Prove that a group of size $\ge18$ people can be assembled from groups of 4 and 7

Question

How can I prove that a group of size $\ge18$ can be assembled from groups of $4$ and $7$ using the well ordering principle?

Well-ordering principle: Every nonempty subset $T$ of $N$ has a least element. That is, there is an $m ∈ T$ such that $m ≤ n$ for all $n ∈ T$.

I have the following unorganized thoughts: I can only prove this by induction not Well Ordering Principle so please let me know how to do that!

Proposition: $P(n)$: If n $\ge18$, there is a group of people of size $n$ made from groups of $4$ or $7$.

$P(0)$: is vacuously true

Inductive step: assume $P(0)...P(n)$ is true. Then $P(n+1)$ must also be true. I don't know how to write this part:

---Unorganized thoughts The rules that I have come up with is:

1. If the size is $\ge 5$ groups of $4$, then you replace by $3$ groups of $7$.

2. If the number cannot be represented in multiples of $5$ groups of $4$, replace one of the groups of $7$ by $2$ groups of $4$

3. Repeat as needed until you get the requested number, which is now made up of groups of $4$ and $7$.

How do I write that as a proof? These are unorganized thoughts.

Please write your steps in detailed order so I can follow. I'm not very proficient in math symbolism so please explain symbols if possible.

Thanks!

Answer 1

Not sure what the well ordering principle is, but here is how I would prove this:

• $A_{18} = \{4,7,7\}$
• $A_{19} = \{4,4,4,7\}$
• $A_{20} = \{4,4,4,4,4\}$
• $A_{21} = \{7,7,7\}$
• $A_{n} = \{4\} \cup A_{n-4}$