Question

sa se arate ca √6+4√2 apartine {a+b√2| a,b apartine Z}.

Answer 1

√6+4√2=√3*√2+4√2=√2(√3+4)=0+(4+√3)√2
a=0∈Z
b=4+√3∉Z

deci√6+4√2 NU APARTINE multimii {a+b√2| a,b apartine Z}.


terxtul nu e acesta
ca cerinta sa fie ADEVARATA
 sau e 6 si nu √6
 sau se cere sa se arate ca NU APARTINE

Answer 2

$\displaystyle Sa~presupunem~ca~exista~a,b \in \mathbb{Z}~astfel~incat~\\ \\ \sqrt{6}+ 4\sqrt{2}=a+b \sqrt{2}. \\ \\ Atunci~\sqrt{6}+ \underbrace{(4-b)}_{k} \sqrt{2}=a. \\ \\ Deci~\sqrt{6}+k \sqrt{2} \in \mathbb{Z}.~Atunci~si~(\sqrt{6}+k \sqrt{2})^2 \in \mathbb{Z} \Leftrightarrow \\ \\ \Leftrightarrow 6+4k \sqrt{3}+2k^2 \in \mathbb{Z} \Rightarrow 4k \sqrt{3} \in \mathbb{Z} \Rightarrow k=0~(caci~k=4-b \in \mathbb{Z}). \\ \\ Rezulta~b=4,~si~deci~a= \sqrt{6} \notin \mathbb{Z}.$

$\displaystyle Prin~urmare~ \sqrt{6}+4 \sqrt{2} \notin \{a+b \sqrt{2}~|~a,b \in \mathbb{Z} \}.$