Identity function

[Papapetros Vaggelis]

Let \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}\) be a strictly increasing function in \(\displaystyle{\mathbb{R}}\) .

If \(\displaystyle{f(r)=r\,,\forall\,r\in\mathbb{Q}}\), then prove that \(\displaystyle{f(x)=x\,,\forall\,x\in\mathbb{R}}\) .

[dr.tasos]

suppose there is one $$ p \in R-Q $$ such that :

$$ f(p) \neq p $$


so for every $$ c>0 $$


the neiborhood $$ (p-c,p+c) $$

contains at least two rational numbers one of them $$ r_{1} \in (p-c,p) $$
and the other one $$ r_{2} \in (p,p+c) $$

we can see that $$ f(r_{1})<f(p)<f(r_{2}) \leftrightarrow r_{1}<f(p)<r_{2} \leftrightarrow r_{1}-p<f(p)-p<r_{2}-p \rightarrow -c<f(p)-p<c \rightarrow |f(p)-p|<c \quad \forall c>0 $$

That means that $$ f(p)-p=0 \Rightarrow f(p)=p $$ which is a contradiction .

We conclude that $$ f(x)=x , \quad \forall x \in R $$