Question

Let G be the set of all real-valued functions on the real line with the binary operation given by pointwise addition of functions: If f, g ∈ G, then f + g is the function whose value at x ∈ R is f (x) + g (x), that is (f + g) (x) = f (x) + g (x). Show that G is a group.

Answer 1

Let's check the axioms:

1)Associativity:

$\forall\ f,g,h\in G: ((f+g)+h)(x)= \\=(f+g)(x)+h(x)=f(x)+g(x)+h(x)= \\=f(x)+(g+h)(x)=(f+(g+h))(x)$

2)Commutativity:

$\forall\ f,g\in G: (f+g)(x)=f(x)+g(x) \\=g(x)+f(x)=(g+f)(x)$

3)Identity element:

$\mathbb{O}\equiv0; \forall\ f\in G: (f+\mathbb{O})(x)= \\= f(x)+\mathbb{O}(x)=f(x)$

4)Inverse element:

$\forall\ f\in G\ \exist (-f): (-f)(x)=-f(x)\\ (f+(-f))(x)=f(x)-f(x)=0=\mathbb{O}(x)$

So it is Abelian group