A more complex question has rarely been asked.

Principia Mathematica took nearly a thousand pages to prove that 1+1=2. It does meander a bit, but had they wanted to prove 1+1=2 alone, it could have done so in 500 pages.

Mathematically speaking, the definition of 1 is:

- There exists a number such that when multiplied upon an element of a specified set, yields the element of the specified set.

It is also defined as:

- 1.0000000000000000000000…
- .9999999999999999999999999…
- as the set of all singletons.
- a singleton is a set with exactly 1 element.

These 4 definitions work in tandem with one another.

For example:

1 = 1 ” id=”MathJax-Element-7-Frame” role=”presentation” tabindex=”0″>

$1=1$Divide both sides by 3.

1 / 3 = 1 / 3 ” id=”MathJax-Element-8-Frame” role=”presentation” tabindex=”0″>

$1/3=1/3$Rewrite.

1 / 3 = .33333333333333333 . . . ” id=”MathJax-Element-9-Frame” role=”presentation” tabindex=”0″>

$1/3=.33333333333333333...$Multiply both sides by 3.

1 = .9999999999999999999 . . . ” id=”MathJax-Element-10-Frame” role=”presentation” tabindex=”0″>

$1=.9999999999999999999...$Similarly:

If

x ” id=”MathJax-Element-11-Frame” role=”presentation” tabindex=”0″>

$x$= .9999999999999999999 . . . ” id=”MathJax-Element-12-Frame” role=”presentation” tabindex=”0″>

$=.9999999999999999999...$10 x = 9.99999999999999999… ” id=”MathJax-Element-13-Frame” role=”presentation” tabindex=”0″>

$10x=\mathrm{9.99999999999999999\dots}$10 x = 9 + .99999999999999 . . . ” id=”MathJax-Element-14-Frame” role=”presentation” tabindex=”0″>

$10x=9+.99999999999999...$10 x = 9 + x ” id=”MathJax-Element-15-Frame” role=”presentation” tabindex=”0″>

$10x=9+x$Simplify by subtracting x from both sides.

9 x = 9 ” id=”MathJax-Element-16-Frame” role=”presentation” tabindex=”0″>

$9x=9$x = 1 ” id=”MathJax-Element-17-Frame” role=”presentation” tabindex=”0″>

$x=1$.99999999999999999999 . . . = 1 ” id=”MathJax-Element-18-Frame” role=”presentation” tabindex=”0″>

$.99999999999999999999...=1$As the set of all singletons, 1 is also THE element that represents the set of all single entities.

That is to say: if you have 7 erasers. What you really have is a set of 7 single entities. The definition of 7 becomes: 1 + 1 + 1 + 1 + 1 + 1 + 1; and not as is commonly believed as: 6 + 1.

- There is an argument for 7 to be defined as 6 + 1, but this argument is a corollary of the Peano Axioms which in turn argues that there exists a set with absolutely nothing in it {} and a set with exactly something in it {x}. More on this later.

The Principia Mathematica uses Peano’s (from the Peano Axioms mentioned earlier) work and notation to expertly slice through the many nuances pertaining to this question.

This is something we will not do; but hopefully, we will also be able to effectively demonstrate why 1 + 1 = 2 in less than 1000 pages.

We will assume these basic principles of number theory:

- There exists a number such that when multiplied to an element of a specific set, yields that element of the specific set.
- There exists a number such that when added to an element of a specific set, yields that element of the specific set.

If we again assume to have only two sets, a set that is empty: {} containing no elements, and a set that is not empty {x} containing an element. We realize that Consequently, we went from nothing {}, to something {x}. This means that {x} is the successor to {}, as the next step up from nothing, is something.

