Is $1+1 = 2$ true in any base? So this may be one of the stupidest questions ever, but is the identity
$$1 + 1 = 2$$
valid in any base?
It seems so, since it's like, for a given base $b$
$$1 = 000\ldots 01 = (0\cdot b^n) + \ldots (1\cdot b^0) = b^0 = 1$$
Which means $1 + 1 = 2$ no matter what base we use.
Is that right? Or is there something I miss? 
 A: It is always valid because $1+1$ is the definition of $2$.  It's not a theorem, it's what we mean by the symbol $2$ to begin with.
A: There are three forces at play here.
First comes our natural understanding of numbers. $2$ is a specific natural, or real, number. We understand it intuitively, and unambiguously. This is the abstraction of the idea of having two apples, two sons, two cats, or two examples for collections with two objects. As such, when we think of the number $2$, it is literally defined to be $1+1$.
Then comes the representation of a number in a particular base. This is now a question of how we dress the abstract number into a slightly more tangible form. Here $2$ is a relatively "bad" example, as most [natural-]bases are large enough so $2$ is just $2$ again. But think of the number ten. In decimal, this is $10$; in trenary this is $101$; in octal, $12$. All of these are different strings which represent the same number. How is this possible? Well, when we change base, we change how we interpret the string of symbols.
And finally comes the purely syntactical evaluation of symbols. This takes the last point and pushes it to $11$.(1) We forget that the symbols even represent the abstract quantity of how many hands a "standard" person should have. We only know that $1$, $2$ and $+$ are symbols in a mathematical language. As such we are in fact allowed to interpret them to mean pretty much anything that we want. You see this with $\pi$, which can denote a homomorphism, a projection, a function, a constant real number; but you see it less often with $2$ or $+$. Nevertheless, we are allowed to interpret those symbols to our liking, and we can easily concoct interpretations where $1+1$ and $2$ are not the same object.
In short, if you consider them as abstract numbers, they are independent of their representations, and then $1+1=2$, always. If you are asking about the interpretation of the symbols as numbers, then the answer is a qualified "usually". If you are asking in general about interpreting the symbols $1,2$ and $+$... then this is a resounding "not enough context".

(1) This can be taken in any base you prefer. Even in base $\omega$.
A: Yes, this is valid in any base in which $1$ and $2$ are both digits (so, with the standard conventions, any base except base $2$).  More generally, a single digit always represents the same number no matter what base you consider it in (as long as it is a valid digit in that base).  So for instance, $3+4=7$ is valid when interpreted in any base (as long as the base is at least $8$, so these are all digits in the base).
To be more precise, we should be clear to distinguish numbers from the sequences of digits we might use to represent them.  Standard notation unfortunately does not make this very clear.  When we write $1+1=2$ normally, what we really mean is "the sum of the number represented by $1$ in decimal notation and the number represented by $1$ in decimal notation is the number represented by $2$ in decimal notation".  So what "$1+1=2$ is valid in any base" really means is "for any base $b>2$, the sum of the number represented by $1$ in base $b$ notation and the number represented by $1$ in base $b$ notation is the number represented by $2$ in base $b$ notation."  This is because, as mentioned above, "the number represented by $1$ in base $b$ notation" is the exact same number as "the number represented by $1$ in decimal notation", and similarly for $2$.
A: ==== 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.
A: No. There are lots of bases in which your expression does not make sense, e.g.


