How do I convince someone that $1+1=2$ may not necessarily be true? Me and my friend were arguing over this "fact" that we all know and hold dear. However, I do know that $1+1=2$ is an axiom. That is why I beg to differ. Neither of us have the required mathematical knowledge to convince each other.
And that is why, we decided to turn to Math Stackexchange for help.
What would be stack's opinion?
 A: It seems that you and your friend lack the mathematical knowledge to handle this delicate point. What is a proof? What is an axiom? What are $1,+,2,=$?
Well, let me try and be concise about things.


*

*A proof is a short sequence of deductions from axioms and assumptions, where at every step we deduce information from our axioms, our assumptions and previously deduced sentences.

*An axiom is simply an assumption.

*$1,+,2,=$ are just letters and symbols. We usually associate $=$ with equality; that is two things are equal if and only if they are the same thing. As for $1,2,+$ we have a natural understanding of what they are but it is important to remember those are just letters which can be used elsewhere (and they are used elsewhere, often).
You want to prove to your friend that $1+1=2$, where those symbols are interpreted as they are naturally perceived. $1$ is the amount of hands attached to a healthy arm of a human being; $2$ is the number of arms attached to a healthy human being; and $+$ is the natural sense of addition. 
From the above, what you want to show, mathematically, is that if you are a healthy human being then you have exactly two hands.
But in mathematics we don't talk about hands and arms. We talk about mathematical objects. We need a suitable framework, and we need axioms to define the properties of these objects. For the sake of the natural numbers which include $1,2,+$ and so on, we can use the Peano Axioms (PA). These axioms are commonly accepted as the definition of the natural numbers in mathematics, so it makes sense to choose them.
I don't want to give a full exposition of PA, so I will only use the part I need from the axioms, the one discussing addition. We have three primary symbols in the language: $0, S, +$. And our axioms are:


*

*For every $x$ and for every $y$, $S(x)=S(y)$ if and only if $x=y$.

*For every $x$ either $x=0$ or there is some $y$ such that $x=S(y)$.

*There is no $x$ such that $S(x)=0$.

*For every $x$ and for every $y$, $x+y=y+x$.

*For every $x$, $x+0=x$.

*For every $x$ and for every $y$, $x+S(y)=S(x+y)$.


This axioms tell us that $S(x)$ is to be thought as $x+1$ (the successor of $x$), and it tells us that addition is commutative and what relations it bears with the successor function.
Now we need to define what are $1$ and $2$. Well, $1$ is a shorthand for $S(0)$ and $2$ is a shorthand for $S(1)$, or $S(S(0))$.
Finally! We can write a proof that $1+1=2$:

  
*
  
*$S(0)+S(0)=S(S(0)+0)$ (by axiom 6).
  
*$S(0)+0 = S(0)$ (by axiom 5).
  
*$S(S(0)+0) = S(S(0))$ (by the second deduction and axiom 1).
  
*$S(0)+S(0) = S(S(0))$ (from the first and third deductions).
  

And that is what we wanted to prove.

Note that the context is quite important. We are free to define the symbols to mean whatever it is we want them to mean. We can easily define a new context, and a new framework in which $1+1\neq 2$. Much like we can invent a whole new language in which Bye is a word for greeting people when you meet them, and Hi is a word for greeting people as they leave.
To see that $1+1\neq2$ in some context, simply define the following axioms:


*

*$1\neq 2$

*For every $x$ and for every $y$, $x+y=x$.


Now we can write a proof that $1+1\neq 2$:


*

*$1+1=1$ (axiom 2 applied for $x=1$).

*$1\neq 2$ (axiom 1).

*$1+1\neq 2$ (from the first and second deductions).



If you read this far, you might also be interested to read these:


*

*How would one be able to prove mathematically that $1+1 = 2$?

*What is the basis for a proof?

*How is a system of axioms different from a system of beliefs?
A: I personally would define the symbol $2$ as $1+1$. 
However, depending on what you assume, this does not ensure $2\ne 0$. What do people usually assume? Often, mathematicians work with the notion of a commutative ring. Think of the integer numbers in the following definition: A commutative ring is a mathematical structure with an addition operator $+$ and a multiplication operator $\times$, containing elements $1$ and $0$ subject to the conditions that


*

*$a + b = b + a$ and $a\times b = b\times a$

*$a + (b + c) = (a + b) + c$ and $a \times (b \times c) = (a \times b)\times c$

*$a + 0 = a$ and $a\times 1 = a$

*For each $a$ there exists $b$ with $a+b=0$, we write $b=-a$

*We have $a\times(b+c)=(a\times b) + (a\times c)$ 


Now the integer numbers clearly satisfy these conditions, and we all have a good idea what $2$ means there. However, consider $\{0,1\}=:\mathbb F_2$. It also satisfies the conditions, if we perform addition and subtraction as usual, but decree that $1+1=0$. Now you have $2=0$ inside $\mathbb F_2$. 
However, note that nothing stops you from limiting yourself to mathematical structures where $1+1\ne 0$. In fact, quite often, one limits himself to structures where $1+\cdots+1\ne 0$ no matter how many times you add $1$ to itself.
In the end, it's all just a matter of definitions. But truly, I would always say $2=1+1$ simply because the symbol $2$ should reasonably be defined as that.
A: Here's an argument to consider. Take the following three propositions:

(A) 1 + 1 = 2
(B) If there is exactly one $F$ and exactly one $G$ (and no $F$ is $G$), then there are exactly two things which are $F$-or-$G$.
(C) $(\exists x)(Fx \land \forall y(Fy \to y = x)$,  $(\exists x)(Gx \land \forall y(Gy \to y = x) \vdash$
$\quad\quad\quad(\exists x)(\exists y)(\neg x = y \land (Fx \lor Gx) \land (Fy \lor Gy) \land (\forall z)(Fz \lor Gz \to (z = x \lor z = y))$

Now, it is a vexed question what exactly the relation is between (A) and (B). But it is very plausible to say that if (A) is indeed interpreted as a statement of ordinary schoolroom arithmetic (and not as a mere de-intepreted string of formal symbols), then so understood, (A) is true iff, for any $F$, $G$, (B) is true. But (B), with numerical quantifiers, can be regimented as (C), with ordinary quantifiers. And (C) is a straight logical theorem of first-order logic.
So (C) is true as a matter of logical necessity, hence (B) is. And it looks a priori that (A)'s truth necessarily goes with the truth of (B) for any $F$, $G$. So it follows that (A), understood the ordinary way, is indeed necessarily true. too.
