Security of one-time-pad without $\bmod $ operation? Assume $r_1,r_2$ are used only once, and picked uniformly at random (each time we want to mask a value) from finite field $\mathbb{F}_p$, where $p$ is a large prime number. Assume $r_i\neq0$ and $b$ is a (non-zero) fixed element of the field. Let $(r_i)^{-1}$ be multiplicative inverse of $r_i$.

Imaging there is a client  $B$ and a server. 
The server does  some homomorphic operation on an encrypted value and then provides $c=E_{pk_{B}}(r_1\cdot b)$ to client $B$. Where the cipchertext space is much larger than $p$.

So the plaintext may overflows in field $\mathbb{F}_p$, in other words  $r_1\cdot b$  may not be in the field $\mathbb{F}_p$. Therefore, there is a rick that client $B$ after decrypting learns something about $b$. So we want to force client $B$, after decrypting the message performs $\bmod p$.

To this end, we compute $c'=E_{pk_{B}}(r_1\cdot b \cdot r_2\cdot (r_2)^{-1})$. 

Question 1: Given $c'$, can client $B$ learn anything about $b$ after decrypting the ciphertext before it performs $\bmod p$ operation?



Simpler version: Given $r_1\cdot b \cdot r_2\cdot (r_2)^{-1}$ without involving $\bmod p$ operation can a semi-honest party learn anything about $b$ ?
Given $c'$, can client $B$ learn anything about $b$ after decrypting the ciphertext before it performs $\bmod p$ operation?
Given $r_1 \cdot b \cdot r_2⋅(r_2)^{−1}$ without involving $\bmod p$ operation can a semi-honest party learn anything about $b$ ?

These would appear to be the same question; it's just that in the first question, $B$ is handed a ciphertext for which he has the decryption key for;.
The answer: obviously, yes.  If $r_1 \cdot b \cdot r_2⋅(r_2)^{−1}$ happens to be relatively prime to a prime $q$ (for example, $q=3$), then the attacker knows that $b$ also is relatively prime to $q$.  This doesn't immediately tell him the value of $b$; however it allows the attacker to deduce what $b$ is not.
If the attacker can factor $r_1 \cdot b \cdot r_2⋅(r_2)^{−1}$, he can then put together a short list of plausible values for $b$; however it is likely that $b$ and either $r_1$ or both of $r_2, r_2^{-1}$ have large enough prime factors to make this intractable (this assumes that $b$ was ultimately chosen in a way that didn't favor smooth numbers).
However, if the attacker is given two such encodings of the same $b$, $r_1 \cdot b \cdot r_2⋅(r_2)^{−1}$ and $r'_1\cdot b \cdot r'_2⋅(r'_2)^{−1}$, then he is in luck; he can compute $\gcd( r_1 \cdot b \cdot r_2⋅(r_2)^{−1}, r'_1\cdot b \cdot r'_2⋅(r'_2)^{−1})$, and that's $r$, multiplied by a (likely small) integer.