Question

If $\left(a,b\right)=d$, prove that role="math" localid="1645880095339" $\left(\frac{a}{d},\frac{b}{d}\right)=1$. [Hint: $a=dr$and $b=ds$for some integers rands (Why?). So $r$and $s$and you must prove that $\left(r,s\right)=1$. Apply Theorem 1.2 to $\left(a,b\right)$and divide the resulting equation by $d$.]

Short answer

It is proved that$\left(\frac{a}{d},\frac{b}{d}\right)=1$.

Step-by-step solution

Theorem

Assume that a and bare any non-zero integers. Let dbe the greatest common divisor of a and b, if and only if dsatisfies the following conditions:

(i) $d|a$and $d|b$

(ii) If$c|a$and$c|b$, then$c|d$.

Prove the part

Assume that $a=d{a}_1$and,$b=d{b}_1$ then prove that localid="1645880573077" $\left(a_1,b_1\right)=1$.

From the stated theorem in step 1, there exist integers $u,v$such that $au+bv=d$.

Use $a=d{a}_1$and $b=d{b}_1$in $au+bv=d$.

$$\begin{gathered} d{a}_{1}u+d{b}_{1}v=d \\ d\left(a_{1}u+b_{1}v\right)=d \\ {a}_{1}u+b_{1}v=1 \end{gathered}$$

Hence, any common divisor of $a_1,b_1$should divide the equation $a_{1}u+b_{1}v=1$, such that it divides 1, which implies that $\left(a_1,b_1\right)=1$.

Then by using $a=d{a}_1$and $b=d{b}_1$into localid="1645881085512" $\left(a_1,b_1\right)=1$, we get:

$\left(\frac{a}{d},\frac{b}{d}\right)=1$

Hence it is proved.