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

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.

For some reason, the comment didn’t come out right:

The first line was supposed to be:

(-n) <= (-x) and (-x) < (-n) + 1.

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.