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

• Assume $x=0$ (real number).
• Assume $x > 0$ is a positive real number.
• 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]