- Assume $x=0$ (real number).
- Then $n=0$ (integer) does the trick: $0\le x < 1,$ because of the its uniqueness.

- Assume $x > 0$ is a positive real number.
- According to the Archimedean principle the set $N:=\{k\in\mathbb N\mid k > x\}$ of natural numbers $k$ greater than $x$ is not empty.
- By the well-ordering principle, $N$ contains a smallest element $m.$
- Note that $m > x > 0,$ therefore $m\ge 1.$
- Set $n=m-1.$
- It follows $n$ is the unique integer with $n\le x < n+1.$

- Assume $x < 0$ is a negative real number.
- Since $-x > 0,$ we have already shown that there is a unique integer $l$ with $l\le -x < l + 1.$
- Therefore, there is a unique integer $-l$ with $-l-1 < x \le -l.$
- If $x = -l,$ then $-l\le x < -l+1$ with a unique $l.$
- Else if $x < -l,$ then $-l-1\le x < l$ with a unique $l.$

q.e.d

| | | | created: 2020-03-01 19:30:44 | modified: 2020-03-01 19:31:03 | by: *bookofproofs* | references: [577]

[577] **Knauer Ulrich**: “Diskrete Strukturen – kurz gefasst”, Spektrum Akademischer Verlag, 2001