# Real Mathematical Proofs & Philosophical Videos

Dear users,

I’ve set up a Pinterest account that you can now explore. If you visit this link https://www.pinterest.com/maths_videos/, you will find plenty of maths proofs that I’ve created – including videos related to the simulation hypothesis, information theory, black holes, space-time and topology.

Hope to see you there! đź™‚

# Logarithmic Proof (4)

Prove that:

$\log _{ a }{ \left( \frac { x }{ p } \right) } =\log _{ a }{ \left( x \right) } -\log _{ a }{ \left( p \right) }$

Say that:

$\log _{ a }{ \left( x \right) } =m\\ \\ \therefore \quad { a }^{ m }=x$

And that:

$\log _{ a }{ \left( p \right) } =n\\ \\ \therefore \quad { a }^{ n }=p$

Therefore:

$\log _{ a }{ \left( \frac { x }{ p } \right) } \\ \\ =\log _{ a }{ \left( \frac { { a }^{ m } }{ { a }^{ n } } \right) } \\ \\ =\log _{ a }{ \left( { a }^{ \left( m-n \right) } \right) } \\ \\ =\left( m-n \right) \log _{ a }{ \left( a \right) } \\ \\ =m\log _{ a }{ \left( a \right) } -n\log _{ a }{ \left( a \right) } \\ \\ =\log _{ a }{ \left( { a }^{ m } \right) -\log _{ a }{ \left( { a }^{ n } \right) } } \\ \\ =\log _{ a }{ \left( x \right) } -\log _{ a }{ \left( p \right) }$

# Logarithmic Proof (3)

Prove that:

$\log _{ a }{ \left( { a }^{ n } \right) =n\log _{ a }{ (a) } }$

Say that:

${ a }^{ n }=p$

Therefore:

$\log _{ a }{ p=n }$

So:

$\log _{ a }{ \left( { a }^{ n } \right) } \\ \\ =\log _{ a }{ \left( p \right) } \\ \\ =\log _{ a }{ \left( p \right) } \cdot 1\\ \\ =\log _{ a }{ \left( p \right) \cdot \log _{ a }{ \left( a \right) } } \\ \\ =n\cdot \log _{ a }{ \left( a \right) }$

# Logarithmic Proof (2)

Prove that:

$\log _{ a }{ (xp)=\log _{ a }{ (x)+\log _{ a }{ (p) } } }$

Say that:

$\log _{ a }{ x } =m\\ \\ \therefore \quad { a }^{ m }=x$

And say that:

$\log _{ a }{ p } =n\\ \\ \therefore \quad { a }^{ n }=p$

Therefore:

$\log _{ a }{ (xp) } \\ \\ =\log _{ a }{ ({ a }^{ m }\cdot { a }^{ n }) } \\ \\ =\log _{ a }{ ({ a }^{ (m+n) }) } \\ \\ =(m+n)\log _{ a }{ (a) } \\ \\ =m\log _{ a }{ (a)+n\log _{ a }{ (a) } } \\ \\ =\log _{ a }{ ({ a }^{ m })+\log _{ a }{ ({ a }^{ n }) } } \\ \\ =\log _{ a }{ (x)+\log _{ a }{ (p) } }$

# Logarithmic Proof (1)

Prove that:

$\log _{ a }{ x=\frac { 1 }{ \log _{ x }{ a } } }$

Proof:

$\log _{ a }{ x } =p\\ \\ { a }^{ p }=x\\ \\ { x }^{ \frac { 1 }{ p } }=a\\ \\ \log _{ x }{ a } =\frac { 1 }{ p } \\ \\ p\log _{ x }{ a } =1\\ \\ p=\frac { 1 }{ \log _{ x }{ a } } \\ \\ As,\quad p=p:\\ \\ \log _{ a }{ x } =\frac { 1 }{ \log _{ x }{ a } }$

# How To Expand (a+b+c)(d+e+f)(g+h+i)

Firstly you have to know what (a+b+c)(d+e+f) is. You can expand this expression using a rectangle:

So you know that: $\left( a+b+c \right) \left( d+e+f \right) =ad+ae+af+bd+be+bf+cd+ce+cf$

