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If you really think about it, is there any proof that does not rely on other facts of addition or multiplication that can solve $2+2$? My question is, is it possible to prove $2+2=4$? If so, an example would be good as well...

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closed as off-topic by Antonio Vargas, William, Claude Leibovici, Andres Caicedo, Asaf Karagila Aug 14 at 12:35

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10  
Yes, see this paper for a proof. –  JimmyK4542 Aug 14 at 3:20
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For this question to make sense you must specify what you mean by $2$, what you mean by $4$, and what you mean by $+$. Before you do that, you did not actually ask a question. –  Ittay Weiss Aug 14 at 3:25
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We can rigorously construct the natural numbers, as well as define what we mean by addition. Statements like $2 + 2 = 4$ can then be proved from those definitions. You may find it helpful to read about the Peano axioms. –  manthanomen Aug 14 at 3:33
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that paper is just a joke –  Gastón Burrull Aug 14 at 3:37
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I'm not sure what this has to do with logic. It's also impossible to tell just from the current question. Such open-ended questions (without saying what is the language and the axioms being used) were discussed before in the "almost the same" 1+1=2 questions: this one and that one and generally see this list. If all those discussions still haven't answered your question, please clarify your question, and I will vote to reopen it. –  Asaf Karagila Aug 14 at 12:39

6 Answers 6

Some of the answers are confusing or even downright misleading, so let me try to set the record straight.

There are many ways of defining $\mathbb{N}.$ In the context of this question, the details aren't that important; what matters is that $\mathbb{N}$ ends up being a set equipped with a distinguished function $S : \mathbb{N} \rightarrow \mathbb{N}$ and a distinguished element $0 \in \mathbb{N}$ subject to a theorem that says "definitions by recursion work." This allows us to prove the existence and uniqueness of a binary operation $+$ on $\mathbb{N}$ satisfying the following specifications. $$n+0 = n, \quad n+S(m) = S(n+m)$$

Now write $4$ as shorthand for $S(S(S(S(0))))$ and write $2$ as shorthand for $S(S(0)).$ Then we have

$$2+2 = 2+S(S(0)) = S(2+S(0)) = S(S(2+0)) = S(S(2)) = S(S(S(S(0)))) = 4$$

Extra Information.

For completeness, here's several ways of defining the naturals.

  1. The algebraic structure $\mathbb{N}$ can be defined as the sole (up to unique isomorphism) model of the Peano Postulates (which are second order).

  2. It can also be defined as the free monounary algebra generated by the singleton set $\{0\}$ (I suggest googling this term if you do not know it).

  3. Set theorists like defining it as the least set $\omega$ such that firstly, $\emptyset \in \omega,$ and secondly, $x \in \omega$ implies $x \cup \{x\} \in \omega$. The entity $\emptyset$ ends up being our $0$, and the function $x \mapsto x \cup \{x\}$ ends up being our successor function.

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You could simply define $+$ as you suggest (as did Peano in the original version of his axioms) or you can prove the existence a unique binary function $+$ with the required properties using the rules of logic and set theory if you have them available. –  Dan Christensen Aug 14 at 4:58
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While this is a good answer in itself, I think it's probably incomprehensible to the OP, who in particular may not know the words "isomorphism," "second order," or "recursion" or be able to read explanations of what a free monounary algebra is. –  Kevin Carlson Aug 14 at 5:14
    
@KevinCarlson, you're right. I've reordered the information to emphasize what really matters. –  goblin Aug 14 at 5:20
    
@KevinCarlson Not to mention the use of notation. I hope the OP knows what $\mathbb{N}.$ means and what $S : \mathbb{N} \rightarrow \mathbb{N}$ means. –  trlkly Aug 14 at 10:13

The usual approach for formally proving that $2+2=4$ is to start from Peano's axioms (which define the set $N$ of natural numbers , $0\in N$ and a successor function on $N$). Using these axioms, along with the rules of logic and set theory, you can formally prove that there exists a unique binary function $+$ such that

$x+0 = x$

$x+(y+1) = (x+y)+1$

where $1$ is the successor of $0$, and $n+1$ is the successor of $n$.

This is a long and tedious process. (Earlier versions of Peano's axioms gave you the above definition to start.)

