# Why is 1+1=2 ?

May 13, 2017

See explanation:

#### Explanation:

Imagine a number line:

You have from $0 - 10$, and you start with $1$. You then $+ 1$, which means to move up one step in the number line. And thus, we have $1 + 1 = 2$

All the best!

May 13, 2017

It depends...

#### Explanation:

It depends on how you start, but one basic answer is that $2$ is the name we give to the number $1 + 1$.

It is much more convenient to have distinct names for numbers than to have to write $1 + 1$ or $1 + 1 + 1 + 1 + 1$, etc.

It really comes down to how you define the concept of numbers at all. Many different approaches have been used, probably the most notorious of which is Whitehead and Russell's "Principia Mathematica" in which they attempted to describe the whole of mathematics based on a very formal approach. It took them until page 379 of the first volume to prove that $1 + 1 = 2$, which they described as "occasionally useful".

More usefully, we can use Peano's axioms for the natural numbers, which can be formulated as follows:

• $0$ is a natural number: $0 \in \mathbb{N}$

• The successor of a natural number is a natural number: $n \in \mathbb{N} \implies \sigma \left(n\right) \in \mathbb{N}$

• $0$ is not the successor of any natural number: $\neg \exists n \in \mathbb{N} : 0 = \sigma \left(n\right)$

• Different natural numbers have different successors: $\forall m , n \in \mathbb{N} , \sigma \left(m\right) = \sigma \left(n\right) \implies m = n$

• (Induction) If a proposition holds for $0$ and for the successor of any natural number for which it holds, then it holds for all natural numbers: $\left(P \left(0\right) \wedge \left(\forall n \in \mathbb{N} , P \left(n\right) \implies P \left(\sigma \left(n\right)\right)\right)\right) \implies \forall n \in \mathbb{N} , P \left(n\right)$

So the natural numbers are: $\left\{0 , \sigma \left(0\right) , \sigma \left(\sigma \left(0\right)\right) , \ldots\right\}$, more commonly known as $\left\{0 , 1 , 2 , \ldots\right\}$

We can define addition recursively by:

• $0 + 0 = 0$

• $\forall m , n \in \mathbb{N} , \sigma \left(m\right) + n = m + \sigma \left(n\right) = \sigma \left(m + n\right)$

Then we find:

$\sigma \left(0\right) + \sigma \left(0\right) = \sigma \left(0 + \sigma \left(0\right)\right) = \sigma \left(\sigma \left(0 + 0\right)\right) = \sigma \left(\sigma \left(0\right)\right)$

That is $1 + 1 = 2$ !