Question

$Fie~a\in~Q*~o~aproximare~~pentru~ \sqrt{3}.Sa~se~demonstreze~ \\ ca~numarul~ \frac{a+3}{a+1} ~este~o~aproximare~mai~buna~a~lui~ \sqrt{3} .$

Answer 1

Am atasat rezolvarea:

Answer 2

$\displaystyle Faptul~ca~x~este~o~aproximatie~mai~buna~decat~y~pentru~c \\ \\ inseamna~ca~x~este~mai~apropiat~de~c. \\ \\ Matematic:~|x-c|\ \textless \ |y-c|. \\ \\ Deci~noi~trebuie~sa~demonstram~inegalitatea~ \\ \\ \left| \frac{a+3}{a+1}- \sqrt{3} \right| \ \textless \ |a- \sqrt{3}|. \\ \\ Aceasta~este~echivalenta~cu~ \left | \frac{(a- \sqrt{3})(1- \sqrt{3})}{a+1} \right| \ \textless \ |a- \sqrt{3}|,~sau, \\ \\ cum~|a- \sqrt{3}|\ \textgreater \ 0,~inegalitatea~devine~|1- \sqrt{3}|\ \textless \ |a+1|. \\ \\ Adica~\sqrt{3}-1\ \textless \ |a+1|.$

$\displaystyle Echivalent~cu~a+1\ \textgreater \ \sqrt{3}-1~SAU~a+1\ \textless \ 1- \sqrt{3}. \\ \\ Prima~afirmatie~este~adevarata,~caci~a \in \mathbb{Q_+};~deci~problema \\ \\ ste~rezolvata.$