Algebra - Piecewise Function Let $p(x)$ be defined on $2 \le x \le 10$ such that$$p(x) = \begin{cases} x + 1 &\quad \lfloor x \rfloor\text{ is prime} \\ p(y) + (x + 1 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}$$where $y$ is the greatest prime factor of $\lfloor x\rfloor.$ Express the range of $p$ in interval notation.
I do not not where to start.
I know at $x \in [2, 3) $, the range is $[3, 4)$. Should I continue with this? Like continuing with $x\in [3,4)$, then $x \in [4, 5)$ and so on?

You are on the right path.. Just a small observation away from the solution. In case you don't get it, I am providing the solution below.
Answer:
The range is $\cup_{p \in S} [p+1, p+2)$ where S is the set of all prime numbers.
To see this, take two cases-

*

*[x] is prime: The largest prime factor would be the number itself. So the range is [p+1,p+2).

*[x] is not prime: [x] can be decomposed into ${p_1}^{a_1}{p_2}^{a_2}...{p_n}^{a_n}$ for this, the range is [$max(p_i)$+1,$max(p_i)$+2).

By 1. we get all [p+1,p+2), where p is prime, in our range. By 2., we remove the possibility of having any [n+1,n+2), where n is not prime, in our range.
Edit: I did not see the constraints put on x. Upon adding the constraints, the answer would include just the prime numbers less than 10. This makes the question kinda trivial though :/
