Proof using continuous f(x+y)=f(x)+f(y)

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SUMMARY

The discussion centers on proving that a continuous function f: ℝ → ℝ, satisfying the functional equation f(x+y) = f(x) + f(y), adheres to the property f(n) = n*f(1) for all natural numbers n. The proof employs mathematical induction, starting with the base case f(0) = 0, which is established by showing that f(0) = f(0 + 0) = f(0) + f(0) leads to f(0) = 0. The inductive step confirms that if f(n) = n*f(1), then f(n+1) = (n+1)*f(1), completing the proof.

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Homework Statement



Let f be a continuous function lR (all real numbers) --> lR such that f(x+y) = f(x) + f (y) for x, y in lR.
prove that f(n) = n*f(1) for all n in lN (all natural numbers)

Homework Equations



f is continuous

also note and prove that f(0) = 0

The Attempt at a Solution



Edit:
I figured out the general proof using induction, assuming that the base case f(0) is true:

Assuming f(n) = nf(1), prove that f(n+1) = (n+1)f(1):

we know f(x+y) =f(x) + f (y)
so f(n+1) =f(n) + f (1)
and according to inductive hypothesis, f(n) = nf(1)
s0 f(n) + f (1)
= nf(1) + f (1)
=(n+1)f(1).

But I still don't understand why the base case f(0) = 0 is true..?
 
Last edited:
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f(0) = f(0 + 0) = f(0) + f(0) => f(0) = 0.
 

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