I was looking through Tao's Analysis I and came across this problem
Prove that $\forall x\in \mathbb R^+,\exists!N\in\mathbb {N}$ s.t. $N\leqslant x<N+1$
I came up with a solution but I used the well-ordering of the natural numbers, which apparently is only treated later on in the book. I am wondering if anyone can suggest how else I should prove it.
My proof: Let $x\in\mathbb R^+$.
Consider the set $S=\{n\in\mathbb N\mid n>x\}$. Then $S\subseteq \mathbb N$. Also $S\neq \emptyset$ by the Archimedean property ($\forall x,\varepsilon\in \mathbb R^+,\exists N\in\mathbb N$ s.t. $N\varepsilon>x$, with $\varepsilon=1$ in this case).
Then by the well-ordering of $\mathbb N$ there exists a minimum $n_0\in S$, which is unique, $\forall n\in S, n_0\leqslant n $. We have $n_0>x>0$ so $n_0\geqslant 1$ which means that $n_0-1$ is defined.
Now $n_0-1\leqslant x$, for if $n_0-1>x$ then $n_0$ wouldn't be the minimal element of $S$.
So take $N=n_0-1$ and we have found a unique (by the uniqueness of the minimal element of $S$) $N\in\mathbb N$ s.t. $N\leqslant x<N+1$. Q.E.D.
As I said earlier, my question is if someone could suggest another method I could use to prove the statement. Furthermore I would be grateful if you point out errors that you find in my proof.
