\( Se\ considera\ functia\ f:\mathbb{R}- \mathbb{Q} \rightarrow \mathbb{R},f(x)=ax+b,a\not=0. \)
\( Sa\ se\ arate\ ca\ functia\ f\ are\ cel\ putin\ o\ valoare\ irationala. \)
Functie gradul I interesanta
-
Laurian Filip
- Site Admin
- Posts: 344
- Joined: Sun Nov 25, 2007 2:34 am
- Location: Bucuresti/Arad
- Contact:
Daca b e rational: problema se reduce la a arata ca \( g(x)=ax \) are cel putin o valoare irationala. Presupunem prin absurd ca nu are.
\(
g(\sqrt 2) \in \mathbb{Q} \)
\( g(\sqrt 2 \cdot \sqrt 3) \in \mathbb{Q} \)
Echivalent cu
\( \sqrt2a \in \mathbb{Q} \)
\( \sqrt2a \cdot sqrt 3 \in \mathbb{Q} \)
Contradictie!
Daca b e irational:
1) a e irational \( \to f(\frac{1}{a})=b+1 \in \mathbb{R}-\mathbb{Q} \)
2) a e rational \( \to f(b)=b(a+1) \in \mathbb{R}-\mathbb{Q} \)
\(
g(\sqrt 2) \in \mathbb{Q} \)
\( g(\sqrt 2 \cdot \sqrt 3) \in \mathbb{Q} \)
Echivalent cu
\( \sqrt2a \in \mathbb{Q} \)
\( \sqrt2a \cdot sqrt 3 \in \mathbb{Q} \)
Contradictie!
Daca b e irational:
1) a e irational \( \to f(\frac{1}{a})=b+1 \in \mathbb{R}-\mathbb{Q} \)
2) a e rational \( \to f(b)=b(a+1) \in \mathbb{R}-\mathbb{Q} \)