Question

Suppose that $a={p_1}^{r_1}{p_2}^{r_2}....{p_k}^{r_k}$and $b={p_1}^{s_1}{p_2}^{s_2}....{p_k}^{s_k}$, where $p_{1,}{p}_{2,...,}{p}_k$are distinct positive primes and each $r_i{s}_i=0$. Prove that $a|b$if and $r_i\le s_i$only if for every i.

Short answer

It is proved that$a|b$ if and only if$s_i\ge r_i$ for every i.

Step-by-step solution

Use the given part

It is given that$a={p_1}^{r_1}{p_2}^{r_2}....{p_k}^{r_k}$ and $b={p_1}^{s_1}{p_2}^{s_2}....{p_k}^{s_k}$.

a and b can be written asfollows:

role="math" localid="1646135507781" $$\begin{gathered} a\prod _{i=1}^k{p_i}^{r_i} \\ b\prod _{i=1}^k{p_i}^{s_i} \end{gathered}$$

Proof part

Divide b by a.

$\begin{array}{rcl}\frac{b}{a}& =& \frac{\prod _{i=1}^k{p_i}^{s_i}}{\prod _{i=1}^k{p_i}^{r_i}}\\ & =& \prod _{i=1}^k\left(\frac{{p_i}^{s_i}}{{p_i}^{r_i}}\right)\\ & =& \prod _{i=1}^k{p_i}^{s_i-r_i}\end{array}$

$\frac{b}{a}$is equivalent to $a|b$.

By the obtained formula $\frac{b}{a}$, it will bean integer if and only if for all$i\le i\le k$ ;then we have$s_i-r_i\ge 0$ or $s_i\ge r_i$. Hence, the given statement is proved.