Prove that $$ \frac{2}{99},\frac{3}{98},...,\frac{97}{4},\frac{98}{3},\frac{99}{2}% $$ are irreducible.
My attempt is: if $a/b$ is irreducible, than $\left( a,b\right) =1$. Now, I choose $a=k,k=2,3,...,99$ and $b=101-k$, hence we need to show that $$ \left( k,101-k\right) =1,~k=2,3,...,99. $$ Since $\left( a,b\right) =\left( b,a\right) $, it is sufficient to show that $\left( k,101-k\right) =1$ for all $k=2,3,...,49$. I know that $$ \left( a,b\right) =\left\{ \begin{array} [c]{cc}% \left( a-b,b\right) , & a>b\\ \left( a,b-a\right) , & a<b \end{array} \right. . $$
If $k=2,3,4,...,49,$ then $101-k>k$ and $$ \left( k,101-k\right) =\left( k,101-2k\right) =... $$ Here is where I don't know how to continue.
More simply, if $k$ and $101-k$ had a common factor $d>1$, would divide $k+101-k=101$. This can't happen, as $101$ is prime.