mersenneforum.org Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

2022-09-28, 07:13   #34
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

3·5·683 Posts

Quote:
 Originally Posted by PhilF Yes, I believe in hydrocarbons, although you're going to have to include some O's with your CH's.
Some -OH with CH, even better...

 2022-10-02, 10:20 #35 MPrimeFinder   "Anonymous" Sep 2022 finding m52 910 Posts (1+1+1+1+1+1+...) is indeed less than (1+2+3+4+5+6+...) In the Gauss formula, there is a pattern regarding powers of 10. 1 - 10 = 11 x 5 = 55 1 - 100 = 101 x 50 = 5050 1 - 1000 = 1001 x 500 = 500500 The pattern: 1 - x = (x+1)*(x/2) = 5 followed by ((log10 of x)-1 zeroes) followed by 5 followed by ((log 10 of x)-1 zeroes) See how there are ((log10 of x)-1)*2 zeroes in the final result? Therefore: 1 - ∞ = (∞+1)*(∞/2) = 5 followed by (∞-1 zeroes) followed by 5 followed by (∞-1 zeroes). Obviously, (1+1+1+1+1+1+...) results in ∞, noticeably less than the result of (1+2+3+4+5+6+...).
2022-10-02, 11:01   #36
paulunderwood

Sep 2002
Database er0rr

29×151 Posts

Quote:
 Originally Posted by MPrimeFinder In the Gauss formula, there is a pattern regarding powers of 10. 1 - 10 = 11 x 5 = 55 1 - 100 = 101 x 50 = 5050 1 - 1000 = 1001 x 500 = 500500 The pattern: 1 - x = (x+1)*(x/2) = 5 followed by ((log10 of x)-1 zeroes) followed by 5 followed by ((log 10 of x)-1 zeroes) See how there are ((log10 of x)-1)*2 zeroes in the final result? Therefore: 1 - ∞ = (∞+1)*(∞/2) = 5 followed by (∞-1 zeroes) followed by 5 followed by (∞-1 zeroes). Obviously, (1+1+1+1+1+1+...) results in ∞, noticeably less than the result of (1+2+3+4+5+6+...).
Another way to see the sums are the same is to use parentheses: (1)+(1+1)+(1+1+1)+(1+1+1+1)....

Last fiddled with by paulunderwood on 2022-10-02 at 11:06

2022-10-02, 11:56   #37
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

29×229 Posts

Quote:
 Originally Posted by MPrimeFinder ... ∞-1 ...
Of course that is perfectly well defined.

2022-10-02, 14:15   #38
Dr Sardonicus

Feb 2017
Nowhere

22×52×61 Posts

Quote:
 Originally Posted by retina Of course that is perfectly well defined.
If you actually make it well-defined, of course, you spoil all the fun. First, you have to be clear about what kind of number you mean by $\infty$. There are cardinal ("how many") numbers and ordinal numbers (the ordinal types of sets with well-orderings) - first, second, third, etc.

I'll assume we're dealing with cardinal numbers, and $\infty$ is the cardinal number $\aleph_{0}$ of the positive integers. Then all the indicated infinities are $\aleph_{0}$.

Finite ordinal numbers are of the form {1, 2, ... n} (positive integers from 1 to n) with the usual linear ordering.

Infinite ordinal numbers are much more complicated. The first infinite ordinal number is the ordinal type of the positive integers with the usual ordering, which may also be viewed as the ordinal type of the set of all finite ordinals, ordered by one being a section (initial segment) of another. The cardinality of this set is not finite, because no ordinal number can be a section of itself. This cardinality is $\aleph_{0}$. The ordinal type is usually denoted $\omega$.

Defining ordinal addition by concatenation, we see that

$1\;+\;\omega\;=\;\omega$, but

$\omega\;+\;1\;\ne\;\omega$.

The ordinal type of the set of all ordinal types which are either finite or of sets with cardinality $\aleph_{0}$ is usually denoted $\Omega$. The cardinality of the set of such ordinals is the next infinite cardinal number after $\aleph_{0}$, denoted $\aleph_{1}$.

Last fiddled with by Dr Sardonicus on 2022-10-02 at 14:41 Reason: Insert omitted word

2022-10-04, 13:57   #39
MPrimeFinder

"Anonymous"
Sep 2022
finding m52

32 Posts

Quote:
 Originally Posted by Dr Sardonicus If you actually make it well-defined, of course, you spoil all the fun. First, you have to be clear about what kind of number you mean by $\infty$. There are cardinal ("how many") numbers and ordinal numbers (the ordinal types of sets with well-orderings) - first, second, third, etc. I'll assume we're dealing with cardinal numbers, and $\infty$ is the cardinal number $\aleph_{0}$ of the positive integers. Then all the indicated infinities are $\aleph_{0}$. Finite ordinal numbers are of the form {1, 2, ... n} (positive integers from 1 to n) with the usual linear ordering. Infinite ordinal numbers are much more complicated. The first infinite ordinal number is the ordinal type of the positive integers with the usual ordering, which may also be viewed as the ordinal type of the set of all finite ordinals, ordered by one being a section (initial segment) of another. The cardinality of this set is not finite, because no ordinal number can be a section of itself. This cardinality is $\aleph_{0}$. The ordinal type is usually denoted $\omega$. Defining ordinal addition by concatenation, we see that $1\;+\;\omega\;=\;\omega$, but $\omega\;+\;1\;\ne\;\omega$. The ordinal type of the set of all ordinal types which are either finite or of sets with cardinality $\aleph_{0}$ is usually denoted $\Omega$. The cardinality of the set of such ordinals is the next infinite cardinal number after $\aleph_{0}$, denoted $\aleph_{1}$.
Ok so what is your "cardinal" solution formula?

Note: See added emphasis in quote.

Last fiddled with by Dr Sardonicus on 2022-10-04 at 14:20 Reason: As indicated

2022-10-09, 02:00   #40
MPrimeFinder

"Anonymous"
Sep 2022
finding m52

32 Posts

Quote:
 Originally Posted by paulunderwood Another way to see the sums are the same is to use parentheses: (1)+(1+1)+(1+1+1)+(1+1+1+1)....
Literally the solution.

 Thread Tools

All times are UTC. The time now is 14:37.

Sun Dec 4 14:37:08 UTC 2022 up 108 days, 12:05, 0 users, load averages: 1.11, 1.11, 0.96

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