# If a/b=c/d, then a/b=(a+c)/(b+d) mathematical proof

In this post I’ll be proving to you that if:

a/b=c/d

Then:

a/b=(a+c)/(b+d)

To prove this, the first thing you need to know is that:

$\frac { a }{ b } =\frac { c }{ d } \\ \\ \therefore \quad ad=bc$

The rest is just mathematical / algebraic trickery… Let me show you…

Proof 1:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab }{ b\left( b+d \right) } +\frac { ad }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { bc }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { c }{ b+d } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 2:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab+bc }{ b\left( b+d \right) } \\ \\ =\frac { b\left( a+c \right) }{ b\left( b+d \right) } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 1 Video:

Proof 2 Video:

# Coded Data Proofs: Mean & Standard Deviation, y=kx+C

Coded Data Proofs (3):

Say y=kx+C and also that:

x={p, q} and y={kp+C, kq+C}

This would mean that:

$\frac { \Sigma y }{ n } =\frac { kp+C+\left\{ kq+C \right\} }{ n } \\ \\ =\frac { kp+kq+nC }{ n } \\ \\ =\frac { k\left( p+q \right) +nC }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } +\frac { nC }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } +C$

And if the above is true:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp+C \right) }^{ 2 }+{ \left( kq+C \right) }^{ 2 } }{ n } \\ \\ =\frac { \left( kp+C \right) \left( kp+C \right) +\left( kq+C \right) \left( kq+C \right) }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+nkpC+{ C }^{ 2 }+\left\{ { k }^{ 2 }{ q }^{ 2 }+nkqC+{ C }^{ 2 } \right\} }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 }+nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) +nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } +\frac { nkC\left( p+q \right) }{ n } +\frac { n{ C }^{ 2 } }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +kC\cdot \Sigma x+{ C }^{ 2 }\\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right)$

Therefore:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ \left( k\cdot \frac { \Sigma x }{ n } +C \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left( k\cdot \frac { \Sigma x }{ n } +C \right) \left( k\cdot \frac { \Sigma x }{ n } +C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left\{ { k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }+nkC\cdot \frac { \Sigma x }{ n } +{ C }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-nkC\cdot \frac { \Sigma x }{ n } -{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-kC\cdot \Sigma x-{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-C\left( k\cdot \Sigma x+C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left\{ \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }$

# Mathematical Programming (Manipulation) Tips

Commutative means that the order of elements contained within an expression do not alter a  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:

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

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Associative 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:

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

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Warning:

Commutative and Associative manipulation should not be used when subtracting  and dividing.