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Question Number 133072 by bemath last updated on 18/Feb/21

Show that (5/(2−(3)^(1/4) )) is in F_2 by expressing the number in form a_1 +b_1 (√k_1 ) where a_1 ,b_1 , k_1 are in F_1

$$\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{5}}{\mathrm{2}−\sqrt[{\mathrm{4}}]{\mathrm{3}}}\:\mathrm{is}\:\mathrm{in}\:\mathrm{F}_{\mathrm{2}} \:\mathrm{by}\:\mathrm{expressing} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{in}\:\mathrm{form}\:{a}_{\mathrm{1}} +{b}_{\mathrm{1}} \sqrt{{k}_{\mathrm{1}} }\:\mathrm{where} \\ $$$${a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,\:{k}_{\mathrm{1}} \:{are}\:{in}\:{F}_{\mathrm{1}} \\ $$

Answered by EDWIN88 last updated on 18/Feb/21

(5/(2−(3)^(1/4) )) ×((2+(3)^(1/4) )/(2+(3)^(1/4) )) = ((10+5 (3)^(1/4) )/(4−(9)^(1/4) )) × ((4+(9)^(1/4) )/(4+(9)^(1/4) )) = ((40+10 (9)^(1/4) +20 (3)^(1/4) +5((27))^(1/4) )/(13)) = ((40+10(√3) +20 (3)^(1/4) +5 (√3) (3)^(1/4) )/(13)) = ((40+10(√3))/(13)) + (((20+5(√3))/(13))) (3)^(1/4) then a_1 =((40+10(√3))/(13)) ; b_1 = ((20+5(√3))/(13)) ; k_1 =(√3)

$$\:\frac{\mathrm{5}}{\mathrm{2}−\sqrt[{\mathrm{4}}]{\mathrm{3}}}\:×\frac{\mathrm{2}+\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{2}+\sqrt[{\mathrm{4}}]{\mathrm{3}}}\:=\:\frac{\mathrm{10}+\mathrm{5}\:\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{4}−\sqrt[{\mathrm{4}}]{\mathrm{9}}}\:×\:\frac{\mathrm{4}+\sqrt[{\mathrm{4}}]{\mathrm{9}}}{\mathrm{4}+\sqrt[{\mathrm{4}}]{\mathrm{9}}} \\ $$$$\:=\:\frac{\mathrm{40}+\mathrm{10}\:\sqrt[{\mathrm{4}}]{\mathrm{9}}\:+\mathrm{20}\:\sqrt[{\mathrm{4}}]{\mathrm{3}}\:+\mathrm{5}\sqrt[{\mathrm{4}}]{\mathrm{27}}}{\mathrm{13}} \\ $$$$=\:\frac{\mathrm{40}+\mathrm{10}\sqrt{\mathrm{3}}\:+\mathrm{20}\:\sqrt[{\mathrm{4}}]{\mathrm{3}}\:+\mathrm{5}\:\sqrt{\mathrm{3}}\:\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{13}} \\ $$$$=\:\frac{\mathrm{40}+\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{13}}\:+\:\left(\frac{\mathrm{20}+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)\:\sqrt[{\mathrm{4}}]{\mathrm{3}} \\ $$$$\mathrm{then}\:{a}_{\mathrm{1}} =\frac{\mathrm{40}+\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{13}}\:;\:{b}_{\mathrm{1}} =\:\frac{\mathrm{20}+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{13}}\:;\:{k}_{\mathrm{1}} =\sqrt{\mathrm{3}}\: \\ $$$$ \\ $$

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