# A What if the (semi) field characteristic of N is not zero?

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1. May 26, 2017

### Cathr

By definition, the characteristic of a field is the smallest number of times one must use the ring's multiplicative identity element (1) in a sum to get the additive identity element (0). Can we use the same rule for the set of natural numbers?

If yes, I found a problem, that has something to do with the Riemann zeta function.

If we calculate the sum for an infinity of identity elements, we obtain the sum of all natural numbers: 1+(1+1)+(1+1+1)+(1+1+1+1)+..., and, if we calculate this sum using the Riemann zeta function, we obtain .

Do it 12 times and add one more multiplicative identity - what we obtain is exactly zero.

Does this make sense? Or it is wrong right from the beginning?

2. May 26, 2017

### Staff: Mentor

Why do you group them like that?
There are methods to assign a finite value to divergent series (e. g. Ramanujan summation), but that value changes if you introduce arbitrary groups like that.
Group it as 1+4+9+16+25+... and the Ramanujan sum is 0.
You cannot "do an infinite sum 12 times in a row and then plus one" in a meaningful way.

3. May 28, 2017

### Cathr

I never really understood why changing the way of grooping numbers changes their sum. For me they just have different intervals after which we visibly insert a "+", but they all contain natural numbers that go to infinity. Or is it a convention, which means that we look at each number as a term? Does it mean that 1+4+9+16+25... is "denser" than 1+2+3+4+5... and therefore it has a bigger partial sum?

4. May 28, 2017

### Staff: Mentor

It is not a proper sum.
Forget Ramanujan summation. It is an exotic method to assign a number to divergent series that you'll never use.