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Let f be a function such that f(x + y) =...

Let f be a function such that f(x + y) = f(x) + f(y) `forall x , y in R`, if f(1) = k, then show that f(n) is equal to nk.

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Since f(x + y) = f(x) + f(y) , we put x = 1, y = 1
`therefore` f(2) = f(1) + f(1) = 2f(1) = 2k
f(3) = f(2) + f(1) = 2k + k = 3k
Proceeding in this way we get f(n) = nk.
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