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.
Let's check the axioms:
1)Associativity:
"\\forall\\ f,g,h\\in G: ((f+g)+h)(x)=\n\\\\=(f+g)(x)+h(x)=f(x)+g(x)+h(x)=\n\\\\=f(x)+(g+h)(x)=(f+(g+h))(x)"
2)Commutativity:
"\\forall\\ f,g\\in G: (f+g)(x)=f(x)+g(x)\n\\\\=g(x)+f(x)=(g+f)(x)"
3)Identity element:
"\\mathbb{O}\\equiv0; \\forall\\ f\\in G: (f+\\mathbb{O})(x)=\n\\\\= f(x)+\\mathbb{O}(x)=f(x)"
4)Inverse element:
"\\forall\\ f\\in G\\ \\exist (-f): (-f)(x)=-f(x)\\\\\n(f+(-f))(x)=f(x)-f(x)=0=\\mathbb{O}(x)"
So it is Abelian group
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