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show that there exists a unique function whose domain is {x∈R} and which is both even and odd.
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my guess is \(f(x)=0\) is both showing it is unique is the hard part
but not really that hard
suppose there is an \(x\) for which \(f(x)=a\neq 0\) then \(f(-x)=a\) since \(f\) is even and also \(f(-x)=-a\) since \(f\) is odd, and therefore \(a=-a\) which shows \(a=0\) contradicting the assumption
that's kinda funny though :P
in a funny sort of way, yes
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