Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ? - Page 4

LaurV · 2022-09-28, 07:13

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...

MPrimeFinder · 2022-10-02, 10:20

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+...).

paulunderwood · 2022-10-02, 11:01

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)....

retina · 2022-10-02, 11:56

Quote:

Originally Posted by MPrimeFinder

... ∞-1 ...

Of course that is perfectly well defined.

Dr Sardonicus · 2022-10-02, 14:15

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 $\aleph_{1}$ .

MPrimeFinder · 2022-10-04, 13:57

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.

MPrimeFinder · 2022-10-09, 02:00

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.

2022-10-02, 10:20	#35
MPrimeFinder "Anonymous" Sep 2022 finding m52 21₁₀ 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+...).