I understand the intuition for how to do this, but I don't know how to formally prove it. For example, I know that if you take (a,b) to be (1,2), then a sample (c,d) could be (1,2) since 1 + 2 = 2 + 1. But how do I formally prove that such numbers c,d exist for an arbitrary c,d such that a,b,c,d are all in the naturals.
Thanks in advance!
On
Well, you can always let $c=a$ and $d=b$. Then $a+d = c+d$.
If you mean a different pair of numbers.
Well, that isn't true. $0 +0$ will have not $\{c,d\} \ne \{0\}$ so that $c + d = 0+ 0 =0$.
Nor is it true for $\{a,b\}=\{0,1\}$ and $a+b = 1$ will only have $\{c,d\} =\{0,1\}$ if $c+d =a+b= 1$.
However if we rule out that $\{a,b\} = \{0,1\}$
Then we can say let $\alpha = \min(a,b)$ and $\beta = \max (a,b)$.
Case 1: $\alpha = 0$. Then $\beta \ne 0, 1$ so $\beta \ge 2$.
Let $c = 1$ and $d = \beta -1$.
Case 2: $\alpha > 0$. Then let $c = \alpha -1$ and $d = \beta + 1$.
You need to be careful with the statement that you are trying to prove. As stated in the title of your question it is false because $c,d$ are chosen before $a,b$. You need $c,d$ to be chosen after $a,b$, so it should be that for any two natural numbers $a,b$ there exists a pair $c,d$ such that $a+d=b+c$. Having stated it correctly, a good way to prove it is an algorithm to find a $c,d$ pair. One such (assuming you accept $0$ as a natural) is
If $a=b, c=d=0$
Else if $a \gt b, d=0, c=a-b$
Else $c=0, d=b-a$