Is there a way to prove that $2+2$ really equals $4$?

Question

In elementary school, one learns that $2+2=4$ by experiment (putting two apples next to two other apples), and maybe also from some addition table to be memorized.

But is there any approach that proves $2+2 = 4$? If so, an example of such a proof would be good.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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: $$\begin{align} 2 + 2 &= 2 + 1' \\ &= (2+1)' \\ &= (2+0')' \\ &= ((2+0)')' \\ &= ((2)')' \\ &= 3'\\ &= 4 \end{align}$$

Answer 5

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.

Answer 6

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.