I think a more interesting question is, given the set of all positive rationals from 0 to 1, say that the number of the elements in this set is n. Then what is 1/n of the sum of all of them? I don't have an answer to that one yet, I'll think about it when I get the time. I know that (1/n) of the sum 1 + 1/2 + 1/3 + ... + 1/n has the limiting value of 0 as n -> ∞.
–
Michael Mendrin
·
3 years, 5 months ago

Log in to reply

@Michael Mendrin
–
Wouldn't this be 1/2 ? We could pair each rational number x (other than 1/2) with (1 - x); then the sum would be lim(n->infinity)[(1/n)(((n-1)/2) + (1/2))] = lim(n->infinity)((1/n)(n/2)) = 1/2.
–
Brian Charlesworth
·
3 years, 5 months ago

Log in to reply

Sure they can all be added up. It's an infinite sum. See harmonic sums, like 1/2 + 1/3 + 1/4 + ... etc. Of course, Gaussian rationals can't be included in this sum because "less than 1" has no meaning for complex numbers.
–
Michael Mendrin
·
3 years, 5 months ago

## Comments

Sort by:

TopNewestI think a more interesting question is, given the set of all positive rationals from 0 to 1, say that the number of the elements in this set is n. Then what is 1/n of the sum of all of them? I don't have an answer to that one yet, I'll think about it when I get the time. I know that (1/n) of the sum 1 + 1/2 + 1/3 + ... + 1/n has the limiting value of 0 as n -> ∞. – Michael Mendrin · 3 years, 5 months ago

Log in to reply

(((n-1)/2) + (1/2))] = lim(n->infinity)((1/n)(n/2)) = 1/2. – Brian Charlesworth · 3 years, 5 months agoLog in to reply

Sure they can all be added up. It's an infinite sum. See harmonic sums, like 1/2 + 1/3 + 1/4 + ... etc. Of course, Gaussian rationals can't be included in this sum because "less than 1" has no meaning for complex numbers. – Michael Mendrin · 3 years, 5 months ago

Log in to reply

– Maharnab Mitra · 3 years, 5 months ago

But how?Log in to reply