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Proof

(related to Corollary: Existence of Unique Integers Exceeding Real Numbers)

• 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 $l.$
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References

Bibliography

1. Knauer Ulrich: "Diskrete Strukturen - kurz gefasst", Spektrum Akademischer Verlag, 2001