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Question Number 142085 by Rexzie last updated on 30/May/21

Proof that 1+3n<n^2 for every positive integer n≥4

$${Proof}\:{that}\:\mathrm{1}+\mathrm{3}{n}<{n}^{\mathrm{2}} \:{for}\:{every}\:{positive}\:{integer}\:{n}\geqslant\mathrm{4} \\ $$

Answered by MJS_new last updated on 26/May/21

it′s wrong for ((3−(√(13)))/2)≤n≤((3+(√(13)))/2)

$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{for}\:\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\leqslant{n}\leqslant\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$

Answered by physicstutes last updated on 26/May/21

Thesame as proving for, 0 < n^2 −3n−1 or n^2 −3n−1 > 0 (n−(3/2))^2 −(9/4)−1 >0 (n−(3/2))^2 −((13)/4)>0 ∀ n ∈ R, (n−(3/2))^2 ≥ 0, but ∀ n ∈R, ⇏ (n−(3/2))^2 −((13)/4)> 0 take the case of the the interval posted above.

$$\mathrm{Thesame}\:\mathrm{as}\:\mathrm{proving}\:\mathrm{for}, \\ $$ $$\:\mathrm{0}\:<\:{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{1}\: \\ $$ $$\mathrm{or}\:{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{1}\:>\:\mathrm{0} \\ $$ $$\:\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{1}\:>\mathrm{0} \\ $$ $$\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{4}}>\mathrm{0} \\ $$ $$\forall\:{n}\:\in\:\mathbb{R},\:\:\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\:\mathrm{0},\:\:\mathrm{but}\:\:\forall\:{n}\:\in\mathbb{R},\:\nRightarrow\:\:\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{4}}>\:\mathrm{0} \\ $$ $$\mathrm{take}\:\mathrm{the}\:\mathrm{case}\:\mathrm{of}\:\mathrm{the}\:\mathrm{the}\:\mathrm{interval}\:\mathrm{posted}\:\mathrm{above}. \\ $$

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