Question

Prove that $a|b$ if and only if $a^2|b^2$.

Short answer

The given statement is proved.

Step-by-step solution

Obtain the a|b if and only if a2|b2 

Consider the equation $a|b$if and only if $a^2|b^2$.

If $a|b\Rightarrow b=at$ for some integer of t , it gives thatlocalid="1652802625910" $b^2=atat=a^2{t}^2$which proves that $a^2|b^2$.

Find proof

Now consider that $a^2|b^2$.

Let’s assume that$a={p_1}^{r_1}{p_2}^{r_2}.....{p_k}^{r_k}$

and $b={p_1}^{s_1}{p_2}^{s_2}.....{p_k}^{s_k}$. Here, $r_i,s_i\ge 0$.

$a^2|b^2$gives${p_1}^{r_1}{p_2}^{r_2}.....{p_k}^{r_k}|{p_1}^{s_1}{p_2}^{s_2}.....{p_k}^{s_k}$ and this indicates that $2s_i\le 2r_i$.

Here, i gives $r_i\le s_i$.

Therefore,

If $a|b\Rightarrow b=at$ for some integer of t , this gives ,$b^2=atat=a^2{t}^2$ which proves that, $a^2|b^2$.

Hence, it is proved that $a|b$if and only if $a^2|b^2$.