Every natural number n greater than or equal to 6 can be written in the form n = 3k +4t for some k,t in N

Question

Prove by strong induction that every natural number $n \geq6$ can be written in the form $n=3k+4t$ for some $k,t \in \mathbb{N}$.

I'm not entirely comfortable with the concept of strong induction. I believe you assume that $P(1), P(2), P(3),... P(n)$ are true to prove $P(n+1)$ is. Now, how does that fit into this problem? Would the base case be where $n=6$?

Answer 1

$P(1)$ through $P(5)$ are vacuously true because $1$ through $5$ are not greater than or equal to $6$. Your base cases are $6,7,8$. Show by example that you can do each of them. Intuitively, you can just add enough $3$s to get to any number. So assume it is true up to $n$, then to do $n+1$ you say that you can do $n-2$ and add a $3$

Answer 2

HINT: Every large enough natural number is of one of the folowwing forms: $n=3k$, $n=3k+(4-3)$ or $n=3k+2(4-3)$.