What is $1 + 2$? The question may seem dumb but how can one prove the answer? I heard there is a proof but don't know where to find it so please help. Thanks in advance.

  • $\begingroup$ If you want to prove something so basic, it's important that you are aware of the exact definitions you are using. So, how do you define "+"? For that matter, how do you define $1,2$ and $3$? $\endgroup$
    – Arthur
    Jun 13, 2015 at 18:04
  • $\begingroup$ To expand on what Arthur is saying, to define the symbol "$+$", you first need to define the symbols "$1$" and "$2$". $\endgroup$ Jun 13, 2015 at 18:06
  • $\begingroup$ See en.wikipedia.org/wiki/Peano_axioms#Addition . $\endgroup$
    – Crostul
    Jun 13, 2015 at 18:07
  • $\begingroup$ Addition is repeated incrementation. $\endgroup$
    – Lucian
    Jun 13, 2015 at 18:55
  • $\begingroup$ You can consider the following definitions as axioms of our numeration system: $2:=1+1$ and $3:=1+2$. ($1+$ can be phrased as "is the successor of".) So there is nothing to be proven. $\endgroup$
    – user65203
    Aug 5, 2016 at 21:22

5 Answers 5


It really depends on what you are calling a proof. The idea is quite universal, but the means to prove things may differ. For example, it took $362$ pages in Russell's Principia Mathematica to prove that $1+1=2$. In this case, they expressed in a quite different background and they also went a little deeper than what I'm about to do here.

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In modern books, we usually use the construction of natural numbers with Peano axioms or systems, which are essentially the same ideas that are on Peano's axioms but expressed in a different form: I've learned this very neat form with a professor of mine, I'll write it very informally so you can get the idea. If you need a deeper understanding, you can consult books on the subject (there is an awesome reference in the end of the answer).

  1. There is a successor function that takes an element on $N$ and connects it on another element of $N$. I'll denote it with $S(n)$. This function is injective. (Curious question: Why does it need to be an injection?)
  2. $0$ is an element on $N$ and it's not the sucessor of any number.
  3. Principle of finite induction: Take a subset $X$ of $N$ and assume $0\in X$, if for every $n$ of $N$ $n\in X$ implies that $S(n)$ is in $X$, then $X=N$.

Sometimes, $3$ seems silly, but it's a very powerful rule that helps you to prove that if something works in a certain pattern, it will work no matter what element of $N$ you are choosing. If you study mathematics closely, you'll notice that there are a lot of things that do seem silly: This is because mathematicians usually choose phrases with profound economy, things must be expressed in less sentences and cover a broad ammount of mathematics. With enough years of mathematics, it's amazing to notice that the silly sentences express truths, sometimes different truths in very different mathematical ideas.

With the given laws I wrote, you "have" the system of natural numbers. But there are a lot of things that need to be proved in order for it to be the same system of natural numbers we're used to. I'll skip all these proofs. So, you take $0$ and (obviously) call it $0$, you take $S(0)$ and call it $1$ and so on. Addition can be defined as follows: Take $m$'s and $n$'s and then you have.

$$\begin{eqnarray*} {m+0}&:=&{m} \\ {m+S(n)}&:=&{S(m+n)} \end{eqnarray*}$$

Example of proof with the rules I gave: $m+0=0+m=m$

At this stage, there are things that seem evident in everyday arithmetic, like commutativity. You may be tempted to say that $m+0=0+m=0$ but that is actually something that needs to be proved and it can be proved only with what I wrote until now, I'll give you a sample proof of this fact. Try to see $m$ as a slot, and whatever you put there, it goes out in the other side of the equality, it doesn't mean that you can commute them. The proof of this very fact is given as follows:

  • Using that first line, you can take $m=0$ and then $0$ goes out in the other side:


  • Using the second line, you can write:


  • And using the result on $(a)$, we can switch the $0+0$ for $0$:


  • Now we need to do:


  • And again, using the result we had on $(b)$, we can switch $0+S(0)$ by $S(0)$:


Do you see that we can continue doing this symbolic manipulation ad infinitum? This is exactly the pattern that I mentioned when I spoke about the principle of finite induction. We can keep switching the inside of the $S(n)$ by what we had before and then, this results holds by induction. We have that $m+0=0+m=0$, we know that the addition of a number with zero commutes. Notice that we still don't know if the addition of numbers different of $0$ commutes, that's something that still needs to be proved.

