So this question in my book looks like it's essentially asking me to prove the ceiling function exists.
This question is slightly different to other things I found in related question because we're asked to prove the existence of the value in the middle of the inequality.
My initial thoughts are to define an $n \in N$ such that $n > x$ and $-n < x$
Following from this one can say $-n < x \leq n < x + 1$.
Now here I just lose any sort of direction and instead think of trying another method defining a set and using the properties of $inf$ and $sup$ to move from there. The tricky thing I encounter as a roadblock is once again the proving of the existence of the $n$ which is different to many related questions.
Any help or solutions are appreciated, ready and rearing to reply :)
You will need the "well ordering property" of the positive integers- every non-empty set of positive integers contains a smallest member. (https://en.wikipedia.org/wiki/Well-ordering_principle)
Given a real number, x, Let A be the set of all positive integers strictly greater than x. By the "Archimedian property" (https://en.wikipedia.org/wiki/Archimedean_property) that set is non-empty. Let m be the smallest member of that set.
Let n= m- 1. Then n< m and since m is the smallest member of A, n is not in A. Therefore, $n\le x$. But n+ 1= m which is in A. n+ 1> x