2 Answers | Add Yours
Suppose such a function `f` exists. I'll give an example of such a function at the end of the post, but even if none existed, we could still assume one did and see what we could say about it.
First we'll show that `f` is odd. Pick any integer `m` and let `f(m)=k,` which should make the next line easier to follow.
`f(-m)=f(f(f(m)))=f(f(k))=-k=-f(m),` so `f` is odd.
Now we'll show that `f` is injective. Suppose that `f(m_1)=f(m_2),` so that `-m_1=f(f(m_1))=f(f(m_2))=-m_2,` which implies that `m_1=m_2.`
Finally, we need to prove surjectivity. Pick any integer `j,` and let `f(-j)=s.` Then `f(s)=f(f(-j))=j,` so `f` is surjective.
An example of such a function is given by `f(0)=0,`
`f(5)=-6,f(6)=5,...,` and so on, where the fact that we want `f(-m)=-f(m)` defines `f` for negative values.
The function f(m) follows the relation f(f(m)) = -m.
If f(m) = -m
f(f(m)) = f(-m) = -(-m) = m
If f(m) = m
f(f(m)) = f(m) = m
The condition f(f(m)) = -m is not true for any function f(m).
We’ve answered 333,958 questions. We can answer yours, too.Ask a question