Question

Fix a positive real number c, and let S = (−c, c) ⊆ R. Consider the formula x ∗ y :=(x + y)/(1 + xy/c^2).

(a)Show that this formula gives a well-defined binary operation on S (I think it is equivalent to say that show the domain of x*y is in (-c,c), but i dont know how to prove that)

(b)this operation makes (S, ∗) into an abelian group. (I have already solved this, you can just ignore)

(c)Explain why the formula does not define a group structure on R. (to have a group structure, need associativity, identity element and inverses, I think these three parts are still right on R, where am I wrong?)

Answer 1

(a) x * y = (x+y)/(1+xy/c²)

We need to prove that x * y is in S = (-c,c) or -c < x*y < c

Now, notice that 1+xy/c² >0 for any x,y in S =(-c,c), so that x * y R

Now -c < x*y < c iff | x * y| < c

| x*y | < c iff | (x+y)/(1+xy/c²) | < c

iff | x+y| <c (1+xy/c²) ...{ as 1+xy/c² >0 }

iff |x+y| < c + xy/c

iff c | x+y| < c² + xy ..{ Multiplying by c on both sides}

Now if x+y >=0 , then | x *y | < c iff (c-x)(c-y) >0 which is true

If x + y <0 then | x*y| <c iff c² + xy>0

Also 0 is an identity element here and -x is inverse

Hence x * y is a binary operation on S

(b) Now take x,y in S

x *y = (x+y)/(1+xy/c²)

Now y *x = (y+x) / (1+yx/c²) = (x+y)/(1+xy/c²) { because x and y are in R and R is abelian under multiplication and adition}

=> y * x = x* y

Hence, S is an abelian group

(c) Not every element has inverse in R to define a group structure