Then you define 2, 3 and 4 such that

$2=1+1$

$3=2+1$

$4=3+1$

Then you have $2+2=2+(1+1)=(2+1)+1=3+1=4$

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By "Peano's axioms" I mean the usual five axioms without $+$, i.e. not the first-order version which is more like Peano's original version of his axioms. I think you mean $1=S(0)$. Perhaps this is the source of your confusion. –  Dan Christensen Aug 14 at 4:36
    
You raise a legitimate point. I did leave out a few details for ease of reading, e.g. the implicit universal quantifiers, and the successor of $x$ being $x+1$, but I believe my definition is essentially correct. –  Dan Christensen Aug 14 at 4:50
    
This was not meant to be a formal proof. (I can supply one if you like, but it is several hundred lines long.) In the context of the OP, it is a stretch to say that my definition is somehow "misleading." –  Dan Christensen Aug 14 at 5:06

We assume the Peano axioms. Specifically:

  1. Zero is a number.
  2. If a is a number, the successor of a is a number.
    (We denote the successor of $x$ as $x'$.)
  3. Zero is not the successor of a number.
  4. Two numbers of which the successors are equal are themselves equal.
  5. (induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

We then define addition recursively as follows:

$$a+0 = a$$ $$a+b' = (a+b)'$$

Now, we will name some numbers. We will denote:

$$\begin{align} 0' &= 1 \\ 1' &= 2 \\ 2' &= 3 \\ 3' &= 4 \end{align}$$

We could keep going, but we only need to be able to denote the numbers $0$ through $4$ (inclusive).

Now, showing that $2+2 = 4$ is a simple application of the recursive formula for addition: $$\begin{align} 2 + 2 &= 2 + 1' \\ &= (2+1)' \\ &= (2+0')' \\ &= ((2+0)')' \\ &= ((2)')' \\ &= 3'\\ &= 4 \end{align}$$

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1  
While (5) is true, it is not quite definitive. I suggest phrasing it as (5*) "if $P(0)$ and $P(x) \rightarrow P(x')$ then $\forall x : P(x)$. The difference can be seen by considering the following structure, let the domain $D$ be $\mathbb{N} \cup \{A, B, C\}$ and successor be extended by defining for $n$ in $\mathbb N$, $n* = n'$, also $A* = B, B* = C, C* = A$. $\,\{\mathbb N, '\}$ satisfies both (5) and (5*), but $\{D, *\}$ satisfies (5) and not (5*) by taking $P(x)$ to be $x \in \mathbb N$. Besides functionality, a purpose of the inductive axiom is to exclude structures like the above. –  DanielV Aug 14 at 7:55

The proposition "2+2 = 4" is a theorem of the Peano arithmetic (the five Peano's axioms). For example: $$1 := 0',$$ $$2 := 1' = 0'',$$ $$3 := 2' = (1')' = 0''',$$ and so on.

Yes, it depends on to what meanings we assign "+" and the numerals.

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The most commonly used method to define sum in $\mathbb{N}$ is derived from Peano Axioms.

$0\in\mathbb{N}$ and $s:\mathbb{N}\to \mathbb{N}\setminus \{0\}$ is a given bijection, we can define $1:=s(0)$, $2:=s(1)$ and so on...

We can define a sum: $n+m:=s^{m}(n)$. In that case $2+2=s(s(2))=s(3)=4$. This sum satisfies the properties which we are accustomed.

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well this kind of proving depands on a set of axiom system of group like Z for example . and these are axioms that been used to prove that 2+2=4

Axiom 1. Algebraic Properties of Z (Properties of + and .)

Properties of Addition

  • A1. Associativity. For every x, y, and z in R, ( x + y ) + z = x + ( y + z ).

  • A2. Commutativity. For every x and y in R, x + y = y + x.

  • A3. Identity. R contains an additive identity, 0, such that for every x in R,
    x+0=x.

  • A4. Additive Inverses. For every x in R, there is an additive inverse, (-x ), in R such that x + (-x ) = 0.

Axiom 2. Order Properties of Z (Properties of <)

  • (i ) Transitive Property. For every a, b, and c in Z, if a < b and b < c, then a < c.

    (ii ) Trichotomy Property. For every a and b in Z, exactly one of the following holds: a =b, a < b, or b < a.

    (iii ) Additive Property. For every a, b, and c in Z, if a < b, then a + c < b + c.

    (iv) Multiplicative Property. For every a, b, and c in Z, if a < b and 0 < c, then ac < bc.

    (v) Order of Identities. 0 < 1.

Axiom 3. Th e Well-Ordering Principle

  • For any integer n, there is a next integer n + 1 that comes immediately after it, with no other integers in between.

with these sets of axioms you can easily prove now that 2+2=4. just folow these steps

1+1=2

(1+1) + 1 =2+1 =3

(1+1+1)+1=3+1=4

hence

(1+1) +(1+1 )=2 +2 =4 done.

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