==== better shorter answer ====

It's not true for base 2 because "2" doesn't exist in base 2.

But changing bases doesn't change what the numbers *are*. It only changes how we represent them. So the real question is does "2" always represent the same number that "1 + 1" represents?

And the answer to that is: Yes, so long as the base is greater than 2 so that it *has* the digit "2".

Any base N will have N digits: {0, 1, 2,........, (N-1)}. "0" is a "null place holder" (sort of, it's a little more complicated than that) and {1,2,3.....,(N-1)} represent the first of the natural numbers. So "2" is always the number we know as 2 if "2" exists. "3" is always the number we know as 3 if "3" exists.... "9" is always the number we know as 9 if "9" exisits. And "A" is always the number we know as 10 if "A" exists" (which it *doesn't* in base 10; ... or should I say "base A"?... It *does* exist in base 11 which has the digits {0,1,2,3,4,5,6,7,8,9,A} which represent the null place holder and the first "ten" natural numbers).

So if "2" and "1" exist. "1 + 1" and "2" always represent the number we know as 2 and the sum we know as 1 + 1.

===== third answer =====

$a + b = c$ where $a,b$ and $c$ are single digit numbers and the base of our numbering system is greater than $c$ will always be true (assuming, of course, that $a + b$ *does* equal $c$).

Whereas if the number system is $N \le c$ or less. $a + b = 1M$ where $M$ is the symbol for $c - N$.

The reason is because digits of a base N system are {0,1,....,N-1} and represent those natural numbers. If $a + b < N$ there *will* be a digit to represent it. If $a+b \ge N$ there won't be.

So for, example $4 + 3 =7$ for all bases that *have* a "$7$". i.e. all bases greater than 7.

So $1+1=2$ is true for all bases that have a "$2$", that is, all bases greater than $2$.

Or to put it a different way:

If $a + b = c< N$ then in base N, $a + b = a*N^0 + b*N^0 = cN^0 = c$. As $c < N$ there is a digit representing c$.

If $a + b = c \ge N; 0 \le a < N; 0 \le b < N$ then in base N $a + b = a*N^0 + b*N^0 = (a+b)*N^0 = c*N^0 = (N + (c - N))*N^0 = N^1 + (c-N)*N^0$. Note that $0 \le c-N < N$ so there is a digit $M = c-N$. So $N^1 + (c-N)*N^0 = 1*N^1 + M*N^0 = 1M$.

==== old answer ====

Um.... 1 and 2 are symbols and mean nothing by themselves.

When we do base arithmetic we share symbols for digits.

Base1 = has one digit-symbol {1}

Base2 = has two digit-symbols {0,1}

Base3 = has three {0,1,2}

Base 4 = has four {0,1,2,3}

...

And so on.

For all bases the symbol "1" means the "base single number.

For all bases that contain the symbol "2", "2" means the "number after 1".

For all bases that contain the symbol "3", "3" means the "number after 2".

And so on.

If "2" is a symbol in the base then, yes, "2" = "the number after the base number" = "the base number plus the base number" = "1+1". That will always be true.

But Base 1 and Base 2 do not have the symbol "2".

In base 2, 1+ 1 = 10.

In base 1, 1 + 1 = 11.

But for all bases > 2 which *do* have the symbol "2". 1 + 1 = 2.