I actually ran into this once not as a joke. I was using some software which had an option that would change the default base it used for all numbers. After typing <command> 16 and fiddling about in hexadecimal I tried switching back to decimal using <command> 10, yet I was still in hex mode. It took me a few more seconds than it should for me to realise what had happened.
It must be the command-line calculator bc
That's possible, it was a long time ago and I can't remember for sure
But every base is not described in the base most commonly written for human reading as "base 10"
Every base is base "one zero"; not every base is base "ten".
Except in Common Lisp, where every base is base 10, but only one base is base 10. .
I laughed at that a lot more than I should have. After a long day, thank you for that.
It can actually represent all integers just fine. It works like any other base, except whereas for example in base 10, positions are multiplied by 10^0 = 1, 10^1 = 10, 10^2 = 100 and so on, in base 1 they are multiplied by 1^0 = 1, 1^1 = 1, 1^2 = 1, and so on. The result is that each digit adds 1 to the number, like tally marks. So we have, for example
111 = 3
11 = 2
1 = 1
= 0
-1 = -1
-11 = -2
-111 = -3
That is wierd.
Base k includes numbers 0,...,k-1. For example 101 is a valid number in base 2 which is 2^2 + 2^1.
Using that logic, base 1 would only contain 0, hence it should not be able to expresss more values that 0.