How do I prove the floor identity $⌊x + n⌋ = ⌊x⌋ + n$ in a more precise way?

Question

I am having trouble understanding the proof provided by the author for the property stated after "Goal:".

Except from the text here is a list of useful properties:

(PROPERTY 1a) $⌊x⌋ = n$ if and only if $n ≤ x < n + 1$

(1b) $⌈x⌉ = n$ if and only if $n − 1 < x ≤ n$

(1c) $⌊x⌋ = n$ if and only if $x − 1 < n ≤ x$

(1d) $⌈x⌉ = n$ if and only if $x ≤ n < x + 1$

(2) $x − 1 < ⌊x⌋ ≤ x ≤ ⌈x⌉ < x + 1$ (3a) $⌊−x⌋ = −⌈x⌉$

(3b) $⌈−x⌉ = −⌊x⌋$

(4a) $⌊x + n⌋ = ⌊x⌋ + n$

(4b) $⌈x + n⌉ = ⌈x⌉ + n$

Goal: As an exercise, prove the property $⌊x + n⌋ = ⌊x⌋ + n$

Proof: We will prove the property using a direct proof. Suppose that $⌊x⌋ = m$, where $m$ is a positive integer. By property (1a), it follows that $m ≤ x < m + 1$. Adding $n$ to all three quantities in this chain of two inequalities shows that $m + n ≤ x + n < m + n + 1$. Using property (1a) again, we see that $⌊x + n⌋ = m + n = ⌊x⌋ + n$. This completes the proof. Proofs of the other properties are left as exercises.

From: Rosen, Kenneth. Discrete Mathematics and Its Applications (p. 159)

My understanding of the property we are tasked with proving is that it is an identity that is stating the proposition $∀x∀n(⌊x + n⌋ = ⌊x⌋ + n)$ where the domains of discourse for $x$ and $n$ are the set of all real numbers and the set of all integers, respectively. With that in mind, since this is a direct proof, shouldn't we start with arbitrary values of the domains (represented by real number $x$ and integer $n$, and NOT JUST POSITIVE INTEGER $n$) and try to show that their properties imply the equation $⌊x + n⌋ = ⌊x⌋ + n$? (Ignoring the positive-n issue, how could we even do this if the only properties we are allowed to assume about $x$ and $n$ are that $x$ is real and $n$ is an integer?) Also, why was he able to start with and assume $⌊x⌋ = m$ ("suppose $⌊x⌋ = m$") out of nowhere? If anyone can provide a version of his proof that is more precise and doesn't skip steps of reasoning or explanations, that would answer my question.

Answer 1

You’re correct in thinking that $m$ should not have been limited to positive integers. However, once the word positive is removed, the proof is fine. In particular, the author is simply assuming that $x$ is a real number and that every real number has a floor. Ignoring positive, which is an error, we should understand his first sentence as being equivalent to the following sentence; where $m$ is an integer is just a reminder to the reader, since the floor function necessarily evaluates to an integer.

Let $x$ be a real number, and let $m=\lfloor x\rfloor$.

Nothing there comes ‘out of nowhere’: we want to show something about all real numbers, so we start with an arbitrary real number $x$. We want to be able to talk easily about its floor, so we give the floor a name, $m$.

The rest of the proof says everything that needs to be said. At most one might add a little explanation after adding $n$, something like this:

Now $m+n$ and $m+n+1$ are consecutive integers, and $$m+n\le x+n<m+n+1\;,$$ so by $(1a)$ we know that $$\lfloor x+n\rfloor=m+n=\lfloor x\rfloor+n\;.$$

But I would do that only when the floor function is first introduced, and then only in a very elementary course — and I say that as one who tends to give more rather than less detail.

Answer 2

First, the statement that $m$ is positive appears to simply be an error in the text and should be ignored (just remove the word "positive" from the proof).

Then, when it says "Suppose that $\lfloor x\rfloor =m$", that is not an assumption but rather a definition of the variable $m$. In other words, they are just defining $m$ as a name for $\lfloor x\rfloor$. Since the floor of a number is by definition an integer, you then know that $m$ is an integer (and so the later steps that use property (1a) are valid).

Answer 3

Let $\lfloor x+n\rfloor=k$

Then by [1a] we have:

$$k \le x+n \lt k+1$$

$$k-n \le x \lt k+1-n$$

Using [1a] again:

$$\lfloor x\rfloor=k-n$$

$$\lfloor x\rfloor+n=k$$