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!!! 😀

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

Surd Problem…

Simplify:

$\sqrt { 1 } \sqrt { 2 } \sqrt { 3 } \sqrt { 4 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 9 } \sqrt { 10 }$

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$\sqrt { 1 } \sqrt { 2 } \sqrt { 3 } \sqrt { 4 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 9 } \sqrt { 10 } \\ \\ =\sqrt { 1 } \sqrt { 4 } \sqrt { 9 } \sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 10 } \\ \\ =1\cdot 2\cdot 3\sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 10 } \\ \\ =6\sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \left( \sqrt { 2 } \sqrt { 3 } \right) \sqrt { 7 } \left( \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \right) \left( \sqrt { 2 } \sqrt { 5 } \right) \\ \\ =6\sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 3 } \sqrt { 3 } \sqrt { 5 } \sqrt { 5 } \sqrt { 7 } \\ \\ =6\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\sqrt { 7 } \\ \\ =720\sqrt { 7 }$

How To Multiply Surds Contained within Fractions

First Scenario:

$\frac { 1 }{ a+\sqrt { b } } =\frac { 1 }{ \left( a+\sqrt { b } \right) } \cdot \frac { \left( a-\sqrt { b } \right) }{ \left( a-\sqrt { b } \right) } \\ \\ =\frac { a-\sqrt { b } }{ { a }^{ 2 }-a\sqrt { b } +a\sqrt { b } -b } =\frac { a-\sqrt { b } }{ { a }^{ 2 }-b }$

Second Scenario:

$\frac { 1 }{ a-\sqrt { b } } =\frac { 1 }{ \left( a-\sqrt { b } \right) } \cdot \frac { \left( a+\sqrt { b } \right) }{ \left( a+\sqrt { b } \right) } \\ \\ =\frac { a+\sqrt { b } }{ { a }^{ 2 }+a\sqrt { b } -a\sqrt { b } -b } =\frac { a+\sqrt { b } }{ { a }^{ 2 }-b }$

Notice that what we’re ultimately doing in both cases is multiplying the surd within a fraction by 1. When the value of the numerator is exactly the same as the value of the denominator in a fraction, what you have is 1.

You should know that 1/1, 2/2, 3/3, (a+b)/(a+b) are all equal to 1.

How To Multiply Surds

In order to multiply surds, you should first know these rules:

${ a }^{ m }\cdot { a }^{ n }={ a }^{ m+n }\\ \\ { a }^{ m }\div { a }^{ n }={ a }^{ m-n }$

You should also know that:

${ a }^{ \frac { 1 }{ 2 } }=\sqrt [ 2 ]{ { a }^{ 1 } } =\sqrt { a }$

So, knowing these rules, what would you get if you multiplied: $\sqrt { 3 } \cdot \sqrt { 3 }$?

Well, $\sqrt { 3 } \cdot \sqrt { 3 } ={ 3 }^{ \frac { 1 }{ 2 } }\cdot { 3 }^{ \frac { 1 }{ 2 } }={ 3 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }={ 3 }^{ 1 }=3$.

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Now how about $\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right)$?

$\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right) =\sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\sqrt { 3 } \cdot \sqrt { 3 } \cdot \left( -1 \right) =3\cdot \left( -1 \right) =-3$

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What about $\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right)$?

$\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right) =\left( -1 \right) \cdot \sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\left( -1 \right) \left( -1 \right) \sqrt { 3 } \sqrt { 3 } =1\cdot 3=3$