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

It is given that:

f: R → R is a continuous function such that:

 ∀  x,y ∈ R

Now, let us assume f(1)=k

Also,

(  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)=0

i.e.

f(0)=0  )

Also,

i.e.

f(2)=f(1)+f(1)         ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

f(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)=mk

Now,

Also,

i.e.  

Then,

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

)

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence such that:

and

( since belongs to Q )

Let f is continuous at x=α

This means that:

This means that:

                       This means that:

                    f(x)=kx for every x∈ R

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )