Radical Notation and Simplifying Them

Introduction

So still trying to get to understand GF(256). Came across this and forgot how interesting it was. Radical notation is a mathematical method for expressing roots of numbers or expressions using the radical symbol √. It represents the inverse operation of exponents (powers), such as finding the square root (√x) or nth root ( n√x ) of a value, which is also equivalent to raising a number to a fractional exponent.

Examples Squares

This was what I found interesting I guess. If memorize the squares of up to four hundred i.e.

2,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400

Then managing radiicals becomes easy So given $\sqrt{18}$, this is actually $\sqrt{2}$$\sqrt{9}$ and $\sqrt{9}$ = 3 so $\sqrt{18}$ = 3 . $\sqrt{2}$
So if we have $\sqrt{75}$ then we have $\sqrt{25}$$\sqrt{3}$ which is 5$\sqrt{3}$
So basically you can split the root by using the product of the roots to simplify the express.
So if we have $\sqrt{80}$ then we have $\sqrt{16}$$\sqrt{5}$ or 4$\sqrt{5}$
So now with variables to
If we have $\sqrt{x^5}$ then this equals 5/2 = 2 remainder 1 to $x^2\sqrt{x^1}$
If we have $\sqrt{x^9}$ then this equals 9/2 = 4 remainder 1 to $x^4\sqrt{x^1}$

Examples Cubes

So the same thing is possible with cubes

1,8,27,64,125,216,343,512,729,1000

So given $\sqrt[3]{54}$ we can see 27 so $\sqrt[3]{27}$$\sqrt[3]{2}$ so 3$\sqrt[3]{2}$
So given $\sqrt[3]{40}$ we can see 8 so $\sqrt[3]{8}$$\sqrt[3]{5}$ so 2$\sqrt[3]{5}$
So now with variables to
If we have $\sqrt[3]{x^8}$ then this equals 8/3 = 2 remainder 2 to $x^2.\sqrt[3]{x^2}$
If we have $\sqrt[3]{x^{1}4}$ then this equals 14/3 = 4 remainder 2 to $x^4.\sqrt[3]{x^2}$

Converting to Exponential Notation and back

If given $\sqrt[5]{x^7}$ then this equals $x^{7/5}$
So given $x^{3/5}$ then this equals $\sqrt[7]{x^2}$
So this was how to manage negative indices.
So given $x^{-7/4}$ we need to invert $\frac{1}{x^{7/4}}$ and then back to rational with $\frac{1}{\sqrt[4]{x^7}}$
And we can do the reverse for negative indices already inverted
So given $\frac{1}{x^{-3/4}}$ we need to invert $x^{3/4}$ and then back to rational with $\sqrt[4]{x^3}$

Simplifying Exponents

Found this tricky with my dyslexia but here goes.
Given $(8{)}^{2/3}$ we can write this as $(8^{1/3}{)}^2$ and this becomes easy.
But the thing I need to remember and forget is
          ${\mathit{x}}^{\mathit{1}/\mathit{3}}$ is the cube root of
          ${\mathit{x}}^{\mathit{1}/\mathit{2}}$ is the square root of

So going back to the problem
          $(8^{1/3}{)}^2$ = $(2{)}^2$ = 4

Absolute Value

The absolute value is the distance from zero neither negative of positive. When simplifying an even root of an even power, the result is the absolute value of the base.
So given
  $\sqrt[2]{x^4}$ we get $(x{)}^{4/2}$ and then $x^2$ and it is even

But if the root is even and the result odd we need to make that result be absolute $|x|$
  $\sqrt[2]{x^{10}}$ = $(x{)}^{10/2}$ = $|x|$
So the point is that this only effects even roots, 5 is not even so this is ok
  $\sqrt[5]{x^{25}}$ = $x^5$
But this is not because the root is even and the result is odd
  $\sqrt[6]{x^{42}}$ = $|x^7|$

Simplifying Rational Exponents

If we have something like
  $3(2x-4{)}^{2/3}+5=17$
To solve this first we move the 5 to the right
  $3(2x-4{)}^{2/3}=12$
Next move the 3 to the right
  $(2x-4{)}^{2/3}=4$
And this is what the question is really about, how to remove the rational exponent. And the answer is raise both sides by the reciprocal
  $[(2x-4{)}^{2/3}{]}^{3/2}=[4{]}^{3/2}$
Now we have
  $(2x-4)=[4{]}^{3/2}$
I found the left to be unreadable but if you remember $x^{1/3}$ is the cube root then $4^{3/2}$ must be $\sqrt[2]{4^3}$ and that = $\sqrt[2]{64}$ with is 8
The other approach which they used a lot was to split the rational exponent to be $(4^{1/2}{)}^3$. We know $x^{1/2}$ is square root and the square root of 4 is 2 so $2^3$ = 8
So the answer is the question
  $(2x-4)=8$
is 2x = 12 and therefore x = 6

Composite Radicals

This is perhaps fairly obvious. If we have $\sqrt[3]{\sqrt[4]{x}}$ we have use the formula which says
  $\sqrt[n]{\sqrt[m]{x}}$ = $\sqrt[mn]{x}$
So
$\sqrt[3]{\sqrt[4]{x}}$ = $\sqrt[12]{x}$

Composite Radicals with different Indices

So just to make this even more fun how do we do this $\sqrt[4]{x^1}.\sqrt[5]{x^1}$. The answer is another formula.
  $\sqrt[m]{x^a}.\sqrt[n]{x^b}$ = $\sqrt[mn]{x^{an+bm}}$

Graphing Radicals

Really liked desmos https://www.desmos.com/calculator. Really easy to use, especially with the robot. So I graphed <math>\sqrt[2]{x}<math> where root was negative and positive and x was negative and positive. I find it soothing to look at graphs