Proof that for all natural numbers $a,b$ if $a>b$ then $a-b$ is natural number

Question

So I am doing exercise 2.4.1 in Bloch's Real numbers and real analysis, and as part of the exercise I need to prove the statement in the title. I made a proof but not sure about it. So I want to ask you to verify it.
Proof: Let $G=\{a\in\mathbb N:a>b \rightarrow (a-b)\in\mathbb N\}$ and $b$ to be any natural number. Suppose $1\notin G$, then it means there is some natural $b$ such that $1>b$ and $(1-b)\notin\Bbb N$. Which contradicts the fact that any natural number is not less than $1$. Then $1\in G$. Suppose that some natural $k\in G$, which means $(k-b)\in\Bbb N$. Then, because of the fact that $\Bbb N$ is inductive $((k-b)+1)\in\Bbb N$, which in turn means that $((k+1)-b)\in\Bbb N$. Which means that $G=\Bbb N$. Hense the theorem is proven.

Answer 1

The idea of doing induction is good, but your attempt fails, sorry.

There is no way to falsify $1\in G$, because there is no natural number less than $1$. Therefore $1\in G$.

Suppose $a\in G$. We want to prove that $a+1\in G$. Let $b<a+1$ be a natural number. There are two cases: $b<a$ or $b=a$, because there is no natural number between $a$ and $a+1$ (I suppose you have already proved this). In the first case $a-b$ is a natural number and $$ (a+1)-b=(a-b)+1 $$ is a natural number, because $a-b$ is. In the case of $b=a$, we have $(a+1)-b=a+1-a=1$, which is a natural number.

Answer 2

If $a>b$, then exists some $p\in\mathbb{N}$ such that $a=b+p$. Then $a-b=p$. Can you go on from this?