Why is this mapping from binary to natural numbers surjective?

Question

In my Lecture we came across something of importance for the Church-Turing-Thesis and i noticed one particular function, which confused me.

Function: Let $bin(x)$ be the injective binary extension without leading zeros. It is not surjective because of the extension without leading zeros.

$$cod := \{0,1\}^* \to \mathbb{N} \hspace{2mm} x \mapsto bin^{-1}(1x)-1$$

Question:

Why is this particular map from the binaries and the natural numbers surjective ? I do not need a intricate proof, but a explanation is enough to wrap my head around this topic.

They also noted that $bin^{-1}$ can be written, because 1x is in the image of $bin$.

Injectivity is given, because $bin$ is injective and on two different strings $x$ and $y$ we get two different decimal numbers out of the function.

Thanks in advance for any awnser on this question, or any comments posted on it.

Answer 1

For any number $n\in\mathbb{N}$, consider $bin(n+1)$. Since $n+1>0$, there must be at least one bit in $bin(n+1)$ in the state of "1".

Its not too hard to see why this immediately implies that there exists some $x\in \{0,1\}^*$ such that $bin(n+1)=1x$ (since we can remove the trailing zeros), and hence $n+1={bin}^{-1}(1x)$, meaning that $n={bin}^{-1}(1x)-1=cod(x)$.

Therefore $cod$ is surjective by definition.