exercitiul 5 va rog

presupunem prin absurd ca fractia ar fi reductibila
adica ar exista un numar k∈N*,k≠1 asa fel incat k|3n+4 si k|5n+7
k|3n+4⇒k|5*(3n+4), k|15n+20
k|5n+7⇒k|3* (5n+7), k|15n+21
atunci k|15n+21-(15n+20)
deci k|1 ⇒k=1
dar noi am presupus k≠1, contradictie, deci presupunerea noastra a fost gresita, deci NU exista k≠1 care sa divida si numaratorul si numitorul⇔fractia esate ireductibila, cerinta

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Аноним Аноним · Answer 2 · 2017-01-20T01:29:12+02:00

[tex]\displaystyle \frac{3n+4}{5n+7} \\ \\ Fie~d=(3n+4,5n+7) \\ \\ \left \{ {{d|(3n+4)\Rightarrow d|5(3n+4)\Rightarrow d|(15n+20)} \atop {d|(5n+7) \Rightarrow d|3(5n+7) \Rightarrow d|(15n+21)}} \right. \Rightarrow \\ \\ \Rightarrow d|[(15n+21)-(15n+20)] \Rightarrow d|(15n+21-15n-20) \Rightarrow \\ \\ \Rightarrow d|(21-20) \Rightarrow d|1 \Rightarrow d=1 \\ \\ (3n+4,5n+7)=1 \\ \\ \Rightarrow \frac{3n+4}{5n+7} ~este~ireductibila[/tex]