On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring
1/3+1/3+1/3 = 1 also 0.9999 recurring
pruve mee rong
And for a similar reason the sum of all of the integer powers of two
from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking about
signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity)
-a = 2 + 4 + 8 + 16 + 32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just ignore 2^(infinity + 1) ... it's a hugemungously massive number!
2S + 1 = S = 1 + 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
Nope.
-a 2S + 1 - 2^(infinity + 1) = S
therefore 2S - S = -1
Nope.
-a 2S - S = 2^(infinity +1) - 1
This forms the basis for how negative integers are written in binary.
Your computer assumes that the representation of -1 as 11111111 goes
on for infinity on the left hand side.
Depsite being one of the most widely used pieces of commercial software, Microsoft Excel spreadsheet is actually hopeless with numbers and in
certain circumstances can be inaccurate due to a decision they made for storing decimal numbers. Apple's Numbers spreadsheet is much better and works properly.
Your Name wrote:
On 2026-02-09 14:26:01 +0000, solar penguin said:
YourCOre right. And it can also be shown another way:
0.9 = 1 - 0.1
0.99 = 1 - 0.01
0.999 = 1 - 0.001
.
.
.
0.9999 recurring = 1 - 0.0000 recurring
Not quote true. That last line still needs a 1 on the end (even though
technically you never actually get there).
0.9999 recurring = 1 - 0.0000...1
For engineers, "close enough is good enough". For mathematicians you
have to be exact. :-)
Methinks this Doctor Who hiatus is gonna be a long one here... we've
already done Bible study, paedophile talk, now it's Maths. What's next
I wonder?
Thank fuck the Olympic men's hockey tournament starts tomorrow
afternoon (15:40 GMT)... that'll surely be better entertainment than
algebra class!
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring
1/3+1/3+1/3 = 1 also 0.9999 recurring
pruve mee rong
And for a similar reason the sum of all of the integer powers of two >>>>> from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking about
signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity)
-a = 2 + 4 + 8 + 16 + 32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just ignore
2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity, two times infinity is still
infinity, and what you wrote down was infinity plus one all two the
power of two which isn't even the same thing as what I originally wrote.
2S + 1 = S = 1 + 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
Nope.
Yes.
-a 2S + 1 - 2^(infinity + 1) = S
therefore 2S - S = -1
Nope.
Yes. 2S - S = S = -1
It's how negative numbers are written in binary and is fully valid in >mathematics and forms a ring and is the basis of 2-adic numbers.
-a 2S - S = 2^(infinity +1) - 1
This forms the basis for how negative integers are written in binary.
Your computer assumes that the representation of -1 as 11111111 goes
on for infinity on the left hand side.
Depsite being one of the most widely used pieces of commercial software,
Microsoft Excel spreadsheet is actually hopeless with numbers and in
certain circumstances can be inaccurate due to a decision they made for
storing decimal numbers. Apple's Numbers spreadsheet is much better and
works properly.
That was 30 years ago and they fixed that but in Excel 95. Try using >LibreOffice Calc if you don't like it.
And you've still not refuted my claim that all of the integers up to >infinity add up to -1/12.
----
The True Doctor https://www.youtube.com/channel/UCngrZwoS0n21IRcXpKO79Lw
"To be woke is to be uninformed which is exactly the opposite of what it >stands for." --William Shatner
On 2026-02-10 11:02:01 +0000, Blueshirt said:
Your Name wrote:
On 2026-02-09 14:26:01 +0000, solar penguin said:
YourCOre right. And it can also be shown another way:
0.9 = 1 - 0.1
0.99 = 1 - 0.01
0.999 = 1 - 0.001
.
.
.
0.9999 recurring = 1 - 0.0000 recurring
Not quote true. That last line still needs a 1 on the end (even though
technically you never actually get there).
0.9999 recurring = 1 - 0.0000...1
For engineers, "close enough is good enough". For mathematicians you
have to be exact. :-)
Methinks this Doctor Who hiatus is gonna be a long one here... we've
already done Bible study, paedophile talk, now it's Maths. What's next
I wonder?
