- #1

mathmari

Gold Member

MHB

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I am looking the following:

Prove that for $m,n\in \mathbb{N}$, $m>n$ it holds that $m-n\in \mathbb{N}$.

Hint: Consider $A=\left \{n\in \mathbb{N}\mid \forall m\in \mathbb{N}, m>n:m-n \in \mathbb{N}\right \}$ and show that $A$ is inductive.

I have done the following:

To show that $A$ is inductive we have to show that $1\in A$ and that if $n\in A$ then also $n+1\in A$, right?

We have that $1\in A$ since $\forall m\in \mathbb{N}$ with $m>1$ we have that $m-1 \in \mathbb{N}$. Can we just say that?

Let $n\in A$. That means that $ \forall m\in \mathbb{N}$ with $m>n$ we have that $m-n \in \mathbb{N}$

We have that $n+1\in A$ since $m-(n+1)=(m-n)+1$ We have that $m-n\in \mathbb{N}$ and the successor is also in $\mathbb{N}$.

Is that correct and complete? :unsure: