Proof that Square Root of Two is Irrational Number

To proof that square root of two is irrational number, we assuming that square root of two is rational number, that is square root of two equals a over b where a and b as integer prime. So a equals b times square root of two or a square equals two times b square. Because a square is two times an integer number, so a square is integer, so that a is integer too. And than we assuming that a is two times c, so the equation is four times c square equals two times b square or two times c square equals b square. So that b square is integer and b is integer too. But it is impossible because a and b is impossible to integer because they are relative prime number. So assumption that square root of two is rational number has brought us to the impossibility and must be annulled. And it is proofed that square root of two is irrationals number.

Abc Formula to Solve a Square Equation

We know that square equation is a times x square plus b times x plus c equals zero. To get the Abc formula we need some step. Step one, we divide all by a, that is x square plus b over a times x plus c over a equals zero. Step two, we need to insert perfect square into that equation. That is x square plus b over a times x plus open bracket b over two of a close bracket square plus c over a equals b over two times a in bracket square. Step three is x plus b over two times a in bracket square equals b square over four times a square in bracket minus c over a equals b square four times a times c all over four times a square. Step four is x plus b over two times a equals about square root of b square four times a times c all over four times a square. Step five is x equals minus b over two times a about one over two times a in bracket times square root of b square four times a times c equals minus b about times square root of b square four times a times c all over two times a. so the formula is:
x equals minus b about times square root of b square four times a times c all over two times a

How to Find Phi

Egypt has found phi about at 200 BC and the value for phi is 3.16. They get that value from the formula of circles area that is square of open bracket diameter times eight over nine close bracket. We know that diameter or d equals two times radius (r), so we get the formula of circles area is open bracket eight over nine in bracket times two of r close bracket square equals sixty four over eighty one in bracket times four of r square equals three point one six times r square. Long time after that Archimedes at 250 BC using phi, but with a different value that is three point one four like which we use now.

Properties of Logarithm

First, a to the power of m in bracket times a to the power of in bracket equals a to the power of m plus n. Second, a to the power of m all over a to the power of n equals a to the power of m minus n. Third, Log b to the base of a equals n, so that b equals a to the power of n. Fourth, Log a to the base of g equals x, so that a equals g to the power of x. Fifth, Log b to the base of g equals y, so that g to the power of y.
Example:
What is the equivalence of Log a times b to the base of g?
Consequence Log a to the base of g equals x, so that a equals g to the power of x, and Log b to the base of g equals y, so that b equals g to the power of y. So a times b equals g to the power of x in bracket times g to the power of y equals g to the power of x plus y. So that Log a times b to the base of g equals Log g to the power of x plus y to the base of g equals x plus y in bracket times Log g to the base of g. If Log g to the base of g is one, so Log a times b to the base of g equals x plus y in bracket times one equals Log a to the base of g plus Log b to the base of g. So:
Log a times b to the base of g equals Log a to the base of g plus Log b to the base of g

What is the equivalence of Log a over b to the base of g?
Consequence a over b equals g to the power of x all over g to the power of y equals g to the power of x minus y. So than Log a over b to the base of g equals Log g to the power of x minus y to the base of g equals x minus y in bracket times Log g to the base of g equals x minus y in bracket times one equals Log a to the base of g minus Log b to the base of g. So:
Log a over b to the base of g equals Log a to the base of g minus Log b to the base of g