Este x rational?
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Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
Este x rational?
Daca \( x\in R \) si \( x^{2003} = x^{2002} + 1 \), atunci \( x \) este rational?
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers
-
Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
Se observa destul de usor ca \( x>1 \).
Daca \( x \in Q \) , atunci \( x= \frac{p}{q} \) unde \( p,q \in Z \) cu \( p>q \) si \( \left\( p,q \right\) =1 \).
Din ipoteza avem: \( \frac {p^{2003}}{q^{2003}} =\frac {p^{2002}}{q^{2002}} + 1 \Rightarrow p^{2003}=q(p^{2002}+q^{2002}) \Rightarrow q|p^{2003} \) contradictie cu \( \left\( p,q \right\) =1 \) \( \Rightarrow x \in {R-Q} \)
Daca \( x \in Q \) , atunci \( x= \frac{p}{q} \) unde \( p,q \in Z \) cu \( p>q \) si \( \left\( p,q \right\) =1 \).
Din ipoteza avem: \( \frac {p^{2003}}{q^{2003}} =\frac {p^{2002}}{q^{2002}} + 1 \Rightarrow p^{2003}=q(p^{2002}+q^{2002}) \Rightarrow q|p^{2003} \) contradictie cu \( \left\( p,q \right\) =1 \) \( \Rightarrow x \in {R-Q} \)
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers