Why does the definition of addition require proofs?

Question

The book is Foundations of Analysis by E. Landau.

I delve into analysis for the first time. Before I would encounter merely a definition of something in a textbook, but now there is something more. In Theorem 4 the author gives a definition of addition. Then he first proves that A) for any fixed $x$ there is at most one possibility to define $x + y$. Then he proves that B) for each $x$ it is possible to define $x + y$.

My question is why does he prove B) that it is possible to define $x + y$? He has already made a definition, so he brought it into existence.

Why does he need proofs here at all? It is a definition, you can define it however you want.

Answer 1

The proof is necessary to ensure that the definition is actually meaningful.

Suppose, for instance, that I define a binary operation $\circledast$ on $\mathbb{Q} \times \mathbb{Q}$ as follows: For any rational numbers $q_1$ and $q_2$, I say that $q_1 \circledast q_2$ is the smallest rational number greater than $q_1 \cdot q_2$. Great! Only problem is, no such smallest rational number exists.

Second example: I define a ternary operation $F$ on $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ as follows: $F(a,b,c)$ is the solution to the equation $a x^2 + bx + c = 0$. Terrific! Only problem here is, sometimes no such solution exists, and sometimes more than one solution exists.

Whenever you define something, it's essential to prove that the thing being defined both exists and is unique. Otherwise the definition is meaningless.

Answer 2

We don't have to simply define addition and hope we get it right. We can actually prove the existence a uniquely defined add function on $N$ with all the required properties.

We can formally construct the $add$ function on the set of natural numbers $N$ by selecting the following subset of $N^3$:

$\forall a,b,c:[(a,b,c)\in add \iff (a,b,c)\in N^3 $

$\land \forall d\subset N^3:[\forall e\in N:(e,0,e)\in d\land \forall e,f,g:[(e,f,g)\in d \implies (e,S(f),S(g))\in d]$

$\implies (a,b,c)\in d]]$

where $S$ is the successor function on $N$ given by Peano's Axioms.

Using Peano's Axioms, we can prove that the set of ordered triples $add$, so defined, is a function such that, using the prefix functional notation, we have:

$\forall a,b\in N: add(a,b)\in N$

$\forall a\in N:add(a,0)=a$

$\forall a,b\in N:add(a,S(b))=S(add(a,b))$

We can also prove that the $add$ function, so defined, is unique. If $add'$ is another function such that we also have:

$\forall a,b\in N: add'(a,b)\in N$

$\forall a\in N:add'(a,0)=a$

$\forall a,b\in N:add'(a,S(b))=S(add'(a,b))$

then we can prove by induction that $\forall a,b\in N: add(a,b)=add'(a,b)$

We then feel justified referring to the add function + (using the infix notation) where

$\forall a,b\in N: a+b\in N$

$\forall a\in N:a+0=a$

$\forall a,b\in N:a+S(b)=S(a+b)$