Thank fuck the Olympic men's hockey tournament starts tomorrow
afternoon (15:40 GMT)... that'll surely be better entertainment than
algebra class!
Nope, "sport sux". I'd far far far rather be in mathematics class. :-p
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
----
Daniel70
Verily, in article <10mhm0p$fsp5$1@dont-email.me>, did daniel47 >@nomail.afraid.org deliver unto us this message:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
What would it be, then?
I'm afraid infinity minus one is still infinity. Otherwise, infinity
plus one couldn't be infinity.
Infinity is a wiggly concept. Have you ever seen that logic puzzle about
the hotel with infinite rooms, which somehow still gets full?
----
The True Melissa - Canal Winchester - Ohio
United States of America - North America - Earth
Solar System - Milky Way - Local Group
Virgo Cluster - Laniakea Supercluster - Cosmos
On 11/02/2026 10:35, Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
Infinity minus one is still infinity. Infinity divided by infinity is >indeterminate, but I can evaluate it to a finite value depending on the >infinities used and the definition of the limit.
----
The True Doctor https://www.youtube.com/channel/UCngrZwoS0n21IRcXpKO79Lw
"To be woke is to be uninformed which is exactly the opposite of what it >stands for." --William Shatner
On 11/02/2026 5:53 am, The True Doctor wrote:
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO,
because there IS at least one larger number!
On 2026-02-10 11:02:01 +0000, Blueshirt said:
Your Name wrote:
On 2026-02-09 14:26:01 +0000, solar penguin said:
.
0.9999 recurring = 1 - 0.0000 recurring
Not quote true. That last line still needs a 1 on the
end (even though technically you never actually get
there). 0.9999 recurring = 1 - 0.0000...1
For engineers, "close enough is good enough". For
mathematicians you have to be exact. :-)
Methinks this Doctor Who hiatus is gonna be a long one
here... we've already done Bible study, paedophile talk,
now it's Maths. What's next I wonder?
Thank fuck the Olympic men's hockey tournament starts
tomorrow afternoon (15:40 GMT)... that'll surely be better
entertainment than algebra class!
Nope, "sport sux". I'd far far far rather be in mathematics
class. :-p
Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO,
because there IS at least one larger number!
Don't give up your day job...
Your Name wrote:
On 2026-02-10 11:02:01 +0000, Blueshirt said:
Your Name wrote:
On 2026-02-09 14:26:01 +0000, solar penguin said:
.
0.9999 recurring = 1 - 0.0000 recurring
Not quote true. That last line still needs a 1 on the
end (even though technically you never actually get
there). 0.9999 recurring = 1 - 0.0000...1
For engineers, "close enough is good enough". For
mathematicians you have to be exact. :-)
Methinks this Doctor Who hiatus is gonna be a long one
here... we've already done Bible study, paedophile talk,
now it's Maths. What's next I wonder?
Thank fuck the Olympic men's hockey tournament starts
tomorrow afternoon (15:40 GMT)... that'll surely be better
entertainment than algebra class!
Nope, "sport sux". I'd far far far rather be in mathematics
class. :-p
Looks like you're in the right place so!
On 11/02/2026 10:35, Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
Infinity minus one is still infinity. Infinity divided by infinity is indeterminate, but I can evaluate it to a finite value depending on
the infinities used and the definition of the limit.
On 2/11/2026 8:31 AM, The True Doctor wrote:
On 11/02/2026 10:35, Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
Infinity minus one is still infinity. Infinity divided by infinity is
indeterminate, but I can evaluate it to a finite value depending on
the infinities used and the definition of the limit.
In mathematics, infinity is a concept describing something unbounded, >limitless, or without end, rather than a specific, countable number. It >represents a quantity larger than any finite number, often used to
describe sets that continue forever (like natural numbers) or limits in >calculus.
Stan
----
This email has been checked for viruses by Avast antivirus software. >www.avast.com
You really are dense Dannyboy.
On 2/11/2026 8:31 AM, The True Doctor wrote:
On 11/02/2026 10:35, Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
Infinity minus one is still infinity. Infinity divided by infinity is
indeterminate, but I can evaluate it to a finite value depending on the
infinities used and the definition of the limit.
