# Demonstrate that function f is bijective and odd if (f(f(m)))=-m, m integer number? f is defined on integer domain with integer range

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### 2 Answers

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(1)=-2,f(2)=1,`

`f(3)=-4,f(4)=3,`

`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).**