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QUESTION

Are the following true or false? Give a proof or a counterexample,
as appropriate.

\begin{description}

\item[(i)]
if gcd$(a,b)=d$, then gcd$(a+d,b)=d$.

\item[(ii)]
If gcd$(a,b)=1$ and $c|a$, then gcd$(c,b)=1$.

\item[(iii)]
If gcd$(a,b)$=gcd$(a,c)=d$, then gcd$(b,c)=d$.

\item[(iv)]
If gcd$(a,b)=1$ and gcd$(a,c)=1$, then gcd$(a,bc)=1$.

\end{description}
(Hint: Corollary 1.5 will help here.)




ANSWER

\begin{description}

\item[(i)]
FALSE: e.g. gcd(10,12)=2, but gcd(10+2,12)=gcde(12,12)=12$\neq2$.

\item[(ii)]
TRUE: Let gcd$(c,b)=d,$ then $d|b$ and $d|c$, and so, since $c|a$,
we have $d|a$ (th.1.3(3)). Thus $d$ is a common divisor of $a$ and
$b$, so $d\leq$gcd$(a,b)=1$. Since, by definition of gcd, we
already know that $d\geq1$, the result follows.

\item[(iii)]
FALSE: e.g. gcd(2,4)=gcd(2,8)=2, but gcd(4,8)=4$\neq$2.

\item[(iv)]
TRUE: gcd$(a,b)$=gcd$(a,c)=1$, so by cor.1.5 we can find integers
$x,\ y,\ u,\ v$ such that $ax+by=1$ and $au+cv=1$. Multiplying
these together and rearranging give:-

$$1=(ax+by)(au+cv)=a(axu+xcv+uby)+bc.yv$$

Thus we have found integers $r=axu+xcv+uby$ and $s=yv$ such that
$1=ar+(bd)s$, and then cor.1.5 gives gcd$(a,bc)=1$ as required.

\end{description}




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