 mersenneforum.org Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?
 Register FAQ Search Today's Posts Mark Forums Read  2022-09-28, 07:13   #34
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

3·23·149 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 3·7 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

3×1,499 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

668910 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

22×1,559 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 . 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 is the cardinal number of the positive integers. Then all the indicated infinities are .

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 . The ordinal type is usually denoted .

Defining ordinal addition by concatenation, we see that

, but

.

The ordinal type of the set of all ordinal types which are either finite or of sets with cardinality is usually denoted . The cardinality of the set of such ordinals is the next infinite cardinal number after , denoted .

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

101012 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 . 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 is the cardinal number of the positive integers. Then all the indicated infinities are . 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 . The ordinal type is usually denoted . Defining ordinal addition by concatenation, we see that , but . The ordinal type of the set of all ordinal types which are either finite or of sets with cardinality is usually denoted . The cardinality of the set of such ordinals is the next infinite cardinal number after , denoted .
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

2110 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 Show Printable Version Email this Page

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