Why is 1+ 1 equal to 2?

When Goldbach wrote to Euler, he put forward two conjectures: (1) Any even number greater than 2 can be decomposed into the sum of two prime numbers; (2) Any odd number greater than 5 can be decomposed into the sum of three prime numbers. Obviously, (2) it is a corollary of unity; (2) It has been proved that vinogradov, a famous mathematician in the former Soviet Union, proved that it is large enough by using the "circle method" and the "triangle sum method" created by himself. This is also the biggest breakthrough of Goldbach's conjecture so far. In the process of proving Goldbach's conjecture, a proposition is put forward: every large enough even number can be expressed as the sum of two numbers with no more than m prime factors and no more than n prime factors. This proposition is abbreviated as "m+n". Obviously, "1+ 1" is the basic proposition of Goldbach's conjecture, and the "three prime numbers theorem" is only a very important inference. In 1973, Chen Jingrun improved the "screening method" and proved that the even number "1+2" can be expressed as the sum of two numbers, one of which is a prime number, and the other is either a prime number or a product of two prime numbers. Chen Jingrun's proof result is called "Chen Theorem", which is the highest record of Goldbach's conjecture so far. The last proof is 1+ 1.

Let me show you a hypothesis:

0, 1 and 2 (such as qv. Quine, Mathematical Logic, Revised Edition. Chapter 6, section 43-44) is defined as follows:

0 := {x: x ={y: ~(y = y)}}

1:= { x:y(yεx . & amp； . x\{y}ε0)}

2:= { x:y(yεx . & amp； . x\{y}ε 1)}

For example, if we take an element from a molecule that belongs to the category of 1, then the molecule will become zero. In other words, 1 is a class composed of all classes with only one element. 〕

Now we generally use the method mainly introduced by von Neumann to define natural numbers. For example:

0:= ∧, 1:= {∧} = {0} =0∪{0},

2:= {∧,{∧}} = {0, 1} = 1∪{ 1}

[∧ is an empty set]

Generally speaking, if we construct a set n, then its successor n * is defined as n ∨{ n}.

In the general axiom system of set theory (such as ZFC), there is an axiom to ensure that this construction process can continue continuously, and all the sets obtained by this construction method can form a set. This axiom is called infinite axiom (of course, we assume that other axioms (such as union axiom) have been established.

Note: Infinite axioms are some so-called illogical axioms. It is these axioms that make some propositions of the logician school represented by Russell impossible in the strictest sense. 〕

Then we can apply the following theorem to define the addition of natural numbers.

Theorem: Life "|N" represents a set of all-natural numbers, so we can uniquely define the mapping A: | NX | n→| n to satisfy the following conditions:

(1) For any element x in |N, we have a (x, 0) = x;

(2) For any element X and Y in |N, we have A(x, y*) = A(x, y)*.

Mapping a is the mapping we use to define addition. We can rewrite the above conditions as follows:

( 1)x+0 = x； (2) x+y* = (x+y)* .

Now, we can prove that "1+ 1 = 2" is as follows:

1+ 1

= 1+0* (because 1:= 0*)

= (1+0)* (according to condition (2))

= 1* (according to the condition (1))

= 2 (because 2:= 1*)

[Note: Strictly speaking, recursion theorem should be used to ensure that the above construction method is appropriate, so I won't repeat it here. ]

1+ 1= 2 "can be said to be a" natural "conclusion drawn by human beings after introducing natural numbers and related operations. However, it was not until the19th century that mathematicians began to establish a strict logical foundation for the analysis based on real number system, and people really examined the basic problems about natural numbers. I believe that the most "classic" proof in this respect should be the one that appeared in the Principles of Mathematics written by Russell and Whitehead.

We can prove that "1+ 1 = 2":

First of all, it can be inferred that:

αε 1 (∑x)(α={x})

βε2 (∑x)(∑y)(β={x，y}。 & amp。 ~(x=y))

ξο 1+ 1(∑x)(∑y)(β= { x } ∨{ y }。 & amp。 ~(x=y))

So for any set γ, we have

γε 1+ 1

(∑x)(∑y)(γ= { x } ∨{ y }。 & amp。 ~(x=y))

(∑x)(∑y)(γ={x，y}。 & amp。 ~(x=y))

γε2

According to zermelo-fraenkel of set theory, we get 1+ 1 = 2.