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

Question

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. $$x=\sqrt4$$ $$x=±4^\frac 12$$ $$(x)^2=(±1)(4^\frac 12)^2$$ $$x^2 = (±1)4^1$$ $$x=±\sqrt4$$

Answer 1

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 $m'^2 = m$,

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

Answer 2

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 $x\ne -2$.

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.