Ytukyg

Note: In this question I am using the radical symbol to denote all roots of a number, not just the positive one.

$$x=sqrt[n] m ≠ x=m^frac 1n $$ when m is an even number

for example:-
$$x=sqrt4$$
$$x=±2$$
or:
$$x=sqrt4$$
$$x=4^frac 12$$
$$(x)^2=(4^frac 12)^2$$
$$x^2 = 4^1$$
$$x=sqrt4$$
$$x=±2$$

but:
$$x=sqrt4$$
$$x=4^frac 12$$
$$x=(2^2)^frac 12$$
$$x=2^frac 22$$
$$x=2^1$$
$$x=2$$

in above example adding ±1 we will have same result as above.

$$x=sqrt4$$
$$x=±4^frac 12$$
$$x=(±1)(2^2)^frac 12$$
$$x=(±1)2^frac 22$$
$$x=(±1)2^1$$
$$x=±2$$
but in second example we will have a negative root that leads to an imaginary number.
$$x=sqrt4$$
$$x=±4^frac 12$$
$$(x)^2=(±1)(4^frac 12)^2$$
$$x^2 = (±1)4^1$$
$$x=±sqrt4$$

I'm sorry but this is pretty much... wrong. Your proposition is not true.

For $x,y>in Bbb R_{>0}$, $x^y$ is defined to be positive. So, as @T.Bongers pointed out, $sqrt{4} = 2$ by definition, even if $x^2 = 4$ has exacly one positive and one negative solution. The convention for the square root is defining that it satisfies $sqrt{x^2} = |x|$ for all $x$. Therefore, since for every non-negative real number $m$ there is some $m' in Bbb R$ such that $m'^2 = m$,

$sqrt{m} = |m'| geq 0$.

First of all $sqrt{n}$ is defined to always be a single value that is never negative.

So if $x=sqrt 4$ then then $x = 2$ and $xne -2$.

You are confusing this with

$x^2 = 4$ then $x =pm sqrt{4}$. But we do not know whether $x = sqrt 4 = 2$ or whether $x = -sqrt{4} = -2$. However we DO know that $sqrt{4} =2$ and $sqrt{4}$ does NOT equal $-2$.

However if we are told $x=sqrt{4}$ then we are being told that $x ne =-sqrt{4}$.

Second, you are treating $pm n$ as though it where a number. It isn't a number. It is statement that there are two possibilities to consider. Either $n$ or $-n$.

But both have to be possible. If we are told $x = 2$ we can't then say $x = pm 2$ because we know $x = -2$ is not possible. (Likewise if we are told $x = -2$ we can't say $x =pm 2$ because we know $x=2$ is not possible.

So if we are told $x^2 = 4$ then there are two possibilities. Either $x = 2$ or $x = -2$ and we don't know which. So we can say $x =pm 2$.

If however we are told $x = sqrt 4$ the we are given one value and we know $x = -sqrt 4$ is not possible.

Third

whe you go from $x = 4^{frac 12}$ to $x^2 = (4^{frac 12})^2$ you must make a mental note that $x > 0$. Squaring adds extranous solutions.

For example. If you have $x = 5$ and then you say $x^2 = 5^2 = 25$ you must remember that $x > 0$. Then when you so $x^2 =25 implies x =pm sqrt{25} = pm 5$ you must add But ALSO $x > 0$ so $x = sqrt{25} = 5$

So to correct everything you've done:

$x = sqrt 4$

$x = 2$

or:

$x = sqrt 4$

$x = 4^{frac 12}$

$x^2 = (4^{frac 12})^2$ (and $x > 0$)

$x^2 = 4^1$

$x = pm sqrt 4$ but $x > 0$ so

$x = sqrt 4$

$x= 2$.

But:

$x = sqrt 4$

$x = 4^{frac 12}$

$x = (2^2)^{frac 12}$

$x = 2^{frac 22}$

$x = 2^1$

$x = 2$

which is basically spinning your wheels in the mud.