If $p, q \in \mathbb{N}=\{1,2,3,\ldots\}$ and $p + q = 2$, then $p = q = 1$.
How do I prove this using the basic properties of Natural Numbers? I could see how exceeding simple it is to solve, but proving something so basic and fundamental involves deep meta-thinking.
On
What exactly do you mean?
Since this question seems that simple and almost selfevident,are you looking for a particular proof or derivation from a formal axiomatic system for natural numbers like set theoretic construction of the natural numbers or Peano Axioms? And what is considered natural numbers here?
The propostion makes only sense if 0 is not included. Otherwise $p=0$ and $q=2$ or $p=2$ and $q=0$ would satisfie the proposition too
The straigthforward way to do it then would just be: $p+q=2$ implies that both p and q are lesser equal $2$(since p,q>0).
So p,q (since natural numbers) must lie in the set $\{1,2\}$
Then consider the binary sum operation on $\{1,2\}$ $$(+):\{1,2\}^2\longrightarrow N,\ (+)(n,m):=n+m$$ The values which $(+)$ takes for the finite number of pairs in $\{1,2\}^2$ are, $$(+)(1,1)=2;\ (+)(2,1)=(+)(1,2)=3;\ (+)(2,2)=4$$
Hence out of all possible pairs in$\{0,1\}^2$ only $(1,1)$ is an preimage of 2 under this operation. q.e.d.
Here is one such proof:
Let $p,q\in \mathbb{N}$. That means, by definition, that $p\geq 1$ and $q\geq 1$. Now, suppose that $p+q=2$. We know that $p+q>p$, since natural numbers are positive, so we have $$2=p+q>p\geq 1$$ That is $2> p\geq 1$, since the only natural number that satisfy the last equality is $1$, then $p=1$. Same argument shows that $q=1$.