# |z_1|/|z_2|=|z_1/z_2| proof (algebraic)

In this post I’ll be proving to you that: $\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right|$

Firstly, I’ll say that: ${ z }_{ 1 }=x+iy,\quad \therefore \quad \left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

And also that: ${ z }_{ 2 }=p+iq,\quad \therefore \quad \left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

If this is the case, then: $\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { x+iy }{ p+iq } \\ \\ =\frac { \left( x+iy \right) }{ \left( p+iq \right) } \cdot \frac { \left( p-iq \right) }{ \left( p-iq \right) } \\ \\ =\frac { px-iqx+ipy-{ i }^{ 2 }qy }{ { p }^{ 2 }-ipq+ipq-{ i }^{ 2 }{ q }^{ 2 } } \\ \\ =\frac { \left( px+qy \right) +i\left( py-qx \right) }{ { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) +i\left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right)$

And as this is in the form: $z=a+ib$

I would have to conclude that: $RHS=\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| \\ \\ =\sqrt { { \left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 }+{ \left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } +\frac { { \left( py-qx \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 }+{ \left( py-qx \right) }^{ 2 } }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) ^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }-2pqxy+{ q }^{ 2 }{ x }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } } \\ \\ =\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =LHS$

Hence, I have my proof.

# 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!!! 😀

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

# Experiments With Surds & Exponentials… $\sqrt { a } \sqrt { a } ={ a }^{ \frac { 1 }{ 2 } }{ a }^{ \frac { 1 }{ 2 } }={ a }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }=a\\ \\ \sqrt [ 3 ]{ a } \sqrt [ 3 ]{ a } \sqrt [ 3 ]{ a } ={ a }^{ \frac { 1 }{ 3 } }{ a }^{ \frac { 1 }{ 3 } }{ a }^{ \frac { 1 }{ 3 } }={ a }^{ \frac { 1 }{ 3 } +\frac { 1 }{ 3 } +\frac { 1 }{ 3 } }=a\\ \\ \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } ={ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }={ a }^{ \frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } }=a$ $\sqrt { \frac { a }{ b } } \sqrt { \frac { a }{ b } } ={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } }={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }=\frac { a }{ b } \\ \\ \sqrt [ 3 ]{ \frac { a }{ b } } \sqrt [ 3 ]{ \frac { a }{ b } } \sqrt [ 3 ]{ \frac { a }{ b } } ={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } +\frac { 1 }{ 3 } +\frac { 1 }{ 3 } }=\frac { a }{ b }$ ${ a }^{ \frac { 1 }{ 2 } }=b\\ \\ \therefore \quad a={ b }^{ 2 }\\ \\ { a }^{ \frac { 3 }{ 4 } }=b\\ \\ \therefore \quad a={ b }^{ \frac { 4 }{ 3 } }\\ \\ { a }^{ \sqrt { m } }=b\\ \\ \therefore \quad a={ b }^{ \frac { 1 }{ \sqrt { m } } }\\ \\ \therefore \quad a={ b }^{ \frac { \sqrt { m } }{ m } }\\ \\ { a }^{ \frac { 1 }{ m+\sqrt { n } } }=b\\ \\ \therefore \quad a={ b }^{ m+\sqrt { n } }$

# Surd Problem…

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

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

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