Distribution of $p_{n+1}$ rough numbers

Question

Is the following argument regarding rough numbers valid? If so, is there a more straight forward way to make the same argument?

Let:

• $p_n$ be the $n$th prime.
• $p_n\#$ be the primorial for $p_n$
• $U_{x,p}$ be the set of integers $u$ where $x < u < x+p\#$ and $\text{gcd}(u,p\#)=1$ where gcd is the greatest common divisor and $x > 1$ is an integer.
• $V_{x,p}$ be the set of integers $v$ where $v \in V_{x,p}$ if $u \in U_{x,p}$ and $v = u - x$.
• $a_i, a_j$ be integers where $p_n < a_i < a_j < p_n\#$ and $\text{gcd}(a_i,p_n\#)=1$ and $\text{gcd}(a_j,p_n\#)=1$

Claim: For all $a_i,a_j$ there exists an integer $w$ where $w \in V_{a_i,p_n}$ but $w \not\in V_{a_j,p_n}$

(1) Assume it is not true.

(2) Let $d = a_j - a_i < p_n\#$

(3) Let $q$ be the lowest prime that does not divide $d$

(4) $q \le p_n$ since $d < p_n\#$

(5) Let $t = \dfrac{p_n\#}{q}$

(6) $\{ a_j, a_j + t, a_j + 2t, \dots, a_j + (q-1)t \}$ forms a complete residue system modulo $q$ with $(q-1)t < p_n\#$

(7) There exists an integer $w$ such that $q | a_j + wt$ and therefore $wt \not\in V_{a_j,p_n}$ and $wt \not\in V_{a_i,p_n}$

(8) But $q \nmid (a_j + wt - d)$ and further, no other prime $\le p_n$ can divide $a_i + \dfrac{p_n\#}{q}$ so we have a contradiction and must reject step(1).

Answer 1

Your proof is basically correct. However, there are a few issues with it, that I'll explain, and also I'll show how to make it a bit more straight forward.

First, your fourth bullet point states:

$V_{x,p}$ be the set of integers $v$ where $v \in V_{x,p}$ if $u \in U_{x,p}$ and $v = u - x$.

The "where $v \in V_{x,p}$" part is redundant since the first part states you're defining the elements of $V$. Also, I find the wording used later somewhat confusing. Instead, I would suggest writing this as something like

$V_{x,p}$ be the set of integers $v$ where $u = x + v$ and $u \in U_{x,p}$

or, even somewhat more simply to avoid defining a new variable of $u$, just say

$V_{x,p}$ be the set of integers $v$ where $x + v \in U_{x,p}$

Next, you try to prove your claim by contradiction. Your step (7) uses this only implicitly, so it's not clear. I suggest you make it explicit instead by doing something like changing the last part of "and $wt \not\in V_{a_i,p_n}$" to ", and step (1) means that $wt \not\in V_{a_i,p_n}$".

However, since you're actually constructing a $w$ such that your claim is true, I believe it's simpler & more appropriate to just prove it directly. Also, one other point is you use $w$ in your claim in one manner and then $w$ later in your (7) in a different way. To help avoid confusion, I'll replace your $w$ in your claim with $y$, so your claim would then be

Claim: For all $a_i,a_j$ there exists an integer $y$ where $y \in V_{a_i,p_n}$ but $y \not\in V_{a_j,p_n}$

You can thus remove your step (1), and your step (7) would become something like

(7) There exists an integer $w$ such that $q | a_j + wt$. Let $y = wt$, so $y \not\in V_{a_j,p_n}$.

Also, your step (8) (note you forgot a factor of $w$ in your "$a_i + \dfrac{p_n\#}{q}$" part) would then become something like

(8) But $q \nmid (a_j + wt - d)$ and further, no other prime $\le p_n$ can divide $a_i + y = a_i + w\left(\dfrac{p_n\#}{q}\right)$, so $y \in V_{a_i,p_n}$. This shows how to construct at least one such integer $y$ stated in the claim, so the claim is always true.

Regarding your question comment, I don't offhand know of any easy way to construct other such differences.