I'm having a bit of a hard time wrapping my head around how the following that I have just learned:

$\sqrt{X^2} = |X|$, and I totally understand why.

But, when expressed as an exponent, doesn't this really just mean the following:

$X^{2/2} = |X|$, if this is the case, and I simplify the rational exponent, I would get:

$X^{1/1}$ or $X^1$, which does not equal $|X|$.

Also, if I apply the following rule of a radical function:

$\sqrt[n]{P^Q} = (\sqrt[n]P)^Q$ where n is the index of the root and $Q$ is the power of the radicand, then this should mean that:

$\sqrt{X^2} = (\sqrt{X})^2$, but the $(\sqrt{X})^2$ does not equal $|X|$ and has a domain where $X > 0$, while the $\sqrt{X^2}$ has a domain equal to all real values for $X$.

Does this mean that when $X$ is raised to an even-numbered power and is the radicand in a radical expression, that one should not simplify the rational exponent or one should not rewrite the radical expression such that the power of the radicand $X$ now lies outside of the root function?

Any replies will be greatly appreciated.

  • 1
    $\begingroup$ Please TeX it up. $\endgroup$ – user305860 Jul 1 '16 at 17:39
  • $\begingroup$ For a lazy person like me it is hard to follow, please tex it up! $\endgroup$ – Babak Miraftab Jul 1 '16 at 17:45

This is due to the fact that there is a slight difference between $\sqrt{x}$ and $x^{1/2}$
I recommend looking up the term "Principle Root", with a basic introduction here.
In essence, for positive numbers there are always two answers to $x^{1/2}$, namely $\pm \sqrt{x}$.
From that, note that the $\sqrt{x}$ function always gives the positive root for a positive argument, and is thus a true function... it outputs one root for each argument $x$. However, the function $x^{1/2}$ outputs two values for each argument $x$, and is thus NOT a true function. Nuances like these are what is messing with your argument!

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any exponent function are defined from $\mathbb{R}^+$ to $\mathbb{R}^+$, because as definition $f(x)=x^{1/p}$ is the inverse (in algebraic sense) of $x^p$ so if $p$ is even we know that $x^p$ is not injective in $\mathbb{R}$ but is bijective in $\mathbb{R}^+$, for general $p$ ($p$ odd) we can define easily $x^p$ in $\mathbb{R}$ as one to one function but the calculus can't be done easily in fact : $$ (-1)=(-1)^1=(-1)^{2/2} $$ but to say that $(x)^{ab}=x^a x^b$ we need to see (as function ) if this is well defined, that mean if $x^a$ is well defined and $x^b$ is too well defined!! so in this case we can't say that $$ (-1)^{2/2}=(-1)^2(-1)^{1/2} $$ because the second term have no sense

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The exponent rules you are referring ( $(a^{b})^c=a^{bc}$ ) work well when the base $a$ is positive, that is, $a>0$. In general, handling exponents of negative numbers should be done with care, precisely because of domain issues of the even-numbered roots.

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