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So I heard of the following way to define naturals in real numbers.
Let's call a subset $A$ of reals to be inductive if and only if:
1)it contains $0$.
2)it contains $k+1$ whenever it contains $k$.
Than, we define real number to be natural if it is contained in every inductive set. Now, can we show that set of all natural real numbers is closed under addition and multiplication, and that $0$ is least element (with real number ordering) ? Or maybe it is possible to prove this properties with other definition of naturals in reals ?

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Let $\mathbb N$ denote the set of elements that are contained in every inductive subset of $\mathbb R$.

It is immediate that $0\in\mathbb N$ and further the set $[0,\infty)$ is inductive so that $\mathbb N\subseteq[0,\infty)$, showing that $0$ serves as least element of $\mathbb N$.

Now define: $$K:=\{n\in\mathbb N\mid \forall k\in\mathbb N[n+k,nk\in\mathbb N]\}$$and prove that $K$ is inductive.

Then $\mathbb N\subseteq K\subseteq\mathbb N$ so that $\mathbb N=K$, and you are ready.

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  • $\begingroup$ Sorry, I am not familiar with this notation $\mathhbb N[n+k,nk\in\mathhbb N]$. Maybe you can call it's name, so I can find it in the internet. $\endgroup$ – Юрій Ярош Jul 28 '18 at 15:35
  • $\begingroup$ In words $K$ is by definition the subset of $\mathbb N$ that contains exactly the elements $n\in\mathbb N$ that satisfy $n+k\in\mathbb N$ and $nk\in\mathbb N$ for every $k\in\mathbb N$. $\endgroup$ – Vera Jul 28 '18 at 15:40
  • $\begingroup$ So could it be written in the form $$K:=\{n\in\mathbb N\mid \forall k\in\mathbb N,n+k,nk\in\mathbb N,\}$$ ? $\endgroup$ – Юрій Ярош Jul 28 '18 at 15:47
  • $\begingroup$ Something like that , yes. $\mathbb N[n+k,nk\in nk\in\mathbb N]$ is not a notation on its own.as you seemed to think. $\endgroup$ – Vera Jul 28 '18 at 15:49

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