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# There is exactly one integer between any real number x and x+1.

Prove that if , then there is exactly one such that .

Proof. By I.3.12, Exercise #4 we know that for there is exactly one such that If , then we already done since .
If , then we have and so . But we already have ; hence, So, is an integer with the desired properties. Uniqueness follows from the uniqueness of our choice of 1. James says:

Hey, looking at your solution gave me an idea for an alternate solution (please excuse the formatting):
From problem 4, for an arbitrary real number (-x), there is only one integer (-n) such that:
(-n) <= (-x) n >= x and x > n – 1 –> x + 1 > n
Thus, there is exactly one n such that x<= n < x + 1.

• James says:

For some reason, the comment didn’t come out right:
The first line was supposed to be:
(-n) <= (-x) and (-x) < (-n) + 1.

• James says:

And then, multiply both by (-1), and get
n>=x and x>n-1, the second of which is x+1>n.
Combined, they are x<=n<x+1.