Question over recursive definitions

Question

Let $E$ be the set of even integers. Then the Base Case: $0\in E$ And the constructor case would be If $n \in E$ then so are $n+2$ and $-n$. This makes sense.

But would the Base case and constructor case be for the set $S$ if $S=\{2^k 3^m 5^n \in N |k,m,n \in N\}$?

Answer 1

Hint

We might have that $k=m=n=0$ which gives us the number $2^03^05^0=1 \in E$.

Now for each number in $S$, if we multiply it by a factor 2, 3 or 5, we get a new number in $S$. How can you formulate this mathematically?

Answer 2

The set of even integers is constructed through addition and subtraction operations. The base case is any element, and $0$ is an example.

$$E = \{0, a \mid a \in \mathbb{E} \implies (a + 2 \in \mathbb{E}) \land (-a \in \mathbb{E})\}$$

1. $E_0 = 0$
2. $E_{n+1}=E_n + 2$
3. $E_{-n} = -E_n$

The set $S$ is constructed by three multiplication operations. The base case is the smallest element of the set, unity.

$$\mathbb{S} = \{1, s \mid s \in \mathbb{S} \implies (2s \in \mathbb{S}) \land (3s\in \mathbb{S}) \land (5s \in \mathbb{S})\}$$

1. $S_0 = 1$
2. $S_{k+1,m,n}=2S_{k,m,n}$
3. $S_{k,m+1,n} = 3S_{k,m,n}$
4. $S_{k,m,n+1}=5 S_{k,m,n}$