Is adding an infinite number of zeroes indeterminate?

Question

Imagine adding an infinite string of zeroes be equal to some number $n$ $$0+0+0+...=n$$

Notice that, since we have an infinite amount of zeroes, the string enclosed is $n$ itself $$0+(0+0+...)=n$$$$0+(n)=n$$

Therefore substituting any number for $n$ would work and the expression $\sum 0=n$ is indeterminate.

This seems very fishy to me. How is it that by adding zeroes indefinitely, you can get to any number you want? $n$ could be anything, whether it be real or imaginary or complex, or is the expression just wrong?

Answer 1

There are two main issues with your reasoning:

Imagine adding an infinite string of zeroes be equal to some number $n$ $$0+0+0+...=n$$

We already have a problem with the first sentence. What does it mean to add something infinitely many times? You cannot just add things infinitely many times, no matter how much you want it. To solve that issue mathematicians invented series. And so given a sequence $a_n=0$ we indeed have a well defined series $\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty 0$ and that is indeed equal to $0$ by simple limit evaluation.

In other situations (who said we are dealing with numbers anyway? it could be the zero of some ring) mathematicians simply apply a convention that $0+0+\cdots = 0$ because $0$ is very special and it makes some concepts easier to write and read, while the convention doesn't break anything at the same time.

Still you cannot just add numbers infinitely many times. You always have to be careful with infinities.

This seems very fishy to me. How is it that by adding zeroes indefinitely, you can get to any number you want? $n$ could be anything

What makes you think that $n$ could be anything? While you are right that any number satisfies $0+n=n$, it is not true that any number satisfies $0+0+0+\cdots = n$.

Here's a simpler example. Say you start with a number $a=5$ and then you observe that $0+a=a$. How does the last equation imply that $a$ is something else than $5$? It does not.

The implication only goes one way, not both.

Answer 2

when you do this step, you are assuming $$(0+0+...)=(n) \rightarrow eq:[1]$$ then if you after get $$0+(n)=n$$ it should meet the previus formulas as well.

So, say n is 7. $0+(7)=7$ is valid equation, but if it doesn't meet the previous formulations 'n' would be a different 'n' from before.

to be sure you are talking about the same 'n' it should meet both equations, and only $0$ could meet them. any other number like 7 only meet the second one not the first one then although both formulas are right they are telling us the restrictions on different 'n' with the same name.