Question

Prove that

(a) $\left(a,b\right)\mathbf{|}\left(a+b,a-b\right)\mathbf{;}$

(b) If $\mathit{a}$is odd and $\mathit{b}$is even, then $\left(a,b\right)\mathbf{=}\left(a+b,a-b\right)\mathbf{;}$

(c) If a and b are odd, then role="math" localid="1647366346855" $\mathbf{2}\left(a,b\right)\mathbf{=}\left(a+b,a-b\right)\mathbf{;}$

Short answer

It is proved that

(a) $\left(a,b\right)|\left(a+b,a-b\right);$

(b) If $a$ is odd and $b$ is even, then $\left(a,b\right)=\left(a+b,a-b\right);$

(c) If $a$ and $b$ are odd, then$2\left(a,b\right)=\left(a+b,a-b\right);$

Step-by-step solution

Prove part (a)

Assume that $d=\left(a,b\right)$. As $d|a,$and $d|b,$this implies that $d|\left(a+b\right),$and $d|\left(a-b\right)$.

Therefore,$d$ is the common divisor of$\left(a+b\right)$ and $\left(a-b\right)$.

As $d$is the divisor of the greatest common divisor, this implies that $d=\left(a,b\right)|\left(a+b,a-b\right)$

Hence, it is proved that$\left(a,b\right)|\left(a+b,a-b\right)$

Prove part (b)

From part (a) $\left(a,b\right)|\left(a+b,a-b\right)$.Assume that $d=\left(a+b,a-b\right)$, this implies that $a+b=dt$and $a-b=du$. Therefore, $2a=d\left(t+u\right)$.

It is given that $a$is odd and $b$is even. This implies that $d$is odd. Therefore, $d|2a,$this implies that $d|a$. Similarly for $2b=d\left(t-u\right);$

Therefore, $d$is the common divisor of $a$and $b$that is, $d|\left(a,b\right).$

Hence,$\left(a,b\right)=\left(a+b,a-b\right)$

Prove part (c)

From part (a)$\left(a,b\right)|\left(a+b,a-b\right)$. Assume that $d=\left(a+b,a-b\right)$, this implies that $\left(a+b\right)=dt$ and $\left(a-b\right)=du$. Therefore,$2a=d\left(t+u\right)$.

It is given that $a$and $b$are odd, this implies that $a+b$and $a-b$are even. Hence $d$is even. Thus $a=\frac{d}{2}\left(t+u\right),$This implies that $\frac{d}{2}|a$.

Similarly, $\frac{d}{2}|b,$so that role="math" localid="1647368976106" $\frac{d}{2}=\frac{\left(a+b,a-b\right)}{2}\left|\left(a,b\right)\right|\left(a+b,a-b\right)$. Therefore, $\left(a,b\right)=\frac{\left(a+b,a-b\right)}{2}$

As, $\left(a,b\right)$is odd and $\left(a+b,a-b\right)$is even, this implies that $\frac{\left(a+b,a-b\right)}{2}=\left(a,b\right)$

$2\left(a,b\right)=\left(a+b,a-b\right)$

Hence it is proved that if $a$and $b$ are odd, then$2\left(a,b\right)=\left(a+b,a-b\right)$