Category Archives: Continued Fractions

Another way to express the golden ratio mathematically

In this post I’m going to be proving that…

\varphi =\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }

So, here I go…

x=\frac { 1+\sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 1+2\sqrt { 5 } +5 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 4 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2 }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 }

Wait for it…

\Rightarrow \quad { x }^{ 2 }-1=1\cdot x\\ \\ \Rightarrow \quad { x }^{ 2 }-1=x\\ \\ \Rightarrow \quad { x }^{ 2 }=x+1\\ \\ \Rightarrow \quad \frac { { x }^{ 2 } }{ x } =\frac { x }{ x } +\frac { 1 }{ x } \\ \\ \Rightarrow \quad x=1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ x } } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } } \\ \\ \therefore \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

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+... }  }  }  }