Proof-that-1-3n-lt-n-2-for-every-positive-integer-n-4-

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

P r o o f t h a t 1 + 3 n < n^{2} f o r e v e r y p o s i t i v e i n t e g e r n ⩾ 4

Answered by MJS_new last updated on 26/May/21

{it}^{'} s wrong for \frac{3 - \sqrt{13}}{2} ⩽ n ⩽ \frac{3 + \sqrt{13}}{2}

Answered by physicstutes last updated on 26/May/21

Thesame as proving for,

0 < n^{2} - 3 n - 1

or n^{2} - 3 n - 1 > 0

{(n - \frac{3}{2})}^{2} - \frac{9}{4} - 1 > 0

{(n - \frac{3}{2})}^{2} - \frac{13}{4} > 0

\forall n \in R, {(n - \frac{3}{2})}^{2} ⩾ 0, but \forall n \in R, ⇏ {(n - \frac{3}{2})}^{2} - \frac{13}{4} > 0

take the case of the the interval posted above .