Next you’d have to multiply (a+b+c)(d+e+f) by (g+h+i) using another rectangle:

And from here you’d figure out that:

$\left( a+b+c \right) \left( d+e+f \right) \left( g+h+i \right) \\ \\ =\left[ \left( a+b+c \right) \left( d+e+f \right) \right] \left( g+h+i \right) \\ \\ =\left[ ad+ae+af+bd+be+bf+cd+ce+cf \right] \left( g+h+i \right) \\ \\ =adg+aeg+afg+bdg+beg+bfg+cdg+ceg+cfg\\ \\ +adh+aeh+afh+bdh+beh+bfh+cdh+ceh+cfh\\ \\ +adi+aei+afi+bdi+bei+bfi+cdi+cei+cfi\\ \\ =ag\left( d+e+f \right) +bg\left( d+e+f \right) +cg\left( d+e+f \right) \\ \\ +ah\left( d+e+f \right) +bh\left( d+e+f \right) +ch\left( d+e+f \right) \\ \\ +ai\left( d+e+f \right) +bi\left( d+e+f \right) +ci\left( d+e+f \right) \\ \\ =\left( d+e+f \right) \left[ ag+bg+cg+ah+bh+ch+ai+bi+ci \right] \\ \\ =\left( d+e+f \right) \left[ g\left( a+b+c \right) +h\left( a+b+c \right) +i\left( a+b+c \right) \right] \\ \\ =\left( d+e+f \right) \left[ \left( a+b+c \right) \left( g+h+i \right) \right] \\ \\ =\left( a+b+c \right) \left( d+e+f \right) \left( g+h+i \right)$

# Even And Odd Functions

An even functionÂ exists when $f\left( x \right) =f\left( -x \right)$ for all values of x.

The graph of anÂ even functionÂ must be symmetrical about the y-axis.

Examples ofÂ even functionsÂ have been listed below:

$f\left( x \right) ={ x }^{ 2 }$

$f\left( x \right) ={ x }^{ 4 }$

$f\left( x \right) =cosx$

———————————————————————–

AnÂ odd function exists when $f\left( x \right) =-f\left( -x \right)$ for all values of x.

Graphs ofÂ odd functionsÂ should have 180 degrees rotational symmetry about the origin (0,0).

Examples ofÂ odd functionsÂ are listed below:

$f\left( x \right) =x$

$f\left( x \right) ={ x }^{ 3 }$

$f\left( x \right) =sinx$

——————————————————————–

Most functions are neither even or odd.

An example of a function that is neither even or odd would be:

$f\left( x \right) ={ x }^{ 2 }+\frac { 1 }{ x }$

# Mathematical Programming (Manipulation) Tips

Commutative means that the order of elements contained within an expression do not alter a Â certain mathematical result.

For example:

$2\cdot 3=3\cdot 2=6\\ \\ 1\cdot 2\cdot 3=2\cdot 1\cdot 3=3\cdot 1\cdot 2=6\\ \\ Or:\\ \\ 1+2=2+1=3\\ \\ 1+2+3=3+2+1=2+1+3=6$

Knowing this, you can say that:

$a\cdot b\cdot c=c\cdot a\cdot b=b\cdot a\cdot c=c\cdot b\cdot a\\ \\ Or:\\ \\ a+b+c=c+a+b=b+a+c=c+b+a$

—————————————————————–

AssociativeÂ means that you can alter the grouping of certain elements in an expression without changing its result.

$\left( 2\cdot 3 \right) \cdot 4=2\cdot \left( 3\cdot 4 \right) =24\\ \\ Or:\\ \\ \left( 2+3 \right) +4=2+\left( 3+4 \right) =9$

Knowing this you could say that:

$\left( a\cdot b \right) \cdot c=a\cdot \left( b\cdot c \right) \\ \\ Or:\\ \\ \left( a+b \right) +c=a+\left( b+c \right)$

——————————————————————

Warning:

Commutative and AssociativeÂ manipulation should not be used when subtracting Â and dividing.

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

——————————————————————————————–

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

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

——————————————————————————————–

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

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