Say a > b >0 . Let gcd(a,b) = d. Now we compute gcd(a - b, b), we claim d is a common divisor. Indeed, d|(a-b) because d|a and d|b. Now we claim that if d' is a larger divisor then d'|(a-b) and d'|b implies d'|(a-b + b) thus d'|a, so d'|a, so d' is a common divisor to a and b so d' <= d.

Same ideal2. if a and b are 2 positive integers show taht gcd(a,b) = gcd(a, a +b)

Use induction.3. show that gcd(fn,fn+1) = 1 for all natural numbers n.

The inductive step if f_n+1 = f_n + f_n-1

But gcd(f_n,f_n-1) = 1 so gcd (f_n+1,f_n)=1 by Euclid's algorithm .