# Golden Ratio Proof

Prove that:

$\frac { a }{ b } =\frac { a+b }{ a } =\frac { 1+\sqrt { 5 } }{ 2 }$

First say that:

$p=\frac { a }{ b }$

$bp=a$

So:

$\frac { a }{ b } =\frac { a+b }{ a } \\ \\ p=\frac { bp+b }{ bp } \\ \\ p=\frac { b\left( p+1 \right) }{ bp } \\ \\ p=\frac { p+1 }{ p } \\ \\ { p }^{ 2 }=p+1\\ \\ { p }^{ 2 }-p=1\\ \\ { \left( p-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=1\\ \\ { \left( p-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4 }{ 4 } +\frac { 1 }{ 4 } \\ \\ { \left( p-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 5 }{ 4 } \\ \\ p-\frac { 1 }{ 2 } =\frac { \sqrt { 5 } }{ \sqrt { 4 } } \\ \\ p-\frac { 1 }{ 2 } =\frac { \sqrt { 5 } }{ 2 } \\ \\ p=\frac { 1 }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \\ \\ p=\frac { 1+\sqrt { 5 } }{ 2 }$

Now:

$\frac { a+b }{ a } \\ \\ =\frac { \left( 1+\sqrt { 5 } \right) +2 }{ \left( 1+\sqrt { 5 } \right) } \\ \\ =\frac { \left( 1+\sqrt { 5 } \right) }{ \left( 1+\sqrt { 5 } \right) } +\frac { 2 }{ \left( 1+\sqrt { 5 } \right) } \cdot \frac { \left( 1-\sqrt { 5 } \right) }{ \left( 1-\sqrt { 5 } \right) } \\ \\ =1+\frac { \left( 2-2\sqrt { 5 } \right) }{ -4 } \\ \\ =\frac { -4 }{ -4 } +\frac { \left( 2-2\sqrt { 5 } \right) }{ -4 } \\ \\ =\frac { -4+\left( 2-2\sqrt { 5 } \right) }{ -4 } \\ \\ =\frac { -4+2-2\sqrt { 5 } }{ -4 } \\ \\ =\frac { -2-2\sqrt { 5 } }{ -4 } \\ \\ =\frac { -2\left( 1+\sqrt { 5 } \right) }{ -2\cdot 2 } \\ \\ =\frac { 1+\sqrt { 5 } }{ 2 }$

# Proving That Two Different Variance Formulas Are Equal To One Another

Prove that:

$\frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

Well you first identify these truths:

$x=\left\{ a,\quad b,\quad c \right\} \\ \\ \therefore \quad \overline { x } =\frac { \Sigma x }{ n } =\frac { a+b+c }{ n } ,\quad \\ \\ \therefore \quad n=3\\ \\ \therefore \quad n\overline { x } =a+b+c\\ \\ \therefore \quad \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ n } .\\ \\ \therefore \quad \Sigma { x }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$

And next:

The standard deviation formula looks like this:

$\sigma =\sqrt { \frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } } =\sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } }$