Do there exist positive integers $a,b$ with $a\bmod p < b\bmod p$ for all primes $p$ 
From this post, if a and b are positive integers with $a\bmod p\leq b\bmod p$ for all primes p (here $a\bmod p$ is the unique integer $r,0\leq r < p$ with $r\equiv a\mod p$) then $a=b.$ So, I came up with the following question:


Do there exist positive integers $a,b$ with $a\bmod p < b\bmod p$ for all primes $p$?

I think the answer is no. Of course only finitely many primes need to be checked to verify if a,b actually satisfy the condition. To solve the problem, it could be useful to use the Sylvester-Schur theorem: there is a prime $p>a$ that divides $b(b+1)\cdots (b+a-1)$ whenever $b> a > 0.$ There's also Bertrand's postulate. Though I'd prefer if there is a simpler approach to this problem.
 A: No, let $a$ and $b$ be two integers; then neither $a$ nor $b$ can be $1$. To see why let $a=1$, and then take on the one hand any prime that divides $b$, and then take on the other hand any prime larger than $b$.
So then assume WLOG  $1<a<b$. Then take on the one hand any prime that divides $a$ and then take on the other hand any prime that divides $b$. [If the inequality $b-a \ge 2$ is satisfied, then taking any prime that divides $b-a$ also suffices.]
To answer your additional question in the comments below, YES, there does exist such infinite families of $a$ and $b$ such that $a \pmod p < b \pmod b$ for all primes $p$ s.t. $(p,b)=1$. Indeed, take $b$ any prime at least $5$ and then set $a=b-1$.

*

*Then for any prime $p<b$ note that $b \pmod p \not = 0$, so [as $a=b-1$]. So it follows that $b \pmod p$ is a positive integer and thus the string of relations $a \pmod p$ $= (b \pmod p )-1$ $< b \pmod p$ hold, for all primes $p < b$.


*For any prime $p>b$, the strict inequality $a \pmod p < b \pmod p$ is obvious.
A: You are right that the answer is no.
If $b=1$, take a prime $p>a$. Then $a\bmod p=a\not\lt 1=b\bmod p$.
Else $b>1$. Take a prime $q\mid b$. Then $0\le a\bmod q \not\lt 0=b\bmod q$.
