Let A=B=$\mathbb{N}$
R is: (a,b)$\in$R iff for some q$\in$Integers a=5q+b WHERE 0$\leq$b<5
Given a relation, show that it's a function.
To Show:
1) $\forall$a$\in$A$\exists$b$\in$B((a,b)$\in$R)
2) $\forall$a$\in$A, there is a unique b$\in$B((a,b)$\in$R)
Proof: Let a$\in$A. Assume (a,b)$\in$R and (a,c)$\in$R where b,c$\in$B.
Do MATH
$\therefore$ b=c. QED
What math should I do to get the full proof. Thanks!
This is not a function when you allow $0 \le b \le 5$. In that case $(5,5)$ and $(5,0)$ are both in $R$.
However if you stipulate that $0 \le b <5$, this is a function, that sends each $a$ to its remainder modulo $5$. So you need to show it's well defined and unique, and both of these proofs will depend on what you already know about numbers.
Unicity comes down to: suppose $(a,b_1), (a,b_2)$ are both in $R$. Then $a = 5q_1 + b_1 = 5q_2 + b_2$, so $5(q_2 - q_1) = b-a$, and then note that 5 cannot divide the difference between two numbers from $\{0,1,2,3,4\}$.