(a) For all positive integers $n$, we have gcd$(2n−1, n) = 1$.
For this, first I tried the Euclidean Algorithm. I divided $2n-1$ by $n$, and got $n-(1/n)$. Then, I didn't know what to do with $1/n$.
I also tried expressing them by a multiple. Say the gcd of them is $d$, so $2n-1=dk$ and $n=dl$ for natural numbers $k,l$. I eventually arrived at $d(2l-k)=1$, and didn't know what to do.
(b) For all positive integers $n$, we have gcd$(4n−2, n) = 2$.
*(b) is solved.