# More ways in which to express the golden ratio

In this blog post I’ll be revealing more ways (4 in fact) in which to express or come up with the value of the golden ratio

Number One: $\varphi =\frac { a+b }{ a } \\ \\ =\frac { a }{ a } +\frac { b }{ a } \\ \\ =1+{ \left( \varphi \right) }^{ -1 }\\ \\ =1+\frac { 1 }{ \varphi }$

Number Two: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad { \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad { \varphi }^{ 2 }-\varphi =1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4 }{ 4 } +\frac { 1 }{ 4 } \\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 5 }{ 4 } \\ \\ \Rightarrow \quad \varphi -\frac { 1 }{ 2 } =\frac { \sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad \varphi =\frac { 1 }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \\ \\ \therefore \quad \varphi =\frac { 1+\sqrt { 5 } }{ 2 }$

Number Three: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } } \\ \\ \therefore \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

Number Four: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ { \Rightarrow \quad \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad \varphi =\sqrt { \varphi +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \varphi +1 } +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \sqrt { \varphi +1 } +1 } +1 } \\ \\ \therefore \quad \varphi =\sqrt { \sqrt { \sqrt { \frac { 1+\sqrt { 5 } }{ 2 } +1 } +1 } +1 } \\ \\$

And check out this calculator trick…

If you’re not satisfied with what I’ve already produced, then you can have a go at proving that… $\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ \\$

Without using the phi (φ) symbol.

Enjoy!!! 😀

# 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! 😀

# THE IRRATIONAL NUMBER VALUE INDEX

Want to derive the value of specific irrational numbers from scratch?

Just click on the surd you’d like to discover more about, and you will be directed to a page which will give you instructions on how to derive its value.

What we haven’t included on this list are the square roots of numbers which can easily be found. We’ve only found the value of surds up to the square root of 11.

 √2 √3 √5 √7 √11 (1+√5)/2

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