Express-28-as-continued-fraction-

Question Number 71518 by TawaTawa last updated on 16/Oct/19

Express \sqrt{28} as continued fraction

Commented by Prithwish sen last updated on 16/Oct/19

\sqrt{28} = 1 + \sqrt{28} - 1

= 1 + \frac{27}{\sqrt{28} + 1} = 1 + \frac{27}{2 + \sqrt{28} - 1}

= 1 + \frac{27}{2 + \frac{27}{\sqrt{28} + 1}} = 1 + \frac{27}{2 + \frac{27}{2 + \frac{27}{2 + \frac{27}{2 + ._{._{. .}}}}}}

please check .

Commented by Prithwish sen last updated on 16/Oct/19

\sqrt{28} = 5 + \sqrt{28} - 5

= 5 + \frac{3}{\sqrt{28} + 5} = 5 + \frac{3}{10 + \sqrt{28} - 5}

= 5 + \frac{3}{10 + \frac{3}{\sqrt{28} + 5}} = 5 + \frac{3}{10 + \frac{3}{10 + \frac{3}{10 + \frac{3}{._{._{.}}}}}}

please check .

Commented by TawaTawa last updated on 16/Oct/19

God bless you sir

Commented by TawaTawa last updated on 16/Oct/19

Commented by TawaTawa last updated on 16/Oct/19

How did wolframe get this sir

Commented by Prithwish sen last updated on 16/Oct/19

let do the back calculation

If x = \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \frac{1}{10 + x}}}}

by calculating we get

x^{2} + 10 x - 3 = 0

\Rightarrow x = - 5 \pm \sqrt{28}

I dont think it will be the correct expression .

Please anyone give some comments .

Commented by mr W last updated on 16/Oct/19

c o r r e c t i s

5 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \frac{1}{10 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \frac{1}{10 + \dots}}}}}}}}

Commented by Prithwish sen last updated on 17/Oct/19

Thank you sir .

Answered by mr W last updated on 16/Oct/19

5 < \sqrt{28} < 6

x_{0} = \sqrt{28}, a_{0} = ⌊ x_{0} ⌋ = 5

x_{1} = \frac{1}{\sqrt{28} - 5} = \frac{\sqrt{28} + 5}{3}, a_{1} = ⌊ x_{1} ⌋ = 3

x_{2} = \frac{1}{\frac{\sqrt{28} + 5}{3} - 3} = \frac{3}{\sqrt{28} - 4} = \frac{\sqrt{28} + 4}{4}, a_{2} = ⌊ x_{2} ⌋ = 2

x_{3} = \frac{1}{\frac{\sqrt{28} + 4}{4} - 2} = \frac{4}{\sqrt{28} - 4} = \frac{\sqrt{28} + 4}{3}, a_{3} = ⌊ x_{3} ⌋ = 3

x_{4} = \frac{1}{\frac{\sqrt{28} + 4}{3} - 3} = \frac{3}{\sqrt{28} - 5} = \sqrt{28} + 5, a_{4} = ⌊ x_{4} ⌋ = 10

x_{5} = \frac{1}{\sqrt{28} + 5 - 10} = \frac{1}{\sqrt{28} - 5} = \frac{\sqrt{28} + 5}{3} = x_{1}, a_{4} = a_{1} = 3

\dots \dots

\Rightarrow \sqrt{28} = 5 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \frac{1}{10 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \frac{1}{10 + \dots}}}}}}}}

a n o t h e r e x a m p l e \sqrt{38}

6 < \sqrt{38} < 7

x_{0} = \sqrt{38}, a_{0} = ⌊ x_{0} ⌋ = 6

x_{1} = \frac{1}{\sqrt{38} - 6} = \frac{\sqrt{38} + 6}{2}, a_{1} = ⌊ x_{1} ⌋ = 6

x_{2} = \frac{1}{\frac{\sqrt{38} + 6}{2} - 6} = \frac{2}{\sqrt{38} - 6} = \sqrt{38} + 6, a_{2} = ⌊ x_{2} ⌋ = 12

x_{3} = \frac{1}{\sqrt{38} + 6 - 12} = \frac{1}{\sqrt{38} - 6} = \frac{\sqrt{38} + 6}{2} = x_{1}, a_{3} = a_{1} = 6

\dots \dots

\Rightarrow \sqrt{38} = 6 + \frac{1}{6 + \frac{1}{12 + \frac{1}{6 + \frac{1}{12 + \dots}}}}

Commented by TawaTawa last updated on 16/Oct/19

God bless you sir . Thanks for your time sir .

Commented by TawaTawa last updated on 17/Oct/19

How can i use it to find the solution to x^{2} - {28 y}^{2} = 1

Commented by mind is power last updated on 17/Oct/19

X^{2} - {28 y}^{2} = 1 ?

Commented by TawaTawa last updated on 17/Oct/19

Yes sir

Commented by mr W last updated on 17/Oct/19

x^{2} - 28 y^{2} = 1 i s a c u r v e, i t h a s i n f i n i t e

s o l u t i o n s (x, y) . t h i s h a s n o t h i n g t o d o

w i t h h o w y o u e x p r e s s \sqrt{28} .

Commented by Prithwish sen last updated on 17/Oct/19

It is a long process . Please see {PELL}^{'} S equation .

Or please give me some time .

Commented by Prithwish sen last updated on 17/Oct/19

x^{2} - 28 y^{2} = 1

now \sqrt{28} = [5, 3, 2, 3, 10]

the expression is periodic from 10

∴ the first 4 convergents are

\frac{5}{1}, \frac{16}{3}, \frac{37}{7}, \frac{127}{24} = \frac{p_{1}}{q_{1}}

thus the equation satisfies for

x = 127 and y = 24

please check

I have the solutions only . For further information

you have to study the {Pell}^{'} s equation .

Commented by TawaTawa last updated on 17/Oct/19

God bless you sir

God bless you sir