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Is it true that $$ \left\lfloor \frac{p_n}{p_i\cdot p_j}\right\rfloor \leq \left\lfloor \frac{p_n}{p_i}\right\rfloor \cdot \ \frac{1}{p_j} $$ for $p_n$ prime, $p_i,p_j$ also primes for $i,j < n$ ? I think like so: $$ \left\lfloor \frac{p_n}{p_i\cdot p_j} \right\rfloor = \left\lfloor \left\lfloor \frac{p_n}{p_i} \right\rfloor \ \cdot\frac{1}{p_j} \right\rfloor \leq \left\lfloor \frac{p_n}{p_i} \right\rfloor \cdot \frac{1}{p_j} $$ Is it true?

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This is a straightforward application of the well-known properties of the floor function:

$$\Big\lfloor \frac{p_n}{p_ip_j}\Big\rfloor=\Big\lfloor\frac{\Big\lfloor\frac{p_n}{p_i}\Big\rfloor}{p_j}\Big\rfloor\le\frac{\Big\lfloor\frac{p_n}{p_i}\Big\rfloor}{p_j}=\Big\lfloor\frac{p_n}{p_i}\Big\rfloor\frac{1}{p_j}$$

The specific properties used are the nested division and that $\lfloor x\rfloor\le x$

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