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**Homework Statement::**Show that for every real number ##x## there is exactly one integer ##N## such that ##N \leq x < N+1##. (This integer is called the integer part of ##x##, and is sometimes denoted ##N = \lfloor x\rfloor##.)

**Relevant Equations::**N/A

I have tried reading the solution given here: https://taoanalysis.wordpress.com/2020/05/06/exercise-5-4-3/

Rules available to us:

trichotomy of order

x<y is defined

We know everything about integers and rational numbers i.e. their laws or algebra and order.

We also know that rational numbers are also real numbers, thus the laws of algebra of rationals also apply to reals.

My thoughts and approach:

- Prove that such an interval exists.
- Prove its uniqueness

- The last part of the solution: The important point is that we have M−1<x<M+2. Now we just have to hunt down x by using order trichotomy.
**Why not just use the order trichotomy from the start?**That is to say, by recursion. We first separate into three cases, where x is an integer, x is not an integer. If x=N where N is an integer, N≤x≤N+1. If x is not an integer, we then have two cases: x is a negative real, or x is a positive real. If x is positive, we test the base case where N=1. If x>1, we proceed, otherwise, we can conclude that 0≤x<1. Suppose x>N is proven. We test the case for N+1. If x>N+1, we proceed. If x<N+1, then N≤x<N+1.

**reduces to a uniqueness problem**. Now, if x<N+1, and x≥N+2, it implies that N+2<N+1, which is impossible, thus there is only a unique interval such that N≤x<N+1.

Is my reasoning correct? I know that it might be less elegant afterall!