Every real number lies between $N$ and $N+1$.

Question

I’m self-studying Tao’s Analysis I and following is a question from it.

Exercise 5.4.3. Show that for every real number $x$ there is exactly one integer $N$ such that $N ≤ x < N +1$. (This integer $N$ is called the integer part of $x$, and is sometimes denoted $N = ⌊x⌋$.)

Uniqueness is easy to show.

I wish to know if my following proof idea for existence is correct. It’d suffice to show this for positive real numbers.

Suppose on the contrary that there existed a positive real number $x$ such that for all positive integers $N$, $x<N+1$ implies $x<N$. Since there exists a positive integer $N_0$ such that $x<N_0=M+1$, where $M=N_0-1<N_0$ is a non-negative integer, and hence, $x<M$. Since $x$ is positive, $M$ is also positive. Hence we have shown that for all positive integers $N_0$, $x<N_0$ implies $x<M$ for some positive integer $M<N_0$, contradicting the principle of infinite descent, since I can create an infinitely descending list of positive integers $n$ satisfying the property $x<n$. Hence the proof completed.

Answer 1

You are going to need to use a basic property of the real numbers: given any real number, x, there exist an integer, N, such that N> x and conversely, given any integer N there exist a real number, x, such that x> N.

Answer 2

I think you have to add some to your proof. As you said let's take $x$ positive. You claim that there exist an integer $N_0 > x$ and this is true (natural number are infinite). Now, by contradiction, since $x<N_0$ we have two cases:

(1) If $N_0=0$, you have a contradiction because $x$ is positive;

(2) If $N_0>0$, then you have $x<N_0-1$ and you can iterate the procedure until you arrive at the case (1) (infinite discent)

Now your proof is completed.

Another approach:

I propose also a more linear approach (in my opinion). Let's take $x$ positive and $\mathbb{N}=\{0,1,2,...\}$. Consider the set \begin{equation} S=\{n\in \mathbb{N} \ | \ x < n+1\} \end{equation} Since $S$ is non empty (natural numbers are infinity), thanks to minimum principle exist the minimum $N\in \mathbb{N}$ of $S$.

Now $x< N+1$ because $N\in S$ and $N\leq x$ because is the minimum of $S$. So $$ N\leq x< N+1 $$