think that $n$ is a natural number and $A$ is the set of the divisors of $n$. Let $f:A \longrightarrow A$ be a function such that if $a \mid b$ then $f(a) \mid f(b)$. Prove that there is a $m$ that $f(m)=m$.
We have learned this kind of questions in the previous class then I cannot do anything.And because of that please give full answers.Thanks.
On
Induction. The base step, for $n=1$, is trivial.
We fix $n$ and assume that the statement is true for numbers lesser than $n$.
If $f(1)=1$ we are done. Otherwise, assume that $d=f(1)>1$. Then $d$ must be a divisor of $n$, because $d\mid f(1)\in A$. The number $d$ also divides $f(a)$ for every $a\in A$.
Let $B$ be the set of the divisors of $n/d$. Define $$f_d(b)=\frac1df(db)$$ for $b\in B$. The function $f_d:A\to A$ is well defined and meets the property. Since $n/d<n$, we can apply the induction hypothesis to say that there is some $m_d$ such that $f_d(m_d)=m_d$, that is, $$m_d=\frac1df(dm_d),$$ QED
Recursive proof:
$n=1$ $f(1)=1$. done.
Suppose $f(n)=n$ done.
Suppose $f(n)=m<n$, for every $a|n, f(a)|m, f(a)$ is an element of $A_m$ the set of divisors of $m$. Let $g$ be the restriction of $g$ to $A_m$, if $a,b\in A_m, a|b, g(a)=f(a)$ divides $f(b)=g(b)$. Recursively, we deduce that there exists $p\in A_m$ such that $g(p)=f(p)=p$.