Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a strictly increasing function in $\mathbb{R}$. If $f(r)=r, \; \forall r \in \mathbb{Q}$ prove that $f(x)=x , \; \forall x \in \mathbb{R}$.
Solution
Let $x$ be an irrational number. If $f(x)<x$ , then we choose a rational $r$ such $f(x)<r<x$ (density of rationals). From the last equation and from the assumption that $f$ is strictly increasing we have that:
$$r = f(r)< f(x)< r$$
leading to a contradiction. Similarly, if we assume $f(x)>x$ we will also be led to a contradiction. The result follows.
The exercise can also be found in mathematica.gr
Solution
Let $x$ be an irrational number. If $f(x)<x$ , then we choose a rational $r$ such $f(x)<r<x$ (density of rationals). From the last equation and from the assumption that $f$ is strictly increasing we have that:
$$r = f(r)< f(x)< r$$
leading to a contradiction. Similarly, if we assume $f(x)>x$ we will also be led to a contradiction. The result follows.
The exercise can also be found in mathematica.gr
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