prove-that-c-5-2-1-3-5-2-1-3-is-a-rational-number-

Question Number 190093 by mnjuly1970 last updated on 27/Mar/23

p r o v e t h a t :

c = {(\sqrt{5} + 2)}^{\frac{1}{3}} - {(\sqrt{5} - 2)}^{\frac{1}{3}}

is a r a t i o n a l number .

Commented by MJS_new last updated on 27/Mar/23

I {don}^{'} t think it is rational

Commented by som(math1967) last updated on 18/May/23

y e s s i r, i t h i n k i t i s n o t r a t i o n a l

Commented by mnjuly1970 last updated on 27/Mar/23

y e s s i r t h a n k s a l o t

i c o r r e t e d i t . \sqrt{5} i n s t e a d o f \sqrt{10}

Answered by MJS_new last updated on 27/Mar/23

\dots = {(2 + \sqrt{5})}^{1 / 3} - {(2 + \sqrt{5})}^{- 1 / 3}

{(2 + \sqrt{5})}^{1 / 3} = \frac{1 + \sqrt{5}}{2}

\Rightarrow c = 1

Answered by som(math1967) last updated on 27/Mar/23

c^{3} = (\sqrt{5} + 2) - (\sqrt{5} - 2) - 3 {(5 - 4)}^{\frac{1}{3}} . c

[c = {(\sqrt{5} + 2)}^{\frac{1}{3}} - {(\sqrt{5} - 2)}^{\frac{1}{3}}]

c^{3} = 4 - 3 c

c^{3} + 3 c - 4 = 0

c^{3} - 1 + 3 (c - 1) = 0

(c - 1) (c^{2} + c + 1 + 3) = 0

(c - 1) (c^{2} + c + 4) = 0

f o r r e a l v a l u e o f c (c^{2} + c + 4) \neq 0

c = 1 r a t i n a l n u m b e r

Commented by mehdee42 last updated on 27/Mar/23

t h a t w a s p e r f e c t .