As such we now have two elements:

Nothing, {}, and something that comes after {}, this something is called the successor, and it is the Successor of nothing.

in written notation we have:

{} and { the Successor of nothing }

Rewritten:

{0, the thing that comes after 0}

Further reworded:

{0, Successor (0) }

Reduced further:

0 , S ( 0 ) ” id=”MathJax-Element-19-Frame” role=”presentation” tabindex=”0″>

$0,S(0)$Where S(0) stands in place of ‘the successor’. Further, we know there are an infinite number of possible Natural numbers, and we get:

{0, Successor of 0, the successor of the successor of 0, the successor of the successor of the successor of 0,…}

Further reduced:

0 , S ( 0 ) , S ( S ( 0 ) ) , S ( S ( S ( 0 ) ) ) , S ( S ( S ( S ( 0 ) ) ) ) , S ( S ( S ( S ( S ( 0 ) ) ) ) , … ” id=”MathJax-Element-20-Frame” role=”presentation” tabindex=”0″>

$0,S(0),S(S(0)),S(S(S(0))),S(S(S(S(0)))),S(S(S(S(S(0)))),\dots $Further explained:

We know that we had nothing, and added something to it, and got something:

Nothing + Something = Successor of nothing.

0 + _ _ = S ( 0 ) ” id=”MathJax-Element-21-Frame” role=”presentation” tabindex=”0″>

$0+\mathrm{\_}\mathrm{\_}=S(0)$We also know that there is nothing closer to 0, than the thing that comes after 0.

0 + S ( 0 ) = S ( 0 ) ” id=”MathJax-Element-22-Frame” role=”presentation” tabindex=”0″>

$0+S(0)=S(0)$This implies that S(0) is the smallest increment possible from natural number to next natural number.

As a consequence, we now have two discovered entities: Something, and Nothing.

Let’s give them names.

We have decided that

- Nothing = 0 .
- 0 = Nothing.
- S(0) is the something that comes after nothing.
- We define a new symbol: 1, to be: 1 = S(0)
- This is to say that 1 IS the symbol that succeeds 0;
- We could have drawn any shape to define the number that succeeds 0; we chose to draw a 1.

0 + S ( 0 ) = S ( 0 ) ” id=”MathJax-Element-23-Frame” role=”presentation” tabindex=”0″>

$0+S(0)=S(0)$0 + 1 = S ( 0 ) ” id=”MathJax-Element-24-Frame” role=”presentation” tabindex=”0″>

$0+1=S(0)$0 + 1 = 1 ” id=”MathJax-Element-25-Frame” role=”presentation” tabindex=”0″>

$0+1=1$0 , 1 , S ( S ( 0 ) ) , S ( S ( S ( 0 ) ) ) , S ( S ( S ( S ( 0 ) ) ) ) , S ( S ( S ( S ( S ( 0 ) ) ) ) , … ” id=”MathJax-Element-26-Frame” role=”presentation” tabindex=”0″>

$0,1,S(S(0)),S(S(S(0))),S(S(S(S(0)))),S(S(S(S(S(0)))),\dots $We now have definitions for 0, and 1. What about a definition for the thing that comes after one? The successor of 1?

As we know S(0) is the smallest increment available, and we are interested in finding S(0)’s successor we investigate:

The successor to the successor of Nothing:

0 + S ( 0 ) = 1 ; 1 + S ( 0 ) = S ( 1 ) ” id=”MathJax-Element-27-Frame” role=”presentation” tabindex=”0″>

$0+S(0)=1;1+S(0)=S(1)$This reads:

The successor of the successor of nothing IS the successor of one

And now… we need a new symbol.

We define the

S ( 1 ) = 2 ” id=”MathJax-Element-28-Frame” role=”presentation” tabindex=”0″>

$S(1)=2$The successor of 1 IS 2.

Thus:

0 + S ( 0 ) = 1 ; 1 + S ( 0 ) = S ( 1 ) = 2 ” id=”MathJax-Element-29-Frame” role=”presentation” tabindex=”0″>

$0+S(0)=1;1+S(0)=S(1)=2$Simplify:

0 + 1 = 1 ; 1 + 1 = S ( 1 ) ; S ( 1 ) = 2. ” id=”MathJax-Element-30-Frame” role=”presentation” tabindex=”0″>

$0+1=1;1+1=S(1);S(1)=2.$Further:

0 + 1 = 1 ; 1 + 1 = 2 ; 2 = 2. ” id=”MathJax-Element-31-Frame” role=”presentation” tabindex=”0″>

$0+1=1;1+1=2;2=2.$1 has many different properties; but all of the properties and their resulting definitions have little to do with why 1 + 1 = 2. And that 1 + 1 = 2 is a byproduct of properties inherent to Natural numbers.