In mathematics, infinity is a concept describing something unbounded, limitless, or without end, rather than a specific, countable number. It represents a quantity larger than any finite number, often used to
describe sets that continue forever (like natural numbers) or limits in calculus.
Stan
Verily, in article <10mhm0p$fsp5$1@dont-email.me>, did daniel47 @nomail.afraid.org deliver unto us this message:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
What would it be, then?
I'm afraid infinity minus one is still infinity. Otherwise, infinity
plus one couldn't be infinity.
Infinity is a wiggly concept. Have you ever seen that logic puzzle about
the hotel with infinite rooms, which somehow still gets full?
The Binky(Word used by paedophiles to indicate their joy of child sexual molestation) Doctor wrote:
You really are dense Dannyboy.Why is everyone ignoring *you*, Binky(Word used by paedophiles
to indicate their joy of child sexual molestation)?!
GUFFAW!
On 2026-02-11 14:53:40 +0000, StanWeiss said:
On 2/11/2026 8:31 AM, The True Doctor wrote:
On 11/02/2026 10:35, Daniel70 wrote:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers >>>>>>>>> of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 + >>>>>> 32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just >>>>>> ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS >>>> at least one larger number!
Infinity minus one is still infinity. Infinity divided by infinity is
indeterminate, but I can evaluate it to a finite value depending on the >>> infinities used and the definition of the limit.
In mathematics, infinity is a concept describing something unbounded,
limitless, or without end, rather than a specific, countable number. It
represents a quantity larger than any finite number, often used to
describe sets that continue forever (like natural numbers) or limits in
calculus.
Stan
Yep. So trying to use "infinity" in any kind of mathematical proof is
simply ridiculous nonsense. It results in more of an 'ideal' rather
than an actual 'fact'.
On 12/02/2026 12:16 am, The True Melissa wrote:
Verily, in article <10mhm0p$fsp5$1@dont-email.me>, did daniel47
@nomail.afraid.org deliver unto us this message:
On 11/02/2026 5:53 am, The True Doctor wrote:
On 09/02/2026 21:35, Your Name wrote:
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:
On 2/9/2026 8:59 AM, The True Doctor wrote:
On 09/02/2026 11:50, Nomen Nescio wrote:
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999
recurring pruve mee rong
And for a similar reason the sum of all of the integer powers
of two from 2^0 to 2^infinity is -1.
Prove me wrong.
If it is a minus integer would depend on, if you are talking
about signed or unsigned integers.
Stan
if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
No it doesn't.
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 +
32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just
ignore 2^(infinity + 1) ... it's a hugemungously massive number!
Infinity plus one is still infinity,
I infinity MINUS one still infinity?? I'm guessing NO, because there IS
at least one larger number!
What would it be, then?
I'm afraid infinity minus one is still infinity. Otherwise, infinity
plus one couldn't be infinity.
Correct .... because it *IS* infinity *PLUS* one!! ;-P (Snigger!! Snigger!!)
--Infinity is a wiggly concept. Have you ever seen that logic puzzle aboutNo can't say I have seen that puzzle.
the hotel with infinite rooms, which somehow still gets full?
--
Daniel70
In article <10mipgp$u3i6$1@dont-email.me>,
Your Name <YourName@YourISP.com> wrote:
On 2026-02-11 14:53:40 +0000, StanWeiss said:So What is 2^infinity ?
On 2/11/2026 8:31 AM, The True Doctor wrote:Yep. So trying to use "infinity" in any kind of mathematical proof is
On 11/02/2026 10:35, Daniel70 wrote:In mathematics, infinity is a concept describing something unbounded,
On 11/02/2026 5:53 am, The True Doctor wrote:Infinity minus one is still infinity. Infinity divided by infinity is
On 09/02/2026 21:35, Your Name wrote:I infinity MINUS one still infinity?? I'm guessing NO, because there IS >>>>> at least one larger number!
