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Real analysis

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Let f be a function defined on all of R that satisfies the additive condition f(x+y)=f(x)+f(y) for all x,y belong to R
a- Show that f(0)=0 and that f(-x)=-f(x) for all x belong to R.

b- Show that if f is continuous at x=0 then f is continuous at every point in R

c- Let k=f(1) show that f f(n)=kn for all n belong to N and then prove that f(z)=kz for all z belong to Z. now prove that f(r)=kr for any rational number r.

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Solution Summary

The solution is a proof regarding continuous functions.

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a. f(0+x)=f(0)+f(x), then we have f(x)=f(0)+f(x). So f(0)=0
f(0)=f(-x+x)=f(-x)+f(x)=0, thus f(-x)=-f(x)
b. Suppose f is continous at 0, then for any e>0, there exists some d>0, such that when |x|<d, ...

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