Prove that `gcd(a+b , a-b)=1` or `gcd(a+b , a-b)=2` if `gcd(a , b)=1`

Check the commutativity and associativity of * on N defined by a*b= gcd(a ,\ b) for all a ,\ b in N .

If x^(2)y+y^(3)=2 then the value of (d^(2)y)/(dx^(2)) at the point (1,1) is (-(a)/(b)) where g.c.d(a, b)=1 then a+b=

Let D = {1, 2, 3, 5, 6, 10, 15, 30}. Define the operations '+', '*' and '"'"' on D as follows a + b = LCM (a, b), a*b = GCD (a, b) and a'=30/a . Then (15'+6)*10 is equal to :

Check for commutative and associative (NN,**) where a**b=gcd(a,b) for all a,binNN .

Check the commutativity and associativity of * on Q defined by a*b=gcd(a,b) for all a,b in N.

If the normal to the ellipse (x)/(18)+(y)/(8)=1 at point (3,2) is ax-by-c=0 .(where gcd(a,b,c)=1) .Then the value of a+b+c ?

The HCF (GCD) of a, b is 12, a, b are positive integers and a gt b gt 12 . The smallest value of (a, b) are respectively