Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR such that f(x) = kx, for every XER

Answer with explanation:It is given that:f: R → R is a continuous function such that:[tex]f(x+y)=f(x)+f(y)------(1)[/tex]  ∀  x,y ∈ RNow, let us assume f(1)=kAlso,f(0)=0(  Since,f(0)=f(0+0)i.e.f(0)=f(0)+f(0)By using property (1)Also,f(0)=2f(0)i.e.2f(0)-f(0)=0i.e.f(0)=0  )Also,f(2)=f(1+1)i.e.f(2)=f(1)+f(1)         ( By using property (1) )i.e.f(2)=2f(1)i.e.f(2)=2kSimilarly for any m ∈ Nf(m)=f(1+1+1+...+1)i.e.f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)i.e.f(m)=mf(1)i.e.f(m)=mkNow,[tex]f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}[/tex]Also,when x∈ Qi.e.  [tex]x=\dfrac{p}{q}[/tex]Then,[tex]f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q[/tex](Now, as we know that:Q is dense in R.so Э x∈ Q' such that Э a seq [tex]<x_n>[/tex] belonging to Q such that:[tex]<x_n>\to x[/tex] )Now, we know that: Q'=RThis means that:Э α ∈ Rsuch that Э sequence [tex]a_n[/tex] such that:[tex]a_n\ belongs\ to\ Q[/tex]and[tex]a_n\to \alpha[/tex][tex]f(a_n)=ka_n[/tex]( since [tex]a_n[/tex] belongs to Q )Let f is continuous at x=αThis means that:[tex]f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha[/tex]This means that:[tex]f(\alpha)=k\alpha[/tex]                        This means that:                     f(x)=kx for every x∈ R