# The Square Root Of 3 Derivation

$x=\sqrt { 3 } \\ \\ { x }^{ 2 }=3\\ \\ { x }^{ 2 }+x=3+x\\ \\ x\left( 1+x \right) =\left( 1+x \right) +2\\ \\ \frac { x\left( 1+x \right) }{ \left( 1+x \right) } =\frac { \left( 1+x \right) }{ \left( 1+x \right) } +\frac { 2 }{ \left( 1+x \right) } \\ \\ x=1+\frac { 2 }{ 1+x } \\ \\ \sqrt { 3 } =1+\frac { 2 }{ 1+x } \\ \\ =1+\frac { 2 }{ 1+x }$

$\\ \\ =1+\frac { 2 }{ 1+\left( 1+\frac { 2 }{ 1+x } \right) } \\ \\ =1+\frac { 2 }{ 2+\frac { 2 }{ 1+x } } \\ \\ =1+\frac { 2 }{ 2+\frac { 2 }{ 1+\left( 1+\frac { 2 }{ 1+x } \right) } } \\ \\ =1+\frac { 2 }{ 2+\frac { 2 }{ 2+\frac { 2 }{ 1+x } } } \\ \\ =1+\frac { 2 }{ 2+\frac { 2 }{ 2+\frac { 2 }{ 1+\left( 1+\frac { 2 }{ 1+x } \right) } } }$

$\\ \\ =1+\frac { 2 }{ 2+\frac { 2 }{ 2+\frac { 2 }{ 2+\frac { 2 }{ 1+x } } } } \\ \\ \therefore \sqrt { 3 } =1+\frac { 2 }{ 2+\frac { 2 }{ 2+\frac { 2 }{ 2+... } } }$