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by adenadel 3206 days ago
This is equivalent to asking what is the average number of digits necessary to represent an arbitrary natural number. If you assume that this is a finite number you can quickly arrive at a contradiction.
1 comments

So there are natural numbers with an infinite number of digits and that is my point. The average can't be larger than the largest number.
The only way to reason about this is as a limit.

lim n->infty sum(1..n)/n.

This limit diverges. Your intuition is breaking down. It might be useful for your understanding to study real analysis where these sorts of questions are handled rigorously.

> necessary to represent an arbitrary natural number

What makes you think this number exists? Infinities can not be added and divided this way.

> The average can't be larger than the largest number.

But it can be equal to it. You mistake is not realizing that infinity minus 1 = infinity.

So the average is "smaller" than the largest, and yet equal to it. Because like I said at the start, you can not just add infinities in the normal way you can manipulate finite numbers.