Question

If $a={p_1}^{r_1}{p_2}^{r_2}...{p_k}^{r_k}$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 $r_i,s_i\ge 0$, then prove that

(a) $\left(a,b\right)={p_1}^{n_1}{p_2}^{n_2}...{p_k}^{n_k}$, where for each i,$n_i=minimumof{r}_i,s_i$ .

(b) role="math" localid="1646214562855" $\left[a,b\right]={p_1}^{t_1}{p_2}^{t_2}...{p_k}^{t_k}$, where for each $t_i=maximumof{r}_i,s_i$.

Short answer

1. It is proved that$\left(a,b\right)={p_1}^{n_1}{p_2}^{n_2}...{p_k}^{n_k}$ for $i,n_i=minimum\left(r_i,s_i\right)$.
2. It is proved that$\left[a,b\right]={p_1}^{t_1}{p_2}^{t_2}...{p_k}^{t_k}$ for $t_i=maximum\left(r_i,s_i\right)$.

Step-by-step solution

Part (a)

It is given that$c|a$ and$c|b$ if and only if role="math" localid="1646215316068" $c=\prod _{i=1}^k{p_i}^{n_i}$with $t_i\le min\left(r_i,s_i\right)$
.

Also, we haverole="math" localid="1646215430592" $\prod _{i=1}^k{p_i}^{t_i}$ with $n_i\le min\left(r_i,s_i\right)$, where it is the greatest number that satisfies the given condition, so it is also the greatest common divisor of a and b.

Part (b) 

As we have $a=\prod _{i=1}^k{p_i}^{r_i}$ and $b=\prod _{i=1}^k{p_i}^{s_i}$.

Find ab and $\left(a,b\right)$.

role="math" localid="1646216178797" $\begin{array}{rcl}ab& =& \left(\prod _{i=1}^k{p_i}^{r_i}\right)\left(\prod _{i=1}^k{p_i}^{s_i}\right)\\ & =& \prod _{i=1}^k\left({p_i}^{r_i}·{p_i}^{s_i}\right)\\ & =& \prod _{i=1}^k{{p_i}^{r_i}}^{+s_i}\end{array}$

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

Now, find $\left[a,b\right]$by using the formula $\left[a,b\right]=\frac{ab}{\left(a,b\right)}$.

$\begin{array}{rcl}\left[a,b\right]& =& \frac{\prod _{i=1}^k{p_i}^{\left(r_i+s_i\right)}}{\prod _{i=1}^k{p_i}^{min\left(r_i,s_i\right)}}\\ & =& \prod _{i=1}^k{p_i}^{max\left(r_i,s_i\right)}\end{array}$

Hence, it is proved that$\left[a,b\right]={p_1}^{t_1}{p_2}^{t_2}...{p_k}^{t_k}$ for $t_i=maximum\left(r_i,s_i\right)$.