MHB Prove $f(x)=Cx$ for All $x$: Functional Equation

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The discussion centers on proving that a continuous function \( f(x) \) satisfying the equation \( f(a+b) = f(a) + f(b) \) for all \( a \) and \( b \) must take the form \( f(x) = Cx \), where \( C = f(1) \). It is established that \( f(x) = Cx \) holds true for rational numbers. The continuity of \( f \) is crucial, as every real number can be approximated by a sequence of rational numbers, allowing the extension of the result to all real numbers. The participants emphasize the significance of continuity in the proof, noting that without it, functions could take on more complex forms. Ultimately, the continuity requirement is essential for ensuring that the functional equation holds for all \( x \).
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Suppose $f(x)$ is continuous for all $x$ and $f(a+b)=f(a)+f(b)$ for all $a$ and $b$. Prove that $f(x)=Cx$, where $C=f(1)$.

I have shown that $f(x)=Cx$ for all rational numbers. How do I use the continuity of $f$ to show it is true for all $x$?
 
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Alexmahone said:
I have shown that $f(x)=Cx$ for all rational numbers.
Having this, remember that rationals are dense in $\mathbb R.$
 
Krizalid said:
Having this, remember that rationals are dense in $\mathbb R.$

Intuitively, I can see that it must be true but I'm having trouble proving it.
 
Alexmahone said:
Intuitively, I can see that it must be true but I'm having trouble proving it.
Every real number is the limit of a sequence of rational numbers.
The function is continuous. What continuity and convergent sequences?
 
Plato said:
Every real number is the limit of a sequence of rational numbers.
The function is continuous. What continuity and convergent sequences?

Got it. Thanks!
 
By the way, if you do not include the requirement that the function be continuous, all f are either of the form f(x)= cx or the graph of y= f(x) is dense in the plane.
 

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