Question Number 165339 by mnjuly1970 last updated on 30/Jan/22
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:#\:{Advanced}\:\:\:{Calculus}\:#\:\:\: \\ $$$$\:\:\:\:\:\mathrm{L}{et}\:,\:\:{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{Q}\:\:{is}\:\:{a}\:\:{continuous} \\ $$$$\:\:\:\:\:\:{function}\:\:.\:{prove}\:{that}\:\:''\:{f}\:''\:{is}\:{a} \\ $$$$\:\:\:\:\:\:\:{constant}\:{function}\:.\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:\:\:\:\ast\:{Adopted}\:{from}\:{mathematical}\:{analysis}\:{book}\:\:\ast\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−−−−−−− \\ $$$$ \\ $$
Answered by mahdipoor last updated on 30/Jan/22
![proof by costradiction: if for ∀a<b , f(a)≠f(b) ∃k∈Q^′ that between f(a) and f(b) ⇒f is continuous function , medium value theorem : ∃c∈[a,b] that f(c)=k∈Q^′ but f:R→Q (and can not f(c)∈Q^′ ) therefor , for ∀a<b , f(a)=f(b) ; this means that f is constant function.](Q165342.png)
$${proof}\:{by}\:{costradiction}: \\ $$$${if}\:{for}\:\forall{a}<{b}\:,\:{f}\left({a}\right)\neq{f}\left({b}\right)\: \\ $$$$\exists{k}\in\mathrm{Q}^{'} \:{that}\:{between}\:{f}\left({a}\right)\:{and}\:{f}\left({b}\right) \\ $$$$\Rightarrow{f}\:{is}\:{continuous}\:{function}\:, \\ $$$${medium}\:{value}\:{theorem}\:: \\ $$$$\exists{c}\in\left[{a},{b}\right]\:{that}\:{f}\left({c}\right)={k}\in\mathrm{Q}^{'} \\ $$$${but}\:{f}:\mathrm{R}\rightarrow\mathrm{Q}\:\left({and}\:{can}\:{not}\:{f}\left({c}\right)\in\mathrm{Q}^{'} \right)\: \\ $$$${therefor}\:,\:{for}\:\forall{a}<{b}\:,\:{f}\left({a}\right)={f}\left({b}\right)\:; \\ $$$${this}\:{means}\:{that}\:{f}\:{is}\:{constant}\:{function}. \\ $$
Commented by mnjuly1970 last updated on 30/Jan/22
$$\:\:{bravo}\:{sir}\:...{zende}\:{bashid} \\ $$
Commented by mahdipoor last updated on 30/Jan/22
$$\left.{payande}\:{bashid}\::\right) \\ $$