Question

Does anyone know the more complicated way of solving for `1` `+` `1`?

Answer 1

We could use Peano's postulates...

Explanation:

We can go back to basics, such as:

• What do we mean by numbers?

• How can we define addition?

The 19th century Italian mathematician Giuseppe Peano presented what are known as the Peano postulates, describing a formal basis for arithmetic of Natural numbers.

In a modern formulation, here are Peano's axioms, describing the basic properties of the Natural numbers:

• `0` is a Natural number.

• If `n` is a Natural number then `sigma(n)` (the successor of `n`) is a Natural number.

• If `m` and `n` are Natural numbers then `m = n` if and only if `sigma(m) = sigma(n)`.

• For any Natural number `n`, `sigma(n) != 0`.

• If `P(n)` is a property such that `P(0)` and for any Natural number `n`, `P(n) => P(sigma(n))` then `P(n)` for all Natural numbers.

Given Peano's axioms, we can define addition of Natural numbers as follows:

• `n + 0 = 0 + n = n` for any Natural number `n`.

• `m + sigma(n) = sigma(m) + n = sigma(m + n)` for any Natural numbers `m, n`.

Then we find:

`sigma(0)+sigma(0) = sigma(0+sigma(0)) = sigma(sigma(0))`

In our notation:

`1+1 = 2`

Answer 2

If you try hard enough anything can be made complicated.

Explanation:

How about this?

Let `k=1+1`

From this it follows that
`color(white)("XXX")kcolor(magneta)(-1)=1+1color(magenta)(-1)`
which can be simplified as
`color(white)("XXX")kcolor(magenta)(-1)=1`
`color(white)("XXXXXX")`critical note at this point:
`color(white)("XXXXXXXXX")`if `k-1 = 1`
`color(white)("XXXXXXXXX")`then `k-1 > 0`
`color(white)("XXXXXXXXX")`and `k > 0`

Squaring both sides
`color(white)("XXX")(k-1)^2=1^2`

`color(white)("XXX")k^2-2k+1=1`

After subtracting `1` from both sides
`color(white)("XXX")k^2-2k+1color(magenta)(-1)=1color(magenta)(-1)`
`color(white)("XXX")k^2-2k=0`

Factoring
`color(white)("XXX")k(k-2)=0`

Which implies
#color(white)("XXX"){: ("either",k=0," or ",k-2=0), (,"not allowed",,rarr k=2), (,"(critical note, above)",,) :}#

Since `k=1+1` (by definition of `k`)
`color(white)("XXX")2=1+1`

Then applying the reflexive property for equality
`color(white)("XXX")1+1=2`