Proving that $1+2=3$.

There are a lot of things that still need to be proved, for example: We would need to prove that the addition of a fixed number with another number has a unique result - How do you know that $m+n=c$ and $c$ is a unique number? How do you know that this sum couldn't give you two answers? It's also possible to prove that with the rules I've written here. I just written these examples and spoke about what needs to be proved for you to feel the drama that is to prove things rigorously. I've given some axioms and then the definition of the addition, to obtain the result that you're looking for, we need the following:


  • From the laws I wrote:


  • And then, we need to go backwards and check that:


  • Remember that $m+0:=0$, that's exactly the result we need here:


  • Now, using this result, we can rewrite:


From this, it follows that $1+2=3$. But from this information alone, you can't say that $2+1=3$ because you still don't know if the addition commutes (here know means that you have deduced something from the rules I wrote). You still need to prove that to prove that $1+2=2+1=3$.

If you study deeper, you'll be able to prove commutativity in general, that is $m+n=n+m$ and then you'll not be bothered to prove every result individually. At the end of the day, this may seem overwhelming, but as I said earlier, do you see that it's possible to build the entire concept of addition with only those two lines plus the axioms for a peano system? For someone starting, it may seem complicated, but as I've shown it's just about symbolic manipulation (everything I did was to switch combinations of symbols for other combinations of symbols).


The Real Numbers: An introduction to set theory and analysis - Stillwell, John $\quad \quad $ On this book you may find about what I wrote, historical facts and much more. It's a very lively read, completely different of the usual dry books in mathematics that expect you to figure it out alone. On the same book, you may found a lot of references to similar books on this subject. In fact, he opens the book speaking about Landau's Foundations of Analysis.

Comprehensive Mathematics for Computer Scientists - Mazzola, Guerino; Milmeister, Gérard; Weissmann, Jody $\quad \quad $ Another interesting fact is that there are different constructions of the natural numbers and definitions of addition, you can see a set theoretic construction in this book. It is very similar to Von Neumann's construction of the numbers, in this case it uses a set theoretic framework. See the link for a brief explanation on them.


The simplest bit of arithmetic that can be proved is that $2+2=4$

the number $1$ is assumed to exist, then the following are definitions of the symbols $2, 3 \text{ and } 4 $

$$ 1+1 \equiv 2$$ $$ 1+2=2+1 \equiv 3$$ $$ 1+3=3+1 \equiv 4$$

now we can prove

$$ 2+2 = 2 + (1 + 1) = (2+1) + 1 = 3+1 =4 $$


The Peano axioms can be augmented with the operations of addition and multiplication and the usual total (linear) ordering on N. The respective functions and relations are constructed in second-order logic, and are shown to be unique using the Peano axioms. Addition

Addition is a function that maps two natural numbers (two elements of N) to another one. It is defined recursively as:

a+0=a a+s(b)=s(a+b)

For example,

a + 1 = a + S(0) = S(a + 0) = S(a).

The structure (N, +) is a commutative semigroup with identity element 0. (N, +) is also a cancellative magma, and thus embeddable in a group. The smallest group embedding N is the integers.


This (and more) is all explained in excruciating detail in Landau's "Foundations of Analysis".

You can buy it here:



One: o. Two: oo. Both together: ooo. Those are three. (This is of course essentially the same answer as Voyska's, but I wanted to put it slightly less formally.)


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