On 2/9/2026 11:26 AM, The True Doctor wrote:Infinity plus one is still infinity,
On 09/02/2026 15:15, StanWeiss wrote:No it doesn't.
On 2/9/2026 8:59 AM, The True Doctor wrote:if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
On 09/02/2026 11:50, Nomen Nescio wrote:If it is a minus integer would depend on, if you are talking >>>>>>>>> about signed or unsigned integers.
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999And for a similar reason the sum of all of the integer powers >>>>>>>>>> of two from 2^0 to 2^infinity is -1.
recurring pruve mee rong
Prove me wrong.
Stan
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 + >>>>>>> 32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just >>>>>>> ignore 2^(infinity + 1) ... it's a hugemungously massive number!
indeterminate, but I can evaluate it to a finite value depending on the >>>> infinities used and the definition of the limit.
limitless, or without end, rather than a specific, countable number. It
represents a quantity larger than any finite number, often used to
describe sets that continue forever (like natural numbers) or limits in
calculus.
Stan
simply ridiculous nonsense. It results in more of an 'ideal' rather
than an actual 'fact'.
The Binky Doctor wrote:
You really are dense Dannyboy.Why is everyone ignoring *you*, Binky?!
GUFFAW!
On 2/11/2026 8:53 PM, The Doctor wrote:
In article <10mipgp$u3i6$1@dont-email.me>,
Your Name <YourName@YourISP.com> wrote:
On 2026-02-11 14:53:40 +0000, StanWeiss said:So What is 2^infinity ?
On 2/11/2026 8:31 AM, The True Doctor wrote:Yep. So trying to use "infinity" in any kind of mathematical proof is
On 11/02/2026 10:35, Daniel70 wrote:In mathematics, infinity is a concept describing something unbounded,
On 11/02/2026 5:53 am, The True Doctor wrote:Infinity minus one is still infinity. Infinity divided by infinity is >>>>> indeterminate, but I can evaluate it to a finite value depending on the >>>>> infinities used and the definition of the limit.
On 09/02/2026 21:35, Your Name wrote:I infinity MINUS one still infinity?? I'm guessing NO, because there IS >>>>>> at least one larger number!
On 2/9/2026 11:26 AM, The True Doctor wrote:
On 09/02/2026 15:15, StanWeiss wrote:No it doesn't.
On 2/9/2026 8:59 AM, The True Doctor wrote:if S = 1 + 2 + 4 + 8 + 16 + ... + 2^infinity
On 09/02/2026 11:50, Nomen Nescio wrote:If it is a minus integer would depend on, if you are talking >>>>>>>>>> about signed or unsigned integers.
look 1/3 = 0.3333 recurring 1/3+1/3+1/3 = 1 also 0.9999 >>>>>>>>>>>> recurring pruve mee rongAnd for a similar reason the sum of all of the integer powers >>>>>>>>>>> of two from 2^0 to 2^infinity is -1.
Prove me wrong.
Stan
then 2S = 2 + 4 + 8 + 16 + 32 + ... + 2^infinity
2S = 2 x (1 + 2 + 4 + 8 + 16 + ... + 2^infinity) = 2 + 4 + 8 + 16 + >>>>>>>> 32 + ... + 2^(infinity + 1)
Even though you can't technically get to infinity, you can't just >>>>>>>> ignore 2^(infinity + 1) ... it's a hugemungously massive number! >>>>>>> Infinity plus one is still infinity,
limitless, or without end, rather than a specific, countable number. It >>>> represents a quantity larger than any finite number, often used to
describe sets that continue forever (like natural numbers) or limits in >>>> calculus.
Stan
simply ridiculous nonsense. It results in more of an 'ideal' rather
than an actual 'fact'.
-aJust another example of you showing-a how clueless you are.
Stan
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This email has been checked for viruses by Avast antivirus software. >www.avast.com
The Binky(Word used by paedophiles to indicate their joy of child sexual molestation) Doctor wrote:
You really are dense Dannyboy.Why is everyone ignoring *you*, Binky(Word used by paedophiles
to indicate their joy of child sexual molestation)?!
GUFFAW!
*Crickets*
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