '4 '8 '16 '32 '64 '64
From this you get
'2 '4 '8 '16 '32 '64 ... = 1
(The Horus eye series had an astronomical function, since a month of 30
days multiplied by '2 '4 '8 '16 '32 '64 yields 29 '2 '32 days or 29 days
12 hours 45 minutes for one lunation of 29 days 12 hours 44 minutes 2.9
seconds, modern average from 1989 AD - mistake of the old value less than
one minute per lunation, or half a day in a lifetime. One eye of the Horus
falcon was the sun, standing for a month of 30 days, his other eye was
the moon, standing for one lunation or synodic month.)
Another infinite series
1 = '1
1 = '1x2 '2
1 = '1x2 '2x3 '3
1 = '1x2 '2x3 '3x4 '4
1 = '1x2 '2x3 '3x4 '4x5 '5
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6
1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 '9x10 '10x11 '11x12 ...
'1x2 '2x3 '5x6 '6x7 '9x10 '10x11 ... = pi/4
Now for 0.999... being 1
1 = 10/10
1 = 9/10 + 10/100
1 = 9/10 + 9/100 + 10/1000
1 = 9/10 + 9/100 + 9/1000 + 10/10000
and so on
1 = 0.999...
Mathematical education should begin with really simple methods, not on
the level of Euclid. Would prevent some people from wasting their life
on kooky notions.
Half a century of studies and work led me to the triangle of language
whose corners are
life with needs and wishes
mathematics, logic of building and maintaining,
based on the formula a = a
art, human measure in a technical world,
based on Goethe's formula all is equal, all unequal ...,
a formula known to artists of all times
Galilei famously wrote that the book of nature is written in the language
of mathematics. Yes, but we can decipher only the first lines on the first
page of the first volume on the first shelf in the first room of the divine
library. God may understand all of nature in mathematical terms, while for us
the equal falls apart into equal unequal. The infinite, equal unequal in
itself, marks the border between the two forms of logic. When we succeed
in taming a part of the infinite on the mathematical side we also increase
the infinite on the other side. For example Georg Cantor's aleph zero of the
rationals and c of the reals revealed infinitely many more and higher alephs
- ever more ever higher infinities. We can't get rid of the other side of
logic. Which is what Goedel proved.
Some people need the illusion of mathematics as a closed and well protected
garden, which might have psychological reasons (holding their life together,
conclusion from long discussions), but they achieve nothing. Progress comes
from taming further parts of the infinite on the mathematical side of logic's
border.
The illusions are sourcing from assuming some illusionary concepts such as infinity concept for a little dirty purpose to make unnecessary theorems or fake results or more precisely approximation theorems, which is - infinity - unreal by its own definition, thus non existing for sure

How many times this was proved here even for a shepherd boy? wonder!
How many times it was said that any number associated with unreal number (infinity by its own basic definition) must be unreal number for sure

How many times it was said that a very little finite integer with finite sequence of digits only and may be in any number system, can fill the size of (say only one million of billions of galaxies), where every one billion digits are assumed stored only in one (mm) cube?, wonder!

But still the top professional mathematician would argue that infinity is still larger than any integer, despite being defined as not being any number, so wonderful!

Finally, how many centuries it would take the human minds that infinity does not exist except in very poor minds, and it was adopted just to justify all those unneeded and illegal talents for sure

And I know that the top professionals would never drop that damn concept, since it represents a Paradise for them, and if they drop it, then nothing would remain for them for sure

So, as many more walks of human life, and with this least and generally unnoticeable or observable category of professional mathematicians, and world of mathematics that is full of corruption for sure

But still people generally would prefer fictions to reality as always as usual for sure

And they would keep running after many Ghosts (they had invented) endlessly

Regards
Bassam King Karzeddin
22th, April, 2017
Franz Gnaedinger
2017-04-24 07:10:05 UTC
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Post by bassam king karzeddin
Finally, how many centuries it would take the human minds that infinity does not exist except in very poor minds, and it was adopted just to justify all those unneeded and illegal talents for sure
Jan Stewart calls mathematics the taming of the infinite. You are dismissing
the infinite as a failed concept. I see the truth in between. The infinite
- either infinitely big or infinitely small - is equal unequal in itself,
marking the border between mathematical logic based on a = a and the wider
logic of all is equal, all unequal ... You can't get rid of the infinite
and the paradoxa it causes, but you can tame parts of the infinite on the
side of mathematical logic. Progress in mathematics comes from there.
Denying the infinite does not solve any problem, it just annihilates
mathematics.

Here a further stairway number, defining 1/2 or '2 in terms of 1/3 or '3

'2 = '2 '6
'2 = '3 '9 '18
'2 = '3 '9 '27 '54
'2 = '3 '9 '27 '81 '162
and so on

'2 = '3 '3x3 '3x3x3 '3x3x3x3 '3x3x3x3x3 '3x3x3x3x3x3 ...
Franz Gnaedinger
2017-04-24 06:51:00 UTC
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Post by Franz Gnaedinger
Half a century of studies and work led me to the triangle of language
whose corners are
life with needs and wishes
mathematics, logic of building and maintaining,
based on the formula a = a
art, human measure in a technical world,
based on Goethe's formula all is equal, all unequal ...,
a formula known to artists of all times
Galilei famously wrote that the book of nature is written in the language
of mathematics. Yes, but we can decipher only the first lines on the first
page of the first volume on the first shelf in the first room of the divine
library. God may understand all of nature in mathematical terms, while for us
the equal falls apart into equal unequal. The infinite, equal unequal in
itself, marks the border between the two forms of logic. When we succeed
in taming a part of the infinite on the mathematical side we also increase
the infinite on the other side. For example Georg Cantor's aleph zero of the
rationals and c of the reals revealed infinitely many more and higher alephs
- ever more ever higher infinities. We can't get rid of the other side of
logic. Which is what Goedel proved.
Some people need the illusion of mathematics as a closed and well protected
garden, which might have psychological reasons (holding their life together,
conclusion from long discussions), but they achieve nothing. Progress comes
from taming further parts of the infinite on the mathematical side of logic's
border.
Now for a geometrical interpretation of the number columns approximating
the square roots of 2 and 4.

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on

Draw a line. Mark two points A and C. Then imagine a point B oscillating
between them

--- A ---------- C ---

if A and C are 1 and 2 respectively, B is 1
if A and C are 2 and 4 respectively, C is 3
if A and C are 5 and 10 respectively, B is 7
if A and C are 12 and 24 respectively, B is 17
if A and C are 29 and 58 respectively, B is 41
if A and C are 99 and 140 respectively, B is 99
and so on

Return to A = 1 and C = 2. The oscillating point B approaches the square
root of 2.

1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
and so on

--- A ---------- C ---

if A and C are 1 and 4 respectively, B is 1
if A and C are 2 and 8 respectively, C is 5
if A and C are 7 and 28 respectively, B is 13
if A and C are 20 and 80 respectively, B is 41
if A and C are 61 and 244 respectively, B is 121
if A and C are 182 and 728 respectively, B is 365
and so on

Return to A = 1 and C = 4. The oscillating point B approaches the square root
of 4 which is 2.

Point B approaches the square root of 2 (first case) and the square root of 4
which is 2 (second case) in the same way. Either both numbers are a mystery
(sqrt2 and 2), or neither of them is.
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