# How to find the square root of 2 using simple algebraic rules

Ever wondered how to discover the value of the square root of 2 using simple algebraic rules?

I’ve been through plenty of maths books and 99.9% of them haven’t demonstrated how this feat can be accomplished.

Today I’m going to share with you the methods which you can use to find the value of irrational numbers as continued fractions, and in the process you will learn how to write the value of the square root of 2 as a continued fraction.

See the mathematics below… ${ x }^{ 2 }=2\\ \\ { x }^{ 2 }+x=2+x\\ \\ x\left( x+1 \right) =\left( x+1 \right) +1\\ \\ \frac { x\left( x+1 \right) }{ \left( x+1 \right) } =\frac { \left( x+1 \right) }{ \left( x+1 \right) } +\frac { 1 }{ \left( x+1 \right) } \\ \\ x=1+\frac { 1 }{ x+1 } \\ \\ x=1+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 }$ $\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } }$ $\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } } \\ \\ x=\sqrt { 2 } \\ \\ \therefore \quad \sqrt { 2 } =1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } }$