If $L$ is recursive then so is $L^*$, and vice versa Given a language $L$ over the alphabet $\{0,1\}$. Let $L^*= \{ w_1w_2...w_n | n \ge 0, w_1,...,w_n \in L\}$.

Prove:


If $L$ is recursive, then $L^*$ is recursive as well.
If $L^*$ is recursive, then $L$ is recursive as well.


For the first proof I have this idea:
If $L$ is recursive, then a turing machine $M_L$ can decide all the words $\{w|w \in L\}$. So $M_L$ can decide $\{w_1,...,w_n | w_1w_2...w_n \in L^*\}$ and appearently it can decide $L^*$ as well. 

For the second proof I tried the same but actually it does not work that way. Any hints and ideas?
The second claim is wrong. Let
$$ L = \{0,1\} \cup \{010^n : \text{the $n$th Turing machine halts}\}. $$
Then $L^* = \{0,1\}^*$ is recursive, but $L$ isn't recursive.

For the first claim, your proof is not convincing. You need to explain how a Turing machine can decide whether a given word $w \in \{0,1\}^*$ can be written as a sequence of words from $L$. One way is to consider all possible partitions into non-empty subwords, and check whether one of them has the property that all subwords are in $L$.