Q-Show-that-the-numbers-3-2-amp-8-cannot-be-terms-of-an-arithmetic-sequence-

Question Number 191525 by mehdee42 last updated on 25/Apr/23

Q : S h o w t h a t t h e n u m b e r s \sqrt{3}, 2 & \sqrt{8} c a n n o t b e t e r m s o f a n a r i t h m e t i c s e q u e n c e .

Answered by mr W last updated on 26/Apr/23

\sqrt{3} < 2 < \sqrt{8}

a s s u m e t h e y a r e t e r m s o f a n A . P .,

t h e n w e h a v e

2 = \sqrt{3} + m d

\sqrt{8} = 2 + n d

w i t h m, n \in N, d \in R .

d = \frac{2 - \sqrt{3}}{m} = \frac{\sqrt{8} - 2}{n}

\Rightarrow 2 (m + n) = m \sqrt{8} + n \sqrt{3}

\Rightarrow 4 {(m + n)}^{2} = 8 m^{2} + 3 n^{2} + 4 m n \sqrt{6}

\Rightarrow \sqrt{6} = \frac{4 {(m + n)}^{2} - 8 m^{2} - 3 n^{2}}{4 m n} = \frac{i n t e g e r}{i n t e g e r}

i . e . \sqrt{6} i s r a t i o n a l, b u t \sqrt{6} i s i n f a c t n o t

r a t i o n a l .

c o n t r a d i c t i o n!

\Rightarrow \sqrt{3}, 2, \sqrt{8} c a n n o t b e t e r m s o f a n A . P .!

Commented by mehdee42 last updated on 27/Apr/23

b e r a v o s i r W

t h e r e s u l t o f c o n t r a d i c t i o n c a n a l s o b e o b t a i n e d f r o m e t h e r e l a t i o n \frac{2 - \sqrt{3}}{m} = \frac{\sqrt{8} - 2}{n} \Rightarrow \frac{2 - \sqrt{3}}{\sqrt{8} - 2} = \frac{m}{n},

b e c a u s e w e h a v e r a t i o n a l n u m b e r o n o n e s i d e a n f d u m b n i m b e r o n t h r o t h e r s i d e

Commented by mehdee42 last updated on 27/Apr/23

a n y w a y, y o u r p r o o f f o r t h i s c o n t r a d i c t i o n i s a p p r i c i a t e d

Commented by mr W last updated on 27/Apr/23

b u t w e c a n n o t g e n e r a l l y s a y t h a t

\frac{i r r a t i o n a l n u m b e r}{i r r a t i o n a l n u m b e r} i s a l s o i r r a t i o n a l .

Commented by mehdee42 last updated on 28/Apr/23

e x a c t l y