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

$$\mathrm{Express}\sqrt{28}\mathrm{as}\mathrm{continued}\mathrm{fraction}$$

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

$$\sqrt{28}=1+\sqrt{28}-1$$$$=1+\frac{27}{\sqrt{28}+1}=1+\frac{27}{2+\sqrt{28}-1}$$$$=1+\frac{27}{2+\frac{27}{\sqrt{28}+1}}=1+\frac{27}{2+\frac{27}{2+\frac{27}{2+\frac{27}{2+{.}_{{.}_{..}}}}}}$$$$\mathbf{please}\mathbf{check}.$$

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

$$\sqrt{28}=5+\sqrt{28}-5$$$$=5+\frac{3}{\sqrt{28}+5}=5+\frac{3}{10+\sqrt{28}-5}$$$$=5+\frac{3}{10+\frac{3}{\sqrt{28}+5}}=5+\frac{3}{10+\frac{3}{10+\frac{3}{10+\frac{3}{{.}_{{.}_{.}}}}}}$$$$\mathbf{please}\mathbf{check}.$$

Commented by TawaTawa last updated on 16/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 16/Oct/19

Commented by TawaTawa last updated on 16/Oct/19

$$\mathrm{How}\mathrm{did}\mathrm{wolframe}\mathrm{get}\mathrm{this}\mathrm{sir}$$

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

$$\mathbf{let}\mathbf{do}\mathbf{the}\mathbf{back}\mathbf{calculation}$$$$\mathbf{If}\mathbf{x}=\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{10+\mathbf{x}}}}}$$$$\mathbf{by}\mathbf{calculating}\mathbf{we}\mathbf{get}$$$${\mathbf{x}}^2+10\mathbf{x}-3=0$$$$\Rightarrow \mathbf{x}=-5\pm \sqrt{28}$$$$\mathbf{I}\mathbf{dont}\mathbf{think}\mathbf{it}\mathbf{will}\mathbf{be}\mathbf{the}\mathbf{correct}\mathbf{expression}.$$$$\mathbf{Please}\mathbf{anyone}\mathbf{give}\mathbf{some}\mathbf{comments}.$$

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

$$correctis$$$$5+\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{10+\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{10+...}}}}}}}}$$

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

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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

$$5<\sqrt{28}<6$$$$x_0=\sqrt{28},a_0=\lfloor x_0\rfloor =5$$$$x_1=\frac{1}{\sqrt{28}-5}=\frac{\sqrt{28}+5}{3},a_1=\lfloor x_1\rfloor =3$$$$x_2=\frac{1}{\frac{\sqrt{28}+5}{3}-3}=\frac{3}{\sqrt{28}-4}=\frac{\sqrt{28}+4}{4},a_2=\lfloor x_2\rfloor =2$$$$x_3=\frac{1}{\frac{\sqrt{28}+4}{4}-2}=\frac{4}{\sqrt{28}-4}=\frac{\sqrt{28}+4}{3},a_3=\lfloor x_3\rfloor =3$$$$x_4=\frac{1}{\frac{\sqrt{28}+4}{3}-3}=\frac{3}{\sqrt{28}-5}=\sqrt{28}+5,a_4=\lfloor x_4\rfloor =10$$$$x_5=\frac{1}{\sqrt{28}+5-10}=\frac{1}{\sqrt{28}-5}=\frac{\sqrt{28}+5}{3}=x_1,a_4=a_1=3$$$$......$$$$\Rightarrow \sqrt{28}=5+\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{10+\frac{1}{3+\frac{1}{2+\frac{1}{3+\frac{1}{10+...}}}}}}}}$$$$$$$$anotherexample\sqrt{38}$$$$6<\sqrt{38}<7$$$$x_0=\sqrt{38},a_0=\lfloor x_0\rfloor =6$$$$x_1=\frac{1}{\sqrt{38}-6}=\frac{\sqrt{38}+6}{2},a_1=\lfloor x_1\rfloor =6$$$$x_2=\frac{1}{\frac{\sqrt{38}+6}{2}-6}=\frac{2}{\sqrt{38}-6}=\sqrt{38}+6,a_2=\lfloor x_2\rfloor =12$$$$x_3=\frac{1}{\sqrt{38}+6-12}=\frac{1}{\sqrt{38}-6}=\frac{\sqrt{38}+6}{2}=x_1,a_3=a_1=6$$$$......$$$$\Rightarrow \sqrt{38}=6+\frac{1}{6+\frac{1}{12+\frac{1}{6+\frac{1}{12+...}}}}$$

Commented by TawaTawa last updated on 16/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by TawaTawa last updated on 17/Oct/19

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

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

$${\mathrm{X}}^2-{28\mathrm{y}}^2=1?$$

Commented by TawaTawa last updated on 17/Oct/19

$$\mathrm{Yes}\mathrm{sir}$$

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

$$x^2-28y^2=1isacurve,ithasinfinite$$$$solutions(x,y).thishasnothingtodo$$$$withhowyouexpress\sqrt{28}.$$

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

$$\mathrm{It}\mathrm{is}\mathrm{a}\mathrm{long}\mathrm{process}.\mathbf{Please}\mathbf{see}{\mathbf{PELL}}^{\prime }\mathbf{S}\mathbf{equation}.$$$$\mathbf{Or}\mathbf{please}\mathbf{give}\mathbf{me}\mathbf{some}\mathbf{time}.$$

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

$${\mathbf{x}}^2-28{\mathbf{y}}^2=1$$$$\mathbf{now}\sqrt{28}=[5,3,2,3,10]$$$$\mathrm{the}\mathrm{expression}\mathrm{is}\mathrm{periodic}\mathrm{from}10$$$$\therefore \mathrm{the}\mathrm{first}4\mathrm{convergents}\mathrm{are}$$$$\frac{5}{1},\frac{16}{3},\frac{37}{7},\frac{127}{24}=\frac{{\mathbf{p}}_1}{{\mathbf{q}}_1}$$$$\mathbf{thus}\mathbf{the}\mathbf{equation}\mathbf{satisfies}\mathbf{for}$$$$\mathbf{x}=127\mathbf{and}\mathbf{y}=24$$$$\mathbf{please}\mathbf{check}$$$$\mathbf{I}\mathbf{have}\mathbf{the}\mathbf{solutions}\mathbf{only}.\mathbf{For}\mathbf{further}\mathbf{information}$$$$\mathbf{you}\mathbf{have}\mathbf{to}\mathbf{study}\mathbf{the}{\mathbf{Pell}}^{\prime }\mathbf{s}\mathbf{equation}.$$

Commented by TawaTawa last updated on 17/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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