Interspersing of integers by reals

Hi everyone I’m asking to myself if the next argument is sound. I know it is costumary to use the well-ordering principle to prove it, but at this point in the book I have not seen yet.

Proposition: For each real number $x$ there exists a unique integer $N$ such that $N\le x<N+1$.


(a) Suppose $x$ is a positive real number and suppose for the sake of contradiction that there is no natural number such that $N\le x<N+1$. This means that whenever $N\le x$ we must have $N+1 \le x$. A simple induction shows that $N+m\le x$ for every natural number $m$. But by the Archimedean Principle we know that there is a natural number $M$ such that $M>x\ge N$. Now we set $m:=M-N$, which implies $N+m = N+(M-N)=M\le x$, a contradiction.

(b) If $x=0$, then $0\le x < 1$ and we’re done.

(c) On the other hand, if $x$ is negative the claim follows from the positive part. Since $-x$ is a positive number and by (a), we have: $M\le -x < M+1$ for at least one natural number. So, $-(M+1)<x \le -M$. If $\,x< -M$ we set $N:=-(M+1)$, but If $\,x = -M$ we put $N=-M$ and we’re done.

To conclude the proof we shall show the uniqueness of the integer. Suppose there are two integers $N$ and $N’$ such that $N\le x<N+1$, $N’\le x<N’+1$ and $N \not= N’$. Without loss of generality we may assume that $N<N’$. So, $\,N+1 \le N’$ and then $x<N’$, a contradiction.

Do you think is correct? Thanks in advance. 🙂

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