*

*Base 1
┌───────────┬───────────┐
│    Base 1 │   Base 10 │
├───────────┼───────────┤
│         1 │         1 │
│        11 │         2 │
│       111 │         3 │
│      1111 │         4 │
│       ... │       ... │
│    1+1=11 │     1+1=2 │
│     1+1=2 │     Error │ Illegal character "2"
└───────────┴───────────┘


*Base 2
┌───────────┬───────────┐
│    Base 2 │   Base 10 │
├───────────┼───────────┤
│         1 │         1 │
│        10 │         2 │
│        11 │         3 │
│       100 │         4 │
│       ... │       ... │
│    1+1=10 │     1+1=2 │
│     1+1=2 │     Error │ Illegal character "2"
└───────────┴───────────┘


*Base φ in standard form (non-unique)
┌───────────┬───────────┐
│    Base φ │   Base 10 │
├───────────┼───────────┤
│         1 │         1 │
│     10.01 │         2 │
│    100.01 │         3 │
│    101.01 │         4 │
│       ... │       ... │
│ 1+1=10.01 │     1+1=2 │
│     1+1=2 │     Error │ Illegal character "2"
└───────────┴───────────┘

Even when restricting to bases which contain the digit "2", it's still not true.
Digits are only representations, and they can represent any value. We usually use base 10 with the digits "0", "1", "2", "3", "4", "5", "6", "7", "8"and "9". So when working with other bases, we reuse these digits if possible, or we borrow latin alphabet letters if we need more, in order to avoid confusion.
But I can define a base 5 positional system with the digits "0", "2", "$", "1", "€". Let's call it "Oriol's system".
┌───────────┬───────────┐
│     Oriol │   Base 10 │
├───────────┼───────────┤
│         2 │         1 │
│         $ │         2 │
│         1 │         3 │
│         € │         4 │
│       ... │       ... │
│     2+2=$ │     1+1=2 │
│     1+1=2 │     3+3=1 │
└───────────┴───────────┘

In that system, the expression 1+1=2 makes sense, but it's false.
Your statement is only true in positional bases whose representations contain the digits "1" and "2", and their meaning is the same as in the decimal system.
A: Integers have their own existence, separate from how we may choose to represent them.
The statement
"(the integer referred to by the decimal symbol 1) + (the integer referred to by the decimal symbol 1) = (the integer referred to by the decimal symbol 2)"
is indeed always true. Whether or not this is expressed in symbols as
"1 + 1 =  2"
depends on how you choose to represent integers.
A: If you put one egg in an empty basket and then put yet an other egg in the same basket, then you have 10 eggs in the basket if the base is 2 or 2 eggs in the basket if the base is greater than 2.
A: Well, the most common definition of $2$ is "the integer that follows $1$". Since the string $a$ for any digit $a$ represents the number $a$, we get that $2=1+1$ no matter the base.
Note that this clearly shows that the definition of $2$ is "algebraic", because you can extend this not only to bases like $2$, $10$, $-5$, $1.875$ or $3+\mathrm{i}$, but also to things like $X+1$ in the ring $(\mathbb{Z}/3\mathbb{Z})[X]$ and similar beasts (yes, you can try to expand polynomials modulo $3$ as $\sum a_i (X+1)^i$ with $a_i$ in a finite set, the question is whether you'll be successful; it should work with $a_i\in\{0,1,2\}$).

Note that one can choose any base and any digit set for the base and start expressing numbers. It does not mean that every combination of a base and a digit set is a good one, since quite often not every real number has a representation, but heck, even in the standard binary system only non-negative numbers can be represented, we need the sign for the negative numbers. So yes, from the generic point of view, $2$ is fine as a digit for the binary system.
A: That's a good question! According to the rules of arithmetic, 1 + 1 = 2 is a true statement about numbers. 
Therefore, any reasonable way of representing numbers—whether base 2 or base 16 or base 10— should allow you to express that true statement.
On the other hand, maybe you're asking about symbols rather than numbers: maybe you're asking whether the symbolic expression 1+1 = 2 is true in any base $b$ (where we interpret 1 and 2 as symbols in base $b$). 
The answer is yes: the statement is true in any base $b>2$. The reason is that for any base $b>2$, the symbol 2 is a meaningful symbol in base $b$; it refers to $2\cdot b^0$. And we have that $b^0 + b^0 = 2\cdot b^0$ by arithmetic— hence the symbolic expression 1 + 1  = 2 is true in any base $b>2$.
