Proving $1 + 1 = 2$

Question

How do you break down the theory of $1 + 1 = 2$? How do you provide a proof, please be precise. This is for one of my discrete math courses and I don't know how this is relevant to the course. And remember, please be detailed and precise.

Answer 1

For a problem like this, it is essential to state what the definitions of $1$ and $2$ are and what set of axioms you are using. Using the Wikipedia Peano axioms we have $1=S(0),2=S(1), 1+1=1+S(0)=S(1+0)=S(1)=2$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal C}\ \mbox{: Cardinal}}$ $$ 0={\cal C}\pars{\emptyset}\,,\quad 1={\cal C}\pars{\braces{0}}\,,\quad 2={\cal C}\pars{\braces{0,1}}={\cal C}\pars{\braces{0,0}} $$

$$ 1 + 1={\cal C}\pars{\braces{0}\cup\braces{0}} = {\cal C}\pars{\braces{0,0}} = 2 $$